Introduction to representation theory
Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, Elena Yudovina
aa r X i v : . [ m a t h . R T ] F e b Introduction to representation theory
Pavel Etingof, Oleg Golberg, Sebastian Hensel,Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena YudovinaFebruary 2, 2011
Contents sl (2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.15 Problems on Lie algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 S n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.13 Proof of Theorem 4.36 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.14 Induced representations for S n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 602.15 The Frobenius character formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.16 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.17 The hook length formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.18 Schur-Weyl duality for gl ( V ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.19 Schur-Weyl duality for GL ( V ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654.20 Schur polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.21 The characters of L λ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.22 Polynomial representations of GL ( V ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.23 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.24 Representations of GL ( F q ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.24.1 Conjugacy classes in GL ( F q ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 684.24.2 1-dimensional representations . . . . . . . . . . . . . . . . . . . . . . . . . . . 704.24.3 Principal series representations . . . . . . . . . . . . . . . . . . . . . . . . . . 714.24.4 Complementary series representations . . . . . . . . . . . . . . . . . . . . . . 734.25 Artin’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.26 Representations of semidirect products . . . . . . . . . . . . . . . . . . . . . . . . . . 76 A , A , A . . . . . . . . . . . . . . . . 815.3 Indecomposable representations of the quiver D . . . . . . . . . . . . . . . . . . . . 835.4 Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.5 Gabriel’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.6 Reflection Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.7 Coxeter elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.8 Proof of Gabriel’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 945.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 INTRODUCTION
Very roughly speaking, representation theory studies symmetry in linear spaces. It is a beautifulmathematical subject which has many applications, ranging from number theory and combinatoricsto geometry, probability theory, quantum mechanics and quantum field theory.Representation theory was born in 1896 in the work of the German mathematician F. G.Frobenius. This work was triggered by a letter to Frobenius by R. Dedekind. In this letter Dedekindmade the following observation: take the multiplication table of a finite group G and turn it into amatrix X G by replacing every entry g of this table by a variable x g . Then the determinant of X G factors into a product of irreducible polynomials in { x g } , each of which occurs with multiplicityequal to its degree. Dedekind checked this surprising fact in a few special cases, but could not proveit in general. So he gave this problem to Frobenius. In order to find a solution of this problem(which we will explain below), Frobenius created representation theory of finite groups. The present lecture notes arose from a representation theory course given by the first author tothe remaining six authors in March 2004 within the framework of the Clay Mathematics InstituteResearch Academy for high school students, and its extended version given by the first author toMIT undergraduate math students in the Fall of 2008. The lectures are supplemented by manyproblems and exercises, which contain a lot of additional material; the more difficult exercises areprovided with hints.The notes cover a number of standard topics in representation theory of groups, Lie algebras, andquivers. We mostly follow [FH], with the exception of the sections discussing quivers, which follow[BGP]. We also recommend the comprehensive textbook [CR]. The notes should be accessible tostudents with a strong background in linear algebra and a basic knowledge of abstract algebra.
Acknowledgements.
The authors are grateful to the Clay Mathematics Institute for hostingthe first version of this course. The first author is very indebted to Victor Ostrik for helping himprepare this course, and thanks Josh Nichols-Barrer and Thomas Lam for helping run the coursein 2004 and for useful comments. He is also very grateful to Darij Grinberg for very careful readingof the text, for many useful comments and corrections, and for suggesting the Exercises in Sections1.10, 2.3, 3.5, 4.9, 4.26, and 6.8. For more on the history of representation theory, see [Cu]. Basic notions of representation theory
In technical terms, representation theory studies representations of associative algebras. Its generalcontent can be very briefly summarized as follows.An associative algebra over a field k is a vector space A over k equipped with an associativebilinear multiplication a, b ab , a, b ∈ A . We will always consider associative algebras with unit,i.e., with an element 1 such that 1 · a = a · a for all a ∈ A . A basic example of an associativealgebra is the algebra End V of linear operators from a vector space V to itself. Other importantexamples include algebras defined by generators and relations, such as group algebras and universalenveloping algebras of Lie algebras.A representation of an associative algebra A (also called a left A -module) is a vector space V equipped with a homomorphism ρ : A → End V , i.e., a linear map preserving the multiplicationand unit.A subrepresentation of a representation V is a subspace U ⊂ V which is invariant under alloperators ρ ( a ), a ∈ A . Also, if V , V are two representations of A then the direct sum V ⊕ V has an obvious structure of a representation of A .A nonzero representation V of A is said to be irreducible if its only subrepresentations are0 and V itself, and indecomposable if it cannot be written as a direct sum of two nonzerosubrepresentations. Obviously, irreducible implies indecomposable, but not vice versa.Typical problems of representation theory are as follows:1. Classify irreducible representations of a given algebra A .2. Classify indecomposable representations of A .3. Do 1 and 2 restricting to finite dimensional representations.As mentioned above, the algebra A is often given to us by generators and relations. Forexample, the universal enveloping algebra U of the Lie algebra sl (2) is generated by h, e, f withdefining relations he − eh = 2 e, hf − f h = − f, ef − f e = h. (1)This means that the problem of finding, say, N -dimensional representations of A reduces to solvinga bunch of nonlinear algebraic equations with respect to a bunch of unknown N by N matrices,for example system (1) with respect to unknown matrices h, e, f .It is really striking that such, at first glance hopelessly complicated, systems of equations canin fact be solved completely by methods of representation theory! For example, we will prove thefollowing theorem. Theorem 1.1.
Let k = C be the field of complex numbers. Then:(i) The algebra U has exactly one irreducible representation V d of each dimension, up to equiv-alence; this representation is realized in the space of homogeneous polynomials of two variables x, y of degree d − , and defined by the formulas ρ ( h ) = x ∂∂x − y ∂∂y , ρ ( e ) = x ∂∂y , ρ ( f ) = y ∂∂x . (ii) Any indecomposable finite dimensional representation of U is irreducible. That is, any finite imensional representation of U is a direct sum of irreducible representations. As another example consider the representation theory of quivers.A quiver is a finite oriented graph Q . A representation of Q over a field k is an assignmentof a k -vector space V i to every vertex i of Q , and of a linear operator A h : V i → V j to every directededge h going from i to j (loops and multiple edges are allowed). We will show that a representationof a quiver Q is the same thing as a representation of a certain algebra P Q called the path algebraof Q . Thus one may ask: what are the indecomposable finite dimensional representations of Q ?More specifically, let us say that Q is of finite type if it has finitely many indecomposablerepresentations.We will prove the following striking theorem, proved by P. Gabriel about 35 years ago: Theorem 1.2.
The finite type property of Q does not depend on the orientation of edges. Theconnected graphs that yield quivers of finite type are given by the following list: • A n : ◦−−◦ · · · ◦−−◦ • D n : ◦−−◦ · · · ◦−−◦|◦ • E : ◦−−◦−−◦−−◦−−◦|◦ • E : ◦−−◦−−◦−−◦−−◦−−◦|◦ • E : ◦−−◦−−◦−−◦−−◦−−◦−−◦|◦ The graphs listed in the theorem are called (simply laced)
Dynkin diagrams . These graphsarise in a multitude of classification problems in mathematics, such as classification of simple Liealgebras, singularities, platonic solids, reflection groups, etc. In fact, if we needed to make contactwith an alien civilization and show them how sophisticated our civilization is, perhaps showingthem Dynkin diagrams would be the best choice!As a final example consider the representation theory of finite groups, which is one of the mostfascinating chapters of representation theory. In this theory, one considers representations of thegroup algebra A = C [ G ] of a finite group G – the algebra with basis a g , g ∈ G and multiplicationlaw a g a h = a gh . We will show that any finite dimensional representation of A is a direct sum ofirreducible representations, i.e., the notions of an irreducible and indecomposable representationare the same for A (Maschke’s theorem). Another striking result discussed below is the Frobeniusdivisibility theorem: the dimension of any irreducible representation of A divides the order of G .Finally, we will show how to use representation theory of finite groups to prove Burnside’s theorem:any finite group of order p a q b , where p, q are primes, is solvable. Note that this theorem does notmention representations, which are used only in its proof; a purely group-theoretical proof of thistheorem (not using representations) exists but is much more difficult!6 .2 Algebras Let us now begin a systematic discussion of representation theory.Let k be a field. Unless stated otherwise, we will always assume that k is algebraically closed,i.e., any nonconstant polynomial with coefficients in k has a root in k . The main example is thefield of complex numbers C , but we will also consider fields of characteristic p , such as the algebraicclosure F p of the finite field F p of p elements. Definition 1.3.
An associative algebra over k is a vector space A over k together with a bilinearmap A × A → A , ( a, b ) ab , such that ( ab ) c = a ( bc ). Definition 1.4.
A unit in an associative algebra A is an element 1 ∈ A such that 1 a = a a . Proposition 1.5.
If a unit exists, it is unique.Proof.
Let 1 , ′ be two units. Then 1 = 11 ′ = 1 ′ .From now on, by an algebra A we will mean an associative algebra with a unit. We will alsoassume that A = 0. Example 1.6.
Here are some examples of algebras over k :1. A = k .2. A = k [ x , ..., x n ] – the algebra of polynomials in variables x , ..., x n .3. A = End V – the algebra of endomorphisms of a vector space V over k (i.e., linear maps, oroperators, from V to itself). The multiplication is given by composition of operators.4. The free algebra A = k h x , ..., x n i . A basis of this algebra consists of words in letters x , ..., x n , and multiplication in this basis is simply concatenation of words.5. The group algebra A = k [ G ] of a group G . Its basis is { a g , g ∈ G } , with multiplication law a g a h = a gh . Definition 1.7.
An algebra A is commutative if ab = ba for all a, b ∈ A .For instance, in the above examples, A is commutative in cases 1 and 2, but not commutative incases 3 (if dim V > n > A is commutative if and only if G is commutative. Definition 1.8.
A homomorphism of algebras f : A → B is a linear map such that f ( xy ) = f ( x ) f ( y ) for all x, y ∈ A , and f (1) = 1. Definition 1.9.
A representation of an algebra A (also called a left A -module) is a vector space V together with a homomorphism of algebras ρ : A → End V .Similarly, a right A -module is a space V equipped with an antihomomorphism ρ : A → End V ;i.e., ρ satisfies ρ ( ab ) = ρ ( b ) ρ ( a ) and ρ (1) = 1.The usual abbreviated notation for ρ ( a ) v is av for a left module and va for the right module.Then the property that ρ is an (anti)homomorphism can be written as a kind of associativity law:( ab ) v = a ( bv ) for left modules, and ( va ) b = v ( ab ) for right modules.Here are some examples of representations. 7 xample 1.10. V = 0.2. V = A , and ρ : A → End A is defined as follows: ρ ( a ) is the operator of left multiplication by a , so that ρ ( a ) b = ab (the usual product). This representation is called the regular representationof A . Similarly, one can equip A with a structure of a right A -module by setting ρ ( a ) b := ba .3. A = k . Then a representation of A is simply a vector space over k .4. A = k h x , ..., x n i . Then a representation of A is just a vector space V over k with a collectionof arbitrary linear operators ρ ( x ) , ..., ρ ( x n ) : V → V (explain why!). Definition 1.11.
A subrepresentation of a representation V of an algebra A is a subspace W ⊂ V which is invariant under all the operators ρ ( a ) : V → V , a ∈ A .For instance, 0 and V are always subrepresentations. Definition 1.12.
A representation V = 0 of A is irreducible (or simple) if the only subrepresenta-tions of V are 0 and V . Definition 1.13.
Let V , V be two representations of an algebra A . A homomorphism (or in-tertwining operator) φ : V → V is a linear operator which commutes with the action of A , i.e., φ ( av ) = aφ ( v ) for any v ∈ V . A homomorphism φ is said to be an isomorphism of representationsif it is an isomorphism of vector spaces. The set (space) of all homomorphisms of representations V → V is denoted by Hom A ( V , V ).Note that if a linear operator φ : V → V is an isomorphism of representations then so is thelinear operator φ − : V → V (check it!).Two representations between which there exists an isomorphism are said to be isomorphic. Forpractical purposes, two isomorphic representations may be regarded as “the same”, although therecould be subtleties related to the fact that an isomorphism between two representations, when itexists, is not unique. Definition 1.14.
Let V , V be representations of an algebra A . Then the space V ⊕ V has anobvious structure of a representation of A , given by a ( v ⊕ v ) = av ⊕ av . Definition 1.15.
A nonzero representation V of an algebra A is said to be indecomposable if it isnot isomorphic to a direct sum of two nonzero representations.It is obvious that an irreducible representation is indecomposable. On the other hand, we willsee below that the converse statement is false in general.One of the main problems of representation theory is to classify irreducible and indecomposablerepresentations of a given algebra up to isomorphism. This problem is usually hard and often canbe solved only partially (say, for finite dimensional representations). Below we will see a numberof examples in which this problem is partially or fully solved for specific algebras.We will now prove our first result – Schur’s lemma. Although it is very easy to prove, it isfundamental in the whole subject of representation theory. Proposition 1.16. (Schur’s lemma) Let V , V be representations of an algebra A over any field F (which need not be algebraically closed). Let φ : V → V be a nonzero homomorphism ofrepresentations. Then:(i) If V is irreducible, φ is injective; ii) If V is irreducible, φ is surjective.Thus, if both V and V are irreducible, φ is an isomorphism.Proof. (i) The kernel K of φ is a subrepresentation of V . Since φ = 0, this subrepresentationcannot be V . So by irreducibility of V we have K = 0.(ii) The image I of φ is a subrepresentation of V . Since φ = 0, this subrepresentation cannotbe 0. So by irreducibility of V we have I = V . Corollary 1.17. (Schur’s lemma for algebraically closed fields) Let V be a finite dimensionalirreducible representation of an algebra A over an algebraically closed field k , and φ : V → V is anintertwining operator. Then φ = λ · Id for some λ ∈ k (a scalar operator). Remark.
Note that this Corollary is false over the field of real numbers: it suffices to take A = C (regarded as an R -algebra), and V = A . Proof.
Let λ be an eigenvalue of φ (a root of the characteristic polynomial of φ ). It exists since k isan algebraically closed field. Then the operator φ − λ Id is an intertwining operator V → V , whichis not an isomorphism (since its determinant is zero). Thus by Proposition 1.16 this operator iszero, hence the result. Corollary 1.18.
Let A be a commutative algebra. Then every irreducible finite dimensional rep-resentation V of A is 1-dimensional. Remark.
Note that a 1-dimensional representation of any algebra is automatically irreducible.
Proof.
Let V be irreducible. For any element a ∈ A , the operator ρ ( a ) : V → V is an intertwiningoperator. Indeed, ρ ( a ) ρ ( b ) v = ρ ( ab ) v = ρ ( ba ) v = ρ ( b ) ρ ( a ) v (the second equality is true since the algebra is commutative). Thus, by Schur’s lemma, ρ ( a ) isa scalar operator for any a ∈ A . Hence every subspace of V is a subrepresentation. But V isirreducible, so 0 and V are the only subspaces of V . This means that dim V = 1 (since V = 0). Example 1.19. A = k . Since representations of A are simply vector spaces, V = A is the onlyirreducible and the only indecomposable representation.2. A = k [ x ]. Since this algebra is commutative, the irreducible representations of A are its1-dimensional representations. As we discussed above, they are defined by a single operator ρ ( x ).In the 1-dimensional case, this is just a number from k . So all the irreducible representations of A are V λ = k , λ ∈ k , in which the action of A defined by ρ ( x ) = λ . Clearly, these representations arepairwise non-isomorphic.The classification of indecomposable representations of k [ x ] is more interesting. To obtain it,recall that any linear operator on a finite dimensional vector space V can be brought to Jordannormal form. More specifically, recall that the Jordan block J λ,n is the operator on k n which inthe standard basis is given by the formulas J λ,n e i = λe i + e i − for i >
1, and J λ,n e = λe . Thenfor any linear operator B : V → V there exists a basis of V such that the matrix of B in this basisis a direct sum of Jordan blocks. This implies that all the indecomposable representations of A are V λ,n = k n , λ ∈ k , with ρ ( x ) = J λ,n . The fact that these representations are indecomposable andpairwise non-isomorphic follows from the Jordan normal form theorem (which in particular saysthat the Jordan normal form of an operator is unique up to permutation of blocks).9his example shows that an indecomposable representation of an algebra need not be irreducible.3. The group algebra A = k [ G ], where G is a group. A representation of A is the same thing asa representation of G , i.e., a vector space V together with a group homomorphism ρ : G → Aut( V ),whre Aut( V ) = GL ( V ) denotes the group of invertible linear maps from the space V to itself. Problem 1.20.
Let V be a nonzero finite dimensional representation of an algebra A . Show thatit has an irreducible subrepresentation. Then show by example that this does not always hold forinfinite dimensional representations. Problem 1.21.
Let A be an algebra over a field k . The center Z ( A ) of A is the set of all elements z ∈ A which commute with all elements of A . For example, if A is commutative then Z ( A ) = A .(a) Show that if V is an irreducible finite dimensional representation of A then any element z ∈ Z ( A ) acts in V by multiplication by some scalar χ V ( z ) . Show that χ V : Z ( A ) → k is ahomomorphism. It is called the central character of V .(b) Show that if V is an indecomposable finite dimensional representation of A then for any z ∈ Z ( A ) , the operator ρ ( z ) by which z acts in V has only one eigenvalue χ V ( z ) , equal to thescalar by which z acts on some irreducible subrepresentation of V . Thus χ V : Z ( A ) → k is ahomomorphism, which is again called the central character of V .(c) Does ρ ( z ) in (b) have to be a scalar operator? Problem 1.22.
Let A be an associative algebra, and V a representation of A . By End A ( V ) onedenotes the algebra of all homomorphisms of representations V → V . Show that End A ( A ) = A op ,the algebra A with opposite multiplication. Problem 1.23.
Prove the following “Infinite dimensional Schur’s lemma” (due to Dixmier): Let A be an algebra over C and V be an irreducible representation of A with at most countable basis.Then any homomorphism of representations φ : V → V is a scalar operator.Hint. By the usual Schur’s lemma, the algebra D := End A ( V ) is an algebra with division.Show that D is at most countably dimensional. Suppose φ is not a scalar, and consider the subfield C ( φ ) ⊂ D . Show that C ( φ ) is a transcendental extension of C . Derive from this that C ( φ ) isuncountably dimensional and obtain a contradiction. A left ideal of an algebra A is a subspace I ⊆ A such that aI ⊆ I for all a ∈ A . Similarly, a rightideal of an algebra A is a subspace I ⊆ A such that Ia ⊆ I for all a ∈ A . A two-sided ideal is asubspace that is both a left and a right ideal.Left ideals are the same as subrepresentations of the regular representation A . Right ideals arethe same as subrepresentations of the regular representation of the opposite algebra A op .Below are some examples of ideals: • If A is any algebra, 0 and A are two-sided ideals. An algebra A is called simple if 0 and A are its only two-sided ideals. • If φ : A → B is a homomorphism of algebras, then ker φ is a two-sided ideal of A . • If S is any subset of an algebra A , then the two-sided ideal generated by S is denoted h S i andis the span of elements of the form asb , where a, b ∈ A and s ∈ S . Similarly we can define h S i ℓ = span { as } and h S i r = span { sb } , the left, respectively right, ideal generated by S .10 .5 Quotients Let A be an algebra and I a two-sided ideal in A . Then A/I is the set of (additive) cosets of I .Let π : A → A/I be the quotient map. We can define multiplication in
A/I by π ( a ) · π ( b ) := π ( ab ) . This is well defined because if π ( a ) = π ( a ′ ) then π ( a ′ b ) = π ( ab + ( a ′ − a ) b ) = π ( ab ) + π (( a ′ − a ) b ) = π ( ab )because ( a ′ − a ) b ∈ Ib ⊆ I = ker π , as I is a right ideal; similarly, if π ( b ) = π ( b ′ ) then π ( ab ′ ) = π ( ab + a ( b ′ − b )) = π ( ab ) + π ( a ( b ′ − b )) = π ( ab )because a ( b ′ − b ) ∈ aI ⊆ I = ker π , as I is also a left ideal. Thus, A/I is an algebra.Similarly, if V is a representation of A , and W ⊂ V is a subrepresentation, then V /W is also arepresentation. Indeed, let π : V → V /W be the quotient map, and set ρ V/W ( a ) π ( x ) := π ( ρ V ( a ) x ).Above we noted that left ideals of A are subrepresentations of the regular representation of A ,and vice versa. Thus, if I is a left ideal in A , then A/I is a representation of A . Problem 1.24.
Let A = k [ x , ..., x n ] and I = A be any ideal in A containing all homogeneouspolynomials of degree ≥ N . Show that A/I is an indecomposable representation of A . Problem 1.25.
Let V = 0 be a representation of A . We say that a vector v ∈ V is cyclic if itgenerates V , i.e., Av = V . A representation admitting a cyclic vector is said to be cyclic. Showthat(a) V is irreducible if and only if all nonzero vectors of V are cyclic.(b) V is cyclic if and only if it is isomorphic to A/I , where I is a left ideal in A .(c) Give an example of an indecomposable representation which is not cyclic.Hint. Let A = C [ x, y ] /I , where I is the ideal spanned by homogeneous polynomials of degree ≥ (so A has a basis , x, y ). Let V = A ∗ be the space of linear functionals on A , with the actionof A given by ( ρ ( a ) f )( b ) = f ( ba ) . Show that V provides such an example. If f , . . . , f m are elements of the free algebra k h x , . . . , x n i , we say that the algebra A := k h x , . . . , x n i / h{ f , . . . , f m }i is generated by x , . . . , x n with defining relations f = 0 , . . . , f m =0.
1. The Weyl algebra, k h x, y i / h yx − xy − i .2. The q -Weyl algebra, generated by x, x − , y, y − with defining relations yx = qxy and xx − = x − x = yy − = y − y = 1. Proposition. (i) A basis for the Weyl algebra A is { x i y j , i, j ≥ } .(ii) A basis for the q-Weyl algebra A q is { x i y j , i, j ∈ Z } .11 roof. (i) First let us show that the elements x i y j are a spanning set for A . To do this, note thatany word in x, y can be ordered to have all the x on the left of the y , at the cost of interchangingsome x and y . Since yx − xy = 1, this will lead to error terms, but these terms will be sums ofmonomials that have a smaller number of letters x, y than the original word. Therefore, continuingthis process, we can order everything and represent any word as a linear combination of x i y j .The proof that x i y j are linearly independent is based on representation theory. Namely, let a bea variable, and E = t a k [ a ][ t, t − ] (here t a is just a formal symbol, so really E = k [ a ][ t, t − ]). Then E is a representation of A with action given by xf = tf and yf = dfdt (where d ( t a + n ) dt := ( a + n ) t a + n − ).Suppose now that we have a nontrivial linear relation P c ij x i y j = 0. Then the operator L = X c ij t i (cid:18) ddt (cid:19) j acts by zero in E . Let us write L as L = r X j =0 Q j ( t ) (cid:18) ddt (cid:19) j , where Q r = 0. Then we have Lt a = r X j =0 Q j ( t ) a ( a − ... ( a − j + 1) t a − j . This must be zero, so we have P rj =0 Q j ( t ) a ( a − ... ( a − j + 1) t − j = 0 in k [ a ][ t, t − ]. Taking theleading term in a , we get Q r ( t ) = 0, a contradiction.(ii) Any word in x, y, x − , y − can be ordered at the cost of multiplying it by a power of q . Thiseasily implies both the spanning property and the linear independence. Remark.
The proof of (i) shows that the Weyl algebra A can be viewed as the algebra ofpolynomial differential operators in one variable t .The proof of (i) also brings up the notion of a faithful representation. Definition.
A representation ρ : A → End V is faithful if ρ is injective.For example, k [ t ] is a faithful representation of the Weyl algebra, if k has characteristic zero(check it!), but not in characteristic p , where ( d/dt ) p Q = 0 for any polynomial Q . However, therepresentation E = t a k [ a ][ t, t − ], as we’ve seen, is faithful in any characteristic. Problem 1.26.
Let A be the Weyl algebra, generated by two elements x, y with the relation yx − xy − . (a) If char k = 0 , what are the finite dimensional representations of A ? What are the two-sidedideals in A ?Hint. For the first question, use the fact that for two square matrices B, C , Tr ( BC ) = Tr ( CB ) .For the second question, show that any nonzero two-sided ideal in A contains a nonzero polynomialin x , and use this to characterize this ideal.Suppose for the rest of the problem that char k = p .(b) What is the center of A ? int. Show that x p and y p are central elements.(c) Find all irreducible finite dimensional representations of A .Hint. Let V be an irreducible finite dimensional representation of A , and v be an eigenvectorof y in V . Show that { v, xv, x v, ..., x p − v } is a basis of V . Problem 1.27.
Let q be a nonzero complex number, and A be the q -Weyl algebra over C generatedby x ± and y ± with defining relations xx − = x − x = 1 , yy − = y − y = 1 , and xy = qyx .(a) What is the center of A for different q ? If q is not a root of unity, what are the two-sidedideals in A ?(b) For which q does this algebra have finite dimensional representations?Hint. Use determinants.(c) Find all finite dimensional irreducible representations of A for such q .Hint. This is similar to part (c) of the previous problem. Definition 1.28. A quiver Q is a directed graph, possibly with self-loops and/or multiple edgesbetween two vertices. Example 1.29. • / / • • o o • O O We denote the set of vertices of the quiver Q as I , and the set of edges as E . For an edge h ∈ E ,let h ′ , h ′′ denote the source and target of h , respectively: • h ′ h / / • h ′′ Definition 1.30.
A representation of a quiver Q is an assignment to each vertex i ∈ I of a vectorspace V i and to each edge h ∈ E of a linear map x h : V h ′ −→ V h ′′ .It turns out that the theory of representations of quivers is a part of the theory of representationsof algebras in the sense that for each quiver Q , there exists a certain algebra P Q , called the pathalgebra of Q , such that a representation of the quiver Q is “the same” as a representation of thealgebra P Q . We shall first define the path algebra of a quiver and then justify our claim thatrepresentations of these two objects are “the same”. Definition 1.31.
The path algebra P Q of a quiver Q is the algebra whose basis is formed byoriented paths in Q , including the trivial paths p i , i ∈ I , corresponding to the vertices of Q , andmultiplication is concatenation of paths: ab is the path obtained by first tracing b and then a . Iftwo paths cannot be concatenated, the product is defined to be zero. Remark 1.32.
It is easy to see that for a finite quiver P i ∈ I p i = 1, so P Q is an algebra with unit. Problem 1.33.
Show that the algebra P Q is generated by p i for i ∈ I and a h for h ∈ E with thedefining relations: . p i = p i , p i p j = 0 for i = j a h p h ′ = a h , a h p j = 0 for j = h ′ p h ′′ a h = a h , p i a h = 0 for i = h ′′ We now justify our statement that a representation of a quiver is the same thing as a represen-tation of the path algebra of a quiver.Let V be a representation of the path algebra P Q . From this representation, we can construct arepresentation of Q as follows: let V i = p i V , and for any edge h , let x h = a h | p h ′ V : p h ′ V −→ p h ′′ V be the operator corresponding to the one-edge path h .Similarly, let ( V i , x h ) be a representation of a quiver Q . From this representation, we canconstruct a representation of the path algebra P Q : let V = L i V i , let p i : V → V i → V be theprojection onto V i , and for any path p = h ...h m let a p = x h ...x h m : V h ′ m → V h ′′ be the compositionof the operators corresponding to the edges occurring in p (and the action of this operator on theother V i is zero).It is clear that the above assignments V ( p i V ) and ( V i ) L i V i are inverses of each other.Thus, we have a bijection between isomorphism classes of representations of the algebra P Q and ofthe quiver Q . Remark 1.34.
In practice, it is generally easier to consider a representation of a quiver as inDefinition 1.30.We lastly define several previous concepts in the context of quivers representations.
Definition 1.35.
A subrepresentation of a representation ( V i , x h ) of a quiver Q is a representation( W i , x ′ h ) where W i ⊆ V i for all i ∈ I and where x h ( W h ′ ) ⊆ W h ′′ and x ′ h = x h | W h ′ : W h ′ −→ W h ′′ forall h ∈ E . Definition 1.36.
The direct sum of two representations ( V i , x h ) and ( W i , y h ) is the representation( V i ⊕ W i , x h ⊕ y h ).As with representations of algebras, a nonzero representation ( V i ) of a quiver Q is said to beirreducible if its only subrepresentations are (0) and ( V i ) itself, and indecomposable if it is notisomorphic to a direct sum of two nonzero representations. Definition 1.37.
Let ( V i , x h ) and ( W i , y h ) be representations of the quiver Q . A homomorphism ϕ : ( V i ) −→ ( W i ) of quiver representations is a collection of maps ϕ i : V i −→ W i such that y h ◦ ϕ h ′ = ϕ h ′′ ◦ x h for all h ∈ E . Problem 1.38.
Let A be a Z + -graded algebra, i.e., A = ⊕ n ≥ A [ n ] , and A [ n ] · A [ m ] ⊂ A [ n + m ] .If A [ n ] is finite dimensional, it is useful to consider the Hilbert series h A ( t ) = P dim A [ n ] t n (thegenerating function of dimensions of A [ n ] ). Often this series converges to a rational function, andthe answer is written in the form of such function. For example, if A = k [ x ] and deg ( x n ) = n then h A ( t ) = 1 + t + t + ... + t n + ... = 11 − t Find the Hilbert series of:(a) A = k [ x , ..., x m ] (where the grading is by degree of polynomials); b) A = k < x , ..., x m > (the grading is by length of words);(c) A is the exterior (=Grassmann) algebra ∧ k [ x , ..., x m ] , generated over some field k by x , ..., x m with the defining relations x i x j + x j x i = 0 and x i = 0 for all i, j (the grading is bydegree).(d) A is the path algebra P Q of a quiver Q (the grading is defined by deg( p i ) = 0 , deg( a h ) = 1 ).Hint. The closed answer is written in terms of the adjacency matrix M Q of Q . Let g be a vector space over a field k , and let [ , ] : g × g −→ g be a skew-symmetric bilinear map.(That is, [ a, a ] = 0, and hence [ a, b ] = − [ b, a ]). Definition 1.39. ( g , [ , ]) is a Lie algebra if [ , ] satisfies the Jacobi identity (cid:2) [ a, b ] , c (cid:3) + (cid:2) [ b, c ] , a (cid:3) + (cid:2) [ c, a ] , b (cid:3) = 0 . (2) Example 1.40.
Some examples of Lie algebras are:1. Any space g with [ , ] = 0 (abelian Lie algebra).2. Any associative algebra A with [ a, b ] = ab − ba .3. Any subspace U of an associative algebra A such that [ a, b ] ∈ U for all a, b ∈ U .4. The space Der( A ) of derivations of an algebra A , i.e. linear maps D : A → A which satisfythe Leibniz rule: D ( ab ) = D ( a ) b + aD ( b ) . Remark 1.41.
Derivations are important because they are the “infinitesimal version” of automor-phisms (i.e., isomorphisms onto itself). For example, assume that g ( t ) is a differentiable family ofautomorphisms of a finite dimensional algebra A over R or C parametrized by t ∈ ( − ǫ, ǫ ) such that g (0) = Id. Then D := g ′ (0) : A → A is a derivation (check it!). Conversely, if D : A → A is aderivation, then e tD is a 1-parameter family of automorphisms (give a proof!).This provides a motivation for the notion of a Lie algebra. Namely, we see that Lie algebrasarise as spaces of infinitesimal automorphisms (=derivations) of associative algebras. In fact, theysimilarly arise as spaces of derivations of any kind of linear algebraic structures, such as Lie algebras,Hopf algebras, etc., and for this reason play a very important role in algebra.Here are a few more concrete examples of Lie algebras:1. R with [ u, v ] = u × v , the cross-product of u and v .2. sl ( n ), the set of n × n matrices with trace 0.For example, sl (2) has the basis e = (cid:18) (cid:19) f = (cid:18) (cid:19) h = (cid:18) − (cid:19) with relations [ h, e ] = 2 e, [ h, f ] = − f, [ e, f ] = h.
15. The Heisenberg Lie algebra H of matrices (cid:16) ∗ ∗ ∗ (cid:17) It has the basis x = y = c = with relations [ y, x ] = c and [ y, c ] = [ x, c ] = 0.4. The algebra aff(1) of matrices ( ∗ ∗ )Its basis consists of X = ( ) and Y = ( ), with [ X, Y ] = Y .5. so ( n ), the space of skew-symmetric n × n matrices, with [ a, b ] = ab − ba . Exercise.
Show that Example 1 is a special case of Example 5 (for n = 3). Definition 1.42.
Let g , g be Lie algebras. A homomorphism ϕ : g −→ g of Lie algebras is alinear map such that ϕ ([ a, b ]) = [ ϕ ( a ) , ϕ ( b )]. Definition 1.43.
A representation of a Lie algebra g is a vector space V with a homomorphismof Lie algebras ρ : g −→ End V . Example 1.44.
Some examples of representations of Lie algebras are:1. V = 0.2. Any vector space V with ρ = 0 (the trivial representation).3. The adjoint representation V = g with ρ ( a )( b ) := [ a, b ] . That this is a representation followsfrom Equation (2). Thus, the meaning of the Jacobi identity is that it is equivalent to theexistence of the adjoint representation.It turns out that a representation of a Lie algebra g is the same thing as a representation of acertain associative algebra U ( g ). Thus, as with quivers, we can view the theory of representationsof Lie algebras as a part of the theory of representations of associative algebras. Definition 1.45.
Let g be a Lie algebra with basis x i and [ , ] defined by [ x i , x j ] = P k c kij x k . The universal enveloping algebra U ( g ) is the associative algebra generated by the x i ’s with thedefining relations x i x j − x j x i = P k c kij x k . Remark.
This is not a very good definition since it depends on the choice of a basis. Later wewill give an equivalent definition which will be basis-independent.
Exercise.
Explain why a representation of a Lie algebra is the same thing as a representationof its universal enveloping algebra.
Example 1.46.
The associative algebra U ( sl (2)) is the algebra generated by e , f , h with relations he − eh = 2 e hf − f h = − f ef − f e = h. Example 1.47.
The algebra U ( H ), where H is the Heisenberg Lie algebra, is the algebra generatedby x , y , c with the relations yx − xy = c yc − cy = 0 xc − cx = 0 . Note that the Weyl algebra is the quotient of U ( H ) by the relation c = 1.16 .10 Tensor products In this subsection we recall the notion of tensor product of vector spaces, which will be extensivelyused below.
Definition 1.48.
The tensor product V ⊗ W of vector spaces V and W over a field k is the quotientof the space V ∗ W whose basis is given by formal symbols v ⊗ w , v ∈ V , w ∈ W , by the subspacespanned by the elements( v + v ) ⊗ w − v ⊗ w − v ⊗ w, v ⊗ ( w + w ) − v ⊗ w − v ⊗ w , av ⊗ w − a ( v ⊗ w ) , v ⊗ aw − a ( v ⊗ w ) , where v ∈ V, w ∈ W, a ∈ k . Exercise.
Show that V ⊗ W can be equivalently defined as the quotient of the free abeliangroup V • W generated by v ⊗ w , v ∈ V, w ∈ W by the subgroup generated by( v + v ) ⊗ w − v ⊗ w − v ⊗ w, v ⊗ ( w + w ) − v ⊗ w − v ⊗ w , av ⊗ w − v ⊗ aw, where v ∈ V, w ∈ W, a ∈ k .The elements v ⊗ w ∈ V ⊗ W , for v ∈ V, w ∈ W are called pure tensors. Note that in general,there are elements of V ⊗ W which are not pure tensors.This allows one to define the tensor product of any number of vector spaces, V ⊗ ... ⊗ V n . Notethat this tensor product is associative, in the sense that ( V ⊗ V ) ⊗ V can be naturally identifiedwith V ⊗ ( V ⊗ V ).In particular, people often consider tensor products of the form V ⊗ n = V ⊗ ... ⊗ V ( n times) fora given vector space V , and, more generally, E := V ⊗ n ⊗ ( V ∗ ) ⊗ m . This space is called the space oftensors of type ( m, n ) on V . For instance, tensors of type (0 ,
1) are vectors, of type (1 ,
0) - linearfunctionals (covectors), of type (1 ,
1) - linear operators, of type (2 ,
0) - bilinear forms, of type (2 , V is finite dimensional with basis e i , i = 1 , ..., N , and e i is the dual basis of V ∗ , then a basisof E is the set of vectors e i ⊗ ... ⊗ e i n ⊗ e j ⊗ ... ⊗ e j m , and a typical element of E is N X i ,...,i n ,j ,...,j m =1 T i ...i n j ...j m e i ⊗ ... ⊗ e i n ⊗ e j ⊗ ... ⊗ e j m , where T is a multidimensional table of numbers.Physicists define a tensor as a collection of such multidimensional tables T B attached to everybasis B in V , which change according to a certain rule when the basis B is changed. Here it isimportant to distinguish upper and lower indices, since lower indices of T correspond to V andupper ones to V ∗ . The physicists don’t write the sum sign, but remember that one should sumover indices that repeat twice - once as an upper index and once as lower. This convention iscalled the Einstein summation , and it also stipulates that if an index appears once, then there isno summation over it, while no index is supposed to appear more than once as an upper index ormore than once as a lower index.One can also define the tensor product of linear maps. Namely, if A : V → V ′ and B : W → W ′ are linear maps, then one can define the linear map A ⊗ B : V ⊗ W → V ′ ⊗ W ′ given by the formula( A ⊗ B )( v ⊗ w ) = Av ⊗ Bw (check that this is well defined!)17he most important properties of tensor products are summarized in the following problem. Problem 1.49. (a) Let U be any k -vector space. Construct a natural bijection between bilinearmaps V × W → U and linear maps V ⊗ W → U .(b) Show that if { v i } is a basis of V and { w j } is a basis of W then { v i ⊗ w j } is a basis of V ⊗ W .(c) Construct a natural isomorphism V ∗ ⊗ W → Hom(
V, W ) in the case when V is finitedimensional (“natural” means that the isomorphism is defined without choosing bases).(d) Let V be a vector space over a field k . Let S n V be the quotient of V ⊗ n ( n -fold tensor productof V ) by the subspace spanned by the tensors T − s ( T ) where T ∈ V ⊗ n , and s is some transposition.Also let ∧ n V be the quotient of V ⊗ n by the subspace spanned by the tensors T such that s ( T ) = T for some transposition s . These spaces are called the n-th symmetric, respectively exterior, powerof V . If { v i } is a basis of V , can you construct a basis of S n V, ∧ n V ? If dim V = m , what are theirdimensions?(e) If k has characteristic zero, find a natural identification of S n V with the space of T ∈ V ⊗ n such that T = sT for all transpositions s , and of ∧ n V with the space of T ∈ V ⊗ n such that T = − sT for all transpositions s .(f ) Let A : V → W be a linear operator. Then we have an operator A ⊗ n : V ⊗ n → W ⊗ n , andits symmetric and exterior powers S n A : S n V → S n W , ∧ n A : ∧ n V → ∧ n W which are defined inan obvious way. Suppose V = W and has dimension N , and assume that the eigenvalues of A are λ , ..., λ N . Find T r ( S n A ) , T r ( ∧ n A ) .(g) Show that ∧ N A = det ( A )Id , and use this equality to give a one-line proof of the fact that det( AB ) = det( A ) det( B ) . Remark.
Note that a similar definition to the above can be used to define the tensor product V ⊗ A W , where A is any ring, V is a right A -module, and W is a left A -module. Namely, V ⊗ A W is the abelian group which is the quotient of the group V • W freely generated by formal symbols v ⊗ w , v ∈ V , w ∈ W , modulo the relations( v + v ) ⊗ w − v ⊗ w − v ⊗ w, v ⊗ ( w + w ) − v ⊗ w − v ⊗ w , va ⊗ w − v ⊗ aw, a ∈ A. Exercise.
Throughout this exercise, we let k be an arbitrary field (not necessarily of charac-teristic zero, and not necessarily algebraically closed).If A and B are two k -algebras, then an ( A, B ) -bimodule will mean a k -vector space V withboth a left A -module structure and a right B -module structure which satisfy ( av ) b = a ( vb ) forany v ∈ V , a ∈ A and b ∈ B . Note that both the notions of ”left A -module” and ”right A -module” are particular cases of the notion of bimodules; namely, a left A -module is the same as an( A, k )-bimodule, and a right A -module is the same as a ( k, A )-bimodule.Let B be a k -algebra, W a left B -module and V a right B -module. We denote by V ⊗ B W the k -vector space ( V ⊗ k W ) / h vb ⊗ w − v ⊗ bw | v ∈ V, w ∈ W, b ∈ B i . We denote the projection ofa pure tensor v ⊗ w (with v ∈ V and w ∈ W ) onto the space V ⊗ B W by v ⊗ B w . (Note that thistensor product V ⊗ B W is the one defined in the Remark after Problem1.49.)If, additionally, A is another k -algebra, and if the right B -module structure on V is part of an( A, B )-bimodule structure, then V ⊗ B W becomes a left A -module by a ( v ⊗ B w ) = av ⊗ B w forany a ∈ A , v ∈ V and w ∈ W . 18imilarly, if C is another k -algebra, and if the left B -module structure on W is part of a ( B, C )-bimodule structure, then V ⊗ B W becomes a right C -module by ( v ⊗ B w ) c = v ⊗ B wc for any c ∈ C , v ∈ V and w ∈ W .If V is an ( A, B )-bimodule and W is a ( B, C )-bimodule, then these two structures on V ⊗ B W can be combined into one ( A, C )-bimodule structure on V ⊗ B W .(a) Let A , B , C , D be four algebras. Let V be an ( A, B )-bimodule, W be a ( B, C )-bimodule,and X a ( C, D )-bimodule. Prove that ( V ⊗ B W ) ⊗ C X ∼ = V ⊗ B ( W ⊗ C X ) as ( A, D )-bimodules.The isomorphism (from left to right) is given by ( v ⊗ B w ) ⊗ C x v ⊗ B ( w ⊗ C x ) for all v ∈ V , w ∈ W and x ∈ X .(b) If A , B , C are three algebras, and if V is an ( A, B )-bimodule and W an ( A, C )-bimodule,then the vector space Hom A ( V, W ) (the space of all left A -linear homomorphisms from V to W )canonically becomes a ( B, C )-bimodule by setting ( bf ) ( v ) = f ( vb ) for all b ∈ B , f ∈ Hom A ( V, W )and v ∈ V and ( f c ) ( v ) = f ( v ) c for all c ∈ C , f ∈ Hom A ( V, W ) and v ∈ V .Let A , B , C , D be four algebras. Let V be a ( B, A )-bimodule, W be a ( C, B )-bimodule, and X a( C, D )-bimodule. Prove that Hom B ( V, Hom C ( W, X )) ∼ = Hom C ( W ⊗ B V, X ) as (
A, D )-bimodules.The isomorphism (from left to right) is given by f ( w ⊗ B v f ( v ) w ) for all v ∈ V , w ∈ W and f ∈ Hom B ( V, Hom C ( W, X )).
The notion of tensor product allows us to give more conceptual (i.e., coordinate free) definitionsof the free algebra, polynomial algebra, exterior algebra, and universal enveloping algebra of a Liealgebra.Namely, given a vector space V , define its tensor algebra T V over a field k to be T V = ⊕ n ≥ V ⊗ n ,with multiplication defined by a · b := a ⊗ b , a ∈ V ⊗ n , b ∈ V ⊗ m . Observe that a choice of a basis x , ..., x N in V defines an isomorphism of T V with the free algebra k < x , ..., x n > .Also, one can make the following definition. Definition 1.50. (i) The symmetric algebra SV of V is the quotient of T V by the ideal generatedby v ⊗ w − w ⊗ v , v, w ∈ V .(ii) The exterior algebra ∧ V of V is the quotient of T V by the ideal generated by v ⊗ v , v ∈ V .(iii) If V is a Lie algebra, the universal enveloping algebra U ( V ) of V is the quotient of T V bythe ideal generated by v ⊗ w − w ⊗ v − [ v, w ], v, w ∈ V .It is easy to see that a choice of a basis x , ..., x N in V identifies SV with the polynomial algebra k [ x , ..., x N ], ∧ V with the exterior algebra ∧ k ( x , ..., x N ), and the universal enveloping algebra U ( V )with one defined previously.Also, it is easy to see that we have decompositions SV = ⊕ n ≥ S n V , ∧ V = ⊕ n ≥ ∧ n V . Problem 1.51.
It is known that if A and B are two polygons of the same area then A can be cutby finitely many straight cuts into pieces from which one can make B. David Hilbert asked in 1900whether it is true for polyhedra in 3 dimensions. In particular, is it true for a cube and a regulartetrahedron of the same volume? he answer is “no”, as was found by Dehn in 1901. The proof is very beautiful. Namely, toany polyhedron A let us attach its “Dehn invariant” D ( A ) in V = R ⊗ ( R / Q ) (the tensor productof Q -vector spaces). Namely, D ( A ) = X a l ( a ) ⊗ β ( a ) π , where a runs over edges of A , and l ( a ) , β ( a ) are the length of a and the angle at a .(a) Show that if you cut A into B and C by a straight cut, then D ( A ) = D ( B ) + D ( C ) .(b) Show that α = arccos(1 / /π is not a rational number.Hint. Assume that α = 2 m/n , for integers m, n . Deduce that roots of the equation x + x − = 2 / are roots of unity of degree n. Conclude that x k + x − k has denominator k and get a contradiction.(c) Using (a) and (b), show that the answer to Hilbert’s question is negative. (Compute theDehn invariant of the regular tetrahedron and the cube). Definition 1.52.
The tensor product of two representations
V, W of a Lie algebra g is the space V ⊗ W with ρ V ⊗ W ( x ) = ρ V ( x ) ⊗ Id + Id ⊗ ρ W ( x ). Definition 1.53.
The dual representation V ∗ to a representation V of a Lie algebra g is the dualspace V ∗ to V with ρ V ∗ ( x ) = − ρ V ( x ) ∗ .It is easy to check that these are indeed representations. Problem 1.54.
Let
V, W, U be finite dimensional representations of a Lie algebra g . Show thatthe space Hom g ( V ⊗ W, U ) is isomorphic to Hom g ( V, U ⊗ W ∗ ) . (Here Hom g := Hom U ( g ) ). sl (2) This subsection is devoted to the representation theory of sl (2), which is of central importance inmany areas of mathematics. It is useful to study this topic by solving the following sequence ofexercises, which every mathematician should do, in one form or another. Problem 1.55.
According to the above, a representation of sl (2) is just a vector space V with atriple of operators E, F, H such that HE − EH = 2 E, HF − F H = − F, EF − F E = H (thecorresponding map ρ is given by ρ ( e ) = E, ρ ( f ) = F , ρ ( h ) = H ).Let V be a finite dimensional representation of sl (2) (the ground field in this problem is C ).(a) Take eigenvalues of H and pick one with the biggest real part. Call it λ . Let ¯ V ( λ ) be thegeneralized eigenspace corresponding to λ . Show that E | ¯ V ( λ ) = 0 .(b) Let W be any representation of sl (2) and w ∈ W be a nonzero vector such that Ew = 0 .For any k > find a polynomial P k ( x ) of degree k such that E k F k w = P k ( H ) w . (First compute EF k w , then use induction in k ).(c) Let v ∈ ¯ V ( λ ) be a generalized eigenvector of H with eigenvalue λ . Show that there exists N > such that F N v = 0 .(d) Show that H is diagonalizable on ¯ V ( λ ) . (Take N to be such that F N = 0 on ¯ V ( λ ) , andcompute E N F N v , v ∈ ¯ V ( λ ) , by (b). Use the fact that P k ( x ) does not have multiple roots). e) Let N v be the smallest N satisfying (c). Show that λ = N v − .(f ) Show that for each N > , there exists a unique up to isomorphism irreducible representationof sl (2) of dimension N . Compute the matrices E, F, H in this representation using a convenientbasis. (For V finite dimensional irreducible take λ as in (a) and v ∈ V ( λ ) an eigenvector of H .Show that v, F v, ..., F λ v is a basis of V , and compute the matrices of the operators E, F, H in thisbasis.)Denote the λ + 1 -dimensional irreducible representation from (f ) by V λ . Below you will showthat any finite dimensional representation is a direct sum of V λ .(g) Show that the operator C = EF + F E + H / (the so-called Casimir operator) commuteswith E, F, H and equals λ ( λ +2)2 Id on V λ .Now it will be easy to prove the direct sum decomposition. Namely, assume the contrary, andlet V be a reducible representation of the smallest dimension, which is not a direct sum of smallerrepresentations.(h) Show that C has only one eigenvalue on V , namely λ ( λ +2)2 for some nonnegative integer λ .(use that the generalized eigenspace decomposition of C must be a decomposition of representations).(i) Show that V has a subrepresentation W = V λ such that V /W = nV λ for some n (use (h)and the fact that V is the smallest which cannot be decomposed).(j) Deduce from (i) that the eigenspace V ( λ ) of H is n + 1 -dimensional. If v , ..., v n +1 is itsbasis, show that F j v i , ≤ i ≤ n + 1 , ≤ j ≤ λ are linearly independent and therefore form a basisof V (establish that if F x = 0 and Hx = µx then Cx = µ ( µ − x and hence µ = − λ ).(k) Define W i = span ( v i , F v i , ..., F λ v i ) . Show that V i are subrepresentations of V and derive acontradiction with the fact that V cannot be decomposed.(l) (Jacobson-Morozov Lemma) Let V be a finite dimensional complex vector space and A : V → V a nilpotent operator. Show that there exists a unique, up to an isomorphism, representation of sl (2) on V such that E = A . (Use the classification of the representations and the Jordan normalform theorem)(m) (Clebsch-Gordan decomposition) Find the decomposition into irreducibles of the represen-tation V λ ⊗ V µ of sl (2) .Hint. For a finite dimensional representation V of sl (2) it is useful to introduce the character χ V ( x ) = T r ( e xH ) , x ∈ C . Show that χ V ⊕ W ( x ) = χ V ( x ) + χ W ( x ) and χ V ⊗ W ( x ) = χ V ( x ) χ W ( x ) .Then compute the character of V λ and of V λ ⊗ V µ and derive the decomposition. This decompositionis of fundamental importance in quantum mechanics.(n) Let V = C M ⊗ C N , and A = J M (0) ⊗ Id N + Id M ⊗ J N (0) , where J n (0) is the Jordan blockof size n with eigenvalue zero (i.e., J n (0) e i = e i − , i = 2 , ..., n , and J n (0) e = 0 ). Find the Jordannormal form of A using (l),(m). Problem 1.56. ( Lie’s Theorem ) The commutant K ( g ) of a Lie algebra g is the linear spanof elements [ x, y ] , x, y ∈ g . This is an ideal in g (i.e., it is a subrepresentation of the adjointrepresentation). A finite dimensional Lie algebra g over a field k is said to be solvable if thereexists n such that K n ( g ) = 0 . Prove the Lie theorem: if k = C and V is a finite dimensionalirreducible representation of a solvable Lie algebra g then V is 1-dimensional. int. Prove the result by induction in dimension. By the induction assumption, K ( g ) has acommon eigenvector v in V , that is there is a linear function χ : K ( g ) → C such that av = χ ( a ) v for any a ∈ K ( g ) . Show that g preserves common eigenspaces of K ( g ) (for this you will need toshow that χ ([ x, a ]) = 0 for x ∈ g and a ∈ K ( g ) . To prove this, consider the smallest vector subspace U containing v and invariant under x . This subspace is invariant under K ( g ) and any a ∈ K ( g ) acts with trace dim( U ) χ ( a ) in this subspace. In particular x, a ]) = dim( U ) χ ([ x, a ]) .). Problem 1.57.
Classify irreducible finite dimensional representations of the two dimensional Liealgebra with basis
X, Y and commutation relation [ X, Y ] = Y . Consider the cases of zero andpositive characteristic. Is the Lie theorem true in positive characteristic? Problem 1.58. (hard!) For any element x of a Lie algebra g let ad ( x ) denote the operator g → g , y [ x, y ] . Consider the Lie algebra g n generated by two elements x, y with the defining relations ad ( x ) ( y ) = ad ( y ) n +1 ( x ) = 0 .(a) Show that the Lie algebras g , g , g are finite dimensional and find their dimensions.(b) (harder!) Show that the Lie algebra g has infinite dimension. Construct explicitly a basisof this algebra. General results of representation theory
Let A be an algebra. Definition 2.1.
A semisimple (or completely reducible) representation of A is a direct sum ofirreducible representations. Example.
Let V be an irreducible representation of A of dimension n . Then Y = End( V ),with action of A by left multiplication, is a semisimple representation of A , isomorphic to nV (thedirect sum of n copies of V ). Indeed, any basis v , ..., v n of V gives rise to an isomorphism ofrepresentations End( V ) → nV , given by x → ( xv , ..., xv n ). Remark.
Note that by Schur’s lemma, any semisimple representation V of A is canonicallyidentified with ⊕ X Hom A ( X, V ) ⊗ X , where X runs over all irreducible representations of A . Indeed,we have a natural map f : ⊕ X Hom(
X, V ) ⊗ X → V , given by g ⊗ x → g ( x ), x ∈ X , g ∈ Hom(
X, V ),and it is easy to verify that this map is an isomorphism.We’ll see now how Schur’s lemma allows us to classify subrepresentations in finite dimensionalsemisimple representations.
Proposition 2.2.
Let V i , ≤ i ≤ m be irreducible finite dimensional pairwise nonisomorphicrepresentations of A , and W be a subrepresentation of V = ⊕ mi =1 n i V i . Then W is isomorphic to ⊕ mi =1 r i V i , r i ≤ n i , and the inclusion φ : W → V is a direct sum of inclusions φ i : r i V i → n i V i givenby multiplication of a row vector of elements of V i (of length r i ) by a certain r i -by- n i matrix X i with linearly independent rows: φ ( v , ..., v r i ) = ( v , ..., v r i ) X i .Proof. The proof is by induction in n := P mi =1 n i . The base of induction ( n = 1) is clear. To performthe induction step, let us assume that W is nonzero, and fix an irreducible subrepresentation P ⊂ W . Such P exists (Problem 1.20). Now, by Schur’s lemma, P is isomorphic to V i for some i ,and the inclusion φ | P : P → V factors through n i V i , and upon identification of P with V i is givenby the formula v ( vq , ..., vq n i ), where q l ∈ k are not all zero.Now note that the group G i = GL n i ( k ) of invertible n i -by- n i matrices over k acts on n i V i by ( v , ..., v n i ) → ( v , ..., v n i ) g i (and by the identity on n j V j , j = i ), and therefore acts on theset of subrepresentations of V , preserving the property we need to establish: namely, under theaction of g i , the matrix X i goes to X i g i , while X j , j = i don’t change. Take g i ∈ G i such that( q , ..., q n i ) g i = (1 , , ..., W g i contains the first summand V i of n i V i (namely, it is P g i ),hence W g i = V i ⊕ W ′ , where W ′ ⊂ n V ⊕ ... ⊕ ( n i − V i ⊕ ... ⊕ n m V m is the kernel of the projectionof W g i to the first summand V i along the other summands. Thus the required statement followsfrom the induction assumption. Remark 2.3.
In Proposition 2.2, it is not important that k is algebraically closed, nor it mattersthat V is finite dimensional. If these assumptions are dropped, the only change needed is that theentries of the matrix X i are no longer in k but in D i = End A ( V i ), which is, as we know, a divisionalgebra. The proof of this generalized version of Proposition 2.2 is the same as before (check it!). Another proof of the existence of P , which does not use the finite dimensionality of V , is by induction in n .Namely, if W itself is not irreducible, let K be the kernel of the projection of W to the first summand V . Then K is a subrepresentation of ( n − V ⊕ ... ⊕ n m V m , which is nonzero since W is not irreducible, so K contains anirreducible subrepresentation by the induction assumption. .2 The density theorem Let A be an algebra over an algebraically closed field k . Corollary 2.4.
Let V be an irreducible finite dimensional representation of A , and v , ..., v n ∈ V be any linearly independent vectors. Then for any w , ..., w n ∈ V there exists an element a ∈ A such that av i = w i .Proof. Assume the contrary. Then the image of the map A → nV given by a → ( av , ..., av n ) is aproper subrepresentation, so by Proposition 2.2 it corresponds to an r -by- n matrix X , r < n . Thus,taking a = 1, we see that there exist vectors u , ..., u r ∈ V such that ( u , ..., u r ) X = ( v , ..., v n ). Let( q , ..., q n ) be a nonzero vector such that X ( q , ..., q n ) T = 0 (it exists because r < n ). Then P q i v i =( u , ..., u r ) X ( q , ..., q n ) T = 0, i.e. P q i v i = 0 - a contradiction with the linear independence of v i . Theorem 2.5. (the Density Theorem). (i) Let V be an irreducible finite dimensional representationof A . Then the map ρ : A → End V is surjective.(ii) Let V = V ⊕ ... ⊕ V r , where V i are irreducible pairwise nonisomorphic finite dimensionalrepresentations of A . Then the map ⊕ ri =1 ρ i : A → ⊕ ri =1 End( V i ) is surjective.Proof. (i) Let B be the image of A in End( V ). We want to show that B = End( V ). Let c ∈ End( V ), v , ..., v n be a basis of V , and w i = cv i . By Corollary 2.4, there exists a ∈ A such that av i = w i .Then a maps to c , so c ∈ B , and we are done.(ii) Let B i be the image of A in End( V i ), and B be the image of A in ⊕ ri =1 End( V i ). Recall that asa representation of A , ⊕ ri =1 End( V i ) is semisimple: it is isomorphic to ⊕ ri =1 d i V i , where d i = dim V i .Then by Proposition 2.2, B = ⊕ i B i . On the other hand, (i) implies that B i = End( V i ). Thus (ii)follows. In this section we consider representations of algebras A = L i Mat d i ( k ) for any field k . Theorem 2.6.
Let A = L ri =1 Mat d i ( k ) . Then the irreducible representations of A are V = k d , . . . , V r = k d r , and any finite dimensional representation of A is a direct sum of copies of V , . . . , V r . In order to prove Theorem 2.6, we shall need the notion of a dual representation.
Definition 2.7. (Dual representation) Let V be a representation of any algebra A . Then thedual representation V ∗ is the representation of the opposite algebra A op (or, equivalently, right A -module) with the action ( f · a )( v ) := f ( av ) . Proof of Theorem 2.6.
First, the given representations are clearly irreducible, as for any v = 0 , w ∈ V i , there exists a ∈ A such that av = w . Next, let X be an n -dimensional representation of A . Then, X ∗ is an n -dimensional representation of A op . But (Mat d i ( k )) op ∼ = Mat d i ( k ) withisomorphism ϕ ( X ) = X T , as ( BC ) T = C T B T . Thus, A ∼ = A op and X ∗ may be viewed as an n -dimensional representation of A . Define φ : A ⊕ · · · ⊕ A | {z } n copies −→ X ∗ φ ( a , . . . , a n ) = a y + · · · + a n y n where { y i } is a basis of X ∗ . φ is clearly surjective, as k ⊂ A . Thus, the dual map φ ∗ : X −→ A n ∗ is injective. But A n ∗ ∼ = A n as representations of A (check it!). Hence, Im φ ∗ ∼ = X is a subrepresen-tation of A n . Next, Mat d i ( k ) = d i V i , so A = ⊕ ri =1 d i V i , A n = ⊕ ri =1 nd i V i , as a representation of A .Hence by Proposition 2.2, X = ⊕ ri =1 m i V i , as desired. Exercise.
The goal of this exercise is to give an alternative proof of Theorem 2.6, not usingany of the previous results of Chapter 2.Let A , A , ... , A n be n algebras with units 1 , 1 , ... , 1 n , respectively. Let A = A ⊕ A ⊕ ... ⊕ A n .Clearly, 1 i j = δ ij i , and the unit of A is 1 = 1 + 1 + ... + 1 n .For every representation V of A , it is easy to see that 1 i V is a representation of A i for every i ∈ { , , ..., n } . Conversely, if V , V , ... , V n are representations of A , A , ... , A n , respectively,then V ⊕ V ⊕ ... ⊕ V n canonically becomes a representation of A (with ( a , a , ..., a n ) ∈ A actingon V ⊕ V ⊕ ... ⊕ V n as ( v , v , ..., v n ) ( a v , a v , ..., a n v n )). (a) Show that a representation V of A is irreducible if and only if 1 i V is an irreducible repre-sentation of A i for exactly one i ∈ { , , ..., n } , while 1 i V = 0 for all the other i . Thus, classify theirreducible representations of A in terms of those of A , A , ... , A n . (b) Let d ∈ N . Show that the only irreducible representation of Mat d ( k ) is k d , and every finitedimensional representation of Mat d ( k ) is a direct sum of copies of k d . Hint:
For every ( i, j ) ∈ { , , ..., d } , let E ij ∈ Mat d ( k ) be the matrix with 1 in the i th row of the j th column and 0’s everywhere else. Let V be a finite dimensional representation of Mat d ( k ). Showthat V = E V ⊕ E V ⊕ ... ⊕ E dd V , and that Φ i : E V → E ii V , v E i v is an isomorphism forevery i ∈ { , , ..., d } . For every v ∈ E V , denote S ( v ) = h E v, E v, ..., E d v i . Prove that S ( v )is a subrepresentation of V isomorphic to k d (as a representation of Mat d ( k )), and that v ∈ S ( v ).Conclude that V = S ( v ) ⊕ S ( v ) ⊕ ... ⊕ S ( v k ), where { v , v , ..., v k } is a basis of E V . (c) Conclude Theorem 2.6.
Let A be an algebra. Let V be a representation of A . A (finite) filtration of V is a sequence ofsubrepresentations 0 = V ⊂ V ⊂ ... ⊂ V n = V . Lemma 2.8.
Any finite dimensional representation V of an algebra A admits a finite filtration V ⊂ V ⊂ ... ⊂ V n = V such that the successive quotients V i /V i − are irreducible.Proof. The proof is by induction in dim( V ). The base is clear, and only the induction step needsto be justified. Pick an irreducible subrepresentation V ⊂ V , and consider the representation U = V /V . Then by the induction assumption U has a filtration 0 = U ⊂ U ⊂ ... ⊂ U n − = U such that U i /U i − are irreducible. Define V i for i ≥ U i − under thetautological projection V → V /V = U . Then 0 = V ⊂ V ⊂ V ⊂ ... ⊂ V n = V is a filtration of V with the desired property. 25 .5 Finite dimensional algebras Definition 2.9.
The radical of a finite dimensional algebra A is the set of all elements of A whichact by 0 in all irreducible representations of A . It is denoted Rad( A ). Proposition 2.10.
Rad ( A ) is a two-sided ideal.Proof. Easy.
Proposition 2.11.
Let A be a finite dimensional algebra.(i) Let I be a nilpotent two-sided ideal in A , i.e., I n = 0 for some n . Then I ⊂ Rad ( A ) .(ii) Rad ( A ) is a nilpotent ideal. Thus, Rad ( A ) is the largest nilpotent two-sided ideal in A .Proof. (i) Let V be an irreducible representation of A . Let v ∈ V . Then Iv ⊂ V is a subrepresen-tation. If Iv = 0 then Iv = V so there is x ∈ I such that xv = v . Then x n = 0, a contradiction.Thus Iv = 0, so I acts by 0 in V and hence I ⊂ Rad( A ).(ii) Let 0 = A ⊂ A ⊂ ... ⊂ A n = A be a filtration of the regular representation of A bysubrepresentations such that A i +1 /A i are irreducible. It exists by Lemma 2.8. Let x ∈ Rad( A ).Then x acts on A i +1 /A i by zero, so x maps A i +1 to A i . This implies that Rad( A ) n = 0, asdesired. Theorem 2.12.
A finite dimensional algebra A has only finitely many irreducible representations V i up to isomorphism, these representations are finite dimensional, and A/ Rad ( A ) ∼ = M i End V i . Proof.
First, for any irreducible representation V of A , and for any nonzero v ∈ V , Av ⊆ V is afinite dimensional subrepresentation of V . (It is finite dimensional as A is finite dimensional.) As V is irreducible and Av = 0, V = Av and V is finite dimensional.Next, suppose we have non-isomorphic irreducible representations V , V , . . . , V r . By Theorem2.5, the homomorphism M i ρ i : A −→ M i End V i is surjective. So r ≤ P i dim End V i ≤ dim A . Thus, A has only finitely many non-isomorphicirreducible representations (at most dim A ).Now, let V , V , . . . , V r be all non-isomorphic irreducible finite dimensional representations of A . By Theorem 2.5, the homomorphism M i ρ i : A −→ M i End V i is surjective. The kernel of this map, by definition, is exactly Rad( A ). Corollary 2.13. P i (dim V i ) ≤ dim A , where the V i ’s are the irreducible representations of A .Proof. As dim End V i = (dim V i ) , Theorem 2.12 implies that dim A − dim Rad( A ) = P i dim End V i = P i (dim V i ) . As dim Rad( A ) ≥ P i (dim V i ) ≤ dim A .26 xample 2.14.
1. Let A = k [ x ] / ( x n ). This algebra has a unique irreducible representation, whichis a 1-dimensional space k , in which x acts by zero. So the radical Rad( A ) is the ideal ( x ).2. Let A be the algebra of upper triangular n by n matrices. It is easy to check that theirreducible representations of A are V i , i = 1 , ..., n , which are 1-dimensional, and any matrix x actsby x ii . So the radical Rad( A ) is the ideal of strictly upper triangular matrices (as it is a nilpotentideal and contains the radical). A similar result holds for block-triangular matrices. Definition 2.15.
A finite dimensional algebra A is said to be semisimple if Rad( A ) = 0. Proposition 2.16.
For a finite dimensional algebra A , the following are equivalent:1. A is semisimple.2. P i (dim V i ) = dim A , where the V i ’s are the irreducible representations of A .3. A ∼ = L i Mat d i ( k ) for some d i .4. Any finite dimensional representation of A is completely reducible (that is, isomorphic to adirect sum of irreducible representations).5. A is a completely reducible representation of A .Proof. As dim A − dim Rad( A ) = P i (dim V i ) , clearly dim A = P i (dim V i ) if and only if Rad( A ) =0. Thus, (1) ⇔ (2).Next, by Theorem 2.12, if Rad( A ) = 0, then clearly A ∼ = L i Mat d i ( k ) for d i = dim V i . Thus,(1) ⇒ (3). Conversely, if A ∼ = L i Mat d i ( k ), then by Theorem 2.6, Rad( A ) = 0, so A is semisimple.Thus (3) ⇒ (1).Next, (3) ⇒ (4) by Theorem 2.6. Clearly (4) ⇒ (5). To see that (5) ⇒ (3), let A = L i n i V i .Consider End A ( A ) (endomorphisms of A as a representation of A ). As the V i ’s are pairwise non-isomorphic, by Schur’s lemma, no copy of V i in A can be mapped to a distinct V j . Also, again bySchur’s lemma, End A ( V i ) = k . Thus, End A ( A ) ∼ = L i Mat n i ( k ). But End A ( A ) ∼ = A op by Problem1.22, so A op ∼ = L i Mat n i ( k ). Thus, A ∼ = ( L i Mat n i ( k )) op = L i Mat n i ( k ), as desired. Let A be an algebra and V a finite-dimensional representation of A with action ρ . Then the character of V is the linear function χ V : A → k given by χ V ( a ) = tr | V ( ρ ( a )) . If [
A, A ] is the span of commutators [ x, y ] := xy − yx over all x, y ∈ A , then [ A, A ] ⊆ ker χ V . Thus,we may view the character as a mapping χ V : A/ [ A, A ] → k . Exercise.
Show that if W ⊂ V are finite dimensional representations of A , then χ V = χ W + χ V/W . Theorem 2.17. (i) Characters of (distinct) irreducible finite-dimensional representations of A arelinearly independent.(ii) If A is a finite-dimensional semisimple algebra, then these characters form a basis of ( A/ [ A, A ]) ∗ . roof. (i) If V , . . . , V r are nonisomorphic irreducible finite-dimensional representations of A , then ρ V ⊕ · · · ⊕ ρ V r : A → End V ⊕ · · · ⊕ End V r is surjective by the density theorem, so χ V , . . . , χ V r arelinearly independent. (Indeed, if P λ i χ V i ( a ) = 0 for all a ∈ A , then P λ i Tr( M i ) = 0 for all M i ∈ End k V i . But each tr( M i ) can range independently over k , so it must be that λ = · · · = λ r = 0.)(ii) First we prove that [Mat d ( k ) , Mat d ( k )] = sl d ( k ), the set of all matrices with trace 0. It isclear that [Mat d ( k ) , Mat d ( k )] ⊆ sl d ( k ). If we denote by E ij the matrix with 1 in the i th row of the j th column and 0’s everywhere else, we have [ E ij , E jm ] = E im for i = m , and [ E i,i +1 , E i +1 ,i ] = E ii − E i +1 ,i +1 . Now { E im }∪{ E ii − E i +1 ,i +1 } forms a basis in sl d ( k ), so indeed [Mat d ( k ) , Mat d ( k )] = sl d ( k ),as claimed.By semisimplicity, we can write A = Mat d ( k ) ⊕ · · · ⊕ Mat d r ( k ) . Then [
A, A ] = sl d ( k ) ⊕ · · · ⊕ sl d r ( k ), and A/ [ A, A ] ∼ = k r . By Theorem 2.6, there are exactly r irreducible representations of A (isomorphic to k d , . . . , k d r , respectively), and therefore r linearly independent characters on the r -dimensional vector space A/ [ A, A ]. Thus, the characters form a basis.
We will now state and prove two important theorems about representations of finite dimensionalalgebras - the Jordan-H¨older theorem and the Krull-Schmidt theorem.
Theorem 2.18. (Jordan-H¨older theorem). Let V be a finite dimensional representation of A ,and V ⊂ V ⊂ ... ⊂ V n = V , V ′ ⊂ ... ⊂ V ′ m = V be filtrations of V , such that therepresentations W i := V i /V i − and W ′ i := V ′ i /V ′ i − are irreducible for all i . Then n = m , and thereexists a permutation σ of , ..., n such that W σ ( i ) is isomorphic to W ′ i .Proof. First proof (for k of characteristic zero). The character of V obviously equals the sumof characters of W i , and also the sum of characters of W ′ i . But by Theorem 2.17, the charac-ters of irreducible representations are linearly independent, so the multiplicity of every irreduciblerepresentation W of A among W i and among W ′ i are the same. This implies the theorem. Second proof (general). The proof is by induction on dim V . The base of induction is clear,so let us prove the induction step. If W = W ′ (as subspaces), we are done, since by the inductionassumption the theorem holds for V /W . So assume W = W ′ . In this case W ∩ W ′ = 0 (as W , W ′ are irreducible), so we have an embedding f : W ⊕ W ′ → V . Let U = V / ( W ⊕ W ′ ), and0 = U ⊂ U ⊂ ... ⊂ U p = U be a filtration of U with simple quotients Z i = U i /U i − (it exists byLemma 2.8). Then we see that:1) V /W has a filtration with successive quotients W ′ , Z , ..., Z p , and another filtration withsuccessive quotients W , ...., W n .2) V /W ′ has a filtration with successive quotients W , Z , ..., Z p , and another filtration withsuccessive quotients W ′ , ...., W ′ n .By the induction assumption, this means that the collection of irreducible representations withmultiplicities W , W ′ , Z , ..., Z p coincides on one hand with W , ..., W n , and on the other hand, with W ′ , ..., W ′ m . We are done.The Jordan-H¨older theorem shows that the number n of terms in a filtration of V with irre-ducible successive quotients does not depend on the choice of a filtration, and depends only on This proof does not work in characteristic p because it only implies that the multiplicities of W i and W ′ i are thesame modulo p , which is not sufficient. In fact, the character of the representation pV , where V is any representation,is zero. . This number is called the length of V . It is easy to see that n is also the maximal length of afiltration of V in which all the inclusions are strict.The sequence of the irreducible representations W , ..., W n enumerated in the order they appearfrom some filtration of V as successive quoteints is called a Jordan-H¨older series of V . Theorem 2.19. (Krull-Schmidt theorem) Any finite dimensional representation of A can be uniquely(up to an isomorphism and order of summands) decomposed into a direct sum of indecomposablerepresentations.Proof. It is clear that a decomposition of V into a direct sum of indecomposable representationsexists, so we just need to prove uniqueness. We will prove it by induction on dim V . Let V = V ⊕ ... ⊕ V m = V ′ ⊕ ... ⊕ V ′ n . Let i s : V s → V , i ′ s : V ′ s → V , p s : V → V s , p ′ s : V → V ′ s be the naturalmaps associated to these decompositions. Let θ s = p i ′ s p ′ s i : V → V . We have P ns =1 θ s = 1. Nowwe need the following lemma. Lemma 2.20.
Let W be a finite dimensional indecomposable representation of A . Then(i) Any homomorphism θ : W → W is either an isomorphism or nilpotent;(ii) If θ s : W → W , s = 1 , ..., n are nilpotent homomorphisms, then so is θ := θ + ... + θ n .Proof. (i) Generalized eigenspaces of θ are subrepresentations of W , and W is their direct sum.Thus, θ can have only one eigenvalue λ . If λ is zero, θ is nilpotent, otherwise it is an isomorphism.(ii) The proof is by induction in n . The base is clear. To make the induction step ( n − n ),assume that θ is not nilpotent. Then by (i) θ is an isomorphism, so P ni =1 θ − θ i = 1. The morphisms θ − θ i are not isomorphisms, so they are nilpotent. Thus 1 − θ − θ n = θ − θ + ... + θ − θ n − is anisomorphism, which is a contradiction with the induction assumption.By the lemma, we find that for some s , θ s must be an isomorphism; we may assume that s = 1. In this case, V ′ = Im( p ′ i ) ⊕ Ker( p i ′ ), so since V ′ is indecomposable, we get that f := p ′ i : V → V ′ and g := p i ′ : V ′ → V are isomorphisms.Let B = ⊕ j> V j , B ′ = ⊕ j> V ′ j ; then we have V = V ⊕ B = V ′ ⊕ B ′ . Consider the map h : B → B ′ defined as a composition of the natural maps B → V → B ′ attached to thesedecompositions. We claim that h is an isomorphism. To show this, it suffices to show that Ker h = 0(as h is a map between spaces of the same dimension). Assume that v ∈ Ker h ⊂ B . Then v ∈ V ′ .On the other hand, the projection of v to V is zero, so gv = 0. Since g is an isomorphism, we get v = 0, as desired.Now by the induction assumption, m = n , and V j ∼ = V ′ σ ( j ) for some permutation σ of 2 , ..., n .The theorem is proved. Exercise.
Let A be the algebra of real-valued continuous functions on R which are periodicwith period 1. Let M be the A -module of continuous functions f on R which are antiperiodic withperiod 1, i.e., f ( x + 1) = − f ( x ).(i) Show that A and M are indecomposable A -modules.(ii) Show that A is not isomorphic to M but A ⊕ A is isomorphic to M ⊕ M .29 emark. Thus, we see that in general, the Krull-Schmidt theorem fails for infinite dimensionalmodules. However, it still holds for modules of finite length , i.e., modules M such that any filtrationof M has length bounded above by a certain constant l = l ( M ). Problem 2.21. Extensions of representations.
Let A be an algebra, and V, W be a pair ofrepresentations of A . We would like to classify representations U of A such that V is a subrepre-sentation of U , and U/V = W . Of course, there is an obvious example U = V ⊕ W , but are thereany others?Suppose we have a representation U as above. As a vector space, it can be (non-uniquely)identified with V ⊕ W , so that for any a ∈ A the corresponding operator ρ U ( a ) has block triangularform ρ U ( a ) = (cid:18) ρ V ( a ) f ( a )0 ρ W ( a ) (cid:19) , where f : A → Hom k ( W, V ) is a linear map.(a) What is the necessary and sufficient condition on f ( a ) under which ρ U ( a ) is a repre-sentation? Maps f satisfying this condition are called (1-)cocycles (of A with coefficients in Hom k ( W, V ) ). They form a vector space denoted Z ( W, V ) .(b) Let X : W → V be a linear map. The coboundary of X , dX , is defined to be the function A → Hom k ( W, V ) given by dX ( a ) = ρ V ( a ) X − Xρ W ( a ) . Show that dX is a cocycle, which vanishes if andonly if X is a homomorphism of representations. Thus coboundaries form a subspace B ( W, V ) ⊂ Z ( W, V ) , which is isomorphic to Hom k ( W, V ) / Hom A ( W, V ) . The quotient Z ( W, V ) /B ( W, V ) isdenoted Ext ( W, V ) .(c) Show that if f, f ′ ∈ Z ( W, V ) and f − f ′ ∈ B ( W, V ) then the corresponding extensions U, U ′ are isomorphic representations of A . Conversely, if φ : U → U ′ is an isomorphism such that φ ( a ) = (cid:18) V ∗ W (cid:19) then f − f ′ ∈ B ( V, W ) . Thus, the space Ext ( W, V ) “classifies” extensions of W by V .(d) Assume that W, V are finite dimensional irreducible representations of A . For any f ∈ Ext ( W, V ) , let U f be the corresponding extension. Show that U f is isomorphic to U f ′ as repre-sentations if and only if f and f ′ are proportional. Thus isomorphism classes (as representations)of nontrivial extensions of W by V (i.e., those not isomorphic to W ⊕ V ) are parametrized by theprojective space P Ext ( W, V ) . In particular, every extension is trivial if and only if Ext ( W, V ) = 0 . Problem 2.22. (a) Let A = C [ x , ..., x n ] , and V a , V b be one-dimensional representations in which x i act by a i and b i , respectively ( a i , b i ∈ C ). Find Ext ( V a , V b ) and classify 2-dimensional repre-sentations of A .(b) Let B be the algebra over C generated by x , ..., x n with the defining relations x i x j = 0 forall i, j . Show that for n > the algebra B has infinitely many non-isomorphic indecomposablerepresentations. Problem 2.23.
Let Q be a quiver without oriented cycles, and P Q the path algebra of Q . Findirreducible representations of P Q and compute Ext between them. Classify 2-dimensional repre-sentations of P Q . roblem 2.24. Let A be an algebra, and V a representation of A . Let ρ : A → End V . A formaldeformation of V is a formal series ˜ ρ = ρ + tρ + ... + t n ρ n + ..., where ρ i : A → End( V ) are linear maps, ρ = ρ , and ˜ ρ ( ab ) = ˜ ρ ( a )˜ ρ ( b ) .If b ( t ) = 1 + b t + b t + ... , where b i ∈ End( V ) , and ˜ ρ is a formal deformation of ρ , then b ˜ ρb − is also a deformation of ρ , which is said to be isomorphic to ˜ ρ .(a) Show that if Ext ( V, V ) = 0 , then any deformation of ρ is trivial, i.e., isomorphic to ρ .(b) Is the converse to (a) true? (consider the algebra of dual numbers A = k [ x ] /x ). Problem 2.25. The Clifford algebra.
Let V be a finite dimensional complex vector spaceequipped with a symmetric bilinear form ( , ) . The Clifford algebra Cl( V ) is the quotient of thetensor algebra T V by the ideal generated by the elements v ⊗ v − ( v, v )1 , v ∈ V . More explicitly, if x i , ≤ i ≤ N is a basis of V and ( x i , x j ) = a ij then Cl( V ) is generated by x i with defining relations x i x j + x j x i = 2 a ij , x i = a ii . Thus, if ( , ) = 0 , Cl( V ) = ∧ V .(i) Show that if ( , ) is nondegenerate then Cl( V ) is semisimple, and has one irreducible repre-sentation of dimension n if dim V = 2 n (so in this case Cl( V ) is a matrix algebra), and two suchrepresentations if dim( V ) = 2 n + 1 (i.e., in this case Cl( V ) is a direct sum of two matrix algebras).Hint. In the even case, pick a basis a , ..., a n , b , ..., b n of V in which ( a i , a j ) = ( b i , b j ) = 0 , ( a i , b j ) = δ ij / , and construct a representation of Cl( V ) on S := ∧ ( a , ..., a n ) in which b i acts as“differentiation” with respect to a i . Show that S is irreducible. In the odd case the situation issimilar, except there should be an additional basis vector c such that ( c, a i ) = ( c, b i ) = 0 , ( c, c ) =1 , and the action of c on S may be defined either by ( − degree or by ( − degree+1 , giving tworepresentations S + , S − (why are they non-isomorphic?). Show that there is no other irreduciblerepresentations by finding a spanning set of Cl( V ) with dim V elements.(ii) Show that Cl( V ) is semisimple if and only if ( , ) is nondegenerate. If ( , ) is degenerate, whatis Cl( V ) / Rad (Cl( V )) ? Let
A, B be algebras. Then A ⊗ B is also an algebra, with multiplication ( a ⊗ b )( a ⊗ b ) = a a ⊗ b b . Exercise.
Show that Mat m ( k ) ⊗ Mat n ( k ) ∼ = Mat mn ( k ).The following theorem describes irreducible finite dimensional representations of A ⊗ B in termsof irreducible finite dimensional representations of A and those of B . Theorem 2.26. (i) Let V be an irreducible finite dimensional representation of A and W anirreducible finite dimensional representation of B . Then V ⊗ W is an irreducible representation of A ⊗ B .(ii) Any irreducible finite dimensional representation M of A ⊗ B has the form (i) for unique V and W . Remark 2.27.
Part (ii) of the theorem typically fails for infinite dimensional representations;e.g. it fails when A is the Weyl algebra in characteristic zero. Part (i) also may fail. E.g. let A = B = V = W = C ( x ). Then (i) fails, as A ⊗ B is not a field.31 roof. (i) By the density theorem, the maps A → End V and B → End W are surjective. Therefore,the map A ⊗ B → End V ⊗ End W = End( V ⊗ W ) is surjective. Thus, V ⊗ W is irreducible.(ii) First we show the existence of V and W . Let A ′ , B ′ be the images of A, B in End M . Then A ′ , B ′ are finite dimensional algebras, and M is a representation of A ′ ⊗ B ′ , so we may assumewithout loss of generality that A and B are finite dimensional.In this case, we claim that Rad( A ⊗ B ) = Rad( A ) ⊗ B + A ⊗ Rad( B ). Indeed, denote the latterby J . Then J is a nilpotent ideal in A ⊗ B , as Rad( A ) and Rad( B ) are nilpotent. On the otherhand, ( A ⊗ B ) /J = ( A/ Rad( A )) ⊗ ( B/ Rad( B )), which is a product of two semisimple algebras,hence semisimple. This implies J ⊃ Rad( A ⊗ B ). Altogether, by Proposition 2.11, we see that J = Rad( A ⊗ B ), proving the claim.Thus, we see that ( A ⊗ B ) / Rad( A ⊗ B ) = A/ Rad( A ) ⊗ B/ Rad( B ) . Now, M is an irreducible representation of ( A ⊗ B ) / Rad( A ⊗ B ), so it is clearly of the form M = V ⊗ W , where V is an irreducible representation of A/ Rad( A ) and W is an irreduciblerepresentation of B/ Rad( B ), and V, W are uniquely determined by M (as all of the algebrasinvolved are direct sums of matrix algebras). 32 Representations of finite groups: basic results
Recall that a representation of a group G over a field k is a k -vector space V together with agroup homomorphism ρ : G → GL ( V ). As we have explained above, a representation of a group G over k is the same thing as a representation of its group algebra k [ G ].In this section, we begin a systematic development of representation theory of finite groups. Theorem 3.1. (Maschke) Let G be a finite group and k a field whose characteristic does not divide | G | . Then:(i) The algebra k [ G ] is semisimple.(ii) There is an isomorphism of algebras ψ : k [ G ] → ⊕ i End V i defined by g
7→ ⊕ i g | V i , where V i are the irreducible representations of G . In particular, this is an isomorphism of representationsof G (where G acts on both sides by left multiplication). Hence, the regular representation k [ G ] decomposes into irreducibles as ⊕ i dim( V i ) V i , and one has | G | = X i dim( V i ) . (the “sum of squares formula”).Proof. By Proposition 2.16, (i) implies (ii), and to prove (i), it is sufficient to show that if V isa finite-dimensional representation of G and W ⊂ V is any subrepresentation, then there exists asubrepresentation W ′ ⊂ V such that V = W ⊕ W ′ as representations.Choose any complement ˆ W of W in V . (Thus V = W ⊕ ˆ W as vector spaces , but not necessarilyas representations .) Let P be the projection along ˆ W onto W , i.e., the operator on V defined by P | W = Id and P | ˆ W = 0. Let P := 1 | G | X g ∈ G ρ ( g ) P ρ ( g − ) , where ρ ( g ) is the action of g on V , and let W ′ = ker P .
Now P | W = Id and P ( V ) ⊆ W , so P = P , so P is a projection along W ′ . Thus, V = W ⊕ W ′ asvector spaces.Moreover, for any h ∈ G and any y ∈ W ′ , P ρ ( h ) y = 1 | G | X g ∈ G ρ ( g ) P ρ ( g − h ) y = 1 | G | X ℓ ∈ G ρ ( hℓ ) P ρ ( ℓ − ) y = ρ ( h ) P y = 0 , so ρ ( h ) y ∈ ker P = W ′ . Thus, W ′ is invariant under the action of G and is therefore a subrepre-sentation of V . Thus, V = W ⊕ W ′ is the desired decomposition into subrepresentations.The converse to Theorem 3.1(i) also holds. Proposition 3.2. If k [ G ] is semisimple, then the characteristic of k does not divide | G | . roof. Write k [ G ] = L ri =1 End V i , where the V i are irreducible representations and V = k is thetrivial one-dimensional representation. Then k [ G ] = k ⊕ r M i =2 End V i = k ⊕ r M i =2 d i V i , where d i = dim V i . By Schur’s Lemma,Hom k [ G ] ( k, k [ G ]) = k ΛHom k [ G ] ( k [ G ] , k ) = kǫ, for nonzero homomorphisms of representations ǫ : k [ G ] → k and Λ : k → k [ G ] unique up to scaling.We can take ǫ such that ǫ ( g ) = 1 for all g ∈ G , and Λ such that Λ(1) = P g ∈ G g . Then ǫ ◦ Λ(1) = ǫ (cid:18)X g ∈ G g (cid:19) = X g ∈ G | G | . If | G | = 0, then Λ has no left inverse, as ( aǫ ) ◦ Λ(1) = 0 for any a ∈ k . This is a contradiction. Example 3.3. If G = Z /p Z and k has characteristic p , then every irreducible representation of G over k is trivial (so k [ Z /p Z ] indeed is not semisimple). Indeed, an irreducible representation of thisgroup is a 1-dimensional space, on which the generator acts by a p -th root of unity, and every p -throot of unity in k equals 1, as x p − x − p over k . Problem 3.4.
Let G be a group of order p n . Show that every irreducible representation of G overa field k of characteristic p is trivial. If V is a finite-dimensional representation of a finite group G , then its character χ V : G → k is defined by the formula χ V ( g ) = tr | V ( ρ ( g )). Obviously, χ V ( g ) is simply the restriction of thecharacter χ V ( a ) of V as a representation of the algebra A = k [ G ] to the basis G ⊂ A , so it carriesexactly the same information. The character is a central or class function : χ V ( g ) depends only onthe conjugacy class of g ; i.e., χ V ( hgh − ) = χ V ( g ). Theorem 3.5.
If the characteristic of k does not divide | G | , characters of irreducible representa-tions of G form a basis in the space F c ( G, k ) of class functions on G .Proof. By the Maschke theorem, k [ G ] is semisimple, so by Theorem 2.17, the characters are linearlyindependent and are a basis of ( A/ [ A, A ]) ∗ , where A = k [ G ]. It suffices to note that, as vectorspaces over k , ( A/ [ A, A ]) ∗ ∼ = { ϕ ∈ Hom k ( k [ G ] , k ) | gh − hg ∈ ker ϕ ∀ g, h ∈ G }∼ = { f ∈ Fun(
G, k ) | f ( gh ) = f ( hg ) ∀ g, h ∈ G } , which is precisely F c ( G, k ). Corollary 3.6.
The number of isomorphism classes of irreducible representations of G equals thenumber of conjugacy classes of G (if | G | 6 = 0 in k ). xercise. Show that if | G | = 0 in k then the number of isomorphism classes of irreduciblerepresentations of G over k is strictly less than the number of conjugacy classes in G .Hint. Let P = P g ∈ G g ∈ k [ G ]. Then P = 0. So P has zero trace in every finite dimensionalrepresentation of G over k . Corollary 3.7.
Any representation of G is determined by its character if k has characteristic 0;namely, χ V = χ W implies V ∼ = W . The following are examples of representations of finite groups over C .1. Finite abelian groups G = Z n × · · · × Z n k . Let G ∨ be the set of irreducible representationsof G . Every element of G forms a conjugacy class, so | G ∨ | = | G | . Recall that all irreduciblerepresentations over C (and algebraically closed fields in general) of commutative algebras andgroups are one-dimensional. Thus, G ∨ is an abelian group: if ρ , ρ : G → C × are irreduciblerepresentations then so are ρ ( g ) ρ ( g ) and ρ ( g ) − . G ∨ is called the dual or character group of G .For given n ≥
1, define ρ : Z n → C × by ρ ( m ) = e πim/n . Then Z ∨ n = { ρ k : k = 0 , . . . , n − } ,so Z ∨ n ∼ = Z n . In general,( G × G × · · · × G n ) ∨ = G ∨ × G ∨ × · · · × G ∨ n , so G ∨ ∼ = G for any finite abelian group G . This isomorphism is, however, noncanonical:the particular decomposition of G as Z n × · · · × Z n k is not unique as far as which elementsof G correspond to Z n , etc. is concerned. On the other hand, G ∼ = ( G ∨ ) ∨ is a canonicalisomorphism, given by ϕ : G → ( G ∨ ) ∨ , where ϕ ( g )( χ ) = χ ( g ).2. The symmetric group S . In S n , conjugacy classes are determined by cycle decompositionsizes: two permutations are conjugate if and only if they have the same number of cyclesof each length. For S , there are 3 conjugacy classes, so there are 3 different irreduciblerepresentations over C . If their dimensions are d , d , d , then d + d + d = 6, so S must havetwo 1-dimensional and one 2-dimensional representations. The 1-dimensional representationsare the trivial representation C + given by ρ ( σ ) = 1 and the sign representation C − given by ρ ( σ ) = ( − σ .The 2-dimensional representation can be visualized as representing the symmetries of theequilateral triangle with vertices 1, 2, 3 at the points (cos 120 ◦ , sin 120 ◦ ), (cos 240 ◦ , sin 240 ◦ ),(1 ,
0) of the coordinate plane, respectively. Thus, for example, ρ ((12)) = (cid:18) − (cid:19) , ρ ((123)) = (cid:18) cos 120 ◦ − sin 120 ◦ sin 120 ◦ cos 120 ◦ (cid:19) . To show that this representation is irreducible, consider any subrepresentation V . V must bethe span of a subset of the eigenvectors of ρ ((12)), which are the nonzero multiples of (1 , , V must also be the span of a subset of the eigenvectors of ρ ((123)), which aredifferent vectors. Thus, V must be either C or 0.3. The quaternion group Q = {± , ± i, ± j, ± k } , with defining relations i = jk = − kj, j = ki = − ik, k = ij = − ji, − i = j = k . { } , {− } , {± i } , {± j } , {± k } , so there are 5 different irreduciblerepresentations, the sum of the squares of whose dimensions is 8, so their dimensions mustbe 1, 1, 1, 1, and 2.The center Z ( Q ) is {± } , and Q /Z ( Q ) ∼ = Z × Z . The four 1-dimensional irreduciblerepresentations of Z × Z can be “pulled back” to Q . That is, if q : Q → Q /Z ( Q ) is thequotient map, and ρ any representation of Q /Z ( Q ), then ρ ◦ q gives a representation of Q .The 2-dimensional representation is V = C , given by ρ ( −
1) = − Id and ρ ( i ) = (cid:18) − (cid:19) , ρ ( j ) = (cid:18) √− −√− (cid:19) , ρ ( k ) = (cid:18) −√− −√− (cid:19) . (3)These are the Pauli matrices, which arise in quantum mechanics. Exercise.
Show that the 2-dimensional irreducible representation of Q can be realized inthe space of functions f : Q → C such that f ( gi ) = √− f ( g ) (the action of G is by rightmultiplication, g ◦ f ( x ) = f ( xg )).4. The symmetric group S . The order of S is 24, and there are 5 conjugacy classes: e, (12) , (123) , (1234) , (12)(34). Thus the sum of the squares of the dimensions of 5 irreduciblerepresentations is 24. As with S , there are two of dimension 1: the trivial and sign repre-sentations, C + and C − . The other three must then have dimensions 2, 3, and 3. Because S ∼ = S / Z × Z , where Z × Z is { e, (12)(34) , (13)(24) , (14)(23) } , the 2-dimensional repre-sentation of S can be pulled back to the 2-dimensional representation of S , which we willcall C .We can consider S as the group of rotations of a cube acting by permuting the interiordiagonals (or, equivalently, on a regular octahedron permuting pairs of opposite faces); thisgives the 3-dimensional representation C .The last 3-dimensional representation is C − , the product of C with the sign representation. C and C − are different, for if g is a transposition, det g | C = 1 while det g | C − = ( − = − C − is by action of S by symmetries (not necessarily rotations)of the regular tetrahedron. Yet another realization of this representation is the space offunctions on the set of 4 elements (on which S acts by permutations) with zero sum ofvalues. If V is a representation of a group G , then V ∗ is also a representation, via ρ V ∗ ( g ) = ( ρ V ( g ) ∗ ) − = ( ρ V ( g ) − ) ∗ = ρ V ( g − ) ∗ . The character is χ V ∗ ( g ) = χ V ( g − ) . We have χ V ( g ) = P λ i , where the λ i are the eigenvalues of g in V . These eigenvalues must beroots of unity because ρ ( g ) | G | = ρ ( g | G | ) = ρ ( e ) = Id . Thus for complex representations χ V ∗ ( g ) = χ V ( g − ) = X λ − i = X λ i = X λ i = χ V ( g ) . In particular, V ∼ = V ∗ as representations (not just as vector spaces) if and only if χ V ( g ) ∈ R for all g ∈ G . 36f V, W are representations of G , then V ⊗ W is also a representation, via ρ V ⊗ W ( g ) = ρ V ( g ) ⊗ ρ W ( g ) . Therefore, χ V ⊗ W ( g ) = χ V ( g ) χ W ( g ).An interesting problem discussed below is to decompose V ⊗ W (for irreducible V, W ) into thedirect sum of irreducible representations.
We define a positive definite Hermitian inner product on F c ( G, C ) (the space of central functions)by ( f , f ) = 1 | G | X g ∈ G f ( g ) f ( g ) . The following theorem says that characters of irreducible representations of G form an orthonormalbasis of F c ( G, C ) under this inner product. Theorem 3.8.
For any representations
V, W ( χ V , χ W ) = dim Hom G ( W, V ) , and ( χ V , χ W ) = (cid:26) , if V ∼ = W, , if V ≇ W if V, W are irreducible.Proof.
By the definition( χ V , χ W ) = 1 | G | X g ∈ G χ V ( g ) χ W ( g ) = 1 | G | X g ∈ G χ V ( g ) χ W ∗ ( g )= 1 | G | X g ∈ G χ V ⊗ W ∗ ( g ) = Tr | V ⊗ W ∗ ( P ) , where P = | G | P g ∈ G g ∈ Z ( C [ G ]) . (Here Z ( C [ G ]) denotes the center of C [ G ]). If X is an irreduciblerepresentation of G then P | X = (cid:26) Id , if X = C , , X = C . Therefore, for any representation X the operator P | X is the G -invariant projector onto the subspace X G of G -invariants in X . Thus,Tr | V ⊗ W ∗ ( P ) = dim Hom G ( C , V ⊗ W ∗ )= dim( V ⊗ W ∗ ) G = dim Hom G ( W, V ) . V of a finitegroup G is irreducible. Indeed, it implies that V is irreducible if and only if ( χ V , χ V ) = 1. Exercise.
Let G be a finite group. Let V i be the irreducible complex representations of G .For every i , let ψ i = dim V i | G | X g ∈ G χ V i ( g ) · g − ∈ C [ G ] . (i) Prove that ψ i acts on V j as the identity if j = i , and as the null map if j = i . (ii) Prove that ψ i are idempotents , i.e., ψ i = ψ i for any i , and ψ i ψ j = 0 for any i = j . Hint: In (i) , notice that ψ i commutes with any element of k [ G ], and thus acts on V j as anintertwining operator. Corollary 1.17 thus yields that ψ i acts on V j as a scalar. Compute thisscalar by taking its trace in V j .Here is another “orthogonality formula” for characters, in which summation is taken over irre-ducible representations rather than group elements. Theorem 3.9.
Let g, h ∈ G , and let Z g denote the centralizer of g in G . Then X V χ V ( g ) χ V ( h ) = (cid:26) | Z g | if g is conjugate to h , otherwisewhere the summation is taken over all irreducible representations of G .Proof. As noted above, χ V ( h ) = χ V ∗ ( h ), so the left hand side equals (using Maschke’s theorem): X V χ V ( g ) χ V ∗ ( h ) = Tr | ⊕ V V ⊗ V ∗ ( g ⊗ ( h ∗ ) − ) =Tr | ⊕ V End V ( x gxh − ) = Tr | C [ G ] ( x gxh − ) . If g and h are not conjugate, this trace is clearly zero, since the matrix of the operator x gxh − in the basis of group elements has zero diagonal entries. On the other hand, if g and h are in thesame conjugacy class, the trace is equal to the number of elements x such that x = gxh − , i.e., theorder of the centralizer Z g of g . We are done. Remark.
Another proof of this result is as follows. Consider the matrix U whose rows arelabeled by irreducible representations of G and columns by conjugacy classes, with entries U V,g = χ V ( g ) / p | Z g | . Note that the conjugacy class of g is G/Z g , thus | G | / | Z g | is the number of elementsconjugate to G . Thus, by Theorem 3.8, the rows of the matrix U are orthonormal. This meansthat U is unitary and hence its columns are also orthonormal, which implies the statement. Definition 3.10.
A unitary finite dimensional representation of a group G is a representation of G on a complex finite dimensional vector space V over C equipped with a G -invariant positive definiteHermitian form ( , ), i.e., such that ρ V ( g ) are unitary operators: ( ρ V ( g ) v, ρ V ( g ) w ) = ( v, w ) . We agree that Hermitian forms are linear in the first argument and antilinear in the second one. heorem 3.11. If G is finite, then any finite dimensional representation of G has a unitarystructure. If the representation is irreducible, this structure is unique up to scaling by a positivereal number.Proof. Take any positive definite form B on V and define another form B as follows: B ( v, w ) = X g ∈ G B ( ρ V ( g ) v, ρ V ( g ) w )Then B is a positive definite Hermitian form on V, and ρ V ( g ) are unitary operators. If V isan irreducible representation and B , B are two positive definite Hermitian forms on V, then B ( v, w ) = B ( Av, w ) for some homomorphism A : V → V (since any positive definite Hermitianform is nondegenerate). By Schur’s lemma, A = λ Id , and clearly λ > V is a finite dimensional representation of a finite group G , thenthe complex conjugate representation V (i.e., the same space V with the same addition and the sameaction of G , but complex conjugate action of scalars) is isomorphic to the dual representation V ∗ .Indeed, a homomorphism of representations V → V ∗ is obviously the same thing as an invariantsesquilinear form on V (i.e. a form additive on both arguments which is linear on the first one andantilinear on the second one), and an isomorphism is the same thing as a nondegenerate invariantsesquilinear form. So one can use a unitary structure on V to define an isomorphism V → V ∗ . Theorem 3.12.
A finite dimensional unitary representation V of any group G is completely re-ducible.Proof. Let W be a subrepresentation of V . Let W ⊥ be the orthogonal complement of W in V under the Hermitian inner product. Then W ⊥ is a subrepresentation of W , and V = W ⊕ W ⊥ .This implies that V is completely reducible.Theorems 3.11 and 3.12 imply Maschke’s theorem for complex representations (Theorem 3.1).Thus, we have obtained a new proof of this theorem over the field of complex numbers. Remark 3.13.
Theorem 3.12 shows that for infinite groups G , a finite dimensional representationmay fail to admit a unitary structure (as there exist finite dimensional representations, e.g. for G = Z , which are indecomposable but not irreducible). Let V be an irreducible representation of a finite group G, and v , v , . . . , v n be an orthonormalbasis of V under the invariant Hermitian form. The matrix elements of V are t Vij ( x ) = ( ρ V ( x ) v i , v j ) . Proposition 3.14. (i) Matrix elements of nonisomorphic irreducible representations are orthog-onal in
Fun( G, C ) under the form ( f, g ) = | G | P x ∈ G f ( x ) g ( x ) . (ii) ( t Vij , t Vi ′ j ′ ) = δ ii ′ δ jj ′ · V Thus, matrix elements of irreducible representations of G form an orthogonal basis of Fun( G, C ) .Proof. Let V and W be two irreducible representations of G. Take { v i } to be an orthonormal basisof V and { w i } to be an orthonormal basis of W under their positive definite invariant Hermitianforms. Let w ∗ i ∈ W ∗ be the linear function on W defined by taking the inner product with39 i : w ∗ i ( u ) = ( u, w i ). Then for x ∈ G we have ( xw ∗ i , w ∗ j ) = ( xw i , w j ). Therefore, putting P = | G | P x ∈ G x, we have( t Vij , t Wi ′ j ′ ) = | G | − X x ∈ G ( xv i , v j )( xw i ′ , w j ′ ) = | G | − X x ∈ G ( xv i , v j )( xw ∗ i ′ , w ∗ j ′ ) = ( P ( v i ⊗ w ∗ i ′ ) , v j ⊗ w ∗ j ′ )If V = W, this is zero, since P projects to the trivial representation, which does not occur in V ⊗ W ∗ . If V = W, we need to consider ( P ( v i ⊗ v ∗ i ′ ) , v j ⊗ v ∗ j ′ ) . We have a G -invariant decomposition V ⊗ V ∗ = C ⊕ L C = span( X v k ⊗ v ∗ k ) L = span a : P k a kk =0 ( X k,l a kl v k ⊗ v ∗ l ) , and P projects to the first summand along the second one. The projection of v i ⊗ v ∗ i ′ to C ⊂ C ⊕ L is thus δ ii ′ dim V X v k ⊗ v ∗ k This shows that ( P ( v i ⊗ v ∗ i ′ ) , v j ⊗ v ∗ j ′ ) = δ ii ′ δ jj ′ dim V which finishes the proof of (i) and (ii). The last statement follows immediately from the sum ofsquares formula. The characters of all the irreducible representations of a finite group can be arranged into a char-acter table, with conjugacy classes of elements as the columns, and characters as the rows. Morespecifically, the first row in a character table lists representatives of conjugacy classes, the secondone the numbers of elements in the conjugacy classes, and the other rows list the values of thecharacters on the conjugacy classes. Due to Theorems 3.8 and 3.9 the rows and columns of acharacter table are orthonormal with respect to the appropriate inner products.Note that in any character table, the row corresponding to the trivial representation consistsof ones, and the column corresponding to the neutral element consists of the dimensions of therepresentations.Here is, for example, the character table of S : S Id (12) (123) C + C − C A , the group of even permutations of 4 items. There are threeone-dimensional representations (as A has a normal subgroup Z ⊕ Z , and A / Z ⊕ Z = Z ).Since there are four conjugacy classes in total, there is one more irreducible representation ofdimension 3 . Finally, the character table is 40 Id (123) (132) (12)(34) C C ǫ ǫ ǫ C ǫ ǫ ǫ C − ǫ = exp( πi ) . The last row can be computed using the orthogonality of rows. Another way to compute thelast row is to note that C is the representation of A by rotations of the regular tetrahedron: inthis case (123) , (132) are the rotations by 120 and 240 around a perpendicular to a face of thetetrahedron, while (12)(34) is the rotation by 180 around an axis perpendicular to two oppositeedges. Example 3.15.
The following three character tables are of Q , S , and A respectively. Q i j k C ++ C + − C − + C −− C S Id (12) (12)(34) (123) (1234) C + C − C C C − A Id (123) (12)(34) (12345) (13245) C C √
52 1 −√ C − −√
52 1+ √ C C Q is obtained from the explicit formula(3) for this representation, or by using the orthogonality.For S , the 2-dimensional irreducible representation is obtained from the 2-dimensional irre-ducible representation of S via the surjective homomorphism S → S , which allows to obtain itscharacter from the character table of S .The character of the 3-dimensional representation C is computed from its geometric realizationby rotations of the cube. Namely, by rotating the cube, S permutes the main diagonals. Thus(12) is the rotation by 180 around an axis that is perpendicular to two opposite edges, (12)(34)41s the rotation by 180 around an axis that is perpendicular to two opposite faces, (123) is therotation around a main diagonal by 120 , and (1234) is the rotation by 90 around an axis that isperpendicular to two opposite faces; this allows us to compute the traces easily, using the fact thatthe trace of a rotation by the angle φ in R is 1 + 2 cos φ . Now the character of C − is found bymultiplying the character of C by the character of the sign representation.Finally, we explain how to obtain the character table of A (even permutations of 5 items). Thegroup A is the group of rotations of the regular icosahedron. Thus it has a 3-dimensional “rotationrepresentation” C , in which (12)(34) is the rotation by 180 around an axis perpendicular to twoopposite edges, (123) is the rotation by 120 around an axis perpendicular to two opposite faces,and (12345), (13254) are the rotations by 72 , respectively 144 , around axes going through twoopposite vertices. The character of this representation is computed from this description in astraightforward way.Another representation of A , which is also 3-dimensional, is C twisted by the automorphismof A given by conjugation by (12) inside S . This representation is denoted by C − . It has thesame character as C , except that the conjugacy classes (12345) and (13245) are interchanged.There are two remaining irreducible representations, and by the sum of squares formula theirdimensions are 4 and 5. So we call them C and C .The representation C is realized on the space of functions on the set { , , , , } with zerosum of values, where A acts by permutations (check that it is irreducible!). The character ofthis representation is equal to the character of the 5-dimensional permutation representation minusthe character of the 1-dimensional trivial representation (constant functions). The former at anelement g equals to the number of items among 1,2,3,4,5 which are fixed by g .The representation C is realized on the space of functions on pairs of opposite vertices of theicosahedron which has zero sum of values (check that it is irreducible!). The character of thisrepresentation is computed similarly to the character of C , or from the orthogonality formula. Character tables allow us to compute the tensor product multiplicities N kij using V i ⊗ V j = X N kij V k , N kij = ( χ i χ j , χ k ) Example 3.16.
The following tables represent computed tensor product multiplicities of irre-ducible representations of S , S , and A respectively. S C + C − C C + C + C − C C − C + C C C + ⊕ C − ⊕ C S C + C − C C C − C + C + C − C C C − C − C + C C − C C C + ⊕ C − ⊕ C C ⊕ C − C ⊕ C − C C + ⊕ C ⊕ C ⊕ C − C − ⊕ C ⊕ C ⊕ C − C − C + ⊕ C ⊕ C ⊕ C − C C C − C C C C C +3 C − C C C C ⊕ C ⊕ C C ⊕ C C − ⊕ C ⊕ C C ⊕ C − ⊕ C ⊕ C C − C ⊕ C ⊕ C C ⊕ C ⊕ C C ⊕ C − ⊕ C ⊕ C C C ⊕ C − ⊕ C ⊕ C ⊕ C C ⊕ C − ⊕ C ⊕ C C C ⊕ C ⊕ C − ⊕ C ⊕ C Problem 3.17.
Let G be the group of symmetries of a regular N-gon (it has 2N elements).(a) Describe all irreducible complex representations of this group (consider the cases of odd andeven N )(b) Let V be the 2-dimensional complex representation of G obtained by complexification of thestandard representation on the real plane (the plane of the polygon). Find the decomposition of V ⊗ V in a direct sum of irreducible representations. Problem 3.18.
Let G be the group of 3 by 3 matrices over F p which are upper triangular and haveones on the diagonal, under multiplication (its order is p ). It is called the Heisenberg group. Forany complex number z such that z p = 1 we define a representation of G on the space V of complexfunctions on F p , by ( ρ f )( x ) = f ( x − , ( ρ f )( x ) = z x f ( x ) . (note that z x makes sense since z p = 1 ).(a) Show that such a representation exists and is unique, and compute ρ ( g ) for all g ∈ G .(b) Denote this representation by R z . Show that R z is irreducible if and only if z = 1 .(c) Classify all 1-dimensional representations of G . Show that R decomposes into a direct sumof 1-dimensional representations, where each of them occurs exactly once.(d) Use (a)-(c) and the “sum of squares” formula to classify all irreducible representations of G . Problem 3.19.
Let V be a finite dimensional complex vector space, and GL ( V ) be the group ofinvertible linear transformations of V . Then S n V and Λ m V ( m ≤ dim ( V ) ) are representations of GL ( V ) in a natural way. Show that they are irreducible representations.Hint: Choose a basis { e i } in V . Find a diagonal element H of GL ( V ) such that ρ ( H ) hasdistinct eigenvalues. (where ρ is one of the above representations). This shows that if W is asubrepresentation, then it is spanned by a subset S of a basis of eigenvectors of ρ ( H ) . Use theinvariance of W under the operators ρ (1 + E ij ) (where E ij is defined by E ij e k = δ jk e i ) for all i = j to show that if the subset S is nonempty, it is necessarily the entire basis. Problem 3.20.
Recall that the adjacency matrix of a graph Γ (without multiple edges) is the matrixin which the ij -th entry is if the vertices i and j are connected with an edge, and zero otherwise.Let Γ be a finite graph whose automorphism group is nonabelian. Show that the adjacency matrixof Γ must have repeated eigenvalues. roblem 3.21. Let I be the set of vertices of a regular icosahedron ( | I | = 12 ). Let Fun( I ) be thespace of complex functions on I . Recall that the group G = A of even permutations of 5 itemsacts on the icosahedron, so we have a 12-dimensional representation of G on Fun( I ) .(a) Decompose this representation in a direct sum of irreducible representations (i.e., find themultiplicities of occurrence of all irreducible representations).(b) Do the same for the representation of G on the space of functions on the set of faces andthe set of edges of the icosahedron. Problem 3.22.
Let F q be a finite field with q elements, and G be the group of nonconstant inho-mogeneous linear transformations, x → ax + b , over F q (i.e., a ∈ F × q , b ∈ F q ). Find all irreduciblecomplex representations of G , and compute their characters. Compute the tensor products of irre-ducible representations.Hint. Let V be the representation of G on the space of functions on F q with sum of all valuesequal to zero. Show that V is an irreducible representation of G . Problem 3.23.
Let G = SU (2) (unitary 2 by 2 matrices with determinant 1), and V = C thestandard 2-dimensional representation of SU (2) . We consider V as a real representation, so it is4-dimensional.(a) Show that V is irreducible (as a real representation).(b) Let H be the subspace of End R ( V ) consisting of endomorphisms of V as a real representation.Show that H is 4-dimensional and closed under multiplication. Show that every nonzero element in H is invertible, i.e., H is an algebra with division.(c) Find a basis , i, j, k of H such that is the unit and i = j = k = − , ij = − ji = k, jk = − kj = i, ki = − ik = j . Thus we have that Q is a subgroup of the group H × of invertible elementsof H under multiplication.The algebra H is called the quaternion algebra.(d) For q = a + bi + cj + dk , a, b, c, d ∈ R , let ¯ q = a − bi − cj − dk , and || q || = q ¯ q = a + b + c + d .Show that q q = ¯ q ¯ q , and || q q || = || q || · || q || .(e) Let G be the group of quaternions of norm 1. Show that this group is isomorphic to SU (2) .(Thus geometrically SU (2) is the 3-dimensional sphere).(f ) Consider the action of G on the space V ⊂ H spanned by i, j, k , by x → qxq − , q ∈ G , x ∈ V . Since this action preserves the norm on V , we have a homomorphism h : SU (2) → SO (3) ,where SO (3) is the group of rotations of the three-dimensional Euclidean space. Show that thishomomorphism is surjective and that its kernel is { , − } . Problem 3.24.
It is known that the classification of finite subgroups of SO (3) is as follows:1) the cyclic group Z /n Z , n ≥ , generated by a rotation by π/n around an axis;2) the dihedral group D n of order n , n ≥ (the group of rotational symmetries in 3-space ofa plane containing a regular n -gon ;3) the group of rotations of the regular tetrahedron ( A ).4) the group of rotations of the cube or regular octahedron ( S ).5) the group of rotations of a regular dodecahedron or icosahedron ( A ). A regular 2-gon is just a line segment. a) Derive this classification.Hint. Let G be a finite subgroup of SO (3) . Consider the action of G on the unit sphere. Apoint of the sphere preserved by some nontrivial element of G is called a pole. Show that everynontrivial element of G fixes a unique pair of opposite poles, and that the subgroup of G fixing aparticular pole P is cyclic, of some order m (called the order of P). Thus the orbit of P has n/m elements, where n = | G | . Now let P , ..., P k be the poles representing all the orbits of G on the setof poles, and m , ..., m k be their orders. By counting nontrivial elements of G , show that − n ) = X i (1 − m i ) . Then find all possible m i and n that can satisfy this equation and classify the corresponding groups.(b) Using this classification, classify finite subgroups of SU (2) (use the homomorphism SU (2) → SO (3) ). Problem 3.25.
Find the characters and tensor products of irreducible complex representations ofthe Heisenberg group from Problem 3.18.
Problem 3.26.
Let G be a finite group, and V a complex representation of G which is faithful,i.e., the corresponding map G → GL ( V ) is injective. Show that any irreducible representation of G occurs inside S n V (and hence inside V ⊗ n ) for some n . Hint. Show that there exists a vector u ∈ V ∗ whose stabilizer in G is 1. Now define the map SV → Fun( G, C ) sending a polynomial f on V ∗ to the function f u on G given by f u ( g ) = f ( gu ).Show that this map is surjective and use this to deduce the desired result. Problem 3.27.
This problem is about an application of representation theory to physics (elasticitytheory). We first describe the physical motivation and then state the mathematical problem.Imagine a material which occupies a certain region U in the physical space V = R (a spacewith a positive definite inner product). Suppose the material is deformed. This means, we haveapplied a diffeomorphism (=change of coordinates) g : U → U ′ . The question in elasticity theoryis how much stress in the material this deformation will cause.For every point P ∈ U , let A P : V → V be defined by A P = dg ( P ) . A P is nondegenerate,so it has a polar decomposition A P = D P O P , where O P is orthogonal and D P is symmetric. Thematrix O P characterizes the rotation part of A P (which clearly produces no stress), and D P isthe distortion part, which actually causes stress. If the deformation is small, D P is close to 1, so D P = 1 + d P , where d P is a small symmetric matrix, i.e., an element of S V . This matrix is calledthe deformation tensor at P .Now we define the stress tensor, which characterizes stress. Let v be a small nonzero vector in V , and σ a small disk perpendicular to v centered at P of area || v || . Let F v be the force with whichthe part of the material on the v -side of σ acts on the part on the opposite side. It is easy to deducefrom Newton’s laws that F v is linear in v , so there exists a linear operator S P : V → V such that F v = S P v . It is called the stress tensor.An elasticity law is an equation S P = f ( d P ) , where f is a function. The simplest such law is alinear law (Hooke’s law): f : S V → End ( V ) is a linear function. In general, such a function isdefined by · parameters, but we will show there are actually only two essential ones – thecompression modulus K and the shearing modulus µ . For this purpose we will use representationtheory. ecall that the group SO (3) of rotations acts on V , so S V , End ( V ) are representations of thisgroup. The laws of physics must be invariant under this group (Galileo transformations), so f mustbe a homomorphism of representations.(a) Show that End ( V ) admits a decomposition R ⊕ V ⊕ W , where R is the trivial representation, V is the standard 3-dimensional representation, and W is a 5-dimensional representation of SO(3).Show that S V = R ⊕ W (b) Show that V and W are irreducible, even after complexification. Deduce using Schur’slemma that S P is always symmetric, and for x ∈ R , y ∈ W one has f ( x + y ) = Kx + µy for somereal numbers K, µ .In fact, it is clear from physics that
K, µ are positive. Physically, the compression modulus K characterises resistance of the material to compression or dilation, while the shearing modulus µ characterizes its resistance to changing the shape of the object without changing its volume. Forinstance, clay (used for sculpting) has a large compression modulus but a small shearing modulus. Representations of finite groups: further results
Suppose that G is a finite group and V is an irreducible representation of G over C . We say that V is- of complex type, if V ≇ V ∗ , - of real type, if V has a nondegenerate symmetric form invariant under G ,- of quaternionic type, if V has a nondegenerate skew form invariant under G. Problem 4.1. (a) Show that
End R [ G ] V is C for V of complex type, Mat ( R ) for V of real type,and H for V of quaternionic type, which motivates the names above.Hint. Show that the complexification V C of V decomposes as V ⊕ V ∗ . Use this to compute thedimension of End R [ G ] V in all three cases. Using the fact that C ⊂ End R [ G ] V , prove the resultin the complex case. In the remaining two cases, let B be the invariant bilinear form on V , and ( , ) the invariant positive Hermitian form (they are defined up to a nonzero complex scalar and apositive real scalar, respectively), and define the operator j : V → V such that B ( v, w ) = ( v, jw ) .Show that j is complex antilinear ( ji = − ij ), and j = λ · Id , where λ is a real number, positive inthe real case and negative in the quaternionic case (if B is renormalized, j multiplies by a nonzerocomplex number z and j by z ¯ z , as j is antilinear). Thus j can be normalized so that j = 1 forthe real case, and j = − in the quaternionic case. Deduce the claim from this.(b) Show that V is of real type if and only if V is the complexification of a representation V R over the field of real numbers. Example 4.2.
For Z /n Z all irreducible representations are of complex type, except the trivial oneand, if n is even, the “sign” representation, m → ( − m , which are of real type. For S all threeirreducible representations C + , C − , C are of real type. For S there are five irreducible representa-tions C + , C − , C , C , C − , which are all of real type. Similarly, all five irreducible representationsof A – C , C , C − , C , C are of real type. As for Q , its one-dimensional representations are ofreal type, and the two-dimensional one is of quaternionic type. Definition 4.3.
The Frobenius-Schur indicator
F S ( V ) of an irreducible representation V is 0 if itis of complex type, 1 if it is of real type, and − Theorem 4.4. (Frobenius-Schur) The number of involutions (=elements of order ≤ ) in G isequal to P V dim( V ) F S ( V ) , i.e., the sum of dimensions of all representations of G of real typeminus the sum of dimensions of its representations of quaternionic type.Proof. Let A : V → V have eigenvalues λ , λ , . . . , λ n . We haveTr | S V ( A ⊗ A ) = X i ≤ j λ i λ j Tr | Λ V ( A ⊗ A ) = X i Assume that all representations of a finite group G are defined over real numbers(i.e., all complex representations of G are obtained by complexifying real representations). Thenthe sum of dimensions of irreducible representations of G equals the number of involutions in G . Exercise. Show that any nontrivial finite group of odd order has an irreducible representationwhich is not defined over R (i.e., not realizable by real matrices). Enumerate the elements of a finite group G as follows: g , g , . . . , g n . Introduce n variables indexedwith the elements of G : x g , x g , . . . , x g n . Definition 4.6. Consider the matrix X G with entries a ij = x g i g j . The determinant of X G is somepolynomial of degree n of x g , x g , . . . , x g n that is called the Frobenius determinant .The following theorem, discovered by Dedekind and proved by Frobenius, became the startingpoint for creation of representation theory (see [Cu]). Theorem 4.7. det X G = r Y j =1 P j ( x ) deg P j for some pairwise non-proportional irreducible polynomials P j ( x ) , where r is the number of conju-gacy classes of G . We will need the following simple lemma. Lemma 4.8. Let Y be an n × n matrix with entries y ij . Then det Y is an irreducible polynomialof { y ij } . Proof. Let Y = t · Id + P ni =1 x i E i,i +1 , where i +1 is computed modulo n , and E i,j are the elementarymatrices. Then det( Y ) = t n − ( − n x ...x n , which is obviously irreducible. Hence det( Y ) isirreducible (since factors of a homogeneous polynomial are homogeneous).Now we are ready to proceed to the proof of Theorem 4.7.48 roof. Let V = C [ G ] be the regular representation of G. Consider the operator-valued polynomial L ( x ) = X g ∈ G x g ρ ( g ) , where ρ ( g ) ∈ End V is induced by g. The action of L ( x ) on an element h ∈ G is L ( x ) h = X g ∈ G x g ρ ( g ) h = X g ∈ G x g gh = X z ∈ G x zh − z So the matrix of the linear operator L ( x ) in the basis g , g , . . . , g n is X G with permuted columnsand hence has the same determinant up to sign.Further, by Maschke’s theorem, we havedet V L ( x ) = r Y i =1 (det V i L ( x )) dim V i , where V i are the irreducible representations of G . We set P i = det V i L ( x ) . Let { e im } be bases of V i and E i,jk ∈ End V i be the matrix units in these bases. Then { E i,jk } is a basis of C [ G ] and L ( x ) | V i = X j,k y i,jk E i,jk , where y i,jk are new coordinates on C [ G ] related to x g by a linear transformation. Then P i ( x ) = det | V i L ( x ) = det( y i,jk )Hence, P i are irreducible (by Lemma 4.8) and not proportional to each other (as they depend ondifferent collections of variables y i,jk ). The theorem is proved. We are now passing to deeper results in representation theory of finite groups. These results requirethe theory of algebraic numbers, which we will now briefly review. Definition 4.9. z ∈ C is an algebraic number (respectively, an algebraic integer ), if z is aroot of a monic polynomial with rational (respectively, integer) coefficients. Definition 4.10. z ∈ C is an algebraic number , (respectively, an algebraic integer ), if z is aneigenvalue of a matrix with rational (respectively, integer) entries. Proposition 4.11. Definitions (4.9) and (4.10) are equivalent.Proof. To show (4.10) ⇒ (4.9), notice that z is a root of the characteristic polynomial of the matrix(a monic polynomial with rational, respectively integer, coefficients).To show (4.9) ⇒ (4.10), suppose z is a root of p ( x ) = x n + a x n − + . . . + a n − x + a n . Then the characteristic polynomial of the following matrix (called the companion matrix ) is p ( x ): 49 . . . − a n . . . − a n − . . . − a n − ...0 0 0 . . . − a . Since z is a root of the characteristic polynomial of this matrix, it is an eigenvalue of this matrix.The set of algebraic numbers is denoted by Q , and the set of algebraic integers by A . Proposition 4.12. (i) A is a ring.(ii) Q is a field. Namely, it is an algebraic closure of the field of rational numbers.Proof. We will be using definition (4.10). Let α be an eigenvalue of A ∈ Mat n ( C )with eigenvector v , let β be an eigenvalue of B ∈ Mat m ( C )with eigenvector w . Then α ± β is an eigenvalue of A ⊗ Id m ± Id n ⊗ B , and αβ is an eigenvalue of A ⊗ B . The corresponding eigenvector is in both cases v ⊗ w . This shows that both A and Q are rings.To show that the latter is a field, it suffices to note that if α = 0 is a root of a polynomial p ( x ) ofdegree d , then α − is a root of x d p (1 /x ). The last statement is easy, since a number α is algebraicif and only if it defines a finite extension of Q . Proposition 4.13. A ∩ Q = Z .Proof. We will be using definition (4.9). Let z be a root of p ( x ) = x n + a x n − + . . . + a n − x + a n , and suppose z = pq ∈ Q , gcd( p, q ) = 1 . Notice that the leading term of p ( x ) will have q n in the denominator, whereas all the other termswill have a lower power of q there. Thus, if q = ± , then p ( z ) / ∈ Z , a contradiction. Thus, z ∈ A ∩ Q ⇒ z ∈ Z . The reverse inclusion follows because n ∈ Z is a root of x − n .Every algebraic number α has a minimal polynomial p ( x ), which is the monic polynomialwith rational coefficients of the smallest degree such that p ( α ) = 0. Any other polynomial q ( x ) withrational coefficients such that q ( α ) = 0 is divisible by p ( x ). Roots of p ( x ) are called the algebraicconjugates of α ; they are roots of any polynomial q with rational coefficients such that q ( α ) = 0.50ote that any algebraic conjugate of an algebraic integer is obviously also an algebraic inte-ger. Therefore, by the Vieta theorem, the minimal polynomial of an algebraic integer has integercoefficients.Below we will need the following lemma: Lemma 4.14. If α , ..., α m are algebraic numbers, then all algebraic conjugates to α + ... + α m are of the form α ′ + ... + α ′ m , where α ′ i are some algebraic conjugates of α i .Proof. It suffices to prove this for two summands. If α i are eigenvalues of rational matrices A i ofsmallest size (i.e., their characteristic polynomials are the minimal polynomials of α i ), then α + α is an eigenvalue of A := A ⊗ Id + Id ⊗ A . Therefore, so is any algebraic conjugate to α + α .But all eigenvalues of A are of the form α ′ + α ′ , so we are done. Problem 4.15. (a) Show that for any finite group G there exists a finite Galois extension K ⊂ C of Q such that any finite dimensional complex representation of G has a basis in which the matricesof the group elements have entries in K .Hint. Consider the representations of G over the field Q of algebraic numbers.(b) Show that if V is an irreducible complex representation of a finite group G of dimension > then there exists g ∈ G such that χ V ( g ) = 0 .Hint: Assume the contrary. Use orthonormality of characters to show that the arithmetic meanof the numbers | χ V ( g ) | for g = 1 is < . Deduce that their product β satisfies < β < .Show that all conjugates of β satisfy the same inequalities (consider the Galois conjugates of therepresentation V , i.e. representations obtained from V by the action of the Galois group of K over Q on the matrices of group elements in the basis from part (a)). Then derive a contradiction. Remark. Here is a modification of this argument, which does not use (a). Let N = | G | . Forany < j < N coprime to N , show that the map g g j is a bijection G → G . Deduce that Q g =1 | χ V ( g j ) | = β . Then show that β ∈ K := Q ( ζ ) , ζ = e πi/N , and does not change under theautomorphism of K given by ζ ζ j . Deduce that β is an integer, and derive a contradiction. Theorem 4.16. Let G be a finite group, and let V be an irreducible representation of G over C .Then dim V divides | G | . Proof. Let C , C , . . . , C n be the conjugacy classes of G . Set λ i = χ V ( g C i ) | C i | dim V , where g C i is a representative of C i . Proposition 4.17. The numbers λ i are algebraic integers for all i .Proof. Let C be a conjugacy class in G , and P = P h ∈ C h . Then P is a central element of Z [ G ], so itacts on V by some scalar λ , which is an algebraic integer (indeed, since Z [ G ] is a finitely generated Z -module, any element of Z [ G ] is integral over Z , i.e., satisfies a monic polynomial equation withinteger coefficients). On the other hand, taking the trace of P in V , we get | C | χ V ( g ) = λ dim V , g ∈ C , so λ = | C | χ V ( g )dim V . 51ow, consider X i λ i χ V ( g C i ) . This is an algebraic integer, since:(i) λ i are algebraic integers by Proposition 4.17,(ii) χ V ( g C i ) is a sum of roots of unity (it is the sum of eigenvalues of the matrix of ρ ( g C i ), andsince g | G | C i = e in G , the eigenvalues of ρ ( g C i ) are roots of unity), and(iii) A is a ring (Proposition 4.12).On the other hand, from the definition of λ i , X C i λ i χ V ( g C i ) = X i | C i | χ V ( g C i ) χ V ( g C i )dim V . Recalling that χ V is a class function, this is equal to X g ∈ G χ V ( g ) χ V ( g )dim V = | G | ( χ V , χ V )dim V . Since V is an irreducible representation, ( χ V , χ V ) = 1 , so X C i λ i χ V ( g C i ) = | G | dim V . Since | G | dim V ∈ Q and P C i λ i χ V ( g C i ) ∈ A , by Proposition 4.13 | G | dim V ∈ Z . Definition 4.18. A group G is called solvable if there exists a series of nested normal subgroups { e } = G ⊳ G ⊳ . . . ⊳ G n = G where G i +1 /G i is abelian for all 1 ≤ i ≤ n − Remark 4.19. Such groups are called solvable because they first arose as Galois groups of poly-nomial equations which are solvable in radicals. Theorem 4.20 (Burnside) . Any group G of order p a q b , where p and q are prime and a, b ≥ , issolvable. This famous result in group theory was proved by the British mathematician William Burnsidein the early 20-th century, using representation theory (see [Cu]). Here is this proof, presented inmodern language.Before proving Burnside’s theorem we will prove several other results which are of independentinterest. Theorem 4.21. Let V be an irreducible representation of a finite group G and let C be a conjugacyclass of G with gcd( | C | , dim( V )) = 1 . Then for any g ∈ C , either χ V ( g ) = 0 or g acts as a scalaron V . Lemma 4.22. If ε , ε . . . ε n are roots of unity such that n ( ε + ε + . . . + ε n ) is an algebraicinteger, then either ε = . . . = ε n or ε + . . . + ε n = 0 .Proof. Let a = n ( ε + . . . + ε n ). If not all ε i are equal, then | a | < 1. Moreover, since any algebraicconjugate of a root of unity is also a root of unity, | a ′ | ≤ a ′ of a . Butthe product of all algebraic conjugates of a is an integer. Since it has absolute value < 1, it mustequal zero. Therefore, a = 0. Proof of theorem 4.21. Let dim V = n . Let ε , ε , . . . ε n be the eigenvalues of ρ V ( g ). They are roots of unity, so χ V ( g ) is an algebraic integer. Also, by Proposition 4.17, n | C | χ V ( g ) is an algebraic integer. Sincegcd( n, | C | ) = 1, there exist integers a, b such that a | C | + bn = 1. This implies that χ V ( g ) n = 1 n ( ε + . . . + ε n ) . is an algebraic integer. Thus, by Lemma 4.22, we get that either ε = . . . = ε n or ε + . . . + ε n = χ V ( g ) = 0. In the first case, since ρ V ( g ) is diagonalizable, it must be scalar. In the second case, χ V ( g ) = 0. The theorem is proved. Theorem 4.23. Let G be a finite group, and let C be a conjugacy class in G of order p k where p is prime and k > . Then G has a proper nontrivial normal subgroup (i.e., G is not simple).Proof. Choose an element g ∈ C . Since g = e , by orthogonality of columns of the character table, X V ∈ Irr G dim V χ V ( g ) = 0 . (4)We can divide Irr G into three parts:1. the trivial representation,2. D , the set of irreducible representations whose dimension is divisible by p , and3. N , the set of non-trivial irreducible representations whose dimension is not divisible by p . Lemma 4.24. There exists V ∈ N such that χ V ( g ) = 0 .Proof. If V ∈ D , the number p dim( V ) χ V ( g ) is an algebraic integer, so a = X V ∈ D p dim( V ) χ V ( g )is an algebraic integer.Now, by (4), we have0 = χ C ( g ) + X V ∈ D dim V χ V ( g ) + X V ∈ N dim V χ V ( g ) = 1 + pa + X V ∈ N dim V χ V ( g ) . This means that the last summand is nonzero. 53ow pick V ∈ N such that χ V ( g ) = 0; it exists by Lemma 4.24. Theorem 4.21 implies that g (and hence any element of C ) acts by a scalar in V . Now let H be the subgroup of G generatedby elements ab − , a, b ∈ C . It is normal and acts trivially in V , so H = G , as V is nontrivial. Also H = 1, since | C | > Proof of Burnside’s theorem. Assume Burnside’s theorem is false. Then there exists a nonsolvable group G of order p a q b . Let G be the smallest such group. Then G is simple, and by Theorem 4.23, it cannot have a conjugacyclass of order p k or q k , k ≥ 1. So the order of any conjugacy class in G is either 1 or is divisibleby pq . Adding the orders of conjugacy classes and equating the sum to p a q b , we see that there hasto be more than one conjugacy class consisting just of one element. So G has a nontrivial center,which gives a contradiction. Theorem 4.25. Let G, H be finite groups, { V i } be the irreducible representations of G over afield k (of any characteristic), and { W j } be the irreducible representations of H over k . Then theirreducible representations of G × H over k are { V i ⊗ W j } .Proof. This follows from Theorem 2.26. Definition 4.26. A virtual representation of a finite group G is an integer linear combination ofirreducible representations of G , V = P n i V i , n i ∈ Z (i.e., n i are not assumed to be nonnegative).The character of V is χ V := P n i χ V i .The following lemma is often very useful (and will be used several times below). Lemma 4.27. Let V be a virtual representation with character χ V . If ( χ V , χ V ) = 1 and χ V (1) > then χ V is a character of an irreducible representation of G .Proof. Let V , V , . . . , V m be the irreducible representations of G , and V = P n i V i . Then byorthonormality of characters, ( χ V , χ V ) = P i n i . So P i n i = 1, meaning that n i = ± i , and n j = 0 for j = i . But χ V (1) > 0, so n i = +1 and we are done. Given a representation V of a group G and a subgroup H ⊂ G , there is a natural way to constructa representation of H . The restricted representation of V to H , Res GH V is the representation givenby the vector space V and the action ρ Res GH V = ρ V | H .There is also a natural, but more complicated way to construct a representation of a group G given a representation V of its subgroup H . Definition 4.28. If G is a group, H ⊂ G , and V is a representation of H , then the inducedrepresentation Ind GH V is the representation of G withInd GH V = { f : G → V | f ( hx ) = ρ V ( h ) f ( x ) ∀ x ∈ G, h ∈ H } g ( f )( x ) = f ( xg ) ∀ g ∈ G . Remark 4.29. In fact, Ind GH V is naturally isomorphic to Hom H ( k [ G ] , V ).Let us check that Ind GH V is indeed a representation: g ( f )( hx ) = f ( hxg ) = ρ V ( h ) f ( xg ) = ρ V ( h ) g ( f )( x ), and g ( g ′ ( f ))( x ) = g ′ ( f )( xg ) = f ( xgg ′ ) =( gg ′ )( f )( x ) for any g, g ′ , x ∈ G and h ∈ H . Remark 4.30. Notice that if we choose a representative x σ from every right H -coset σ of G , thenany f ∈ Ind GH V is uniquely determined by { f ( x σ ) } .Because of this, dim(Ind GH V ) = dim V · | G || H | . Problem 4.31. Check that if K ⊂ H ⊂ G are groups and V a representation of K then Ind GH Ind HK V is isomorphic to Ind GK V . Exercise. Let K ⊂ G be finite groups, and χ : K → C ∗ be a homomorphism. Let C χ be thecorresponding 1-dimensional representation of K . Let e χ = 1 | K | X g ∈ K χ ( g ) − g ∈ C [ K ]be the idempotent corresponding to χ . Show that the G -representation Ind GK C χ is naturally iso-morphic to C [ G ] e χ (with G acting by left multiplication). Let us now compute the character χ of Ind GH V . In each right coset σ ∈ H \ G , choose a representative x σ . Theorem 4.32. (The Mackey formula) One has χ ( g ) = X σ ∈ H \ G : x σ gx − σ ∈ H χ V ( x σ gx − σ ) . Remark. If the characteristic of the ground field k is relatively prime to | H | , then this formulacan be written as χ ( g ) = 1 | H | X x ∈ G : xgx − ∈ H χ V ( xgx − ) . Proof. For a right H -coset σ of G , let us define V σ = { f ∈ Ind GH V | f ( g ) = 0 ∀ g σ } . Then one has Ind GH V = M σ V σ , and so χ ( g ) = X σ χ σ ( g ) , χ σ ( g ) is the trace of the diagonal block of ρ ( g ) corresponding to V σ .Since g ( σ ) = σg is a right H -coset for any right H -coset σ , χ σ ( g ) = 0 if σ = σg .Now assume that σ = σg . Then x σ g = hx σ where h = x σ gx − σ ∈ H . Consider the vector spacehomomorphism α : V σ → V with α ( f ) = f ( x σ ). Since f ∈ V σ is uniquely determined by f ( x σ ), α is an isomorphism. We have α ( gf ) = g ( f )( x σ ) = f ( x σ g ) = f ( hx σ ) = ρ V ( h ) f ( x σ ) = hα ( f ) , and gf = α − hα ( f ). This means that χ σ ( g ) = χ V ( h ). Therefore χ ( g ) = X σ ∈ H \ G,σg = σ χ V ( x σ gx − σ ) . A very important result about induced representations is the Frobenius Reciprocity Theorem whichconnects the operations Ind and Res. Theorem 4.33. (Frobenius Reciprocity)Let H ⊂ G be groups, V be a representation of G and W a representation of H . ThenHom G ( V, Ind GH W ) is naturally isomorphic to Hom H (Res GH V, W ). Proof. Let E = Hom G ( V, Ind GH W ) and E ′ = Hom H (Res GH V, W ). Define F : E → E ′ and F ′ : E ′ → E as follows: F ( α ) v = ( αv )( e ) for any α ∈ E and ( F ′ ( β ) v )( x ) = β ( xv ) for any β ∈ E ′ .In order to check that F and F ′ are well defined and inverse to each other, we need to checkthe following five statements.Let α ∈ E , β ∈ E ′ , v ∈ V , and x, g ∈ G .(a) F ( α ) is an H -homomorphism, i.e., F ( α ) hv = hF ( α ) v .Indeed, F ( α ) hv = ( αhv )( e ) = ( hαv )( e ) = ( αv )( he ) = ( αv )( eh ) = h · ( αv )( e ) = hF ( α ) v .(b) F ′ ( β ) v ∈ Ind GH W , i.e., ( F ′ ( β ) v )( hx ) = h ( F ′ ( β ) v )( x ).Indeed, ( F ′ ( β ) v )( hx ) = β ( hxv ) = hβ ( xv ) = h ( F ′ ( β ) v )( x ).(c) F ′ ( β ) is a G -homomorphism, i.e. F ′ ( β ) gv = g ( F ′ ( β ) v ).Indeed, ( F ′ ( β ) gv )( x ) = β ( xgv ) = ( F ′ ( β ) v )( xg ) = ( g ( F ′ ( β ) v ))( x ).(d) F ◦ F ′ = Id E ′ .This holds since F ( F ′ ( β )) v = ( F ′ ( β ) v )( e ) = β ( v ).(e) F ′ ◦ F = Id E , i.e., ( F ′ ( F ( α )) v )( x ) = ( αv )( x ).Indeed, ( F ′ ( F ( α )) v )( x ) = F ( αxv ) = ( αxv )( e ) = ( xαv )( e ) = ( αv )( x ), and we are done. Exercise. The purpose of this exercise is to understand the notions of restricted and inducedrepresentations as part of a more advanced framework. This framework is the notion of tensorproducts over k -algebras (which generalizes the tensor product over k which we defined in Definition56.48). In particular, this understanding will lead us to a new proof of the Frobenius reciprocityand to some analogies between induction and restriction.Throughout this exercise, we will use the notation and results of the Exercise in Section 1.10.Let G be a finite group and H ⊂ G a subgroup. We consider k [ G ] as a ( k [ H ] , k [ G ])-bimodule(both module structures are given by multiplication inside k [ G ]). We denote this bimodule by k [ G ] . On the other hand, we can also consider k [ G ] as a ( k [ G ] , k [ H ])-bimodule (again, bothmodule structures are given by multiplication). We denote this bimodule by k [ G ] .(a) Let V be a representation of G . Then, V is a left k [ G ]-module, thus a ( k [ G ] , k )-bimodule.Thus, the tensor product k [ G ] ⊗ k [ G ] V is a ( k [ H ] , k )-bimodule, i. e., a left k [ H ]-module. Provethat this tensor product is isomorphic to Res GH V as a left k [ H ]-module. The isomorphism Res GH V → k [ G ] ⊗ k [ G ] V is given by v ⊗ k [ G ] v for every v ∈ Res GH V .(b) Let W be a representation of H . Then, W is a left k [ H ]-module, thus a ( k [ H ] , k )-bimodule. Then, Ind GH W ∼ = Hom H ( k [ G ] , W ), according to Remark 4.30. In other words, Ind GH W ∼ =Hom k [ H ] ( k [ G ] , W ). Now, use part (b) of the Exercise in Section 1.10 to conclude Theorem 4.33.(c) Let V be a representation of G . Then, V is a left k [ G ]-module, thus a ( k [ G ] , k )-bimodule.Prove that not only k [ G ] ⊗ k [ G ] V , but also Hom k [ G ] ( k [ G ] , V ) is isomorphic to Res GH V as a left k [ H ]-module. The isomorphism Hom k [ G ] ( k [ G ] , V ) → Res GH V is given by f f (1) for every f ∈ Hom k [ G ] ( k [ G ] , V ).(d) Let W be a representation of H . Then, W is a left k [ H ]-module, thus a ( k [ H ] , k )-bimodule. Show that Ind GH W is not only isomorphic to Hom k [ H ] ( k [ G ] , W ), but also isomorphic to k [ G ] ⊗ k [ H ] W . The isomorphism Hom k [ H ] ( k [ G ] , W ) → k [ G ] ⊗ k [ H ] W is given by f P g ∈ P g − ⊗ k [ H ] f ( g ) for every f ∈ Hom k [ H ] ( k [ G ] , W ), where P is a set of distinct representatives for the right H -cosets in G . (This isomorphism is independent of the choice of representatives.)(e) Let V be a representation of G and W a representation of H . Use (b) to prove thatHom G (cid:0) Ind GH W, V (cid:1) is naturally isomorphic to Hom H (cid:0) W, Res GH V (cid:1) .(f) Let V be a representation of H . Prove that Ind GH ( V ∗ ) ∼ = (cid:0) Ind GH V (cid:1) ∗ as representations of G . [Hint: Write Ind GH V as k [ G ] ⊗ k [ H ] V and write Ind GH ( V ∗ ) as Hom k [ H ] ( k [ G ] , V ∗ ). Provethat the map Hom k [ H ] ( k [ G ] , V ∗ ) × (cid:0) Ind GH ( V ∗ ) (cid:1) → k given by (cid:0) f, (cid:0) x ⊗ k [ H ] v (cid:1)(cid:1) ( f ( Sx )) ( v ) isa nondegenerate G -invariant bilinear form, where S : k [ G ] → k [ G ] is the linear map defined by Sg = g − for every g ∈ G .] Here are some examples of induced representations (we use the notation for representations fromthe character tables).1. Let G = S , H = Z . Using the Frobenius reciprocity, we obtain: Ind GH C + = C ⊕ C + ,Ind GH C − = C ⊕ C − .2. Let G = S , H = Z . Then we obtain Ind GH C + = C + ⊕ C − , Ind GH C ǫ = Ind GH C ǫ = C .3. Let G = S , H = S . Then Ind GH C + = C + ⊕ C − , Ind GH C − = C − ⊕ C , Ind GH C = C ⊕ C − ⊕ C . Problem 4.34. Compute the decomposition into irreducibles of all the representations of A in-duced from all the irreducible representations of a) Z (b) Z (c) Z (d) A (e) Z × Z S n In this subsection we give a description of the representations of the symmetric group S n for any n . Definition 4.35. A partition λ of n is a representation of n in the form n = λ + λ + ... + λ p ,where λ i are positive integers, and λ i ≥ λ i +1 .To such λ we will attach a Young diagram Y λ , which is the union of rectangles − i ≤ y ≤ − i +1,0 ≤ x ≤ λ i in the coordinate plane, for i = 1 , ..., p . Clearly, Y λ is a collection of n unit squares. A Young tableau corresponding to Y λ is the result of filling the numbers 1 , ..., n into the squares of Y λ in some way (without repetitions). For example, we will consider the Young tableau T λ obtainedby filling in the numbers in the increasing order, left to right, top to bottom.We can define two subgroups of S n corresponding to T λ :1. The row subgroup P λ : the subgroup which maps every element of { , ..., n } into an elementstanding in the same row in T λ .2. The column subgroup Q λ : the subgroup which maps every element of { , ..., n } into anelement standing in the same column in T λ .Clearly, P λ ∩ Q λ = { } .Define the Young projectors: a λ := 1 | P λ | X g ∈ P λ g,b λ := 1 | Q λ | X g ∈ Q λ ( − g g, where ( − g denotes the sign of the permutation g . Set c λ = a λ b λ . Since P λ ∩ Q λ = { } , thiselement is nonzero.The irreducible representations of S n are described by the following theorem. Theorem 4.36. The subspace V λ := C [ S n ] c λ of C [ S n ] is an irreducible representation of S n underleft multiplication. Every irreducible representation of S n is isomorphic to V λ for a unique λ . The modules V λ are called the Specht modules .The proof of this theorem is given in the next subsection. Example 4.37. For the partition λ = ( n ), P λ = S n , Q λ = { } , so c λ is the symmetrizer, and hence V λ is the trivialrepresentation. 58or the partition λ = (1 , ..., Q λ = S n , P λ = { } , so c λ is the antisymmetrizer, and hence V λ isthe sign representation. n = 3. For λ = (2 , V λ = C . n = 4. For λ = (2 , V λ = C ; for λ = (3 , V λ = C − ; for λ = (2 , , V λ = C . Corollary 4.38. All irreducible representations of S n can be given by matrices with rational entries. Problem 4.39. Find the sum of dimensions of all irreducible representations of the symmetricgroup S n .Hint. Show that all irreducible representations of S n are real, i.e., admit a nondegenerateinvariant symmetric form. Then use the Frobenius-Schur theorem. Lemma 4.40. Let x ∈ C [ S n ] . Then a λ xb λ = ℓ λ ( x ) c λ , where ℓ λ is a linear function.Proof. If g ∈ P λ Q λ , then g has a unique representation as pq , p ∈ P λ , q ∈ Q λ , so a λ gb λ = ( − q c λ .Thus, to prove the required statement, we need to show that if g is a permutation which is not in P λ Q λ then a λ gb λ = 0.To show this, it is sufficient to find a transposition t such that t ∈ P λ and g − tg ∈ Q λ ; then a λ gb λ = a λ tgb λ = a λ g ( g − tg ) b λ = − a λ gb λ , so a λ gb λ = 0. In other words, we have to find two elements i, j standing in the same row in thetableau T = T λ , and in the same column in the tableau T ′ = gT (where gT is the tableau of thesame shape as T obtained by permuting the entries of T by the permutation g ). Thus, it suffices toshow that if such a pair does not exist, then g ∈ P λ Q λ , i.e., there exists p ∈ P λ , q ′ ∈ Q ′ λ := gQ λ g − such that pT = q ′ T ′ (so that g = pq − , q = g − q ′ g ∈ Q λ ).Any two elements in the first row of T must be in different columns of T ′ , so there exists q ′ ∈ Q ′ λ which moves all these elements to the first row. So there is p ∈ P λ such that p T and q ′ T ′ havethe same first row. Now do the same procedure with the second row, finding elements p , q ′ suchthat p p T and q ′ q ′ T ′ have the same first two rows. Continuing so, we will construct the desiredelements p, q ′ . The lemma is proved.Let us introduce the lexicographic ordering on partitions: λ > µ if the first nonvanishing λ i − µ i is positive. Lemma 4.41. If λ > µ then a λ C [ S n ] b µ = 0 .Proof. Similarly to the previous lemma, it suffices to show that for any g ∈ S n there exists atransposition t ∈ P λ such that g − tg ∈ Q µ . Let T = T λ and T ′ = gT µ . We claim that there aretwo integers which are in the same row of T and the same column of T ′ . Indeed, if λ > µ , this isclear by the pigeonhole principle (already for the first row). Otherwise, if λ = µ , like in the proofof the previous lemma, we can find elements p ∈ P λ , q ′ ∈ gQ µ g − such that p T and q ′ T ′ have thesame first row, and repeat the argument for the second row, and so on. Eventually, having done i − λ i > µ i , which means that some two elements of the i -th row of the firsttableau are in the same column of the second tableau, completing the proof.59 emma 4.42. c λ is proportional to an idempotent. Namely, c λ = n !dim V λ c λ .Proof. Lemma 4.40 implies that c λ is proportional to c λ . Also, it is easy to see that the trace of c λ in the regular representation is n ! (as the coefficient of the identity element in c λ is 1). Thisimplies the statement. Lemma 4.43. Let A be an algebra and e be an idempotent in A . Then for any left A -module M ,one has Hom A ( Ae, M ) ∼ = eM (namely, x ∈ eM corresponds to f x : Ae → M given by f x ( a ) = ax , a ∈ Ae ).Proof. Note that 1 − e is also an idempotent in A . Thus the statement immediately follows fromthe fact that Hom A ( A, M ) ∼ = M and the decomposition A = Ae ⊕ A (1 − e ).Now we are ready to prove Theorem 4.36. Let λ ≥ µ . Then by Lemmas 4.42, 4.43Hom S n ( V λ , V µ ) = Hom S n ( C [ S n ] c λ , C [ S n ] c µ ) = c λ C [ S n ] c µ . The latter space is zero for λ > µ by Lemma 4.41, and 1-dimensional if λ = µ by Lemmas 4.40and 4.42. Therefore, V λ are irreducible, and V λ is not isomorphic to V µ if λ = µ . Since the numberof partitions equals the number of conjugacy classes in S n , the representations V λ exhaust all theirreducible representations of S n . The theorem is proved. S n Denote by U λ the representation Ind S n P λ C . It is easy to see that U λ can be alternatively defined as U λ = C [ S n ] a λ . Proposition 4.44. Hom ( U λ , V µ ) = 0 for µ < λ , and dim Hom ( U λ , V λ ) = 1 . Thus, U λ = ⊕ µ ≥ λ K µλ V µ , where K µλ are nonnegative integers and K λλ = 1 . Definition 4.45. The integers K µλ are called the Kostka numbers . Proof. By Lemmas 4.42 and 4.43,Hom( U λ , V µ ) = Hom( C [ S n ] a λ , C [ S n ] a µ b µ ) = a λ C [ S n ] a µ b µ , and the result follows from Lemmas 4.40 and 4.41.Now let us compute the character of U λ . Let C i be the conjugacy class in S n having i l cyclesof length l for all l ≥ i is a shorthand notation for ( i , ..., i l , ... )). Also let x , ..., x N bevariables, and let H m ( x ) = X i x mi be the power sum polynomials. Theorem 4.46. Let N ≥ p (where p is the number of parts of λ ). Then χ U λ ( C i ) is the coefficient of x λ := Q x λ j j in the polynomial Y m ≥ H m ( x ) i m . If j > p , we define λ j to be zero. roof. The proof is obtained easily from the Mackey formula. Namely, χ U λ ( C i ) is the number ofelements x ∈ S n such that xgx − ∈ P λ (for a representative g ∈ C i ), divided by | P λ | . The order of P λ is Q i λ i !, and the number of elements x such that xgx − ∈ P λ is the number of elements in P λ conjugate to g (i.e. | C i ∩ P λ | ) times the order of the centralizer Z g of g (which is n ! / | C i | ). Thus, χ U λ ( C i ) = | Z g | Q j λ j ! | C i ∩ P λ | . Now, it is easy to see that the centralizer Z g of g is isomorphic to Q m S i m ⋉ ( Z /m Z ) i m , so | Z g | = Y m m i m i m ! , and we get χ U λ ( C i ) = Q m m i m i m ! Q j λ j ! | C i ∩ P λ | . Now, since P λ = Q j S λ j , we have | C i ∩ P λ | = X r Y j ≥ λ j ! Q m ≥ m r jm r jm ! , where r = ( r jm ) runs over all collections of nonnegative integers such that X m mr jm = λ j , X j r jm = i m . Indeed, an element of C i that is in P λ would define an ordered partition of each λ j into parts(namely, cycle lengths), with m occuring r jm times, such that the total (over all j ) number of timeseach part m occurs is i m . Thus we get χ U λ ( C i ) = X r Y m i m ! Q j r jm !But this is exactly the coefficient of x λ in Y m ≥ ( x m + ... + x mN ) i m ( r jm is the number of times we take x mj ). Let ∆( x ) = Q ≤ i Let N ≥ p . Then χ V λ ( C i ) is the coefficient of x λ + ρ := Q x λ j + N − jj in the polyno-mial ∆( x ) Y m ≥ H m ( x ) i m . Remark. Here is an equivalent formulation of Theorem 4.47: χ V λ ( C i ) is the coefficient of x λ in the (Laurent) polynomial Y i Lemma 4.48. Q i Multiply both sides by Q i,j ( z i − y j ). Then the right hand side must vanish on the hyperplanes z i = z j and y i = y j (i.e., be divisible by ∆( z )∆( y )), and is a homogeneous polynomial of degree N ( N − z N − y N and then setting z N = y N .Now setting in the lemma z i = 1 /x i , we get Corollary 4.49. (Cauchy identity) R ( x, y ) = det( 11 − x i y j ) = X σ ∈ S N Q Nj =1 (1 − x j y σ ( j ) ) . x λ + ρ y λ + ρ is 1. Indeed, if σ = 1 is a permu-tation in S N , the coefficient of this monomial in Q (1 − x j y σ ( j ) ) is obviously zero. Remark. For partitions λ and µ of n , let us say that λ (cid:22) µ or µ (cid:23) λ if µ − λ is a sum ofvectors of the form e i − e j , i < j (called positive roots). This is a partial order, and µ (cid:23) λ implies µ ≥ λ . It follows from Theorem 4.47 and its proof that χ λ = ⊕ µ (cid:23) λ e K µλ χ U µ . This implies that the Kostka numbers K µλ vanish unless µ (cid:23) λ . In the following problems, we do not make a distinction between Young diagrams and partitions. Problem 4.50. For a Young diagram µ , let A ( µ ) be the set of Young diagrams obtained by addinga square to µ , and R ( µ ) be the set of Young diagrams obtained by removing a square from µ .(a) Show that Res S n S n − V µ = ⊕ λ ∈ R ( µ ) V λ . (b) Show that Ind S n S n − V µ = ⊕ λ ∈ A ( µ ) V λ . Problem 4.51. The content c ( λ ) of a Young diagram λ is the sum P j P λ j i =1 ( i − j ) . Let C = P i The formula follows from formula (5). Namely, note that l ! Q Theorem 4.54. Let A , B be two subalgebras of the algebra End E of endomorphisms of a finitedimensional vector space E , such that A is semisimple, and B = End A E . Then:(i) A = End B E (i.e., the centralizer of the centralizer of A is A );(ii) B is semisimple;(iii) as a representation of A ⊗ B , E decomposes as E = ⊕ i ∈ I V i ⊗ W i , where V i are all theirreducible representations of A , and W i are all the irreducible representations of B . In particular,we have a natural bijection between irreducible representations of A and B .Proof. Since A is semisimple, we have a natural decomposition E = ⊕ i ∈ I V i ⊗ W i , where W i :=Hom A ( V i , E ), and A = ⊕ i End V i . Therefore, by Schur’s lemma, B = End A ( E ) is naturally identi-fied with ⊕ i End( W i ). This implies all the statements of the theorem.We will now apply Theorem 4.54 to the following situation: E = V ⊗ n , where V is a finitedimensional vector space over a field of characteristic zero, and A is the image of C [ S n ] in End E .Let us now characterize the algebra B . Let gl ( V ) be End V regarded as a Lie algebra with operation ab − ba . Theorem 4.55. The algebra B = End A E is the image of the universal enveloping algebra U ( gl ( V )) under its natural action on E . In other words, B is generated by elements of the form ∆ n ( b ) := b ⊗ ⊗ ... ⊗ ⊗ b ⊗ ... ⊗ ... + 1 ⊗ ⊗ ... ⊗ b,b ∈ gl ( V ) .Proof. Clearly, the image of U ( gl ( V )) is contained in B , so we just need to show that any elementof B is contained in the image of U ( gl ( V )). By definition, B = S n End V , so the result follows frompart (ii) of the following lemma. 64 emma 4.56. Let k be a field of characteristic zero.(i) For any finite dimensional vector space U over k , the space S n U is spanned by elements ofthe form u ⊗ ... ⊗ u , u ∈ U .(ii) For any algebra A over k , the algebra S n A is generated by elements ∆ n ( a ) , a ∈ A .Proof. (i) The space S n U is an irreducible representation of GL ( U ) (Problem 3.19). The subspacespanned by u ⊗ ... ⊗ u is a nonzero subrepresentation, so it must be everything.(ii) By the fundamental theorem on symmetric functions, there exists a polynomial P withrational coefficients such that P ( H ( x ) , ..., H n ( x )) = x ...x n (where x = ( x , ..., x n )). Then P (∆ n ( a ) , ∆ n ( a ) , ..., ∆ n ( a n )) = a ⊗ ... ⊗ a. The rest follows from (i).Now, the algebra A is semisimple by Maschke’s theorem, so the double centralizer theoremapplies, and we get the following result, which goes under the name “Schur-Weyl duality”. Theorem 4.57. (i) The image A of C [ S n ] and the image B of U ( gl ( V )) in End( V ⊗ n ) are central-izers of each other.(ii) Both A and B are semisimple. In particular, V ⊗ n is a semisimple gl ( V ) -module.(iii) We have a decomposition of A ⊗ B -modules V ⊗ n = ⊕ λ V λ ⊗ L λ , where the summationis taken over partitions of n , V λ are Specht modules for S n , and L λ are some distinct irreduciblerepresentations of gl ( V ) or zero. GL ( V ) The Schur-Weyl duality for the Lie algebra gl ( V ) implies a similar statement for the group GL ( V ). Proposition 4.58. The image of GL ( V ) in End( V ⊗ n ) spans B .Proof. Denote the span of g ⊗ n , g ∈ GL ( V ), by B ′ . Let b ∈ End V be any element.We claim that B ′ contains b ⊗ n . Indeed, for all values of t but finitely many, t · Id + b is invertible,so ( t · Id + b ) ⊗ n belongs to B ′ . This implies that this is true for all t , in particular for t = 0, since( t · Id + b ) ⊗ n is a polynomial in t .The rest follows from Lemma 4.56. Corollary 4.59. As a representation of S n × GL ( V ) , V ⊗ n decomposes as ⊕ λ V λ ⊗ L λ , where L λ = Hom S n ( V λ , V ⊗ n ) are distinct irreducible representations of GL ( V ) or zero. Example 4.60. If λ = ( n ) then L λ = S n V , and if λ = (1 n ) ( n copies of 1) then L λ = ∧ n V . It wasshown in Problem 3.19 that these representations are indeed irreducible (except that ∧ n V is zeroif n > dim V ). 65 .20 Schur polynomials Let λ = ( λ , ..., λ p ) be a partition of n , and N ≥ p . Let D λ ( x ) = X s ∈ S N ( − s N Y j =1 x λ j + N − js ( j ) = det( x λ j + N − ji ) . Define the polynomials S λ ( x ) := D λ ( x ) D ( x )(clearly D ( x ) is just ∆( x )). It is easy to see that these are indeed polynomials, as D λ is an-tisymmetric and therefore must be divisible by ∆. The polynomials S λ are called the Schurpolynomials . Proposition 4.61. Y m ( x m + ... + x mN ) i m = X λ : p ≤ N χ λ ( C i ) S λ ( x ) . Proof. The identity follows from the Frobenius character formula and the antisymmetry of∆( x ) Y m ( x m + ... + x mN ) i m . Certain special values of Schur polynomials are of importance. Namely, we have Proposition 4.62. S λ (1 , z, z , ..., z N − ) = Y ≤ i 1) = Y ≤ i The first identity is obtained from the definition using the Vandermonde determinant. Thesecond identity follows from the first one by setting z = 1. L λ Proposition 4.61 allows us to calculate the characters of the representations L λ .Namely, let dim V = N , g ∈ GL ( V ), and x , ..., x N be the eigenvalues of g on V . To computethe character χ L λ ( g ), let us calculate Tr V ⊗ n ( g ⊗ n s ), where s ∈ S n . If s ∈ C i , we easily get that thistrace equals Y m Tr( g m ) i m = Y m H m ( x ) i m . On the other hand, by the Schur-Weyl dualityTr V ⊗ n ( g ⊗ n s ) = X λ χ λ ( C i )Tr L λ ( g ) . Comparing this to Proposition 4.61 and using linear independence of columns of the character tableof S n , we obtain 66 heorem 4.63. (Weyl character formula) The representation L λ is zero if and only if N < p ,where p is the number of parts of λ . If N ≥ p , the character of L λ is the Schur polynomial S λ ( x ) .Therefore, the dimension of L λ is given by the formula dim L λ = Y ≤ i The representation L λ +1 N (where N = (1 , , ..., ∈ Z N ) is isomorphic to L λ ⊗ ∧ N V .Proof. Indeed, L λ ⊗ ∧ N V ⊂ V ⊗ n ⊗ ∧ N V ⊂ V ⊗ n + N , and the only component of V ⊗ n + N that hasthe same character as L λ ⊗ ∧ N V is L λ +1 N . This implies the statement. GL ( V ) Definition 4.65. We say that a finite dimensional representation Y of GL ( V ) is polynomial (or algebraic , or rational ) if its matrix elements are polynomial functions of the entries of g, g − , g ∈ GL ( V ) (i.e., belong to k [ g ij ][1 / det( g )]).For example, V ⊗ n and hence all L λ are polynomial. Also define L λ − r · N := L λ ⊗ ( ∧ N V ∗ ) ⊗ r (thisdefinition makes sense by Proposition 4.64). This is also a polynomial representation. Thus wehave attached a unique irreducible polynomial representation L λ of GL ( V ) = GL N to any sequence( λ , ..., λ N ) of integers (not necessarily positive) such that λ ≥ ... ≥ λ N . This sequence is calledthe highest weight of L λ . Theorem 4.66. (i) Every finite dimensional polynomial representation of GL ( V ) is completelyreducible, and decomposes into summands of the form L λ (which are pairwise non-isomorphic).(ii) (the Peter-Weyl theorem for GL ( V ) ). Let R be the algebra of polynomial functions on GL ( V ) . Then as a representation of GL ( V ) × GL ( V ) (with action ( ρ ( g, h ) φ )( x ) = φ ( g − xh ) , g, h, x ∈ GL ( V ) , φ ∈ R ), R decomposes as R = ⊕ λ L ∗ λ ⊗ L λ , where the summation runs over all λ .Proof. (i) Let Y be a polynomial representation of GL ( V ). We have an embedding ξ : Y → Y ⊗ R given by ( u, ξ ( v ))( g ) := u ( gv ), u ∈ V ∗ . It is easy to see that ξ is a homomorphism of representations(where the action of GL ( V ) on the first component of Y ⊗ R is trivial). Thus, it suffices to provethe theorem for a subrepresentation Y ⊂ R m . Now, every element of R is a polynomial of g ij times a nonpositive power of det( g ). Thus, R is a quotient of a direct sum of representations of theform S r ( V ⊗ V ∗ ) ⊗ ( ∧ N V ∗ ) ⊗ s . So we may assume that Y is contained in a quotient of a (finite)direct sum of such representations. As V ∗ = ∧ N − V ⊗ ∧ N V ∗ , Y is contained in a direct sum ofrepresentations of the form V ⊗ n ⊗ ( ∧ N V ∗ ) ⊗ s , and we are done.(ii) Let Y be a polynomial representation of GL ( V ), and let us regard R as a representationof GL ( V ) via ( ρ ( h ) φ )( x ) = φ ( xh ). Then Hom GL ( V ) ( Y, R ) is the space of polynomial functionson GL ( V ) with values in Y ∗ , which are GL ( V )-equivariant. This space is naturally identified67ith Y ∗ . Taking into account the proof of (i), we deduce that R has the required decomposition,which is compatible with the second action of GL ( V ) (by left multiplications). This implies thestatement.Note that the Peter-Weyl theorem generalizes Maschke’s theorem for finite group, one of whoseforms states that the space of complex functions Fun( G, C ) on a finite group G as a representationof G × G decomposes as ⊕ V ∈ Irrep(G) V ∗ ⊗ V . Remark 4.67. Since the Lie algebra sl ( V ) of traceless operators on V is a quotient of gl ( V ) byscalars, the above results extend in a straightforward manner to representations of the Lie algebra sl ( V ). Similarly, the results for GL ( V ) extend to the case of the group SL ( V ) of operators withdeterminant 1. The only difference is that in this case the representations L λ and L λ +1 m areisomorphic, so the irreducible representations are parametrized by integer sequences λ ≥ ... ≥ λ N up to a simultaneous shift by a constant.In particular, one can show that any finite dimensional representation of sl ( V ) is completelyreducible, and any irreducible one is of the form L λ (we will not do this here). For dim V = 2 onethen recovers the representation theory of sl (2) studied in Problem 1.55. Problem 4.68. (a) Show that the S n -representation V ′ λ := C [ S n ] b λ a λ is isomorphic to V λ .Hint. Define S n -homomorphisms f : V λ → V ′ λ and g : V ′ λ → V λ by the formulas f ( x ) = xa λ and g ( y ) = yb λ , and show that they are inverse to each other up to a nonzero scalar.(b) Let φ : C [ S n ] → C [ S n ] be the automorphism sending s to ( − s s for any permutation s .Show that φ maps any representation V of S n to V ⊗ C − . Show also that φ ( C [ S n ] a ) = C [ S n ] φ ( a ) ,for a ∈ C [ S n ] . Use (a) to deduce that V λ ⊗ C − = V λ ∗ , where λ ∗ is the conjugate partition to λ ,obtained by reflecting the Young diagram of λ . Problem 4.69. Let R k,N be the algebra of polynomials on the space of k -tuples of complex N by N matrices X , ..., X k , invariant under simultaneous conjugation. An example of an element of R k,N is the function T w := Tr( w ( X , ..., X k )) , where w is any finite word on a k -letter alphabet. Showthat R k,N is generated by the elements T w .Hint. Consider invariant functions that are of degree d i in each X i , and realize this space asa tensor product ⊗ i S d i ( V ⊗ V ∗ ) . Then embed this tensor product into ( V ⊗ V ∗ ) ⊗ N = End( V ) ⊗ n ,and use the Schur-Weyl duality to get the result. GL ( F q ) GL ( F q )Let F q be a finite field of size q of characteristic other than 2, and G = GL ( F q ). Then | G | = ( q − q − q ) , since the first column of an invertible 2 by 2 matrix must be non-zero and the second column maynot be a multiple of the first one. Factoring, | GL ( F q ) | = q ( q + 1)( q − . G .To begin, let us find the conjugacy classes in GL ( F q ).Representatives Number of elements in a conjugacyclass Number of classesScalar (cid:0) x x (cid:1) q − x )Parabolic (cid:0) x x (cid:1) q − (cid:0) t u t (cid:1) , t =0) q − x )Hyperbolic (cid:0) x y (cid:1) , y = x q + q (elements that commute withthis one are of the form (cid:0) t u (cid:1) , t, u =0) ( q − q − 2) ( x, y = 0and x = y )Elliptic (cid:0) x εyy x (cid:1) , x ∈ F q , y ∈ F × q , ε ∈ F q \ F q (characteris-tic polynomial over F q is irre-ducible) q − q (the reason will be describedbelow) q ( q − 1) (matrices with y and − y are conjugate)More on the conjugacy class of elliptic matrices: these are the matrices whose characteristicpolynomial is irreducible over F q and which therefore don’t have eigenvalues in F q . Let A be sucha matrix, and consider a quadratic extension of F q , F q ( √ ε ) , ε ∈ F q \ F q . Over this field, A will have eigenvalues α = α + √ εα and α = α − √ εα , with corresponding eigenvectors v, v ( Av = αv, Av = αv ) . Choose a basis { e = v + v, e = √ ε ( v − v ) } . In this basis, the matrix A will have the form (cid:18) α εα α α (cid:19) , justifying the description of representative elements of this conjugacy class.In the basis { v, v } , matrices that commute with A will have the form (cid:18) λ λ (cid:19) , for all λ ∈ F × q , so the number of such matrices is q − 1. 69 .24.2 1-dimensional representations First, we describe the 1-dimensional representations of G . Proposition 4.70. [ G, G ] = SL ( F q ) .Proof. Clearly, det( xyx − y − ) = 1 , so [ G, G ] ⊆ SL ( F q ) . To show the converse, it suffices to show that the matrices (cid:18) (cid:19) , (cid:18) a a − (cid:19) , (cid:18) (cid:19) are commutators (as such matrices generate SL ( F q ) . ) Clearly, by using transposition, it sufficesto show that only the first two matrices are commutators. But it is easy to see that the matrix (cid:18) (cid:19) is the commutator of the matrices A = (cid:18) / 20 1 (cid:19) , B = (cid:18) − (cid:19) , while the matrix (cid:18) a a − (cid:19) is the commutator of the matrices A = (cid:18) a 00 1 (cid:19) , B = (cid:18) (cid:19) , This completes the proof.Therefore, G/ [ G, G ] ∼ = F × q via g → det( g ) . The one-dimensional representations of G thus have the form ρ ( g ) = ξ (cid:0) det( g ) (cid:1) , where ξ is a homomorphism ξ : F × q → C × ;so there are q − C ξ . .24.3 Principal series representations Let B ⊂ G, B = { (cid:18) ∗ ∗ ∗ (cid:19) } (the set of upper triangular matrices); then | B | = ( q − q, [ B, B ] = U = { (cid:18) ∗ (cid:19) } , and B/ [ B, B ] ∼ = F × q × F × q (the isomorphism maps an element of B to its two diagonal entries).Let λ : B → C × be a homomorphism defined by λ (cid:18) a b c (cid:19) = λ ( a ) λ ( c ),for some pair of homomorphisms λ , λ : F × q → C × . Define V λ ,λ = Ind GB C λ , where C λ is the 1-dimensional representation of B in which B acts by λ. We havedim( V λ ,λ ) = | G || B | = q + 1 . Theorem 4.71. λ = λ ⇒ V λ ,λ is irreducible.2. λ = λ = µ ⇒ V λ ,λ = C µ ⊕ W µ , where W µ is a q -dimensional irreducible representation of G .3. W µ ∼ = W ν if and only if µ = ν ; V λ ,λ ∼ = V λ ′ ,λ ′ if and only if { λ , λ } = { λ ′ , λ ′ } (in thesecond case, λ = λ , λ ′ = λ ′ ).Proof. From the Mackey formula, we have tr V λ ,λ ( g ) = 1 | B | X a ∈ G, aga − ∈ B λ ( aga − ) . If g = (cid:18) x x (cid:19) , the expression on the right evaluates to λ ( g ) | G || B | = λ ( x ) λ ( x ) (cid:0) q + 1 (cid:1) . If g = (cid:18) x x (cid:19) , λ ( g ) · , since here aga − ∈ B ⇒ a ∈ B. If g = (cid:18) x y (cid:19) , the expression evaluates to (cid:0) λ ( x ) λ ( y ) + λ ( y ) λ ( x ) (cid:1) · , since here aga − ∈ B ⇒ a ∈ B or a is an element of B multiplied by the transposition matrix.If g = (cid:18) x εyy x (cid:19) , x = y the expression on the right evaluates to 0 because matrices of this type don’t have eigenvalues over F q (and thus cannot be conjugated into B ). From the definition, λ i ( x )( i = 1 , 2) is a root of unity,so | G |h χ V λ ,λ , χ V λ ,λ i = ( q + 1) ( q − 1) + ( q − q − q + q ) ( q − q − q + q ) X x = y λ ( x ) λ ( y ) λ ( y ) λ ( x ) . The last two summands come from the expansion | a + b | = | a | + | b | + ab + ab. If λ = λ = µ, the last term is equal to ( q + q )( q − q − , and the total in this case is( q + 1)( q − q + 1) + ( q − 1) + 2 q ( q − q + 1)( q − q ( q − 1) = 2 | G | , so h χ V λ ,λ , χ V λ ,λ i = 2 . Clearly, C µ ⊆ Ind GB C µ,µ , since Hom G ( C µ , Ind GB C µ,µ ) = Hom B ( C µ , C µ ) = C (Theorem 4.33).Therefore, Ind GB C µ,µ = C µ ⊕ W µ ; W µ is irreducible; and the character of W µ is different for distinctvalues of µ, proving that W µ are distinct. 72f λ = λ , let z = xy − , then the last term of the summation is( q + q ) X x = y λ ( z ) λ ( z ) = ( q + q ) X x ; z =1 λ λ ( z ) = ( q + q )( q − X z =1 λ λ ( z ) . Since X z ∈ F × q λ λ ( z ) = 0 , because the sum of all roots of unity of a given order m > − ( q + q )( q − X z =1 λ λ (1) = − ( q + q )( q − . The difference between this case and the case of λ = λ is equal to − ( q + q )[( q − q − 1) + ( q − | G | , so this is an irreducible representation by Lemma 4.27.To prove the third assertion of the theorem, we look at the characters on hyperbolic elementsand note that the function λ ( x ) λ ( y ) + λ ( y ) λ ( x )determines λ , λ up to permutation. Let F q ⊃ F q be a quadratic extension F q ( √ ε ) , ε ∈ F q \ F q . We regard this as a 2-dimensional vectorspace over F q ; then G is the group of linear transformations of F q over F q . Let K ⊂ G be the cyclicgroup of multiplications by elements of F × q ,K = { (cid:18) x εyy x (cid:19) } , | K | = q − . For ν : K → C × a homomorphism, let Y ν = Ind GK C ν . This representation, of course, is very reducible. Let us compute its character, using the Mackeyformula. We get χ (cid:18) x x (cid:19) = q ( q − ν ( x ); χ ( A ) = 0 for A parabolic or hyperbolic; χ (cid:18) x εyy x (cid:19) = ν (cid:18) x εyy x (cid:19) + ν (cid:18) x εyy x (cid:19) q . The last assertion holds because if we regard the matrix as an element of F q , conjugation is anautomorphism of F q over F q , but the only nontrivial automorphism of F q over F q is the q th powermap. 73e thus have Ind GK C ν q ∼ = Ind GK C ν because they have the same character. Therefore, for ν q = ν we get q ( q − 1) representations.Next, we look at the following tensor product: W ⊗ V α, , where 1 is the trivial character and W is defined as in the previous section. The character of thisrepresentation is χ (cid:18) x x (cid:19) = q ( q + 1) α ( x ); χ ( A ) = 0 for A parabolic or elliptic; χ (cid:18) x y (cid:19) = α ( x ) + α ( y ) . Thus the ”virtual representation” W ⊗ V α, − V α, − Ind GK C ν , where α is the restriction of ν to scalars, has character χ (cid:18) x x (cid:19) = ( q − α ( x ); χ (cid:18) x x (cid:19) = − α ( x ); χ (cid:18) x y (cid:19) = 0; χ (cid:18) x εyy x (cid:19) = − ν (cid:18) x εyy x (cid:19) − ν q (cid:18) x εyy x (cid:19) . In all that follows, we will have ν q = ν. The following two lemmas will establish that the inner product of this character with itself isequal to 1, that its value at 1 is positive. As we know from Lemma 4.27, these two properties implythat it is the character of an irreducible representation of G . Lemma 4.72. Let χ be the character of the ”virtual representation” defined above. Then h χ, χ i = 1 and χ (1) > . Proof. χ (1) = q ( q + 1) − ( q + 1) − q ( q − 1) = q − > . We now compute the inner product h χ, χ i . Since α is a root of unity, this will be equal to1( q − q ( q + 1) (cid:2) ( q − · ( q − · q − · · ( q − q ( q − · X ζ elliptic ( ν ( ζ )+ ν q ( ζ ))( ν ( ζ ) + ν q ( ζ )) (cid:3) ν is also a root of unity, the last term of the expression evaluates to X ζ elliptic (2 + ν q − ( ζ ) + ν − q ( ζ )) . Let’s evaluate the last summand.Since F × q is cyclic and ν q = ν , X ζ ∈ F × q ν q − ( ζ ) = X ζ ∈ F × q ν − q ( ζ ) = 0 . Therefore, X ζ elliptic ( ν q − ( ζ ) + ν − q ( ζ )) = − X ζ ∈ F × q ( ν q − ( ζ ) + ν − q ( ζ )) = − q − 1) =since F × q is cyclic of order q − . Therefore, h χ, χ i = 1( q − q ( q + 1) (cid:0) ( q − · ( q − · q − · · ( q − q ( q − · (2( q − q ) − q − (cid:1) = 1 . We have now shown that for any ν with ν q = ν the representation Y ν with the same characteras W ⊗ V α, − V α, − Ind GK C ν exists and is irreducible. These characters are distinct for distinct pairs ( α, ν ) (up to switch ν → ν q ), so there are q ( q − such representations, each of dimension q − q − G, q ( q − principal series repre-sentations, and q ( q − complementary series representations, for a total of q − G. This implies that we have in fact found all irreduciblerepresentations of GL ( F q ). Theorem 4.73. Let X be a conjugation-invariant system of subgroups of a finite group G . Thentwo conditions are equivalent:(i) Any element of G belongs to a subgroup H ∈ X .(ii) The character of any irreducible representation of G belongs to the Q -span of characters ofinduced representations Ind GH V , where H ∈ X and V is an irreducible representation of H . Remark. Statement (ii) of Theorem 4.73 is equivalent to the same statement with Q -spanreplaced by C -span. Indeed, consider the matrix whose columns consist of the coefficients of thedecomposition of Ind GH V (for various H, V ) with respect to the irreducible representations of G .Then both statements are equivalent to the condition that the rows of this matrix are linearlyindependent. 75 roof. Proof that (ii) implies (i). Assume that g ∈ G does not belong to any of the subgroups H ∈ X . Then, since X is conjugation invariant, it cannot be conjugated into such a subgroup.Hence by the Mackey formula, χ Ind GH ( V ) ( g ) = 0 for all H ∈ X and V . So by (ii), for any irreduciblerepresentation W of G , χ W ( g ) = 0. But irreducible characters span the space of class functions, soany class function vanishes on g , which is a contradiction.Proof that (i) implies (ii). Let U be a virtual representation of G over C (i.e., a linear combina-tion of irreducible representations with nonzero integer coefficients) such that ( χ U , χ Ind GH V ) = 0 forall H, V . So by Frobenius reciprocity, ( χ U | H , χ V ) = 0. This means that χ U vanishes on H for any H ∈ X . Hence by (i), χ U is identically zero. This implies (ii) (because of the above remark). Corollary 4.74. Any irreducible character of a finite group is a rational linear combination ofinduced characters from its cyclic subgroups. Let G, A be groups and φ : G → Aut( A ) be a homomorphism. For a ∈ A , denote φ ( g ) a by g ( a ).The semidirect product G ⋉ A is defined to be the product A × G with multiplication law( a , g )( a , g ) = ( a g ( a ) , g g ) . Clearly, G and A are subgroups of G ⋉ A in a natural way.We would like to study irreducible complex representations of G ⋉ A . For simplicity, let us doit when A is abelian.In this case, irreducible representations of A are 1-dimensional and form the character group A ∨ , which carries an action of G . Let O be an orbit of this action, x ∈ O a chosen element,and G x the stabilizer of x in G . Let U be an irreducible representation of G x . Then we define arepresentation V ( O,U ) of G ⋉ A as follows.As a representation of G , we set V ( O,x,U ) = Ind GG x U = { f : G → U | f ( hg ) = hf ( g ) , h ∈ G x } . Next, we introduce an additional action of A on this space by ( af )( g ) = x ( g ( a )) f ( g ). Then it’seasy to check that these two actions combine into an action of G ⋉ A . Also, it is clear that thisrepresentation does not really depend on the choice of x , in the following sense. Let x, y ∈ O ,and g ∈ G be such that gx = y , and let g ( U ) be the representation of G y obtained from therepresentation U of G x by the action of g . Then V ( O,x,U ) is (naturally) isomorphic to V ( O,y,g ( U )) .Thus we will denote V ( O,x,U ) by V ( O,U ) (remembering, however, that x has been fixed). Theorem 4.75. (i) The representations V ( O,U ) are irreducible.(ii) They are pairwise nonisomorphic.(iii) They form a complete set of irreducible representations of G ⋉ A .(iv) The character of V = V ( O,U ) is given by the Mackey-type formula χ V ( a, g ) = 1 | G x | X h ∈ G : hgh − ∈ G x x ( h ( a )) χ U ( hgh − ) . roof. (i) Let us decompose V = V ( O,U ) as an A -module. Then we get V = ⊕ y ∈ O V y , where V y = { v ∈ V ( O,U ) | av = ( y, a ) v, a ∈ A } . (Equivalently, V y = { v ∈ V ( O,U ) | v ( g ) = 0 unless gy = x } ). So if W ⊂ V is a subrepresentation, then W = ⊕ y ∈ O W y , where W y ⊂ V y . Now, V y is arepresentation of G y , which goes to U under any isomorphism G y → G x determined by g ∈ G mapping x to y . Hence, V y is irreducible over G y , so W y = 0 or W y = V y for each y . Also, if hy = z then hW y = W z , so either W y = 0 for all y or W y = V y for all y , as desired.(ii) The orbit O is determined by the A -module structure of V , and the representation U bythe structure of V x as a G x -module.(iii) We have X U,O dim V U,O ) = X U,O | O | (dim U ) = X O | O | | G x | = X O | O || G/G x || G x | = | G | X O | O | = | G || A ∨ | = | G ⋉ A | . (iv) The proof is essentially the same as that of the Mackey formula. Exercise. Redo Problems 3.17(a), 3.18, 3.22 using Theorem 4.75. Exercise. Deduce parts (i)-(iii) of Theorem 4.75 from part (iv).77 Quiver Representations Problem 5.1. Field embeddings. Recall that k ( y , ..., y m ) denotes the field of rational functionsof y , ..., y m over a field k . Let f : k [ x , ..., x n ] → k ( y , ..., y m ) be an injective k -algebra homomor-phism. Show that m ≥ n . (Look at the growth of dimensions of the spaces W N of polynomials ofdegree N in x i and their images under f as N → ∞ ). Deduce that if f : k ( x , ..., x n ) → k ( y , ..., y m ) is a field embedding, then m ≥ n . Problem 5.2. Some algebraic geometry. Let k be an algebraically closed field, and G = GL n ( k ) . Let V be a polynomial representationof G . Show that if G has finitely many orbits on V then dim( V ) ≤ n . Namely:(a) Let x , ..., x N be linear coordinates on V . Let us say that a subset X of V is Zariski denseif any polynomial f ( x , ..., x N ) which vanishes on X is zero (coefficientwise). Show that if G hasfinitely many orbits on V then G has at least one Zariski dense orbit on V .(b) Use (a) to construct a field embedding k ( x , ..., x N ) → k ( g pq ) , then use Problem 5.1.(c) generalize the result of this problem to the case when G = GL n ( k ) × ... × GL n m ( k ) . Problem 5.3. Dynkin diagrams. Let Γ be a graph, i.e., a finite set of points (vertices) connected with a certain number of edges(we allow multiple edges). We assume that Γ is connected (any vertex can be connected to anyother by a path of edges) and has no self-loops (edges from a vertex to itself ). Suppose the verticesof Γ are labeled by integers , ..., N . Then one can assign to Γ an N × N matrix R Γ = ( r ij ) , where r ij is the number of edges connecting vertices i and j . This matrix is obviously symmetric, and iscalled the adjacency matrix. Define the matrix A Γ = 2 I − R Γ , where I is the identity matrix. Main definition: Γ is said to be a Dynkin diagram if the quadratic from on R N with matrix A Γ is positive definite.Dynkin diagrams appear in many areas of mathematics (singularity theory, Lie algebras, rep-resentation theory, algebraic geometry, mathematical physics, etc.) In this problem you will get acomplete classification of Dynkin diagrams. Namely, you will prove Theorem. Γ is a Dynkin diagram if and only if it is one on the following graphs: • A n : ◦−−◦ · · · ◦−−◦ • D n : ◦−−◦ · · · ◦−−◦|◦ • E : ◦−−◦−−◦−−◦−−◦|◦ E : ◦−−◦−−◦−−◦−−◦−−◦|◦ • E : ◦−−◦−−◦−−◦−−◦−−◦−−◦|◦ (a) Compute the determinant of A Γ where Γ = A N , D N . (Use the row decomposition rule, andwrite down a recursive equation for it). Deduce by Sylvester criterion that A N , D N are Dynkindiagrams. (b) Compute the determinants of A Γ for E , E , E (use row decomposition and reduce to (a)).Show they are Dynkin diagrams.(c) Show that if Γ is a Dynkin diagram, it cannot have cycles. For this, show that det ( A Γ ) = 0 for a graph Γ below (show that the sum of rows is 0). Thus Γ has to be a tree.(d) Show that if Γ is a Dynkin diagram, it cannot have vertices with 4 or more incoming edges,and that Γ can have no more than one vertex with 3 incoming edges. For this, show that det ( A Γ ) = 0 for a graph Γ below: 11 112 2 (e) Show that det ( A Γ ) = 0 for all graphs Γ below: Recall the Sylvester criterion: a symmetric real matrix is positive definite if and only if all its upper left cornerprincipal minors are positive. The Sylvester criterion says that a symmetric bilinear form ( , ) on R N is positive definite if and only if for any k ≤ N , det ≤ i,j ≤ k ( e i , e j ) > Please ignore the numerical labels; they will be relevant for Problem 5.5 below. (f ) Deduce from (a)-(e) the classification theorem for Dynkin diagrams.(g) A (simply laced) affine Dynkin diagram is a connected graph without self-loops such that thequadratic form defined by A Γ is positive semidefinite. Classify affine Dynkin diagrams. (Show thatthey are exactly the forbidden diagrams from (c)-(e)). Problem 5.4. Let Q be a quiver with set of vertices D . We say that Q is of finite type if ithas finitely many indecomposable representations. Let b ij be the number of edges from i to j in Q ( i, j ∈ D ).There is the following remarkable theorem, proved by P. Gabriel in early seventies. Theorem. A connected quiver Q is of finite type if and only if the corresponding unorientedgraph (i.e., with directions of arrows forgotten) is a Dynkin diagram.In this problem you will prove the “only if ” direction of this theorem (i.e., why other quiversare NOT of finite type).(a) Show that if Q is of finite type then for any rational numbers x i ≥ which are not simul-taneously zero, one has q ( x , ..., x N ) > , where q ( x , ..., x N ) := X i ∈ D x i − X i,j ∈ D b ij x i x j . Hint. It suffices to check the result for integers: x i = n i . First assume that n i ≥ , and considerthe space W of representations V of Q such that dim V i = n i . Show that the group Q i GL n i ( k ) actswith finitely many orbits on W ⊕ k , and use Problem 5.2 to derive the inequality. Then deduce theresult in the case when n i are arbitrary integers.(b) Deduce that q is a positive definite quadratic form.Hint. Use the fact that Q is dense in R .(c) Show that a quiver of finite type can have no self-loops. Then, using Problem 5.3, deducethe theorem. Problem 5.5. Let G = 1 be a finite subgroup of SU (2) , and V be the 2-dimensional representationof G coming from its embedding into SU (2) . Let V i , i ∈ I , be all the irreducible representations of G . Let r ij be the multiplicity of V i in V ⊗ V j . (a) Show that r ij = r ji .(b) The McKay graph of G , M ( G ), is the graph whose vertices are labeled by i ∈ I , and i isconnected to j by r ij edges. Show that M ( G ) is connected. (Use Problem 3.26)(c) Show that M ( G ) is an affine Dynkin graph (one of the “forbidden” graphs in Problem 5.3).For this, show that the matrix a ij = 2 δ ij − r ij is positive semidefinite but not definite, and useProblem 5.3.Hint. Let f = P x i χ V i , where χ V i be the characters of V i . Show directly that ((2 − χ V ) f, f ) ≥ M ( G ) has no self-loops, by using that if G is not cyclicthen G contains the central element − Id ∈ SU (2).80d) Which groups from Problem 3.24 correspond to which diagrams?(e) Using the McKay graph, find the dimensions of irreducible representations of all finite G ⊂ SU (2) (namely, show that they are the numbers labeling the vertices of the affine Dynkindiagrams on our pictures). Compare with the results on subgroups of SO (3) we obtained inProblem 3.24. A , A , A We have seen that a central question about representations of quivers is whether a certain connectedquiver has only finitely many indecomposable representations. In the previous subsection it is shownthat only those quivers whose underlying undirected graph is a Dynkin diagram may have thisproperty. To see if they actually do have this property, we first explicitly decompose representationsof certain easy quivers. Remark 5.6. By an object of the type 1 / / / / / / Example 5.7 ( A ) . The quiver A consists of a single vertex and has no edges. Since a repre-sentation of this quiver is just a single vector space, the only indecomposable representation is theground field (=a one-dimensional space). Example 5.8 ( A ) . The quiver A consists of two vertices connected by a single edge. • / / • A representation of this quiver consists of two vector spaces V, W and an operator A : V → W . • V A / / • W To decompose this representation, we first let V ′ be a complement to the kernel of A in V andlet W ′ be a complement to the image of A in W . Then we can decompose the representation asfollows • V A / / • W = • ker A / / • ⊕ • V ′ / / A ∼ • Im A ⊕ • / / • W ′ The first summand is a multiple of the object 1 / / / / / / A has three indecomposable representations, namely1 / / , / / / / . Example 5.9 ( A ) . The quiver A consists of three vertices and two connections between them.So we have to choose between two possible orientations. • / / • / / • or • / / • • o o 1. We first look at the orientation • / / • / / • . • V A / / • W B / / • Y . Like in Example 5.8 we first split away • ker A / / • / / • . This object is a multiple of 1 / / / / Y ′ be a complement of Im B in Y .Then we can also split away • / / • / / • Y ′ which is a multiple of the object 0 / / / / A is injective and the map B is surjective (we rename the spaces to simplify notation): • V (cid:31) (cid:127) A / / • W B / / / / • Y . Next, let X = ker( B ◦ A ) and let X ′ be a complement of X in V . Let W ′ be a complementof A ( X ) in W such that A ( X ′ ) ⊂ W ′ . Then we get • V (cid:31) (cid:127) A / / • W B / / / / • Y = • X A / / • A ( X ) B / / • ⊕ • X ′ (cid:31) (cid:127) A / / • W ′ B / / / / • Y The first of these summands is a multiple of 1 / / ∼ / / A is injective, B is surjective and furthermore ker( B ◦ A ) = 0.To simplify notation, we redefine V = X ′ , W = W ′ . Next we let X = Im( B ◦ A ) and let X ′ be a complement of X in Y . Furthermore, let W ′ = B − ( X ′ ). Then W ′ is a complement of A ( V ) in W . This yields the decomposition • V (cid:31) (cid:127) A / / • W B / / / / • Y = • V ∼ A / / • A ( V ) ∼ B / / • X ⊕ • / / • W ′ B / / / / • X ′ Here, the first summand is a multiple of 1 / / ∼ / / ∼ B ,the second summand can be decomposed into multiples of 0 / / / / ∼ / / / / . So, on the whole, this quiver has six indecomposable representations:1 / / / / , / / / / , / / ∼ / / , / / ∼ / / ∼ , / / / / ∼ , / / / / 02. Now we look at the orientation • / / • • o o . Very similarly to the other orientation, we can split away objects of the type1 / / o o , / / o o which results in a situation where both A and B are injective: • V (cid:31) (cid:127) A / / • W o o B ? _ • Y . 82y identifying V and Y as subspaces of W , this leads to the problem of classifying pairs ofsubspaces of a given space W up to isomorphism (the pair of subspaces problem ). To doso, we first choose a complement W ′ of V ∩ Y in W , and set V ′ = W ′ ∩ V , Y ′ = W ′ ∩ Y .Then we can decompose the representation as follows: • V (cid:31) (cid:127) / / • W o o ? _ • Y = • V ′ (cid:31) (cid:127) / / • W ′ o o ? _ • Y ′ ⊕ • V ∩ Y / / ∼ • V ∩ Y • o o ∼ V ∩ Y . The second summand is a multiple of the object 1 / / ∼ o o ∼ . We go on decomposing thefirst summand. Again, to simplify notation, we let V = V ′ , W = W ′ , Y = Y ′ . We can now assume that V ∩ Y = 0. Next, let W ′ be a complement of V ⊕ Y in W . Thenwe get • V (cid:31) (cid:127) / / • W o o ? _ • Y = • V (cid:31) (cid:127) / / • V ⊕ Y o o ? _ • Y ⊕ • / / • W ′ • o o The second of these summands is a multiple of the indecomposable object 0 / / o o .The first summand can be further decomposed as follows: • V (cid:31) (cid:127) / / • V ⊕ Y o o ? _ • Y = • V / / ∼ • V • o o ⊕ • / / • Y • Y o o ∼ These summands are multiples of1 / / o o , / / o o So - like in the other orientation - we get 6 indecomposable representations of A :1 / / o o , / / o o , ∼ / / ∼ o o , / / o o , / / o o , / / o o D As a last - slightly more complicated - example we consider the quiver D . Example 5.10 ( D ) . We restrict ourselves to the orientation • / / • • o o • O O . So a representation of this quiver looks like • V A / / • V • V A o o • V A O O A , A , A . Moreprecisely, we split away the representations • ker A / / • • o o • O O • / / • • o o • ker A O O • / / • • ker A o o • O O These representations are multiples of the indecomposable objects • / / • • o o • O O • / / • • o o • O O • / / • • o o • O O So we get to a situation where all of the maps A , A , A are injective. • V (cid:31) (cid:127) A / / • V • V ? _ A o o • V (cid:31) ? A O O As in 2, we can then identify the spaces V , V , V with subspaces of V . So we get to the triple ofsubspaces problem of classifying a triple of subspaces of a given space V .The next step is to split away a multiple of • / / • • o o • O O to reach a situation where V + V + V = V. By letting Y = V ∩ V ∩ V , choosing a complement V ′ of Y in V , and setting V ′ i = V ′ ∩ V i , i = 1 , , 3, we can decompose this representation into • V ′ (cid:31) (cid:127) / / • V ′ • V ′ ? _ o o • V ′ (cid:31) ? O O ⊕ • Y ∼ / / • Y • Y ∼ o o • Y O O O(cid:15) The last summand is a multiple of the indecomposable representation • ∼ / / • • ∼ o o • O O O(cid:15) 84o - considering the first summand and renaming the spaces to simplify notation - we are in asituation where V = V + V + V , V ∩ V ∩ V = 0 . As a next step, we let Y = V ∩ V and we choose a complement V ′ of Y in V such that V ⊂ V ′ ,and set V ′ = V ′ ∩ V , V ′ = V ′ ∩ V . This yields the decomposition • V (cid:31) (cid:127) / / • V • V ? _ o o • V (cid:31) ? O O = • V ′ (cid:31) (cid:127) / / • V ′ • V ? _ o o • V ′ (cid:31) ? O O ⊕ • Y ∼ / / • Y • o o • Y O O O(cid:15) The second summand is a multiple of the indecomposable object • ∼ / / • • o o • O O O(cid:15) . In the resulting situation we have V ∩ V = 0. Similarly we can split away multiples of • ∼ / / • • ∼ o o • O O and • / / • • ∼ o o • O O O(cid:15) to reach a situation where the spaces V , V , V do not intersect pairwise V ∩ V = V ∩ V = V ∩ V = 0 . If V * V ⊕ V we let Y = V ∩ ( V ⊕ V ). We let V ′ be a complement of Y in V . Since then V ′ ∩ ( V ⊕ V ) = 0, we can select a complement V ′ of V ′ in V which contains V ⊕ V . This givesus the decomposition • V (cid:31) (cid:127) / / • V • V ? _ o o • V (cid:31) ? O O = • V ′ ∼ / / • V ′ • o o • O O ⊕ • Y (cid:31) (cid:127) / / • V ′ • V ? _ o o • V (cid:31) ? O O The first of these summands is a multiple of • ∼ / / • • o o • O O By splitting these away we get to a situation where V ⊆ V ⊕ V . Similarly, we can split awayobjects of the type • / / • • o o • O O O(cid:15) and • / / • • ∼ o o • O O to reach a situation in which the following conditions hold85. V + V + V = V. V ∩ V = 0 , V ∩ V = 0 , V ∩ V = 0 . V ⊆ V ⊕ V , V ⊆ V ⊕ V , V ⊆ V ⊕ V . But this implies that V ⊕ V = V ⊕ V = V ⊕ V = V. So we get dim V = dim V = dim V = n and dim V = 2 n. Since V ⊆ V ⊕ V we can write every element of V in the form x ∈ V , x = ( x , x ) , x ∈ V , x ∈ V . We then can define the projections B : V → V , ( x , x ) x ,B : V → V , ( x , x ) x . Since V ∩ V = 0 , V ∩ V = 0, these maps have to be injective and therefore are isomorphisms. Wethen define the isomorphism A = B ◦ B − : V → V . Let e , . . . , e n be a basis for V . Then we get V = C e ⊕ C e ⊕ · · · ⊕ C e n V = C Ae ⊕ C Ae ⊕ · · · ⊕ C Ae n V = C ( e + Ae ) ⊕ C ( e + Ae ) ⊕ · · · ⊕ C ( e n + Ae n ) . So we can think of V as the graph of an isomorphism A : V → V . From this we obtain thedecomposition • V (cid:31) (cid:127) / / • V • V ? _ o o • V (cid:31) ? O O = n M j =1 • C (1 , (cid:31) (cid:127) / / • C • C (1 , ? _ o o • C (0 , (cid:31) ? O O These correspond to the indecomposable object • / / • • o o • O O Thus the quiver D with the selected orientation has 12 indecomposable objects. If one were toexplicitly decompose representations for the other possible orientations, one would also find 12indecomposable objects.It appears as if the number of indecomposable representations does not depend on the orienta-tion of the edges, and indeed - Gabriel’s theorem will generalize this observation.86 .4 Roots From now on, let Γ be a fixed graph of type A n , D n , E , E , E . We denote the adjacency matrixof Γ by R Γ . Definition 5.11 (Cartan Matrix) . We define the Cartan matrix as A Γ = 2Id − R Γ . On the lattice Z n (or the space R n ) we then define an inner product B ( x, y ) = x T A Γ y corresponding to the graph Γ. Lemma 5.12. B is positive definite.2. B ( x, x ) takes only even values for x ∈ Z n .Proof. 1. This follows by definition, since Γ is a Dynkin diagram.2. By the definition of the Cartan matrix we get B ( x, x ) = x T A Γ x = X i,j x i a ij x j = 2 X i x i + X i,j, i = j x i a ij x j = 2 X i x i + 2 · X i A root with respect to a certain positive inner product is a shortest (with respectto this inner product), nonzero vector in Z n .So for the inner product B , a root is a nonzero vector x ∈ Z n such that B ( x, x ) = 2 . Remark 5.14. There can be only finitely many roots, since all of them have to lie in some ball. Definition 5.15. We call vectors of the form α i = (0 , . . . , i − th z}|{ , . . . , simple roots .The α i naturally form a basis of the lattice Z n . Lemma 5.16. Let α be a root, α = P ni =1 k i α i . Then either k i ≥ for all i or k i ≤ for all i .Proof. Assume the contrary, i.e., k i > k j < 0. Without loss of generality, we can also assumethat k s = 0 for all s between i and j . We can identify the indices i, j with vertices of the graph Γ. • • i ǫ • i ′ • • j • •• ǫ be the edge connecting i with the next vertex towards j and i ′ be the vertex on the otherend of ǫ . We then let Γ , Γ be the graphs obtained from Γ by removing ǫ . Since Γ is supposedto be a Dynkin diagram - and therefore has no cycles or loops - both Γ and Γ will be connectedgraphs, which are not connected to each other. • • i Γ • • • j • •• Γ Then we have i ∈ Γ , j ∈ Γ . We define β = X m ∈ Γ k m α m , γ = X m ∈ Γ k m α m . With this choice we get α = β + γ. Since k i > , k j < β = 0 , γ = 0 and therefore B ( β, β ) ≥ , B ( γ, γ ) ≥ . Furthermore, B ( β, γ ) = − k i k i ′ , since Γ , Γ are only connected at ǫ . But this has to be a nonnegative number, since k i > k i ′ ≤ 0. This yields B ( α, α ) = B ( β + γ, β + γ ) = B ( β, β ) | {z } ≥ +2 B ( β, γ ) | {z } ≥ + B ( γ, γ ) | {z } ≥ ≥ . But this is a contradiction, since α was assumed to be a root. Definition 5.17. We call a root α = P i k i α i a positive root if all k i ≥ 0. A root for which k i ≤ i is called a negative root. Remark 5.18. Lemma 5.16 states that every root is either positive or negative. Example 5.19. 1. Let Γ be of the type A N − . Then the lattice L = Z N − can be realized asa subgroup of the lattice Z N by letting L ⊆ Z N be the subgroup of all vectors ( x , . . . , x N )such that X i x i = 0 . The vectors α = (1 , − , , . . . , α = (0 , , − , , . . . , α N − = (0 , . . . , , , − L . Furthermore, the standard inner product( x, y ) = X x i y i Z N restricts to the inner product B given by Γ on L , since it takes the same values on thebasis vectors: ( α i , α i ) = 2( α i , α j ) = (cid:26) − i, j adjacent0 otherwiseThis means that vectors of the form(0 , . . . , , , , . . . , , − , , . . . , 0) = α i + α i +1 + · · · + α j − and (0 , . . . , , − , , . . . , , , , . . . , 0) = − ( α i + α i +1 + · · · + α j − )are the roots of L . Therefore the number of positive roots in L equals N ( N − . 2. As a fact we also state the number of positive roots in the other Dynkin diagrams: D N N ( N − E 36 roots E 63 roots E 120 roots Definition 5.20. Let α ∈ Z n be a positive root. The reflection s α is defined by the formula s α ( v ) = v − B ( v, α ) α. We denote s α i by s i and call these simple reflections . Remark 5.21. As a linear operator of R n , s α fixes any vector orthogonal to α and s α ( α ) = − α Therefore s α is the reflection at the hyperplane orthogonal to α , and in particular fixes B . The s i generate a subgroup W ⊆ O ( R n ), which is called the Weyl group of Γ. Since for every w ∈ W , w ( α i ) is a root, and since there are only finitely many roots, W has to be finite. Definition 5.22. Let Q be a quiver with any labeling 1 , . . . , n of the vertices. Let V = ( V , . . . , V n )be a representation of Q . We then call d ( V ) = (dim V , . . . , dim V n )the dimension vector of this representation.We are now able to formulate Gabriel’s theorem using roots. Theorem 5.23 (Gabriel’s theorem) . Let Q be a quiver of type A n , D n , E , E , E . Then Q hasfinitely many indecomposable representations. Namely, the dimension vector of any indecomposablerepresentation is a positive root (with respect to B Γ ) and for any positive root α there is exactlyone indecomposable representation with dimension vector α . .6 Reflection Functors Definition 5.24. Let Q be any quiver. We call a vertex i ∈ Q a sink if all edges connected to i point towards i . / / • i o o O O We call a vertex i ∈ Q a source if all edges connected to i point away from i . • i o o / / (cid:15) (cid:15) Definition 5.25. Let Q be any quiver and i ∈ Q be a sink (a source). Then we let Q i be thequiver obtained from Q by reversing all arrows pointing into (pointing out of) i .We are now able to define the reflection functors (also called Coxeter functors ). Definition 5.26. Let Q be a quiver, i ∈ Q be a sink. Let V be a representation of Q . Then wedefine the reflection functor F + i : Rep Q → Rep Q i by the rule F + i ( V ) k = V k if k = iF + i ( V ) i = ker ϕ : M j → i V j → V i . Also, all maps stay the same but those now pointing out of i ; these are replaced by compositionsof the inclusion of ker ϕ into ⊕ V j with the projections ⊕ V j → V k . Definition 5.27. Let Q be a quiver, i ∈ Q be a source. Let V be a representation of Q . Let ψ bethe canonical map ψ : V i → M i → j V j . Then we define the reflection functor F − i : Rep Q → Rep Q i by the rule F − i ( V ) k = V k if k = iF − i ( V ) i = Coker ( ψ ) = M i → j V j / Im ψ. Again, all maps stay the same but those now pointing into i ; these are replaced by the compositionsof the inclusions V k → ⊕ i → j V j with the natural map ⊕ V j → ⊕ V j / Im ψ . Proposition 5.28. Let Q be a quiver, V an indecomposable representation of Q . . Let i ∈ Q be a sink. Then either dim V i = 1 , dim V j = 0 for j = i or ϕ : M j → i V j → V i is surjective.2. Let i ∈ Q be a source. Then either dim V i = 1 , dim V j = 0 for j = i or ψ : V i → M i → j V j is injective.Proof. 1. Choose a complement W of Im ϕ . Then we get V = • / / • W • o o • O O ⊕ V ′ Since V is indecomposable, one of these summands has to be zero. If the first summand iszero, then ϕ has to be surjective. If the second summand is zero, then the first one has to beof the desired form, because else we could write it as a direct sum of several objects of thetype • / / • • o o • O O which is impossible, since V was supposed to be indecomposable.2. Follows similarly by splitting away the kernel of ψ . Proposition 5.29. Let Q be a quiver, V be a representation of Q .1. If ϕ : M j → i V j → V i is surjective, then F − i F + i V = V. 2. If ψ : V i → M i → j V j is injective, then F + i F − i V = V. Proof. In the following proof, we will always mean by i → j that i points into j in the originalquiver Q . We only establish the first statement and we also restrict ourselves to showing that thespaces of V and F − i F + i V are the same. It is enough to do so for the i -th space. Let ϕ : M j → i V j → V i 91e surjective and let K = ker ϕ. When applying F + i , the space V i gets replaced by K . Furthermore, let ψ : K → M j → i V j . After applying F − i , K gets replaced by K ′ = M j → i V j / (Im ψ ) . But Im ψ = K and therefore K ′ = M j → i V j / ker( ϕ : M j → i V j → V i ) = Im( ϕ : M j → i V j → V i )by the homomorphism theorem. Since ϕ was assumed to be surjective, we get K ′ = V i . Proposition 5.30. Let Q be a quiver, and V be an indecomposable representation of Q . Then F + i V and F − i V (whenever defined) are either indecomposable or 0.Proof. We prove the proposition for F + i V - the case F − i V follows similarly. By Proposition 5.28 itfollows that either ϕ : M j → i V j → V i is surjective or dim V i = 1 , dim V j = 0 , j = i . In the last case F + i V = 0 . So we can assume that ϕ is surjective. In this case, assume that F + i V is decomposable as F + i V = X ⊕ Y with X, Y = 0. But F + i V is injective at i , since the maps are canonical projections, whose directsum is the tautological embedding. Therefore X and Y also have to be injective at i and hence (by5.29) F + i F − i X = X, F + i F − i Y = Y In particular F − i X = 0 , F − i Y = 0 . Therefore V = F − i F + i V = F − i X ⊕ F − i Y which is a contradiction, since V was assumed to be indecomposable. So we can infer that F + i V is indecomposable. 92 roposition 5.31. Let Q be a quiver and V a representation of Q .1. Let i ∈ Q be a sink and let V be surjective at i . Then d ( F + i V ) = s i ( d ( V )) . 2. Let i ∈ Q be a source and let V be injective at i . Then d ( F − i V ) = s i ( d ( V )) . Proof. We only prove the first statement, the second one follows similarly. Let i ∈ Q be a sink andlet ϕ : M j → i V j → V i be surjective. Let K = ker ϕ . Thendim K = X j → i dim V j − dim V i . Therefore we get (cid:0) d ( F + i V ) − d ( V ) (cid:1) i = X j → i dim V j − V i = − B ( d ( V ) , α i )and (cid:0) d ( F + i V ) − d ( V ) (cid:1) j = 0 , j = i. This implies d ( F + i V ) − d ( V ) = − B ( d ( V ) , α i ) α i ⇔ d ( F + i V ) = d ( V ) − B ( d ( V ) , α i ) α i = s i ( d ( V )) . Definition 5.32. Let Q be a quiver and let Γ be the underlying graph. Fix any labeling 1 , . . . , n of the vertices of Γ. Then the Coxeter element c of Q corresponding to this labeling is defined as c = s s . . . s n . Lemma 5.33. Let β = X i k i α i with k i ≥ for all i but not all k i = 0 . Then there is N ∈ N , such that c N β has at least one strictly negative coefficient. roof. c belongs to a finite group W . So there is M ∈ N , such that c M = 1 . We claim that 1 + c + c + · · · + c M − = 0as operators on R n . This implies what we need, since β has at least one strictly positive coefficient,so one of the elements cβ, c β, . . . , c M − β must have at least one strictly negative one. Furthermore, it is enough to show that 1 is not aneigenvalue for c , since (1 + c + c + · · · + c M − ) v = w = 0 ⇒ cw = c (cid:0) c + c + · · · + c M − (cid:1) v = ( c + c + c + · · · + c M − + 1) v = w. Assume the contrary, i.e., 1 is a eigenvalue of c and let v be a corresponding eigenvector. cv = v ⇒ s . . . s n v = v ⇔ s . . . s n v = s v. But since s i only changes the i -th coordinate of v , we get s v = v and s . . . s n v = v. Repeating the same procedure, we get s i v = v for all i . But this means B ( v, α i ) = 0 . for all i , and since B is nondegenerate, we get v = 0. But this is a contradiction, since v is aneigenvector. Let V be an indecomposable representation of Q . We introduce a fixed labeling 1 , . . . n on Q , suchthat i < j if one can reach j from i . This is possible, since we can assign the highest label to anysink, remove this sink from the quiver, assign the next highest label to a sink of the remainingquiver and so on. This way we create a labeling of the desired kind.We now consider the sequence V (0) = V, V (1) = F + n V, V (2) = F + n − F + n V, . . . This sequence is well defined because of the selected labeling: n has to be a sink of Q , n − Q n (where Q n is obtained from Q by reversing all the arrows at the vertex r ) andso on. Furthermore, we note that V ( n ) is a representation of Q again, since every arrow has beenreversed twice (since we applied a reflection functor to every vertex). This implies that we candefine V ( n +1) = F + n V ( n ) , . . . and continue the sequence to infinity. 94 heorem 5.34. There is m ∈ N , such that d (cid:16) V ( m ) (cid:17) = α p for some p .Proof. If V ( i ) is surjective at the appropriate vertex k , then d (cid:16) V ( i +1) (cid:17) = d (cid:16) F + k V ( i ) (cid:17) = s k d (cid:16) V ( i ) (cid:17) . This implies, that if V (0) , . . . , V ( i − are surjective at the appropriate vertices, then d (cid:16) V ( i ) (cid:17) = . . . s n − s n d ( V ) . By Lemma 5.33 this cannot continue indefinitely - since d (cid:0) V ( i ) (cid:1) may not have any negative entries.Let i be smallest number such that V ( i ) is not surjective at the appropriate vertex. By Proposition5.30 it is indecomposable. So, by Proposition 5.28, we get d ( V ( i ) ) = α p for some p .We are now able to prove Gabriel’s theorem. Namely, we get the following corollaries. Corollary 5.35. Let Q be a quiver, V be any indecomposable representation. Then d ( V ) is apositive root.Proof. By Theorem 5.34 s i . . . s i m ( d ( V )) = α p . Since the s i preserve B , we get B ( d ( V ) , d ( V )) = B ( α p , α p ) = 2 . Corollary 5.36. Let V, V ′ be indecomposable representations of Q such that d ( V ) = d ( V ′ ) . Then V and V ′ are isomorphic.Proof. Let i be such that d (cid:16) V ( i ) (cid:17) = α p . Then we also get d (cid:0) V ′ ( i ) (cid:1) = α p . So V ′ ( i ) = V ( i ) =: V i . Furthermore we have V ( i ) = F + k . . . F + n − F + n V (0) V ′ ( i ) = F + k . . . F + n − F + n V ′ (0) . But both V ( i − , . . . , V (0) and V ′ ( i − , . . . , V ′ (0) have to be surjective at the appropriate vertices.This implies F − n F − n − . . . F − k V i = (cid:26) F − n F − n − . . . F − k F + k . . . F + n − F + n V (0) = V (0) = VF − n F − n − . . . F − k F + k . . . F + n − F + n V ′ (0) = V ′ (0) = V ′ Corollary 5.37. For every positive root α , there is an indecomposable representation V with d ( V ) = α. Proof. Consider the sequence s n α, s n − s n α, . . . Consider the first element of this sequence which is a negative root (this has to happen by Lemma5.33) and look at one step before that, calling this element β . So β is a positive root and s i β is anegative root for some i . But since the s i only change one coordinate, we get β = α i and ( s q . . . s n − s n ) α = α i . We let C ( i ) be the representation having dimension vector α i . Then we define V = F − n F − n − . . . F − q C ( i ) . This is an indecomposable representation and d ( V ) = α. Example 5.38. Let us demonstrate by example how reflection functors work. Consider the quiver D with the orientation of all arrows towards the node (which is labeled by 4). Start with the1-dimensional representation V α sitting at the 4-th vertex. Apply to V α the functor F − F − F − .This yields F − F − F − V α = V α + α + α + α . Now applying F − we get F − F − F − F − V α = V α + α + α +2 α . Note that this is exactly the inclusion of 3 lines into the plane, which is the most complicatedindecomposable representation of the D quiver. Problem 5.39. Let Q n be the cyclic quiver of length n , i.e., n vertices connected by n oriented edgesforming a cycle. Obviously, the classification of indecomposable representations of Q is given bythe Jordan normal form theorem. Obtain a similar classification of indecomposable representationsof Q . In other words, classify pairs of linear operators A : V → W and B : W → V up toisomorphism. Namely:(a) Consider the following pairs (for n ≥ ):1) E n,λ : V = W = C n , A is the Jordan block of size n with eigenvalue λ , B = 1 ( λ ∈ C ).2) E n, ∞ : is obtained from E n, by exchanging V with W and A with B . ) H n : V = C n with basis v i , W = C n − with basis w i , Av i = w i , Bw i = v i +1 for i < n , and Av n = 0 .4) K n is obtained from H n by exchanging V with W and A with B .Show that these are indecomposable and pairwise nonisomorphic.(b) Show that if E is a representation of Q such that AB is not nilpotent, then E = E ′ ⊕ E ′′ ,where E ′′ = E n,λ for some λ = 0 .(c) Consider the case when AB is nilpotent, and consider the operator X on V ⊕ W givenby X ( v, w ) = ( Bw, Av ) . Show that X is nilpotent, and admits a basis consisting of chains (i.e.,sequences u, Xu, X u, ...X l − u where X l u = 0 ) which are compatible with the direct sum decompo-sition (i.e., for every chain u ∈ V or u ∈ W ). Deduce that (1)-(4) are the only indecomposablerepresentations of Q .(d)(harder!) generalize this classification to the Kronecker quiver, which has two vertices and and two edges both going from to .(e)(still harder!) can you generalize this classification to Q n , n > , with any orientation? Problem 5.40. Let L ⊂ Z be the lattice of vectors where the coordinates are either all integersor all half-integers (but not integers), and the sum of all coordinates is an even integer.(a) Let α i = e i − e i +1 , i = 1 , ..., , α = e + e , α = − / P i =1 e i . Show that α i are a basisof L (over Z ).(b) Show that roots in L (under the usual inner product) form a root system of type E (computethe inner products of α i ).(c) Show that the E and E lattices can be obtained as the sets of vectors in the E lattice L where the first two, respectively three, coordinates (in the basis e i ) are equal.(d) Show that E , E , E have 72,126,240 roots, respectively (enumerate types of roots in termsof the presentations in the basis e i , and count the roots of each type). Problem 5.41. Let V α be the indecomposable representation of a Dynkin quiver Q which corre-sponds to a positive root α . For instance, if α i is a simple root, then V α i has a 1-dimensional spaceat i and 0 everywhere else.(a) Show that if i is a source then Ext ( V, V α i ) = 0 for any representation V of Q , and if i isa sink, then Ext ( V α i , V ) = 0 .(b) Given an orientation of the quiver, find a Jordan-H¨older series of V α for that orientation. Introduction to categories We have now seen many examples of representation theories and of operations with representations(direct sum, tensor product, induction, restriction, reflection functors, etc.) A context in which onecan systematically talk about this is provided by Category Theory .Category theory was founded by Saunders MacLane and Samuel Eilenberg around 1940. It is afairly abstract theory which seemingly has no content, for which reason it was christened “abstractnonsense”. Nevertheless, it is a very flexible and powerful language, which has become totallyindispensable in many areas of mathematics, such as algebraic geometry, topology, representationtheory, and many others.We will now give a very short introduction to Category theory, highlighting its relevance to thetopics in representation theory we have discussed. For a serious acquaintance with category theory,the reader should use the classical book [McL]. Definition 6.1. A category C is the following data:(i) a class of objects Ob ( C );(ii) for every objects X, Y ∈ Ob ( C ), the class Hom C ( X, Y ) = Hom( X, Y ) of morphisms (orarrows) from X, Y (for f ∈ Hom( X, Y ), one may write f : X → Y );(iii) For any objects X, Y, Z ∈ Ob ( C ), a composition map Hom( Y, Z ) × Hom( X, Y ) → Hom( X, Z ),( f, g ) f ◦ g ,which satisfy the following axioms:1. The composition is associative, i.e., ( f ◦ g ) ◦ h = f ◦ ( g ◦ h );2. For each X ∈ Ob ( C ), there is a morphism 1 X ∈ Hom( X, X ), called the unit morphism, suchthat 1 X ◦ f = f and g ◦ X = g for any f, g for which compositions make sense. Remark. We will write X ∈ C instead of X ∈ Ob ( C ). Example 6.2. 1. The category Sets of sets (morphisms are arbitrary maps).2. The categories Groups , Rings (morphisms are homomorphisms).3. The category Vect k of vector spaces over a field k (morphisms are linear maps).4. The category Rep( A ) of representations of an algebra A (morphisms are homomorphisms ofrepresentations).5. The category of topological spaces (morphisms are continuous maps).6. The homotopy category of topological spaces (morphisms are homotopy classes of continuousmaps). Important remark. Unfortunately, one cannot simplify this definition by replacing the word“class” by the much more familiar word “set”. Indeed, this would rule out the important Example 1,as it is well known that there is no set of all sets, and working with such a set leads to contradictions.The precise definition of a class and the precise distinction between a class and a set is the subjectof set theory, and cannot be discussed here. Luckily, for most practical purposes (in particular, inthese notes), this distinction is not essential. 98e also mention that in many examples, including examples 1-6, the word “class” in (ii) canbe replaced by “set”. Categories with this property (that Hom( X, Y ) is a set for any X, Y ) arecalled locally small; many categories that we encounter are of this kind.Sometimes the collection Hom( X, Y ) of morphisms from X to Y in a given locally small category C is not just a set but has some additional structure (say, the structure of an abelian group, or avector space over some field). In this case one says that C is enriched over another category D (which is a monoidal category, i.e., has a product operation and a unit object under this product, e.g.the category of abelian groups or vector spaces with the tensor product operation). This means thatfor each X, Y ∈ C , Hom( X, Y ) is an object of D , and the composition Hom( Y, Z ) × Hom( X, Y ) → Hom( X, Z ) is a morphism in D . E.g., if D is the category of vector spaces, this means that thecomposition is bilinear, i.e. gives rise to a linear map Hom( Y, Z ) ⊗ Hom( X, Y ) → Hom( X, Z ). Fora more detailed discussion of this, we refer the reader to [McL]. Example. The category Rep( A ) of representations of a k -algebra A is enriched over thecategory of k -vector spaces. Definition 6.3. A full subcategory of a category C is a category C ′ whose objects are a subclassof objects of C , and Hom C ′ ( X, Y ) = Hom C ( X, Y ). Example. The category AbelianGroups is a full subcategory of the category Groups . We would like to define arrows between categories. Such arrows are called functors . Definition 6.4. A functor F : C → D between categories C and D is(i) a map F : Ob ( C ) → Ob ( D );(ii) for each X, Y ∈ C , a map F = F X,Y : Hom( X, Y ) → Hom( F ( X ) , F ( Y )) which preservescompositions and identity morphisms.Note that functors can be composed in an obvious way. Also, any category has the identityfunctor. Example 6.5. 1. A (locally small) category C with one object X is the same thing as a monoid.A functor between such categories is a homomorphism of monoids.2. Forgetful functors Groups → Sets , Rings → AbelianGroups .3. The opposite category of a given category is the same category with the order of arrows andcompositions reversed. Then V V ∗ is a functor Vect k Vect opk .4. The Hom functors: If C is a locally small category then we have the functor C → Sets givenby Y Hom( X, Y ) and C op → Sets given by Y Hom( Y, X ).5. The assignment X Fun( X, Z ) is a functor Sets → Rings op .6. Let Q be a quiver. Consider the category C ( Q ) whose objects are the vertices and morphismsare oriented paths between them. Then functors from C ( Q ) to Vect k are representations of Q over k . 7. Let K ⊂ G be groups. Then we have the induction functor Ind GK : Rep( K ) → Rep( G ), andRes GK : Rep( G ) → Rep( K ). 99. We have an obvious notion of the Cartesian product of categories (obtained by taking theCartesian products of the classes of objects and morphisms of the factors). The functors of directsum and tensor product are then functors Vect k × Vect k → Vect k . Also the operations V V ⊗ n , V S n V , V 7→ ∧ n V are functors on Vect k . More generally, if π is a representation of S n , wehave functors V Hom S n ( π, V ⊗ n ). Such functors (for irreducible π ) are called the Schur functors.They are labeled by Young diagrams.9. The reflection functors F ± i : Rep( Q ) → Rep( ¯ Q i ) are functors between representation cate-gories of quivers. One of the important features of functors between categories which distinguishes them from usualmaps or functions is that the functors between two given categories themselves form a category,i.e., one can define a nontrivial notion of a morphism between two functors. Definition 6.6. Let C , D be categories and F, G : C → D be functors between them. A morphism a : F → G (also called a natural transformation or a functorial morphism) is a collection ofmorphisms a X : F ( X ) → G ( X ) labeled by the objects X of C , which is functorial in X , i.e., forany morphism f : X → Y (for X, Y ∈ C ) one has a Y ◦ F ( f ) = G ( f ) ◦ a X .A morphism a : F → G is an isomorphism if there is another morphism a − : G → F such that a ◦ a − and a − ◦ a are the identities. The set of morphisms from F to G is denoted by Hom( F, G ). Example 6.7. 1. Let FVect k be the category of finite dimensional vector spaces over k . Then thefunctors id and ∗∗ on this category are isomorphic. The isomorphism is defined by the standardmaps a V : V → V ∗∗ given by a V ( u )( f ) = f ( u ), u ∈ V , f ∈ V ∗ . But these two functors are notisomorphic on the category of all vector spaces Vect k , since for an infinite dimensional vector space V , V is not isomorphic to V ∗∗ .2. Let FVect ′ k be the category of finite dimensional k -vector spaces, where the morphismsare the isomorphisms. We have a functor F from this category to itself sending any space V to V ∗ and any morphism a to ( a ∗ ) − . This functor satisfies the property that V is isomorphic to F ( V ) for any V , but it is not isomorphic to the identity functor. This is because the isomorphism V → F ( V ) = V ∗ cannot be chosen to be compatible with the action of GL ( V ), as V is notisomorphic to V ∗ as a representation of GL ( V ).3. Let A be an algebra over a field k , and F : A − mod → Vect k be the forgetful functor.Then as follows from Problem 1.22, End F = Hom( F, F ) = A .4. The set of endomorphisms of the identity functor on the category A − mod is the center of A (check it!). When two algebraic or geometric objects are isomorphic, it is usually not a good idea to say thatthey are equal (i.e., literally the same). The reason is that such objects are usually equal in manydifferent ways, i.e., there are many ways to pick an isomorphism, but by saying that the objects areequal we are misleading the reader or listener into thinking that we are providing a certain choiceof the identification, which we actually do not do. A vivid example of this is a finite dimensionalvector space V and its dual space V ∗ . 100or this reason in category theory, one most of the time tries to avoid saying that two objectsor two functors are equal. In particular, this applies to the definition of isomorphism of categories.Namely, the naive notion of isomorphism of categories is defined in the obvious way: a functor F : C → D is an isomorphism if there exists F − : D → C such that F ◦ F − and F − ◦ F are equalto the identity functors. But this definition is not very useful. We might suspect so since we haveused the word “equal” for objects of a category (namely, functors) which we are not supposed todo. And in fact here is an example of two categories which are “the same for all practical purposes”but are not isomorphic; it demonstrates the deficiency of our definition.Namely, let C be the simplest possible category: Ob ( C ) consists of one object X , withHom( X, X ) = { X } . Also, let C have two objects X, Y and 4 morphisms: 1 X , Y , a : X → Y and b : Y → X . So we must have a ◦ b = 1 Y , b ◦ a = 1 X .It is easy to check that for any category D , there is a natural bijection between the collectionsof isomorphism classes of functors C → D and C → D (both are identified with the collection ofisomorphism classes of objects of D ). This is what we mean by saying that C and C are “the samefor all practical purposes”. Nevertheless they are not isomorphic, since C has one object, and C has two objects (even though these two objects are isomorphic to each other).This shows that we should adopt a more flexible and less restrictive notion of isomorphism ofcategories. This is accomplished by the definition of an equivalence of categories.Definition 6.8. A functor F : C → D is an equivalence of categories if there exists F ′ : D → C such that F ◦ F ′ and F ′ ◦ F are isomorphic to the identity functors.In this situation, F ′ is said to be a quasi-inverse to F .In particular, the above categories C and C are equivalent (check it!).Also, the category FSet of finite sets is equivalent to the category whose objects are nonneg-ative integers, and morphisms are given by Hom( m, n ) = Maps( { , ..., m } , { , ..., n } ). Are thesecategories isomorphic? The answer to this question depends on whether you believe that thereis only one finite set with a given number of elements, or that there are many of those. It seemsbetter to think that there are many (without asking “how many”), so that isomorphic sets need notbe literally equal, but this is really a matter of choice. In any case, this is not really a reasonablequestion; the answer to this question is irrelevant for any practical purpose, and thinking about itwill give you nothing but a headache. A fundamental notion in category theory is that of a representable functor . Namely, let C be a(locally small) category, and F : C → Sets be a functor. We say that F is representable if thereexists an object X ∈ C such that F is isomorphic to the functor Hom( X, ?). More precisely, if weare given such an object X , together with an isomorphism ξ : F ∼ = Hom( X, ?), we say that thefunctor F is represented by X (using ξ ).In a similar way, one can talk about representable functors from C op to Sets . Namely, onecalls such a functor representable if it is of the form Hom(? , X ) for some object X ∈ C , up to anisomorphism.Not every functor is representable, but if a representing object X exists, then it is unique.Namely, we have the following lemma. 101 emma 6.9. (The Yoneda Lemma) If a functor F is represented by an object X , then X is uniqueup to a unique isomorphism. I.e., if X, Y are two objects in C , then for any isomorphism of functors φ : Hom ( X, ?) → Hom ( Y, ?) there is a unique isomorphism a φ : X → Y inducing φ .Proof. (Sketch) One sets a φ = φ − Y (1 Y ), and shows that it is invertible by constructing the inverse,which is a − φ = φ X (1 X ). It remains to show that the composition both ways is the identity, whichwe will omit here. This establishes the existence of a φ . Its uniqueness is verified in a straightforwardmanner. Remark. In a similar way, if a category C is enriched over another category D (say, the categoryof abelian groups or vector spaces), one can define the notion of a representable functor from C to D . Example 6.10. Let A be an algebra. Then the forgetful functor to vector spaces on the categoryof left A -modules is representable, and the representing object is the free rank 1 module (=theregular representation) M = A . But if A is infinite dimensional, and we restrict attention to thecategory of finite dimensional modules, then the forgetful functor, in general, is not representable(this is so, for example, if A is the algebra of complex functions on Z which are zero at all pointsbut finitely many). Another fundamental notion in category theory is the notion of adjoint functors . Definition 6.11. Functors F : C → D and G : D → C are said to be a pair of adjoint functors if forany X ∈ C , Y ∈ D we are given an isomorphism ξ XY : Hom C ( F ( X ) , Y ) → Hom D ( X, G ( Y )) which isfunctorial in X and Y ; in other words, if we are given an isomorphism of functors Hom( F (?) , ?) → Hom(? , G (?)) ( C × D → Sets ). In this situation, we say that F is left adjoint to G and G is rightadjoint to F .Not every functor has a left or right adjoint, but if it does, it is unique and can be constructedcanonically (i.e., if we somehow found two such functors, then there is a canonical isomorphismbetween them). This follows easily from the Yoneda lemma, as if F, G are a pair of adjoint functorsthen F ( X ) represents the functor Y Hom( X, G ( Y )), and G ( Y ) represents the functor X Hom( F ( X ) , Y ). Remark 6.12. The terminology “left and right adjoint functors” is motivated by the analogybetween categories and inner product spaces. More specifically, we have the following useful dic-tionary between category theory and linear algebra, which helps understand better many notionsof category theory. 102ictionary between category theory and linear algebraCategory C Vector space V with a nondegenerate inner productThe set of morphisms Hom( X, Y ) Inner product ( x, y ) on V (maybe nonsymmetric)Opposite category C op Same space V with reversed inner productThe category Sets The ground field k Full subcategory in C Nondegenerate subspace in V Functor F : C → D Linear operator f : V → W Functor F : C → Sets Linear functional f ∈ V ∗ = Hom( V, k )Representable functor Linear functional f ∈ V ∗ given by f ( v ) = ( u, v ), u ∈ V Yoneda lemma Nondegeneracy of the inner product (on both sides)Not all functors are representable If dim V = ∞ , not ∀ f ∈ V ∗ , f ( v ) = ( u, v )Left and right adjoint functors Left and right adjoint operatorsAdjoint functors don’t always exist Adjoint operators may not exist if dim V = ∞ If they do, they are unique If they do, they are uniqueLeft and right adjoints may not coincide The inner product may be nonsymmetric Example 6.13. 1. Let V be a finite dimensional representation of a group G or a Lie algebra g .Then the left and right adjoint to the functor V ⊗ on the category of representations of G is thefunctor V ∗ ⊗ .2. The functor Res GK is left adjoint to Ind GK . This is nothing but the statement of the Frobeniusreciprocity.3. Let Assoc k be the category of associative unital algebras, and Lie k the category of Liealgebras over some field k . We have a functor L : Assoc k → Lie k , which attaches to an associativealgebra the same space regarded as a Lie algebra, with bracket [ a, b ] = ab − ba . Then the functor L has a left adjoint, which is the functor U of taking the universal enveloping algebra of a Lie algebra.4. We have the functor GL : Assoc k → Groups , given by A GL ( A ) = A × . This functorhas a left adjoint, which is the functor G k [ G ], the group algebra of G .5. The left adjoint to the forgetful functor Assoc k → Vect k is the functor of tensor algebra: V T V . Also, if we denote by Comm k the category of commutative algebras, then the left adjointto the forgetful functor Comm k → Vect k is the functor of the symmetric algebra: V SV .One can give many more examples, spanning many fields. These examples show that adjointfunctors are ubiquitous in mathematics. The type of categories that most often appears in representation theory is abelian categories .The standard definition of an abelian category is rather long, so we will not give it here, referringthe reader to the textbook [Fr]; rather, we will use as the definition what is really the statement ofthe Freyd-Mitchell theorem: Definition 6.14. An abelian category is a category (enriched over the category of abelian groups),which is equivalent to a full subcategory C of the category A -mod of left modules over a ring A ,closed under taking finite direct sums, as well as kernels, cokernels, and images of morphisms.We see from this definition that in an abelian category, Hom( X, Y ) is an abelian group for each X, Y , compositions are group homomorphisms with respect to each argument, there is the zero ob-ject, the notion of an injective morphism (monomorphism) and surjective morphism (epimorphism),and every morphism has a kernel, a cokernel, and an image.103 xample 6.15. The category of modules over an algebra A and the category of finite dimensionalmodules over A are abelian categories. Remark 6.16. The good thing about Definition 6.14 is that it allows us to visualize objects,morphisms, kernels, and cokernels in terms of classical algebra. But the definition also has a bigdrawback, which is that even if C is the whole category A -mod, the ring A is not determined by C .In particular, two different rings can have equivalent categories of modules (such rings are called Morita equivalent ). Actually, it is worse than that: for many important abelian categories thereis no natural (or even manageable) ring A at all. This is why people prefer to use the standarddefinition, which is free from this drawback, even though it is more abstract.We say that an abelian category C is k - linear if the groups Hom C ( X, Y ) are equipped witha structure of a vector space over k , and composition maps are k -linear in each argument. Inparticular, the categories in Example 6.15 are k -linear. Definition 6.17. A sequence of objects and morphisms X → X → ... → X n +1 in an abelian category is said to be a complex if the composition of any two consecutive arrowsis zero. The cohomology of this complex is H i = Ker ( d i ) / Im( d i − ), where d i : X i → X i +1 (thusthe cohomology is defined for 1 ≤ i ≤ n ). The complex is said to be exact in the i -th term if H i = 0, and is said to be an exact sequence if it is exact in all terms. A short exact sequence is an exact sequence of the form 0 → X → Y → Z → . Clearly, 0 → X → Y → Z → X → Y is injective, Y → Z is surjective, and the induced map Y /X → Z is an isomorphism. Definition 6.18. A functor F between two abelian categories is additive if it induces homomor-phisms on Hom groups. Also, for k -linear categories one says that F is k -linear if it induces k -linearmaps between Hom spaces.It is easy to show that if F is an additive functor, then F ( X ⊕ Y ) is canonically isomorphic to F ( X ) ⊕ F ( Y ). Example 6.19. The functors Ind GK , Res GK , Hom G ( V, ?) in the theory of group representations overa field k are additive and k -linear. Definition 6.20. An additive functor F : C → D between abelian categories is left exact if forany exact sequence 0 → X → Y → Z, the sequence 0 → F ( X ) → F ( Y ) → F ( Z )is exact. F is right exact if for any exact sequence X → Y → Z → , the sequence F ( X ) → F ( Y ) → F ( Z ) → F is exact if it is both left and right exact.104 efinition 6.21. An abelian category C is semisimple if any short exact sequence in this categorysplits, i.e., is isomorphic to a sequence0 → X → X ⊕ Y → Y → Example 6.22. The category of representations of a finite group G over a field of characteristicnot dividing | G | (or 0) is semisimple.Note that in a semisimple category, any additive functor is automatically exact on both sides. Example 6.23. (i) The functors Ind GK , Res GK are exact.(ii) The functor Hom( X, ?) is left exact, but not necessarily right exact. To see that it need notbe right exact, it suffices to consider the exact sequence0 → Z → Z → Z / Z → , and apply the functor Hom( Z / Z , ?).(iii) The functor X ⊗ A for a right A -module X (on the category of left A -modules) is right exact,but not necessarily left exact. To see this, it suffices to tensor multiply the above exact sequenceby Z / Z . Exercise. Show that if ( F, G ) is a pair of adjoint additive functors between abelian categories,then F is right exact and G is left exact. Exercise. (a) Let Q be a quiver and i ∈ Q a source. Let V be a representation of Q , and W arepresentation of Q i (the quiver obtained from Q by reversing arrows at the vertex i ). Prove thatthere is a natural isomorphism between Hom (cid:0) F − i V, W (cid:1) and Hom (cid:0) V, F + i W (cid:1) . In other words, thefunctor F + i is right adjoint to F − i .(b) Deduce that the functor F + i is left exact, and F − i is right exact.105 Structure of finite dimensional algebras In this section we return to studying the structure of finite dimensional algebras. Throughout thesection, we work over an algebraically closed field k (of any characteristic). Let A be an algebra, and P be a left A -module. Theorem 7.1. The following properties of P are equivalent:(i) If α : M → N is a surjective morphism, and ν : P → N any morphism, then there exists amorphism µ : P → M such that α ◦ µ = ν .(ii) Any surjective morphism α : M → P splits, i.e., there exists µ : P → M such that α ◦ µ = id .(iii) There exists another A -module Q such that P ⊕ Q is a free A -module, i.e., a direct sum ofcopies of A .(iv) The functor Hom A ( P, ?) on the category of A -modules is exact.Proof. To prove that (i) implies (ii), take N = P . To prove that (ii) implies (iii), take M to be free(this can always be done since any module is a quotient of a free module). To prove that (iii) implies(iv), note that the functor Hom A ( P, ?) is exact if P is free (as Hom A ( A, N ) = N ), so the statementfollows, as if the direct sum of two complexes is exact, then each of them is exact. To prove that(iv) implies (i), let K be the kernel of the map α , and apply the exact functor Hom A ( P, ?) to theexact sequence 0 → K → M → N → . Definition 7.2. A module satisfying any of the conditions (i)-(iv) of Theorem 7.1 is said to be projective . Let A be a ring, and I ⊂ A a nilpotent ideal. Proposition 7.3. Let e ∈ A/I be an idempotent, i.e., e = e . There exists an idempotent e ∈ A which is a lift of e (i.e., it projects to e under the reduction modulo I ). This idempotent is uniqueup to conjugation by an element of I .Proof. Let us first establish the statement in the case when I = 0. Note that in this case I is aleft and right module over A/I . Let e ∗ be any lift of e to A . Then e ∗ − e ∗ = a ∈ I , and e a = ae .We look for e in the form e = e ∗ + b , b ∈ I . The equation for b is e b + be − b = a .Set b = (2 e − a . Then e b + be − b = 2 e a − (2 e − a = a, so e is an idempotent. To classify other solutions, set e ′ = e + c . For e ′ to be an idempotent, wemust have ec + ce − c = 0. This is equivalent to saying that ece = 0 and (1 − e ) c (1 − e ) = 0, so c = ec (1 − e ) + (1 − e ) ce = [ e, [ e, c ]]. Hence e ′ = (1 + [ c, e ]) e (1 + [ c, e ]) − .106ow, in the general case, we prove by induction in k that there exists a lift e k of e to A/I k +1 ,and it is unique up to conjugation by an element of 1 + I k (this is sufficient as I is nilpotent).Assume it is true for k = m − 1, and let us prove it for k = m . So we have an idempotent e m − ∈ A/I m , and we have to lift it to A/I m +1 . But ( I m ) = 0 in A/I m +1 , so we are done. Definition 7.4. A complete system of orthogonal idempotents in a unital algebra B is a collectionof elements e , ..., e n ∈ B such that e i e j = δ ij e i , and P ni =1 e i = 1. Corollary 7.5.