aa r X i v : . [ m a t h . C O ] A ug THE STABILIZER OF IMMANANTS
KE YE
Abstract.
Immanants are homogeneous polynomials of degree n in n variables asso-ciated to the irreducible representations of the symmetric group S n of n elements. Wedescribe immanants as trivial S n modules and show that any homogeneous polynomialof degree n on the space of n × n matrices preserved up to scalar by left and right actionby diagonal matrices and conjugation by permutation matrices is a linear combinationof immanants. Building on works of M. Ant´onia Duffner [ ? ] and Coelho, M. Purifica¸c˜ao[ ? ], we prove that for n ≥ π = (4 , , , S n ) ⋉ T ( GL n × GL n ) ⋉Z ,where T ( GL n × GL n ) is the group consisting of pairs of n × n diagonal matrices withthe product of determinants 1, acting by left and right matrix multiplication, ∆( S n ) isthe diagonal of S n × S n , acting by conjugation, ( S n is the group of symmetric group.)and Z acts by sending a matrix to its transpose. Based on the work of Coelho, M.Purifica¸c˜ao and Duffner, M. Ant´onia [ ? ], we also prove that for n ≥ S n ) ⋉ T ( GL n × GL n ) ⋉Z . Introduction
D.E. Littlewood [ ? ] defined polynomials of degree n in n variables generalizing thenotion of determinant and permanent, called immanants, and are defined as follows: Definition 1.1.
For any partition π ⊢ n , define a polynomial of degree n in matrixvariables ( x ij ) n × n associated to π as follows: P π := X σ ∈ S n χ π ( σ ) n Y i =1 x iσ ( i ) This polynomial is called the immanant associated to π . Example 1.2. If π = (1 , , ...,
1) then P π is exactly the determinant of the matrix ( x ij ) n × n . Example 1.3. If π = ( n ) then P π is the permanent P σ ∈ S n Q ni =1 x iσ ( i ) . Remark . Let E and F be C n . Since immanants are homogeneous polynomials ofdegree n in n variables, we can identify them as elements in S n ( E ⊗ F ) (Identify thespace E ⊗ F with the space of n × n matrices). The space S n ( E ⊗ F ) is a representationof GL ( E ⊗ F ), in particular, it is a representation of GL ( E ) × GL ( F ) ⊂ GL ( E ⊗ F ).So we can use the representation theory of GL ( E ) × GL ( F ) to study immanants. The Date : Nov 2010. explicit expression of an immanant P π in S n ( E ⊗ F ) is: X σ ∈ S n χ π ( σ ) n Y i =1 e i ⊗ f σ ( i ) , where Q is interpreted as the symmetric tensor product.In section 3 we remark that immanants can be defined as trivial S n modules (Proposi-tion 3.3). Duffner, M. Ant´onia found the system of equations determining the stabilizerof immanants (except determinant and permanent) for n ≥ ? ] in year 1994. 2 yearslater, Coelho, M. Purifica¸c˜ao proved in [ ? ] that if the system of equations in [ ? ] has asolution, then permutations τ and τ in the system must be the same. Building on worksof Duffner and Coelho, We prove the main results Theorem 1.5 and Theorem 1.7 of thispaper in section 4. Theorem 1.5.
Let π be a partition of n ≥ such that π = (1 , ..., , (4 , , , or ( n ) , thenthe identity component of the stabilizer of the immanant P π is ∆( S n ) ⋉ T ( GL n × GL n ) ⋉Z , where T ( GL n × GL n ) is the group consisting of pairs of n × n diagonal matriceswith the product of determinants , acting by left and right matrix multiplication, ∆( S n ) is the diagonal of S n × S n , acting by conjugation, ( S n is the group of symmetric group.)and Z acts by sending a matrix to its transpose. Remark . It is well-known that Theorem 1.5 is true for permanent as well, but ourproof does not recover this case.
Theorem 1.7.
Let n ≥ and let π be a partition of n which is not symmetric, that is, π is not equal to its transpose, and π = (1 , ..., or ( n ) , then the stabilizer of the immanantis ∆( S n ) ⋉ T ( GL n × GL n ) ⋉ Z . Remark . One can compute directly from the system of equations determined byDuffner, M. Ant´onia for the case n = 4 and π = (2 ,
2) and see that in this case, Theorem1.7 fails, since there will be many additional components. For example, C = e − e − e e − − e − e e − e stabilizes the immanant P (2 , , but it’s not in the identity component.2. Notations and preliminaries (1) E and F are n -dimensional complex vector spaces.(2) { e i } ni =1 and { f i } ni =1 are fixed basis of E and F , respectively.(3) S n is the symmetric group on n elements. Given σ ∈ S n , we can express σ as disjoint product of cycles, and we can denote the conjugacy class of σ by(1 i i ...n i n ), meaning that σ is a disjoint product of i i i n n -cycles. Sometimes we might use ( k , k , ..., k p ) (where n ≥ k ≥ k ≥ ... ≥ k p ≥ P pi =1 k i = n ) to indicate the cycle type of σ . This notation means that σ contains a k -cycle, a k -cycle . . . and a k p -cycle. HE STABILIZER OF IMMANANTS 3 (4) π ⊢ n is a partition of n , we can write π = ( π , ..., π k ) with π ≥ ... ≥ π k .(5) π ′ is the conjugate partition of π .(6) M π (sometimes we might use [ π ] as well) is the irreducible representation of thesymmetric group S n corresponding to the partition π .(7) S π ( E ) is the irreducible representation of GL ( E ) corresponding to π .(8) χ π is the character of M π .(9) Let V be a representation of SL ( E ), then the weight-zero-subspace of V is denotedas V .(10) For S π ( E ), we have a realization in the tensor product E ⊗ n via the young sym-metrizer c π : S π ( E ) ∼ = c π ( E ⊗ n ),where π is any young tableau of shape π .(11) The action of S n on E is the action of the Weyl group of SL ( E ), that is, thepermutation representation on C n .(12) Fix a young tableau π , define P ¯ π := { σ ∈ S n | σ preserves each row of ¯ π } and Q ¯ π := { σ ∈ S n | σ preserves each column of ¯ π } (13) For a polynomial P in variables ( x ij ) n × n , denote the stabilizer of P in GL ( E ⊗ F )by G ( P ). 3. The description of immanants as modules
Consider the action of T ( E ) × T ( F ) on immanants, where T ( E ), T ( F ) are maximal tori(diagonal matrices) of SL ( E ), SL ( F ), respectively. For any ( A, B ) ∈ T ( E ) × T ( F ), A = a . . . a . . . . . . a n , B = b . . . b . . . . . . b n For the immanant P π = P σ ∈ S n χ π ( σ ) Q ni =1 x iσ ( i ) , the action of ( A, B ) on P π is given by( A, B ) .P π := X σ ∈ S n χ π ( σ ) n Y i =1 a i b σ ( i ) x iσ ( i ) = P π . That is, immanants are in the SL ( E ) × SL ( F ) weight-zero-subspace of S n ( E ⊗ F ). Onthe other hand, the decomposition of S n ( E ⊗ F ) as GL ( E ) × GL ( F )-modules: S n ( E ⊗ F ) = X λ ⊢ n S λ ( E ) ⊗ S λ ( F ) . Accordingly, we have a decomposition of the weight-zero-subspace:( S n ( E ⊗ F )) = X λ ⊢ n ( S λ ( E )) ⊗ ( S λ ( F )) . Proposition 3.1.
For λ ⊢ n , ( S λ ( E )) ∼ = M λ as S n -modules. Proof.
See [ ? ] page 272. (cid:3) Thus we can identify ( S λ ( E )) ⊗ ( S λ ( F )) with M λ ⊗ M λ as an S n × S n module. Also,the diagonal ∆( S n ) of S n × S n is isomorphic to S n , so M λ ⊗ M λ is a S n -module. M λ ⊗ M λ KE YE is an irreducible S n × S n module, but it is reducible as and S n -module, so that we candecompose it.Consider the action of S n on S n ( E ⊗ F ), let σ ∈ S n , then: σ.x ij = x σ ( i ) σ ( j ) . So immanants are invariant under the action of S n , hence are contained in the isotypiccomponent of the trivial S n representation of L λ ⊢ n ( S λ ( E )) ⊗ ( S λ ( F )) = L λ ⊢ n M λ ⊗ M λ . Proposition 3.2.
As a S n module, M λ ⊗ M λ contains only one copy of trivial represen-tation. Proof.
Denote the character of σ ∈ S n on M λ ⊗ M λ by χ ( σ ), and let χ trivial be the characterof the trivial representation. From the general theory of characters, it suffices to showthat the inner product ( χ, χ trivial ) = 1. First, the character χ ( σ, τ ) of ( σ, τ ) ∈ S n × S n on the module M λ ⊗ M λ is χ λ ( σ ) χ λ ( τ ). So in particular, the character χ of M λ ⊗ M λ on σ is ( χ λ ( σ )) . Next, ( χ, χ trivial ) = 1 n ! ( X σ ∈ S n χ ( σ ) χ trivial ( σ ))= 1 n ! X σ ∈ S n ( χ λ ( σ )) =( χ λ , χ λ )=1since χ trivial ( σ ) = 1 , ∀ σ ∈ S n . (cid:3) By the above proposition, M λ ⊗ M λ = C λ ⊕ · · · , where C λ means the unique copyof trivial representation in M λ ⊗ M λ , and dots means other components in this module.Hence P π ∈ M λ ⊢ n C λ . We can further locate immanants:
Proposition 3.3.
Let P π be the immanant associated to the partition π ⊢ n . Assume C π is the unique copy of the trivial S n -representation contained in ( S π ( E )) ⊗ ( S π ( F )) .Then: P π ∈ C π . Before proving this proposition, we remark that it gives an equivalent definition ofthe immanant: P π is the element of the trivial representation C π of M π ⊗ M π such that P π ( Id ) = dim ([ π ]). For more information about this definition, see for example, [ ? ]. Example 3.4. If π = (1 , ..., S π E = V n E , which is already a 1 dimensionalvector space. If π = ( n ), then S π E = S n E , in which there’s only one (up to scale) weightzero vector e ◦ · · · ◦ e n . Proof.
Fix a partition π ⊢ n , we want to show that the immanant P π is in C π , but we knowthat P π is in the weight-zero-subspace ( S n ( E ⊗ F )) = ⊕ λ ⊢ n ( S λ ( E )) ⊗ ( S λ ( F )) .Since( S λ ( E )) ⊗ ( S λ ( F )) ⊂ S λ ( E ) ⊗ S λ ( F ), it suffices to show that P π ∈ S π ( E ) ⊗ S π ( F ). Thenit suffices to show that for any young symmetrizer c ¯ λ not of the shape π , c ¯ λ ⊗ c ¯ λ ( P π ) = 0. HE STABILIZER OF IMMANANTS 5
It suffices to check that 1 ⊗ c ¯ λ ( P π ) = 0, since c ¯ λ ⊗ c ¯ λ = ( c ¯ λ ⊗ ◦ (1 ⊗ c ¯ λ ). Express P π asan element in S n ( E ⊗ F ): P π = X σ ∈ S n X τ ∈ S n χ π ( σ )( ⊗ ni =1 e τ ( i ) ) ⊗ ( ⊗ ni =1 f σ ◦ τ ( i ) )The young symmetrizer c ¯ λ = P p ∈ P λ , q ∈ Q λ sgn ( q ) p q . So1 ⊗ c λ ( P π ) = X τ ∈ S n ⊗ ni =1 e τ ( i ) ⊗ ( X p ∈ P λ q ∈ Q λ σ ∈ S n χ π ( σ ) sgn ( q ) ⊗ ni =1 f σ · τ · q · p ( i ) )In the above expression, c ¯ λ acts on f i ’s. Now it suffices to show that: X p ∈ P λ q ∈ Q λ σ ∈ S n χ π ( σ ) sgn ( q ) n O i =1 f σ · τ · q · p ( i ) = 0 , ∀ τ ∈ S n . For any τ ∈ S n , X p ∈ P λ q ∈ Q λ α ∈ S n χ π ( α · τ − ) sgn ( q ) n O i =1 f α · q · p ( i ) = X γ ∈ S n X p ∈ P λ q ∈ Q λ α ∈ S n α · q · p = γ χ π ( α · τ − ) sgn ( q ) n O i =1 f γ ( i ) = X γ ∈ S n [ X p ∈ P λ q ∈ Q λ χ π ( γ · p − · q − · τ − ) sgn ( q )] n O i =1 f γ ( i ) . Let σ , τ ∈ S n , σ · τ = σ · τ · σ − · σ , so σ · τ = τ ′ · σ , where τ ′ is conjugate to τ in S n by σ . Therefore, we can rewrite the previous equation as: X γ ∈ S n [ X p ∈ P λ q ∈ Q λ χ π ( τ − · γ · p − · q − ) sgn ( q )] n O i =1 f γ ( i ) . Therefore, it suffices to show: X p ∈ P ¯ λ q ∈ Q ¯ λ χ π ( γ · p · q ) sgn ( q ) = 0 , ∀ γ ∈ S n This equality holds because the left hand side is the trace of γ · c ¯ λ as an operator on thespace C S n · c ¯ π ,the group algebra of S n , which is a realization of M π in C S n , but thisoperator is in fact zero: ∀ σ ∈ S n , γ · c ¯ λ · σ · c ¯ π = γ · σ · c ¯ λ ′ · c ¯ π = 0 , where ¯ λ ′ is the young tableau of shape λ ′ which is conjugate to ¯ λ by σ . This implies that c ¯ λ and c ¯ π are of different type, hence c ¯ λ · c ¯ π = 0, in particular, the trace of this operatoris 0. Therefore, P π ∈ S π ( E ) ⊗ S π ( F ). (cid:3) KE YE
Corollary 3.5.
If a homogeneous polynomial Q of degree n is preserved by the group ∆( S n ) × T ( E ) × T ( F ) , then Q is a linear combination of immanants. Furthermore,immanants are linearly independent and form a basis of the space of all homogeneousdegree n polynomials preserved by ∆( S n ) × T ( E ) × T ( F )) . Proof. If Q is preserved by S n × T ( GL ( E ) × GL ( F )), then Q is in L λ ⊢ n C λ . By theproposition, immanants form a basis of L λ ⊢ n C λ , the corollary follows. (cid:3) The stabilizer of immanant
Next, we study the stabilizer of immanants in the group GL ( E ⊗ F ). Example 4.1. : For π = (1 , , ...,
1) and π = ( n ), G ( P π ) are well-known: If π =(1 , , ..., G ( P π ) = S ( GL ( E ) × GL ( F )) ⋉ Z , and if π = ( n ),then G ( P π ) = S n × S n × T ( GL ( E ) × GL ( F )) ⋉Z , where S ( GL ( E ) × GL ( F )) is a subgroup of GL ( E ) × GL ( F )consisting of pairs ( A, B ) with det ( A ) det ( B ) = 1, T ( GL ( E ) × GL ( F )) is the pair of di-agonal matrices with the product of determinants 1, and ⋉Z means that we are allowedto take the transpose of matrices. For the stabilizer of determinant, see G.Frobenius [ ? ].For the stabilizer of permanent, see Botta [ ? ].Assume C = ( c ij ) and X = ( x ij ) are n × n matrices. Denote the torus action of C on X by C ∗ X = Y , where Y = ( y ij ) is an n × n matrix with entry y ij = c ij x ij . Note thatthe torus action is just the action of the diagonal matrices in End ( E ⊗ F ) on the vectorspace E ⊗ F . To find G ( P π ), we need the following result from [ ? ]: Theorem 4.2 ([ ? ]) . Assume n ≥ , π = (1 , , ..., and ( n ) . A linear transformation T ∈ GL ( E ⊗ F ) preserves the immanant P π iff T ∈ T ( GL ( E ⊗ F )) ⋉ S n ⋉ S n ⋉ Z , andsatisfies the relation: χ π ( σ ) n Y i =1 c iσ ( i ) = χ π ( τ στ − ) , where σ runs over all elements in S n , T ( GL ( E ⊗ F )) is the torus of GL ( E ⊗ F ) , actingby the torus action described above, S n is the symmetric group in n elements, acting byleft and right multiplication, and Z sending a matrix to its transpose. sketch. Step 1: Let π be a fixed partition of n . Define a subset A of the set M n ( C ) of n by n matrices as follows, X ( n − := { A ∈ M n ( C ) : degree of P π ( xA + B ) , for every B ∈ M n ( C ) } . Geometrically, X ( n − is the most singular locus of IM π = 0. If A is in X ( n − , and T preserves P π (it can be shown that T is invertible), then we have that T ( A ) ∈ X ( n − , since the preserver of the hypersurface IM π = 0 will preserve the mostsingular locus as well.Step 2: Characterize the set X ( n − . To do this, first define a subset R i (resp. R i ) of M n ( C ), consisting of matrices that have nonzero entries only in i -th row (resp. column).Then one proves that A ∈ X ( n − if and only if it is in one of the forms:(1) R i or R i for some i .(2) The nonzero elements are in the 2 × A [ i, h | i, h ], and χ π ( σ ) a ii a hh + χ π ( τ ) a ih a hi = 0 HE STABILIZER OF IMMANANTS 7 for every σ and τ satisfying σ ( i ) = i , τ ( h ) = h and τ = σ ( ih ).(3) π = (2 , , ...,
1) and there are complementary sets of indices { i , ..., i p } , { j , ..., j q } such that the nonzero elements are in A [ i , ..., i p | j , ..., j q ] and the rank of A [ i , ..., i p | j , ..., j q ] is one.(4) π = ( n − , A [ u, v | r, s ], andthe permanent of this submatrix is zero.Step 3: Characterize T by sets R i and R j . (cid:3) We will start from this theorem. From this theorem, we know that G ( P π ) is containedin the group S n × S n × C n ⋉ Z , and subject to the relation in Theorem 4.2. Remark . In the equation in the Theorem 4.2, c ij = 0 , ∀ i, j n , since thestabilizer of P π is a group.Now instead of considering n parameters, we can consider n ! parameters, i.e , considerthe stabilizer of immanant in the bigger monoid S n × S n × C n ! ⋉ Z . We can ignore the Z -part of this monoid. The action of the monoid on the weight zero space of S n ( E ⊗ F )spanned by monomials x σ (1) x σ (2) · · · x nσ ( n ) , σ ∈ S n is:( τ , τ , ( c σ ) σ ∈ S n ) · ( x σ (1) x σ (2) · · · x nσ ( n ) ) = c σ x τ (1) τ σ (1) x τ (2) τ σ (2) · · · x τ ( n ) τ σ ( n ) . Proposition 4.4.
The stabilizer of P π in S n × S n × C n ! is determined by equations (1) c τ − τ σ χ π ( τ − τ σ ) = χ π ( σ ) , ∀ σ ∈ S n Proof.
The action of this monoid on P π is:( τ , τ , ( c σ ) σ ∈ S n ) · P π = X σ ∈ S n χ π ( σ ) c σ n Y i =1 x τ ( i ) ,τ σ ( i ) = X σ ∈ S n χ π ( σ ) c σ n Y i =1 x i,τ στ − ( i ) = X σ ∈ S n χ π ( τ − στ ) c τ − στ n Y i =1 x i,σ ( i ) = X σ ∈ S n χ π ( τ − τ ( τ − στ )) c τ − τ ( τ − στ ) n Y i =1 x i,σ ( i ) If ( τ , τ , ( c σ )) stabilizes P π , then χ π ( τ − τ ( τ − στ )) c τ − τ ( τ − στ ) = χ π ( σ ) = χ π ( τ − στ ) , ∀ σ ∈ S n . Therefore, we have: c τ − τ σ χ π ( τ − τ σ ) = χ π ( σ ) , ∀ σ ∈ S n . (cid:3) Our next task is to find τ and τ , such that the equation (1) has solution for ( c σ ) σ ∈ S n .For convenience, in the equation (1), set τ − τ = τ , so we get a new equation:(2) c τσ χ π ( τ σ ) = χ π ( σ ) , ∀ σ ∈ S n KE YE
Lemma 4.5.
If the equation (2) has a solution then τ ∈ S n satisfies: (1) If χ π ( σ ) = 0 , then χ π ( τ σ ) = 0 ; (2) If χ π ( σ ) = 0 , then χ π ( τ σ ) = 0 ; Proof.
Clear. (cid:3)
Definition 4.6. : For a fixed partition π ⊢ n , define: P := { σ ∈ S n | χ π ( σ ) = 0 } ,Q := { σ ∈ S n | χ π ( σ ) = 0 } .G := ( ∩ σ ∈ P P σ ) ∩ ( ∩ σ ∈ Q Qσ ) . Lemma 4.7.
The equation (2) has a solution, then τ ∈ G . Proof.
It suffices to show that the 2 conditions in Lemma 4.5 imply τ ∈ G . If τ satisfiesconditions 1 and 2, then τ σ ∈ P, ∀ σ ∈ P ; τ σ ′ ∈ Q, ∀ σ ′ ∈ Q, therefore τ ∈ P σ − , ∀ σ ∈ P ; τ ∈ Qσ ′ − , ∀ σ ′ ∈ Q, so τ ∈ G. (cid:3) Example 4.8.
We can compute G directly for small n . If n = 3, then we have threerepresentations M (3) , M (1 , , , M (2 , , G = S , A , A , respectively. If n = 4, thenwe have five representations M (4) , M (1 , , , , M , , M (2 , , , M (2 , , Here G = S , A , { (1) , (12)(34) , (13)(24) , (14)(23) } , { (1) , (12)(34) , (13)(24) , (14)(23) } , A , respectively. Ifthe partition π = (1 , , ..., G = A n . If the partition π = ( n ), then G = S n . Notethat in these examples, G is a normal subgroup. In fact, this holds in general.The following proposition is due to Coelho, M. Purifica¸c˜ao in [ ? ], using Murnagham-Nakayama Rule. One can give a different proof using Frobenius character formula forcycles (see, for example, [ ? ]). Proposition 4.9 ([ ? ]) . For any n ≥ and partition π ⊢ n , G is a normal subgroup of S n . Moreover, if π = (1 , , ..., or ( n ) , then G is the trivial subgroup (1) of S n . Sketch.
It is easy to show that G is a normal subgroup of S n . And then one can prove G = S n by computing character χ π . Then assume G = A n , one can show that Q = A n and P = S n − A n . If such a partition π exists, then it must be symmetric. But then onecan construct cycles σ contained in A n case by case (using Murnagham-Nakayama ruleor Frobenius’s character formula) such that χ π ( σ ) = 0. It contradicts that Q = A n . (cid:3) Let’s return to the equation (see Theorem 4.2): χ π ( σ ) n Y i =1 c iσ ( i ) = χ π ( τ στ − ) ∀ σ ∈ S n HE STABILIZER OF IMMANANTS 9
By Proposition 4.9, we can set τ = τ in the above equation then we have equations for c σ ’s:(3) n Y i =1 c iσ ( i ) = 1 ∀ σ ∈ S n , with χ π ( σ ) = 0 . So elements in G ( P π ) can be expressed as triples ( τ, τ, ( c ij )) where matrices ( c ij ) is deter-mined by equation(3). Remark . The coefficients of those linear equations are n × n permutation matrices.If we ignore the restriction χ π ( σ ) = 0,then we get all n × n permutation matrices. Lemma 4.11.
The permutation matrices span a linear space of dimension ( n − + 1 in M at n × n ∼ = C n . Proof.
Consider the action of S n on C n by permuting the entries, so σ ∈ S n is an elementin End ( C n ), corresponding to a permutation matrix, and vice versa. Now as S n modules, C n ∼ = M ( n − , ⊕ C , where C is the trivial representation of S n . So we have a decompositionof vector spaces: End ( C n ) ∼ = End ( M ( n − , ⊕ C ) ∼ = End ( M ( n − , ) ⊕ End ( C ) ⊕ Hom ( M ( n − , , C ) ⊕ Hom ( C , M ( n − , ) . Since C and M ( n − , are S n modules, S n ֒ → End ( M ( n − , ) ⊕ End ( C ). Note that dim ( End ( M ( n − , ) ⊕ End ( C )) = ( n − + 1, so it suffices to show that S n will span End ( M ( n − , ) ⊕ End ( C ), but this is not hard to see because we have an algebra isomor-phism: C [ S n ] ∼ = M λ ⊢ n End ( M λ )Hence, we have: C [ S n ] ։ End ( M ( n − , ) ⊕ End ( C ) . (cid:3) Remark . Lemma 4.11 shows that the dimension of the stabilizer of an immanant isat least 2 n −
2. We will show next that for any partition π of n ≥ , ...,
1) and( n ), the dimension of the stabilizer G ( P π ) is exactly 2 n − G ( P π ). Since G ( P π ) ⊂ GL ( E ⊗ F ),the Lie algebra of G ( P π ) is a subalgebra of gl ( E ⊗ F ). We have the decompsition of gl ( E ⊗ F ) gl ( E ⊗ F ) = End ( E ⊗ F ) ∼ = ( E ⊗ F ) ∗ ⊗ ( E ⊗ F ) ∼ = E ∗ ⊗ E ⊗ F ∗ ⊗ F ∼ = ( sl R ( E ) ⊕ Id E ⊕ T ( E )) ⊗ ( sl R ( F ) ⊕ Id F ⊕ T ( F )) ∼ = sl R ( E ) ⊗ sl R ( F ) ⊕ sl R ( E ) ⊗ Id F ⊕ Id E ⊗ sl R ( F ) ⊕ Id E ⊗ Id F ⊕ T ( E ) ⊗ T ( F ) ⊕ T ( E ) ⊗ Id F ⊕ Id E ⊗ T ( F ) Where sl R ( E ) is the root space of sl ( E ), T ( E ) is the torus of sl ( E ), and Id E is thespace spanned by identity matrix. The similar notation is for F . We will show that theLie algebra of { C ∈ M n × n | P π ( C ∗ X ) = P π ( X ) } is T ( E ) ⊗ Id F ⊕ Id E ⊗ T ( F ). Let { e i | i = 1 , ..., n } be a fixed basis of E and { α i | i = 1 , ..., n } be the dual basis. Let H E i = α ⊗ e − α i ⊗ e i . Then { H i | i = 2 , ..., n } is a basis of T ( E ). We use H F for F and define A ij = H E i ⊗ H F j for all i ≥
2, ≥ A ij on variable x pq . C i,jp,q := A ij ( x pq ) = ( δ p − δ ip )( δ q − δ jq ) C i,jp,q = , p = q = 1 − , p = i, q = 1 − , p = 1 , q = j , p = i, q = j , otherwise (4)Equation (3) implies that the matrices C = ( c ij ) that stabilize P π is contained in thetorus of GL ( E ⊗ F ), hence the Lie algebra of the set of such matrices is contained in thetorus of gl ( E ⊗ F ), that is, it is contained in t := Id E ⊗ Id F ⊕ T ( E ) ⊗ T ( F ) ⊕ T ( E ) ⊗ Id F ⊕ Id E ⊗ T ( F ). Now let L be an element of t , then L can be expressed as the linearcombination of A ij ’s and Id E ⊗ Id F . Hence: L = ( aId E ⊗ Id F + X i,j> a ij A ij )for some a , a ij ∈ C .Then L ( x pq ) = ( a + P i,j> a ij ) x , p = q = 1( a − P i> a iq ) x q , p = 1 , q = 1( a − P j> a pj ) x p , p = 1 , q = 1( a + a pq ) x pq , p = 1 , q = 1(5)Now for a permutation σ ∈ S n , the action of L on the monomial x σ (1) x σ (2) ... x nσ ( n ) is:if σ (1) = 1,(6) L ( n Y p =1 x pσ ( p ) ) = ( na + X i,j> a ij + n X p =2 a pσ ( p ) ) n Y p =1 x pσ ( p ) . if σ (1) = 1 and σ ( k ) = 1,(7) L ( n Y p =1 x pσ ( p ) ) = ( na + X p =1 ,k a pσ ( p ) − X i> a iσ (1) − X j> a kj ) n Y p =1 x pσ ( p ) . Lemma 4.13.
For any solution of the system of linear equations a ij + a jk + a km = a ik + a kj + a jm , where { i, j, k, m } = { , , , } (8) a ij + a jk = a ij ′ + a j ′ k , where { i, j, k, j ′ } = { , , , } (9) HE STABILIZER OF IMMANANTS 11
There exists a number λ such that for any permutation µ of the set { , , , } moving l elements, (10) X i =2 µ ( i ) = i a iµ ( i ) = lλ Proof. check by solving this linear system. (cid:3)
Lemma 4.14. let n ≥ be an integer, π be a fixed partition of n , which is not (1 , ..., or ( n ) . Assume that there exists a permutation(so a conjugacy class) τ ∈ S n such that: (1) χ π ( τ ) = 0 ; (2) τ contains a cycle moving at least numbers; (3) τ fixes at least number.Also assume that L ( P π ) = 0 . Then under the above assumptions, a = a ij = 0 for all i, j > . Proof. L ( P π ) = 0 means that L ( Q np =1 x pσ ( p ) ) for all σ ∈ S n such that χ π ( σ ) = 0. Considerpermutations (2345 ... ) ... ( ... ) and (2435 ... ) ... ( ... )(all cycles are the same except the firstone, and for the first cycle, all numbers are the same except the first 4), from formula (6),we have:(11) na + X i,j> a ij + a + a + a + E = 0(12) na + X i,j> a ij + a + a + a + E = 0for some linear combination E of a ij ’s. Thus(13) a + a + a = a + a + a Similarly,(14) a ij + a jk + a km = a ik + a kj + a jm , for i, j, k, m distinct . Next, consider permutations (1234 ...k ) ... ( ... ) and (1254 ...k ) ... ( ... ), again, from formula (7),we obtain:(15) na + a + a + E ′ − X i> a i, − X j> a k,j = 0(16) na + a + a + E ′ − X i> a i, − X j> a k,j = 0Hence,(17) a + a = a + a and thus(18) a ij + a jk = a ij ′ + a j ′ k ,for all i, j, j ′ , k distinct . Now for 2 ≤ i < j < k < m ≤ n , we have system of linear equations of the same form asLemma 4.13. So we have relations: X m ≥ p ≥ iµ ( p ) = p a pµ ( p ) = lλ ijkm where µ is a permutation of the set { i, j, k, m } , l is the number of elements moved by µ ,and λ ijkm is a constant number.It is easy to see that λ ijkm is the same for different choices of the set { ≤ i < j < k 2. For this purpose, consider τ =(243 ...k ) ... ( ... ) and τ = (253 ...k ) ... ( ... ), then na + P i,j> a ij + a + ... + a k + a + (sum of a ii ’s for i = 5 fixed by τ ) = 0(20) na + P i,j> a ij + a + ... + a k + a + (sum of a ii ’s for i = 4 fixed by τ ) = 0(21)Combine these two equations and equation (19) to obtain a = a . The same argument implies that a ii = a for all n ≥ i ≥ 2. Now we have: X i,j> a ij = X
1) + l ) λ + (3( n − − l ) a = 0where l is the number of elements moved by σ . Let σ = (123 ...p ... ( ... ) and σ =(143 ...p ... ( ... ). Then formula (7) gives:0 = na + ( a + ... + a p ) + ˜ E − P i> a i − P j> a j (24) 0 = na + ( a + ... + a p ) + ˜ E − P i> a i − P j> a j (25)Note that ˜ E comes from the product of disjoint cycles in σ and σ except the first one,so they are indeed the same, and if we assume that σ moves l ′ elements, and the firstcycle in σ moves r elements, then ˜ E = ( l ′ − r ) λ + ( n − l ′ ) a . On the other hand, a + ... + a p = a + ... + a p + a − a = rλ − a a + ... + a p = a + ... + a p + a − a = rλ − a HE STABILIZER OF IMMANANTS 13 Equations (24) and (25) gives:(26) na + ( l ′ + 2 n − λ + ( n − l ′ + 2) a = 0Now equations (22), (23), and (26) imply that λ = a = a = 0. From equation (24), wehave: a = X j> a j + X i> a i similarly, a k = X j> a kj + X i> a i for all n ≥ k ≥ X i> a i = ( n − X i> a i + X i,j> a ij = ( n − X i> a i . Hence P i> a i = 0. For the same reason P j> a j = 0, therefore a = 0. By the sameargument, a ij = 0 for all n ≥ i = j ≥ 2, and this completes the proof of the lemma. (cid:3) Lemma 4.15. If n ≥ , then for any partition λ of n , except λ = (3 , , , , λ = (4 , , and λ = (4 , , , , there exists a permutation τ ∈ S n satisfying conditions (1), (2), (3)in Lemma 4.14. Proof. Write λ = ( λ , λ , ..., λ p ) where λ ≥ λ ≥ ...λ p ≥ P pi =1 λ i = n . Withoutloose of generality, we may assume p ≥ λ , otherwise, we can consider the conjugate λ ′ of λ . There exists a largest integer m such that the Young diagram of λ contains an m × m square.Now we will construct τ using the Murnagham-Nakayama Rule (See [ ? ])case by case:(1) If m = 1 then λ is a hook:( λ , , ..., p > λ and λ ≥ 4. Take τ = ( p − , n − p +1 ) then χ λ ( τ ) = 0 by theMurnagham-Nakayama Rule. In this case, n ≥ p > λ and λ = 1. This case is trivial.(c) p > λ and λ = 2 or 3. τ = (4 , n − ) will work.(d) p = λ . Take τ = ( p − , n − p +1 ) if p ≥ τ = ( p − , n − p +2 ) if p ≥ n ≥ p = λ = 5. τ exists by checking the character tables.(2) If m ≥ 2, let ξ be the length of the longest skew hook contained in the youngdiagram of λ . Then take τ = ( ξ, n − ξ ). (cid:3) proof of Theorem 1.5 . For the case n = 5, one can check directly. By Lemma 4.14and lemma 4.15, we know that for n ≥ π not equal to (3 , , , 1) and (4 , , , { C ∈ M n × n | P π ( C ∗ X ) = P π ( X ) } is T ( E ) ⊗ Id F ⊕ Id E ⊗ T ( F ), so theidentity component of { C ∈ M n × n | P π ( C ∗ X ) = P π ( X ) } is T ( GL ( E ) × GL ( F )), andhence the identity component of G ( P π ) is ∆( S n ) ⋉ T ( GL ( E ) × GL ( F )) ⋉Z . For cases π = (3 , , , π = (4 , , 1) the statement is true by Theorem 1.7. (cid:3) By investigating the equation (3), we can give a sufficient condition for the stabilizerof P π to be ∆( S n ) ⋉ T ( GL ( E ) × GL ( F )) ⋉Z as follows: Lemma 4.16. Let π be a partition of n which is not (1 , ..., or ( n ) . Assume thatthere exist permutations σ , τ ∈ S n and an integer p ≥ , such that χ π (( i ...i p ) σ ) =0 , χ π ((1 i ...i p ) σ ) = 0 , χ π ( τ ) = 0 and χ π (( ij ) τ ) = 0 , where ( i ...i p ) and (1 i ...i p ) arecycles disjoint from σ , and ( ij ) is disjoint from τ . Then the stabilizer of P π is ∆( S n ) ⋉ T ( GL ( E ) × GL ( F )) ⋉Z . Proof. For convenience, we will show for the case σ = (1) and p = 2, the other cases aresimilar. In equations 3, let σ = ( ij ) and (1 ij ), where 1 < i, j n and i = j . Then(27) c ji c ij c Y k =1 ,i,j c kk = 1(28) c ji c j c i Y k =1 ,i,j c kk = 1So c ij = c i c j c . Similarly, the existence of τ will give the relation c ii = c i c i c for all1 ≤ i ≤ n . Set A = a . . . a . . . . . . a n , B = b . . . b . . . . . . b n Where a ii = c i and b jj = c j c . So we have C ∗ X = AXB, with det ( AB ) = 1 . (cid:3) The following two propositions guarantee the existence of permutations satisfied con-ditions in Lemma (4.16). Proposition 4.17. Let n ≥ , and π be a non-symmetric partition of n , then there existsnonnegative integers k ,..., k r such that k + ... + k r = n − , such that | ( χ π ( τ )) | = 1 ,where τ is a permutation of type ( k ,..., k r , ) or ( k ,..., k r , . Proof. See Proposition (3 . ? ]. (cid:3) Proposition 4.18. Let n > and π be a non-symmetric partition of n , then there existsnonnegative integers k , ..., k r and q with k r > , q ≥ , and k + ... + k r + q = n , suchthat χ π ( σ ) = 0 , where σ ∈ S is of type ( k , ..., k r , q ) or ( k , ..., k r + 1 , q − ) . Proof. See Proposition (3 . ? ]. (cid:3) proof of Theorem 1.7 . Since n ≥ 5, by propositions 4.17 and 4.18, there exist permu-tations satisfying conditions in Lemma 4.16, then the theorem follows. (cid:3) Acknowledgement I would like to thank the anonymous referee for his/her valuable comments. E-mail address ::