aa r X i v : . [ m a t h . R T ] J u l ALGEBRAIC RELATIONS VIA A MONTE CARLO SIMULATION
ALISON ELAINE BECKERABSTRACT. The conjugation action of the complex orthogonal group on the polynomial functions on n × n matrices gives rise to a graded algebra of invariant polynomials. A spanning set of this algebra is inbijective correspondence to a set of unlabeled, cyclic graphs with directed edges equivalent under dihedralsymmetries. When the degree of the invariants is n + 1, we show that the dimension of the space ofrelations between the invariants grows linearly in n . Furthermore, we present two methods to obtain abasis of the space of relations. First, we construct a basis using an idempotent of the group algebrareferred to as Young symmetrizers, but this quickly becomes computationally expensive as n increases.Thus, we propose a more computationally efficient method for this problem by repeatedly generatingrandom matrices using a Monte Carlo algorithm. The goal of this work is to understand the relations between invariants of the conjugationaction of the complex orthogonal group, O n ( C ), on the polynomial functions on n × n matri-ces, P ( M n ). We present results about the dimension of the space of relations, denoted RE L ,and further discuss two methods for obtaining a basis of this space. First, we construct abasis of relations by using Young symmetrizers, however, this can be computationally expen-sive. Thus we propose a more efficient method to determine a basis of
RE L by repeatedlygenerating random matrices via a Monte Carlo algorithm.Throughout this paper we let GL n denote the complex general linear group, and we let O n denote the complex orthogonal group. In Section 3 we discuss our main results about thedimension of RE L when the degree of the polynomial invariants is equal to n + 1. Again,these are the invariants under the conjugation action of O n on P ( M n ), g · f ( x ) = f ( g T xg )where x ∈ M n , f ∈ P ( M n ) and g ∈ O n .We have that P d ( M n ), the homogeneous polynomials of degree d , are a finite dimensionalrepresentation of O n . Therefore, we have a graded structure on P ( M n ), P ( M n ) = M d ≥ P d ( M n )and thus a graded algebra of invariant polynomials, P ( M n ) O n . It is shown in [7] that thisalgebra is generated by functions of the form T r ( x a ( x T ) a x a ( x T ) a · · · x a m − ( x T ) a m ) for x ∈ M n . Products of these functions span the set of invariants, however, they are not linearlyindependent if the degree of the polynomials is greater than n .There are no relations between invariants in what we define as the stable range , wherethe degree of the invariant polynomials is smaller than n [12]. The first time relations arise isjust outside of the stable range, where the degree of the polynomials is n + 1. The following1heorem gives the dimension of the space of relations between the polynomial invariants, RE L n +1 , which we see grows linearly in n . Theorem.
Let n be a positive integer. The dimension of the space of relations, RE L n +1 ,between the degree n + 1 invariants of the O n conjugation action on P ( M n ) is equal to dim ( RE L n +1 ) = n n even n + 32 n oddIn order to prove Theorem 1, we rely on correspondences between the elements of P ( M n ) O n and several different spaces. In Section 2 we discuss a useful bijection from [12] be-tween the invariant polynomials and Necklace diagrams, which are unlabeled, cyclic graphsembedded in the plane.Next, we present two ways to determine a basis of RE L . In Section 4 we explicitlydescribe the relations using idempotents of the group algebra, C [ S n ], called Young sym-metrizers. However, the size of these elements grows exponentially in n , and thus it becomescomputationally overwhelming to obtain a basis using this method as n increases.Therefore, in Chapter 5, we propose a new approach to explicitly define a basis of therelations that is more efficient and does not rely on Young symmetrizers. This algorithmis designed using a Monte Carlo simulation that repeatedly generates random matrices togive numerical values to the invariants, and we solve a linear system to recover a basis ofrelations.All of our code is written in Python and Sage [9] and run on the SageMath cloud servers. . n invariant polynomials on matrices. Consider the complex general linear group, GL n , and the polynomial functions on n × n matrices, P ( M n ). There is a conjugation actionof GL n on P ( M n ) defined by g · f ( x ) = f ( g − xg )for g ∈ GL n , x ∈ M n , and f ∈ P ( M n ).We can see that T r ( x ) is invariant under this action, since g · T r ( x ) = T r ( g − xg ) = T r ( gg − x ) = T r ( x )It can be shown [7] that T r ( x k ) is invariant for k ∈ N , k ≤ n , and these functionsgenerate the invariant algebra P ( M n ) GL n . Furthermore, there are no relations between theseinvariants. Thus, we restrict the adjoint action of GL n to the action of a subgroup of GL n ,the complex orthogonal group O n , on P ( M n ) defined by g · f ( x ) = f ( g T xg )2he homogeneous polynomials of degree d , denoted P d ( M n ), are a finite dimensional repre-sentation of O n . Therefore, we have a graded algebra structure, P ( M n ) = M d P d ( M n )and a similar grading of the algebra of polynomial invariants, P ( M n ) O n .The polynomial invariants under the GL n action are also invariant under the orthogonalgroup, but there are additional invariants under the conjugation action of O n . It is shownin [7], that P ( M n ) O n is generated by traces of monomials in x and x T , T r ( x a ( x T ) a x a ( x T ) a · · · x a m − ( x T ) a m )for x ∈ M n , a i ∈ Z + .Products of these polynomial functions span the set of invariants. When the degree ofthe invariants is less than or equal to n , there are no relations between the polynomials. Thisis called the stable range, defined below. Definition 1.
For d, n ∈ N , we define the ordered pair ( d, n ) to be in the stable range if d ≤ n . If the degree, d , of the polynomials is greater than the dimension of the defining repre-sentation, the spanning set of invariants is not linearly independent. We look at the firstoccurrence of relations between invariants, which happens just outside of the stable rangewhere the invariant polynomials have degree d = n + 1. There is a useful correspondencebetween these polynomials and unlabeled cyclic graphs called Necklace diagrams. . There is a bijective correspondence between unlabeled cyclicgraphs with oriented edges and the polynomial generators of the invariant ring P ( M n ) O n detailed in [12]. These graphs are called Necklace diagrams, defined as follows. Definition 2.
Let m ∈ N . We embed an unlabeled, directed, cyclic graph with m nodes inthe plane and centered at the origin; each edge is given an orientation of clockwise or coun-terclockwise determined by choosing an arbitrary edge, E , and noting the direction traveledto subsequent edges.These graphs, called Necklace diagrams and denoted N m , have the following structure: • An edge may join a node to itself • At most two edges may join two different nodesFurthermore, diagrams are considered equivalent under dihedral symmetries of rotation andreflection.
Let D d denote the set of all (not necessarily connected) Necklace diagrams in which eachconnected component is a Necklace diagram N m i , and there are d total nodes in the diagram.That is, define D d = { N m i | N m i is a Necklace diagram with m i nodes, and X i m i = d } . emark. The combinatorics presented here regarding the construction of Necklace diagramsis also seen in [1]. Here, we allow a diagram with d nodes to consist of possibly disconnectedcomponents. In [1], the author considers the number of connected diagrams with no propersubpattern.In the stable range where d ≤ n , the dimension of the space of invariants under the O n conjugation action is exactly equal to the number of Necklace diagrams with d total nodes,that is, | D d | = dim P d ( M n ) O n . As a result, there are no relations between the invariants in this setting. Thus, we presenta new result about the dimension of the invariant space outside of the stable range, whererelations between the invariants arise. To aid in this discussion, it is useful to note a corre-spondence between Necklace diagrams and a set of fixed-point free involutions.By definition, the set of all fixed-point free involutions on a set A is the set of all trans-positions of elements in A . Let I n denote the set of all fixed-point free involutions on the set { , , . . . , n } . Then by [12], we have a bijective correspondenceΘ : D d −→ I d between the set of Necklace diagrams with d total nodes and involutions without fixed pointson { , , . . . , d } . Thus, we are free to consider the degree d elements of P ( M n ) O n as products of traces of x and x T , or as Necklace diagrams, or as fixed-point free involutions of the set { , , . . . , d } .In the next section, we describe an important correspondence between these fixed-point freeinvolutions, I d , and a set of double cosets of the symmetric group. Consider the symmetricgroup S n , and note the following inclusion:∆ S n ⊆ S n × S n ֒ → S n (2.1)where ∆ S n denotes the diagonally embedded copy of S n in S n × S n , that is,∆ S n = { ( σ, σ ) | σ ∈ S n } Additionally, we consider S n × S n where the first copy of S n is composed of permutations ofthe set { , , . . . , n } and the second copy of S n is permutations of the set { n +1 , n +2 , . . . , n } .Then we have an embedding of S n × S n into S n , and thus we can consider ∆ S n and S n × S n as subgroups of S n .We define the following element of S n using disjoint cycle notation: τ = (1 2)(3 4) · · · ( i i + 1) · · · (2 n − n )and we define H n as the centralizer of τ in the group S n , H n = { σ ∈ S n | στ = τ σ } . τ is also an element of I n . Now, we can construct a set of doublecosets of the group S n as follows:(∆ S n ) \ S n /H n = { ( S n ) σH n | σ ∈ S n } There is a bijection, [12], between these double cosets and elements of the invariant algebra P ( M n ) O n . We note that this construction of invariants is used in our code to find theelements of P ( M n ) O n .In the following sections our goal is to understand the dimension of the space of relationsbetween the elements of P ( M n ) O n ; in order to do this we rely on the combinatorial objectcalled Littlewood-Richardson numbers. There are several equivalent definitions of Littlewood-Richardson numbers, for this work we consider them as the coefficients that arise from in-ducing a representation from a subgroup of a symmetric group to the entire symmetricgroup.
Definition 3.
Let µ and ν be partitions of positive integers n and m , respectively. Thenthe module Y µ ⊗ Y ν is naturally an ( S n × S m ) -module. Thus we can induce to an S n + m representation, Ind S n + m S n × S m Y µ ⊗ Y ν = M λ ⊢ n + m c λµν Y λ (2.2) where the coefficients, c λµν are Littlewood-Richardson numbers . Furthermore, the Littlewood-Richardson rule states that the coefficients c λµν , count thenumber of skew semi-standard Young tableaux [2] of shape λ/µ with weight ν , with theadditional restriction that the concatenation of the reversed rows is a lattice word.It can be shown using character theory that c λµν = 0 when µ, ν ⊆ λ . That is, theYoung diagrams of µ and ν must fit inside the Young diagram of λ in order to have nonzerocoefficients.We are now ready to state and prove a theorem about the dimension of the space ofrelations between the polynomial invariants under the conjugation action of the complexorthogonal group on P ( M n ). Under the conjugation action of O n on P ( M n ), we know that there are no relations betweenthe invariants in the stable range, where the degree of the polynomials is less than or equalto n . Therefore, we analyze a space outside of the stable range where there are relations.In this section, we present a result about the dimension of the space relations between theinvariant polynomials of degree d = n + 1. We start by giving an example. Example:
Let n = 2. We consider invariant polynomials under the action of O ; weknow that there are no relations between the invariants if d ≤
2. Thus we take one stepoutside of the stable range and let d = 3, where we consider cubic invariant polynomials5nder the action of O on P ( M ). Here, dim ( P ( M )) O = 5The space consists of the following cubic invariants, T r ( x ) , T r ( x x T ) , T r ( x ) T r ( x ) , T r ( xx T ) T r ( x ) , T r ( x ) In the language of Section 2, these polynomials correspond to the set of Necklace diagramswith 3 total nodes and oriented edges.
T r ( x ) T r ( x x T ) T r ( x ) T r ( x ) T r ( xx T ) T r ( x ) T r ( x ) Figure 3.1:
Bijection between Necklace diagrams D and elements of P ( M ) O . These polynomials span the set of invariants, however, they are not linearly independent.There are actually two relations,
T r ( x ) − T r ( x ) T r ( x ) + 2 T r ( x ) = 02 T r ( x x T ) − T r ( xx T ) T r ( x ) + T r ( x ) − T r ( x ) T r ( x ) = 0The primary result of this section describes how many relations there are between thedegree n + 1 invariants under the action of O n . We start by giving some set up for the maintheorem. n invariants. Consider the follow-ing inclusion map from the square n -dimensional matrices to the square ( n + 1)-dimensionalmatrices: M n ֒ → M n +1 x ֒ → (cid:18) x
00 0 (cid:19) n +1 Then we have a surjection between the polynomial spaces: P [ M n +1 ] ։ P [ M n ]6here we restrict the ( n + 1) × ( n + 1) dimensional matrix down to an n × n dimensionalmatrix. Recall, in Section 2 we discuss the invariants of this space under the conjugationaction of the complex orthogonal group O n . Under this action we have a surjection betweenthe invariant rings: P [ M n +1 ] O n +1 ։ P [ M n ] O n (3.1) Remark:
We note that P ( M n ) is a C -algebra of polynomial functions on M n with agraded structure: P ( M n ) = M d P d ( M n )where P d ( M n ) denotes the subspace of homogeneous degree d polynomials, which are a finitedimensional representation of O n . Thus we also have a graded structure on the invariantalgebras, P ( M n ) O n = M d P d ( M n ) O n Therefore if we fix the degree, d = n + 1, of the invariant polynomials, we have a surjectionbetween the algebras P n +1 ( M n +1 ) O n +1 ։ P n +1 ( M n ) O n (3.2)Note that when d = 1, we have that P ( M n ) = M ∗ n , which leads to the following identi-fication: P ( M n ) = M d P d ( M n ) = M d S d ( M ∗ n ) = S ( M ∗ n )where S ( M ∗ n ) is the symmetric algebra of the dual space, M ∗ n , of polynomials on M n . Thenif we again fix the degree d = n + 1, we see that: P n +1 ( M n ) = S n +1 ( M ∗ n ) = S n +1 ( M n ) ∗ Furthermore, as we are working over C , we have that S n +1 ( M n ) ≃ [ ⊗ n +1 ( M n )] ∆ S n +1 . Notethat the symmetric tensors are invariant under the natural permutation action of the sym-metric group on the tensor factors. Thus map 3.2 can be written as the surjective map:[ ⊗ n +1 ( M n +1 )] O n +1 × ∆ S n +1 ։ [ ⊗ n +1 ( M n )] O n × ∆ S n +1 (3.3)where ∆ S n +1 denotes the diagonally embedded copy of S n +1 in S n +1 × S n +1 .Next, we use the following decomposition of M n in the above map 3.3. Consider the Gl n × Gl n action on M n given by: ( g, h ) · x = gxh T for x ∈ M n and ( g, h ) ∈ Gl n × Gl n . Restricting to the diagonal Gl n action on this spacegives: ( g, g ) · x = gxg T and thus under this action, we have a decomposition of M n as follows, M n ≃ C n ⊗ C n (3.4)7e focus on the O n decomposition of M n under the adjoint action of Gl n , where x → gxg − .Under this action we have that M n ≃ ( C n ) ∗ ⊗ C However, consider the map: C n → ( C n ) ∗ v ϕ : C n → C w → v · w where v, w ∈ C n and v · w is the usual dot product. Since the dot product is invariant underthe O n action, we have that as an O n - representation, C n ≃ ( C n ) ∗ . Thus, we are free todecompose M n as in 3.4 using this property that O n is self-dual. Remark on the Brauer algebra:
Before we symmetrize and consider the ∆ S n +1 action, we can decompose the even dimensional tensor space using the dual, ( C n ) ∗ , whichgives the following, [( ⊗ n +1 C n ) ∗ ⊗ ( ⊗ n +1 C n )] O n ∼ = End O n ( ⊗ n +1 C n )where the endomorphism group is defined to be the Brauer algebra [4]. If we consider the O n action on the tensor space ⊗ k C n instead of the traditional general linear group action,then the Brauer algebra replaces the symmetric group in the decomposition of the space viaSchur-Weyl duality. A recent discussion can be found in [5].Now, returning to our main thread, using the decomposition of M n described above, wecan write map 3.3 as: ζ : [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1 ։ [ ⊗ n +1) C n ] O n × ∆ S n +1 (3.5)We see that in the domain of this map, [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1 , we are in the stable rangewhere the degree of the invariants is equal to the dimension of the defining representation.In the codomain, [ ⊗ n +1) C n ] O n × ∆ S n +1 , is where relations arise. Thus, the kernel, RE L , ofthis map (0) → RE L → [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1 → [ ⊗ n +1) C n ] O n × ∆ S n +1 is exactly the space of relations between the degree n + 1 invariants. Definition 4.
Let n ∈ N , and let ∆ S n +1 denote the diagonally embedded copy of the sym-metric group S n +1 in S n +1 × S n +1 . Then the kernel of the map, ζ : [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1 ։ [ ⊗ n +1) C n ] O n × ∆ S n +1 denoted RE L , is defined to be the space of relations between the invariants of the conjugationaction of the complex orthogonal group on P ( M n ) . Our goal is to understand
RE L as a vector space in order to compute its dimension, anddetermine a basis. The following theorem describes the dimension of this vector space whenwe are just outside of the stable range, that is, when the degree of the invariants is n + 1.8 heorem 1. Let n be a positive integer. The dimension of the space of relations, RE L n +1 ,between the degree n + 1 invariants of the O n conjugation action on P ( M n ) is equal to dim ( RE L n +1 ) = n n even n + 32 n oddIn order to determine the dimension of RE L n +1 , the proof will proceed as follows. First,we consider the projection: (cid:2) ⊗ n +1 ( M n ) (cid:3) O n −→ P n +1 ( M n ) O n (3.6)where when we project to polynomial space, the tensors become symmetric. The relationsin the polynomial algebra pull back into the tensor algebra, and they form an irreduciblerepresentation of S n +1) corresponding to the partition [ n + 1 , n + 1]. When this partition isrestricted to S n +1 × S n +1 , there is a multiplicity-free decomposition into irreducible represen-tations Y α ⊗ Y α corresponding to size n + 1 diagrams with at most two parts. Thus, to each α there exists a polynomial relation which corresponds to the S n +1 invariant where we embed S n +1 into S n +1 × S n +1 diagonally. This ∆ S n +1 is the symmetric group that symmetrizes togo from the tensor algebra to polynomial space.The linear growth of RE L n +1 is shown in the highlighted diagonal of the following table:Dimension of RE L d \ n n , and the rows by the degree of the polynomialinvariants in P n +1 ( M n ) O n . The data in this table again shows that there are no relationswhen we are in the stable range where the degree of the invariants is less than or equal to n ,and it illustrates the linear behavior of the dimension of the space of relations. The values inthis table are all generated using code we have written to compute the number of invariantsin each scenario. Proof of Theorem 1 : Recall we are determining the dimension of the kernel,
RE L n +1 , RE L n +1 → [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1
9e have the following decomposition into irreducible representations via Schur-Weyl Duality: ⊗ n +1) C n +1 ≃ M µ : µ ⊢ n +1) ℓ ( µ ) ≤ n +1 F µn +1 ⊗ Y µ n +1) (3.7)where the F µn +1 and Y µ n +1) are irreducible representations of Gl n +1 and S n +1) , respectively,which are associated to Young Diagram µ with 2( n + 1) boxes and number of nonzero rows ≤ n + 1. We compute the O n +1 invariants,[ ⊗ n +1) C n +1 ] O n +1 ≃ M µ h F µn +1 ⊗ Y µ n +1) i O n +1 ≃ M µ ( F µn +1 ) O n +1 ⊗ Y µ n +1) By the Cartan-Helgason Theorem, we have that dim ( F µn +1 ) O n +1 is nonzero and equal toone only when the corresponding tableaux µ has all even parts. Thus, we let µ = 2 λ : M µ : µ =2 λµ ⊢ n +1) ℓ ( µ ) ≤ n +1 ( F µn +1 ) O n +1 ⊗ Y µ n +1) = M λ : 2 λ ⊢ n +1) ℓ (2 λ ) ≤ n +1 ( F λn +1 ) O n +1 ⊗ Y λ n +1) (3.8)This space consists of all Young diagrams 2 λ , where 2 λ ⊢ n + 1), and ℓ (2 λ ) ≤ n + 1.We know from [12] that this space of invariants corresponds to the set of fixed-point freeinvolutions, thus we have that dim M λ : 2 λ ⊢ n +1) ℓ (2 λ ) ≤ n +1 ( F λn +1 ) O n +1 ⊗ Y λ n +1) = (2( n + 1))!2 n +1 ( n + 1)! (3.9) Remark:
The Robinson-Schensted correspondence associates a pair of standard Youngtableaux, (
P, Q ) to a permutation. It is shown, [10], that if the permutation is an involution,then P = Q . Furthermore, due to a result of Schutzenberger [11], we have that the fixed-point free involutions correspond to Young diagrams with all even rows. Thus, the spacedescribed above in Equation 3.9 consists of all standard Young tableaux of shape 2 λ , whichagain have all even rows.Recall the kernel contained in this space, RE L ⊂ M λ : 2 λ ⊢ n +1) ℓ (2 λ ) ≤ n +1 ( F λn +1 ) O n +1 ⊗ Y λ n +1) is the space of relations between the invariants. Once again, we know that relations do notexist in the stable range. Therefore, relations occur when we violate the inequality ℓ (2 λ ) ≤ n .As such, RE L is the space of irreducible representations corresponding to the Young10iagrams in our space where ℓ (2 λ ) > n , RE L = M λ : 2 λ ⊢ n +1) ℓ (2 λ ) >n ( F λn +1 ) O n +1 ⊗ Y λ n +1) ⊂ M λ : 2 λ ⊢ n +1) ℓ (2 λ ) ≤ n +1 ( F λn +1 ) O n +1 ⊗ Y λ n +1) since L λ ⊢ n +1) ℓ (2 λ ) >n ( F λn +1 ) O n +1 ⊗ Y λ n +1) is the kernel of the map: ζ : [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1 ։ [ ⊗ n +1) C n ] O n × ∆ S n +1 . Now, since we have the restriction that the length of λ must be greater than n , we have onlyone option for the Young diagram, that is, ℓ (2 λ ) = n + 1, n + 1 =: 2 λ ...Thus we define Young diagram 2 λ as the Young diagram pictured above, with twocolumns and n + 1 rows. Here, we have that the dimension of the irreducible representation Y λ n +1) is equal to the number of standard Young tableaux of the column shape ( n + 1) × dim M λ : 2 λ ⊢ n +1) ℓ (2 λ ) >n ( F λn +1 ) O n +1 ⊗ Y λ n +1) = C n +1 where C n +1 is a Catalan number.The irreducible representation Y λ n +1) corresponds to partition [ n + 1 , n + 1], and wewant to restrict this partition to S n +1 × S n +1 . As dicussed in Section 3, we can inducerepresentations in the following way: Ind S n +1) S n +1 × S n +1 Y αn +1 ⊗ Y βn +1 = M λ : 2 λ ⊢ n +1) c λαβ Y λ n +1) (3.10)Where the coefficients c λαβ are the Littlewood-Richardson numbers discussed in Section 3.3.These coefficients count the number of skew semi-standard Young tableaux of shape 2 λ/α with weight β .By Frobenius reciprocity for finite groups, we can restate Equation 3.10 in terms ofrestricting the representation, Res S n +1) S n +1 × S n +1 Y λ n +1) = M α ⊢ n +1 β ⊢ n +1 c λαβ Y αn +1 ⊗ Y βn +1 M λ : 2 λ ⊢ n +1) ℓ (2 λ )= n +1 ( F λn +1 ) O n +1 ⊗ Y λ n +1) = M λ : 2 λ ⊢ n +1) α ⊢ n +1 β ⊢ n +1 ( F λn +1 ) O n +1 ⊗ c λαβ ( Y αn +1 ⊗ Y βn +1 ) ∆ S n +1 The Littlewood-Richardson rule tells us c λαβ = 0 when the Young diagrams of α and β fitinside the Young diagram of 2 λ . Furthermore, since | λ | = 2( n + 1) and | α | = n + 1 and | β | = n + 1 we must have that α = β .Alternately, we determine that α = β by considering homomorphisms between the twoirreducible representations. We have that as representations, the S n +1 are self dual, and thuswe can view the tensor of irreducible representations as an endomorphism group: Y αn +1 ⊗ Y βn +1 ∼ = End ( Y αn +1 , Y βn +1 )By Schur’s Lemma we see that there are no nonzero homomorphisms between distinct irre-ducible representations, and thus we must have that α = β : M λ : 2 λ ⊢ n +1) α ⊢ n +1 β ⊢ n +1 c λαβ Y αn +1 ⊗ Y βn +1 = M λ : 2 λ ⊢ n +1) α ⊢ n +1 c λαα Y αn +1 ⊗ Y αn +1 Now, the Littlewood-Richardson coefficients c λαα correspond to the number of semi-standardfillings of tableaux of skew-shape 2 λ/α of weight α . Thus, since 2 λ has two columns of length n + 1, it is clear that we must have that each coefficient is equal to 1.Now, we recall that we are working in the space of relations, RE L , and thus we havethat
RE L n +1 = M α : α ⊢ n +1 c λαα =12 λ ⊢ n +1) ℓ (2 λ )= n +1 ( Y αn +1 ⊗ Y αn +1 ) ∆ S n +1 So, to each α there exists a polynomial relation which corresponds to the irreduciblerepresentation Y αn +1 .The number of α that satisfy this is precisely the dimension of the space of relations, dim ( RE L n +1 ) = n n even n + 32 n odd (3.11)We proceed by induction on n , the degree of the invariants.Case I: Let n be a positive, even integer; n = 2 m for some m ∈ N . Base case: Let n = 2.Then we are considering the dimension of the space of relations between the elements of P ( M ) O . Thus we look at all partitions, α ⊢
3. We are concerned with the specific α thatgive relations in our space; these are the ones in which the Young diagrams correspondingto α fit exactly inside the column shaped Young diagram (2 , , oung diagram (2 , , Below we show all of the possible partitions of α and their corresponding Young diagrams: ↔ (1 , , ↔ (2 , ↔ (3)Clearly, only partitions (1 , ,
1) and (2 ,
1) will fit appropriately inside the Young diagramof column shape (2 , , n = 2, we have the dimension of the space of relationsis equal to 2 = 22 + 1.Induction step: We assume the proposition holds for even integer n = k , that is, for pos-itive even integer k , the dimension of the space of relations between elements of P k +1 ( M k ) O k is equal to k k + 2.Thus, consider the following two partitions α ⊢ k + 2 + 1 = k + 3 and their correspondingYoung diagrams: k + 3 ··· ↔ (1 + 1 + · · · + 1 | {z } k +3 times ) ··· ↔ (2 + 1 + · · · + 1 | {z } k +1 times ) Then, it is clear that the Young diagram corresponding to partition (1+ · · · +1) fits insidethe diagram (2 , . . . ,
2) of 2( k + 1) boxes. Additionally, we know from our assumption thatthere are k k + 1 that satisfy our condition on the corresponding Youngdiagrams. Thus the total number of partitions that work is k k + 22 + 1 . Case II: Let n be an odd, positive integer; n = 2 m + 1 for some m ∈ N . Base case: Let n = 1. Then we are considering the dimension of the space of relations between the elementsof P ( M ) O . Thus we look at all partitions, α ⊢
2. We are again concerned with only the13 which correspond to relations in our space, that is, the α in which their correspondingYoung diagrams fit exactly into the column shaped Young diagram (2 , Young diagram (2 , Below we show all the possible partitions of α and the corresponding Young diagrams: ↔ (1 , ↔ (2)Both partitions (1 ,
1) and (2) fit appropriately inside the column shape (2 , n = 1, we have the dimension of the space of relations is equal to 2 = 1 + 32 .Induction step: We suppose the proposition holds for odd integer n = k . Thereforewe assume for odd integer k , the dimension of the space of relations between elements of P k +1 ( M k ) O k is equal to k + 32 . We show this proposition holds for k + 2.Similar to Case I, we consider the following two partitions of k + 2 + 1 = k + 3:1 + 1 + · · · + 1 | {z } k +3 times · · · + 1 | {z } k +1 times Then, we know from our assumption that there are k + 32 partitions of the k + 1 that satisfyour conditions. Thus the total number of partitions that work is k + 32 + 1 = k + 3 + 22 . which thus concludes the proof. RE L
In the previous subsection we state and prove a theorem regarding the dimension of thespace
RE L n +1 . Here, we present a method to determine a basis for this space of relationsby relying on a construction of elements from the group algebra C [ S n +1) ], called Youngsymmetrizers. The proceeding section presents a short explanation of Young symmetrizers,and states several concepts that we will use to obtain a basis of the space of relations.14 .1 A discussion of Young symmetrizers. We create a standard Young tableau byfilling the Young diagram with the numbers 1 , . . . , n such that the rows and columns strictlyincrease. Then, we can define the following: P λ = { σ ∈ S n | σ preserves each row of λ } Q λ = { σ ∈ S n | σ preserves each column of λ } These subgroups of S n define elements a λ and b λ in the group algebra, C [ S n ]: a λ := X σ ∈ P λ e σ b λ := X σ ∈ Q λ sgn( σ ) e σ where e σ denotes the unit vector corresponding to σ . By construction, the elements a λ and b λ are idempotents in the group algebra. They do not commute, however, their product isalso idempotent, and is defined as the Young Symmetrizer [ ? ], [ ? ]. Definition 5.
The Young symmetrizer corresponding to Young diagram λ is defined as y λ := a λ · b λ . We note that any element, d , of C [ S n ] gives an invariant subspace, C [ S n ] d , of C [ S n ].However, the image of a Young symmetrizer (by right multiplication on C [ S n ]) is an invariantsubspace which is irreducible under the action of C [ S n ], and unique for each partition λ [3].The following theorem tells us that the subspaces C [ S n ] y λ given by the Young symmetriz-ers are, in fact, irreducible representations of S n , and every irreducible representation of S n is of this form. Theorem 2.
Given S n , let λ be a partition of n . Define Y λ as the subspace of C [ S n ] spannedby the Young symmetrizer y λ . Then: • Y λ is an irreducible representation of S n • If λ, µ are distinct partitions of n , then Y λ ≇ Y µ • The Y λ account for all irreducible representations of S n . Proof of this theorem can be found in Fulton and Harris [3]. Because we can constructthe irreducible representations of S n in this way, we revisit our decomposition from Section3 and construct a basis of RE L n +1 . We show in Section 3 that thespace of relations between invariants under the conjugation action of O n on P ( M n ), RE L n +1 is, RE L n +1 = M α : α ⊢ n +1 c λαα =12 λ : 2 λ ⊢ n +1) ℓ (2 λ )= n +1 ( Y αn +1 ⊗ Y αn +1 ) ∆ S n +1 M α : α ⊢ n +1 c λαα =12 λ ⊢ n +1) ℓ (2 λ )= n +1 ( Y αn +1 ⊗ Y αn +1 ) ∆ S n +1 ≃ M λ : 2 λ ⊢ n +1) ℓ (2 λ )= n +1 ( F λn +1 ) O n +1 ⊗ Y λ n +1) . For each partition α of n +1, there exists a polynomial relation, and this relation correspondsto the Y αn +1 .Furthermore, recall that the dimension of the irreducible representation Y λ n +1) is equalto a Catalan number, the dimension of the vector space. A natural basis is in one-to-onecorrespondence with the standard Young tableau with two columns of length n + 1. Thesetableaux are all the diagrams n + 1 =: 2 λ ...with a standard filling of the numbers [1 , , . . . , n ] such that each row and column strictlyincreases. Let T λ := { T | T is a standard tableau of shape 2 λ . } . Then we know that each T in the set T λ corresponds to a relation between the elements of P n +1 ( M n ) O n . Additionally, since each element of T λ is a standard tableau, it correspondsto an element of C [ S n +1) ] called a Young symmetrizer. We denote this element y T .Recall in Theorem 1 we determine the dimension of the space of relations between degree n + 1 invariants under the conjugation action of O n on polynomials on n × n matrices. Byaveraging over ∆ S n +1 , we have the projection,[ ⊗ n +1 ( M n )] O n → P n +1 ( M n ) O n between the space of orthogonally invariant tensors and the polynomial invariants under the O n action. We discussed the following maps in Section 2, S n +1) → S n +1) /H n +1 → ∆ S n +1 \ S n +1) /H n +1 . Recall, the cosets S n +1) /H n +1 are in one-to-one correspondence with the set I n +1) , thefixed-point free involutions on S n +1) . Furthermore, the double cosets in the above map arein bijective correspondence to a basis of the degree n + 1 polynomial invariants under theorthogonal group action on P ( M n ).Thus we have the following invariant subspaces of the full group algebra: C [ S n +1) ] ∆ S n +1 × H n +1 ֒ → C [ S n +1) ] H n +1 ֒ → C [ S n +1) ] (4.1)where elements of C [ S n +1) ] ∆ S n +1 × H n +1 are linear combinations of the permutations thatcorrespond to the polynomial invariants, P n +1 ( M n ) O n .16hen, we consider the following maps: C [ S n +1) ] ∆ S n +1 × H n +1 E −→ [ ⊗ n +1 M n ] O n × ∆ S n +1 → P n +1 [ M n ] O n where E takes elements of the group algebra invariant under the left ∆ S n +1 action and theright H n +1 action into endomorphisms on tensors.So, for arbitrary degree of invariants d , we have the projection: C [ S d ] ∆ S d × H d ։ P d ( M n ) O n where if d = n + 1, there exists a nonzero kernel which corresponds to the relations betweendegree n + 1 invariants.Thus we take a Young symmetrizer and conjugate τ = (12)(34) · · · (2( n + 1) − n + 1))by each of its terms in order to write it as a linear combination of fixed point free involutions.We can then determine which double coset, ∆ S n +1 \ S n +1) /H n +1 , each term is in. Thus, werewrite the Young symmetrizer using coset representatives for each of its terms. We denotethis by f y T , so that f y T ∈ C [ S n +1) ] ∆ S n +1 × H n +1 Now, the f y T form a spanning set for the relations between the degree n + 1 invariants. Inorder to find a basis of relations, we look for a subspace of this vector space with dimensiondictated by Theorem 1.We use Python and Sage to write the code for finding relations via this method of Youngsymmetrizers. Our code runs on the Sagemath cloud with 11 GB disk space, 5 GB ofRAM and 1 core. However, the size of the Young symmetrizers grows exponentially and thecalculations quickly become too RAM intensive as we increase the degree of the invariants.For example, in [ ⊗ M ] O there are 42 Young symmetrizers, each with 460 ,
800 components;the calculation to find the relations using the method discussed here takes a little over 43hours to compute on this server.In the next chapter, we discuss a much faster method for finding the relations betweenthe invariants under the O n action on P ( M n ). In this chapter we introduce a new method for determining a basis for the space of relations,
RE L n +1 . We want to avoid the lengthy calculations involved in computing relations usingYoung symmetrizers, discussed in Section 4. We know that the kernel of the map ζ : [ ⊗ n +1) C n +1 ] O n +1 × ∆ S n +1 ։ [ ⊗ n +1) C n ] O n × ∆ S n +1 is exactly our space RE L n +1 and consists of relations between the degree n + 1 invariantsunder the complex orthogonal group action on P ( M n ).We showed in Chapter 3 that P ( M n ) O n , the algebra of invariant polynomials under theconjugation action of O n , is generated by elements of the form T r ( x a ( x T ) a x a ( x T ) a · · · x a M )for matrix x ∈ M n . Products of the above polynomials form a spanning set of the invariants,17nd they are not linearly independent when the degree of the monomials is greater than n .We want to find a basis of RE L using elements of the invariant algebra, P n +1 ( M n ) O n ,and avoid computations with Young symmetrizers. Determining the relations between thesepolynomials requires solving the nonlinear equations: k X i =1 y i T r ( x a i ( x T ) a i · · · ) = 0 (5.1)where x ∈ M n , k = dim ( P n +1 ( M n ) O n ), (recall that this is the number of double cosets S n +1 \ S n +1) /H n +1 ), and the y i are constant coefficients in C .Solving these nonlinear equations can be computationally challenging. However, we in-troduce a method for finding relations that allows us to instead solve a k × k linear systemof equations via a Monte Carlo algorithm.Each of the invariants T r ( x a i ( x T ) a i · · · ) are constructed using a matrix x ∈ M n ( C ). Bydefinition, T r ( x a i ( x T ) a i · · · ) ∈ C Therefore, if we randomly generate a matrix x ∈ M n we can compute a numerical value foreach of the k invariants in P n +1 ( M n ) O n . Each equation then becomes linear in the variables y i : k X i =1 y i T r ( x a i ( x T ) a i · · · ) | {z } ∈ C = 0Thus, if we generate k random matrices in M n , and compute the values of each invariant in P n +1 ( M n ) O n , this gives k different linear equations.Let x , x , . . . x k denote the k randomly generated matrices in M n ( C ). We can then solvethe k × k linear system for the y i , y T r ( x a ( x T ) a · · · ) + y T r ( x a ( x T ) a · · · ) + · · · + y k T r ( x a k ( x T ) a k · · · ) = 0 (1) y T r ( x a ( x T ) a · · · ) + y T r ( x a ( x T ) a · · · ) + · · · + y k T r ( x a k ( x T ) a k · · · ) = 0 (2)... y T r ( x a k ( x Tk ) a · · · ) + y T r ( x a k ( x Tk ) a · · · ) + · · · + y k T r ( x a k k ( x Tk ) a k · · · ) = 0 (k)and the solution is exactly the relations in the space RE L n +1 . Example:
We revisit the space P ( M ) O of degree 3 invariants under the conjugationaction of O on M . We know that the dimension of the invariant space is 5, recall we havethe following invariants: T r ( x ) , T r ( x x T ) , T r ( x ) T r ( x ) , T r ( xx T ) T r ( x ) , T r ( x ) We construct a basis for this space using the Monte Carlo method described above. Therelations can be described as solutions to the following the nonlinear equations in the invari-18nts: y T r ( x ) + y T r ( x x T ) + y T r ( x ) T r ( x ) + y T r ( xx T ) T r ( x ) + y T r ( x ) = 0where the y i ’s are constant coefficients.In order to avoid solving a complicated nonlinear system, we first repeatedly generaterandom complex 2 × y T r ( x ) + y T r ( x x T ) + y T r ( x ) T r ( x ) + y T r ( xx T ) T r ( x ) + y T r ( x ) = 0 (1) y T r ( x ) + y T r ( x x T ) + y T r ( x ) T r ( x ) + y T r ( xx T ) T r ( x ) + y T r ( x ) = 0 (2)... y T r ( x ) + y T r ( x x T ) + y T r ( x ) T r ( x ) + y T r ( xx T ) T r ( x ) + y T r ( x ) = 0 (5)where each equation uses a different randomly generated 2 × y i with coefficientsin C .We then run code to solve this 5 × RE L , T r ( x ) − T r ( x ) T r ( x ) + 2 T r ( x ) = 02 T r ( x x T ) − T r ( xx T ) T r ( x ) + T r ( x ) − T r ( x ) T r ( x ) = 0Again, we use Python and Sage to write the code for finding relations via this MonteCarlo simulation method. Our code runs on the Sagemath cloud with 11 GB disk space, 5GB of RAM and 1 core. However, this method is much faster at finding relations than themethod of Young symmetrizers. For example, in the case of [ ⊗ M ] O which took over 43hours to compute relations via Young symmetrizers, our new method takes just a bit over 2minutes. REFERENCES [1] K.M. Brucks,
MSS Sequences, Colorings of Necklaces, and Periodic Points of f ( z ) = z −
2, Advances in Applied Mathematics, (1987), 434-445.[2] W. Fulton, Young Tableaux , Cambridge University Press, Cambridge, UK 1997.[3] W. Fulton and J. Harris,
Representation theory, A first course , Graduate Texts in Math-ematics , Springer-Verlag, New York, 1991.[4] R. Goodman and N.R. Wallach,
Symmetry, Representations, and Invariants , GraduateTexts in Mathematics, Springer (2009).[5] M. Kim, D. Koo,
Polynomial invariants on matrices and partition, Brauer algebra ,arXiv:2006.14812 [math.RA], 2020 196] D.E. Littlewood and A.R. Richardson,
Group Characters and Algebra , The Royal Society, (1934), 721-730.[7] C. Procesi,
The invariant theory of n × n matrices , Adv. in Math. (1976), 306-381[8] C. Procesi and H. Kraft, Classical Invariant Theory, A Primer , Preliminary Version,1996.[9]
SageMath, the Sage Mathematics Software System , The Sage Developers, .[10] C. Schensted,
Longest increasing and decreasing subsequences , Canadian Journal ofMathematics, (1961), 179-191.[11] M. Sch¨utzenerger, La correspondance de Robinson , Lecture Notes in Mathematics,
Springer-Verlag, New York (1977), 59-113.[12] J. Willenbring, A Stable Range for Dimensions of Homogeneous O(n)-Invariant Polyno-mials on the n × n Matrices, J. Algebra 242 (2) (2001) 691-708 MRI1848965 (2002f:13013).