harmonics for twisted steenrod operators
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Harmonics for Twisted Steenrod Operators. Fran cois Bergeron LaCIM, Universit e du Qu ebec ` a Montr eal with: Adriano Garsia and Nolan Wallach FPSAC August 2010 S n -Harmonic Polynomials Solutions of the system ( 1 + . . . +


  1. Harmonics for Twisted Steenrod Operators. Fran¸ cois Bergeron LaCIM, Universit´ e du Qu´ ebec ` a Montr´ eal with: Adriano Garsia and Nolan Wallach FPSAC August 2010

  2. S n -Harmonic Polynomials Solutions of the system ( ∂ 1 + . . . + ∂ n ) f ( x 1 , . . . , x n ) = 0 ( ∂ 2 1 + . . . + ∂ 2 n ) f ( x 1 , . . . , x n ) = 0 . . . ( ∂ n 1 + . . . + ∂ n n ) f ( x 1 , . . . , x n ) = 0 ∂ are said to be S n -harmonic polynomials . Here ∂ i := ∂x i . They can be characterized as all the linear combinations of � ∂ k 1 1 ∂ k 2 2 · · · ∂ k 2 ( x i − x j ) n i<j 2 FPSAC, August 2010

  3. Graded S n -Module The space H S n of harmonic polynomials for the symmetric group is a graded S n -module , i.e.: � H S n ≃ π d ( H S n ) , d ≥ 0 where π d is the linear projection sending polynomials to there degree d homogeneous component. Recall that the group S n acts on polynomials in n variables x = x 1 , x 2 , . . . , x n by permuting variables: σ · x i = x σ ( i ) . 3 FPSAC, August 2010

  4. Graded Irreducible Decomposition From the now classical decomposition Q [ x ] ≃ Q [ x ] S n ⊗ H S n , as graded S n -modules, we get � � � Q [ x ] ≃ e µ V τ , where: ℓ ( µ ) ≤ n λ ⊢ n sh( τ )= λ V τ is some copy of an irreducible representation of S n , of (1) Frobenius characteristic s λ , in the homogeneous component, in H S n , of degree equal to the cocharge , co( τ ), of τ . (2) The indices τ run over the set of standard tableaux of shape λ . (3) Finally, e µ denotes the elementary polynomials in the variables x , considered as linear operator on Q [ x ]. 4 FPSAC, August 2010

  5. Hilbert Series Recall that we have � � n � 1 dim( π d ( Q [ x ])) t d = 1 − t d ≥ 0 and the previous decomposition gives � � n n � � � 1 1 n λ t co( τ ) = 1 − t 1 − t i i =1 λ ⊢ n sh( τ )= λ with n λ equal to the number of standard tableaux of shape λ . Recall that � ( i − 1) λ i . n ( λ ) := i is the smallest possible value for the cocharge of a standard tableaux of shape λ . 5 FPSAC, August 2010

  6. Twisted Steenrod Operators and Associated Harmonics n � q x i ∂ k +1 + ∂ k D k ; q := H n ; q := { f | D k ; q f = 0 , ∀ k ≥ 1 } i i i =1 n � � H x := { f | � � x i ∂ k +1 D k := D k f = 0 , ∀ k ≥ 1 } i i =1 n � � H x := { f | � � x i ∂ k +1 + ( k + 1) ∂ k D k f = 0 , ∀ k ≥ 1 } D k := i i i =1 n � a i ∂ k D k := H n := { f | D k f = 0 , ∀ k ≥ 1 } i i =1 All of these spaces are homogeneous, and we write � � H n ; q ( t ) , H n ( t ) , H n ( t ) , and H n ( t ) , for the respective Hilbert series. 6 FPSAC, August 2010

  7. Conjecture of Hivert-Thi´ ery Conjecture (HT) . The space H n ; q is isomorphic, as a graded S n -module, to the space of S n -harmonic polynomials, for “generic” values of q . In fact H n ; q = { f | D 1; q f = 0 , D 2; q f = 0 } and since [ D k ; q , D j ; q ] = q ( k − j ) D k + j ; q 7 FPSAC, August 2010

  8. General Conjecture For the general operators n � b i x i ∂ 2 D 1 := i + a i ∂ i , and i =1 n � d i x i ∂ 3 i + c i ∂ 2 := D 2 i , i =1 set H n = { f | D 1 f = 0 , and D 2 f = 0 } . Then Conjecture (B) . There is a graded space isomorphism between the space H n and the space of S n -harmonic polynomials, for “generic” values of a i , b i , c i , and d i . 8 FPSAC, August 2010

  9. Dual Point of View For the scalar product on Q [ x ] defined by  if , x a = x b ,  a !  � x a , x b � :=   0 otherwise , for two monomials x a and x b (in vector notation), with a ! standing for a 1 ! a 2 ! · · · a n !, we easily check that i ∂ j i f, g � = � f, x j � x k i ∂ k i g � thus we get the dual operators n � q x k +1 D ∗ ∂ i + x k k ; q = i i i =1 9 FPSAC, August 2010

  10. Hit Polynomials Following R. Wood we say that a polynomial is hit , for the operators D ∗ k ; q , if it can be expressed in the form � D ∗ f ( x ) = k ; q g k ( x ) , k ≥ 1 for some polynomials g k . We write C n ; q for the graded quotient of the space of polynomial by the subspace of hit-polynomials for the operators D ∗ k ; q . Likewise, we write � � C n , C n and for the spaces respectively associated to the operators n n � � � � x k +1 x k +1 ∂ i + ( k + 1) x k D ∗ D ∗ k := and k := i . ∂ i , i i i =1 i =1 10 FPSAC, August 2010

  11. Wood’s Conjecture Conjecture (W) . The space � C n contains a copy of the regular representation spanned by the monomials e x x a , 0 ≤ a i < i, with a = ( a 1 , . . . , a n ) , with e x = x 1 x 2 · · · x n . In fact, we will see that the entire space can be described as follows. Conjecture (BGW) . The space � C n affords the basis e y y a , with a = ( a 1 , . . . , a k ) , 0 ≤ a i < i, with e y = y 1 y 2 · · · y k , k varying from 0 to n , and y varying in all k -subsets of x . Clearly the spaces C n ; q , � C n , � C n are respectively isomorphic, as graded S n -modules, to the spaces H n ; q , � H n , � H n . 11 FPSAC, August 2010

  12. Example For the space � C 3 we have the basis 1 , x 1 , x 2 , x 3 , x 1 x 2 , x 1 x 2 2 , x 1 x 3 , x 1 x 2 3 , x 2 x 3 , x 2 x 2 3 , x 1 x 2 x 3 , x 1 x 2 2 x 3 , x 1 x 2 x 2 3 , x 1 x 2 2 x 2 3 , x 1 x 2 x 3 3 , x 1 x 2 2 x 3 3 . Modulo the conjecture, the associated Hilbert series is � n � n � � t k [ k ] t ! . H n ( t ) = k k =0 12 FPSAC, August 2010

  13. Graded Frobenius Characteristic Recall that the graded Frobenius characteristic of an invariant homogeneous subspace � V = V d , d ≥ 0 of Q [ x ] is � � t d 1 χ V d ( σ ) p λ ( σ ) F V ( t ) := n ! d ≥ 0 σ ∈ S n Since the associated operators are symmetric, the spaces H n ; q , � H n , � H n are invariant homogeneous spaces, we have corresponding � � F n ; q ( t ) , F n ( t ) , and F n ( t ) , graded Frobenius characteristics. 13 FPSAC, August 2010

  14. First Results We have the following Theorem ( 1). If q is considered as a formal parameter, then the space H x ; q is isomorphic, as a graded S n -module, to a submodule of the S n -harmonics. Theorem ( 2). Let the Hilbert series of H x ; q be � c d,n t d , H x ; q ( t ) = d ≥ 0 then � � c d,n = [ n ] t ! ∀ d ≤ n. t d , � Theorem ( 3). The space of tilde-harmonics has the direct sum decomposition � � e y � H x = H y . y ⊆ x 14 FPSAC, August 2010

  15. Proof of Theorem 3 Decompose f in Q [ x ] in the form � f = e y f y y ⊆ x with f y in Q [ y ]. Then one checks that f is in � H x if and only if all f y are chosen to lie in � H y , using the operator identity D k e x = e x � � D k . In other words, we get � e y � H y = � H x , y ⊆ x thus finishing the proof. 15 FPSAC, August 2010

  16. Implication for the Frobenius It follows from this proof that the graded Frobenius characteristic of � H a x is given by the symmetric function n � t k � � F a ( t ) = F a ( t ) h n − k . k =0 Here a stands for the characteristic function for selection of some subset of indices for which we set � { f | � H a := D k f = 0 , if a ( k ) = 0 } , and x � { f | � H a := D k f = 0 , if a ( k ) = 0 } . x 16 FPSAC, August 2010

  17. Kernel of D k The Hilbert series of the kernel of n � b i x i ∂ k +1 + a i ∂ k D k := I i is i =1 � � n − 1 1 (1 + t + . . . + t k − 1 ) , 1 − t and we have an explicit description of it. For k = 1, the elements of the kernel take the form � c r ( y r + Ψ 1 ( y r )) , f = r where, setting x = x n , a = a n , b = b n and y = x 1 , . . . , x n − 1 ; we have � ( − 1) m D m x m 1 ( g ) Ψ 1 ( g ) := m ! . [ a ; b ] m m ≥ 1 17 FPSAC, August 2010

  18. One Generic Case We have the following Theorem ( 4). For all choices of a i ’s such that � a k � = 0 , ∀ K ⊆ { 1 , . . . , n } , K � = ∅ , k ∈ K the Hilbert series of the space n � a i ∂ m + j { f | 1 ≤ j ≤ n } , f = 0 , for i i =1 is � m + n � [ n ]! t n t for all m ≥ 0 . 18 FPSAC, August 2010

  19. Proof of Theorem 4 To prove the theorem, we use the fact that Proposition. Polynomials θ 1 ( x ) , θ 2 ( x ) , . . . , θ n ( x ) form a regular sequence in Q [ x ] if and only if the system of equations θ 1 ( x ) = 0 , θ 2 ( x ) = 0 , . . . , θ n ( x ) = 0 has, for x ∈ Q n , the unique solution x 1 = 0 , x 2 = 0 , . . . , x n = 0 . For our case, we formulate this in the format       x m +1 x m +1 x m +1 0 . . . a 1 1 2 n             x m +2 x m +2 x m +2 0  . . .   a 2    1 2 n       = . . . . .       ... . . . . .       . . . . .       x m + n x m + n x m + n . . . a n 0 1 2 n 19 FPSAC, August 2010

  20. An Hyperplane Arrangement For the hyperplane arrangement � � � � = 0 , a k K ⊆{ 1 ,...,n } k ∈ K K � = ∅ the number of chambers are 1 , 2 , 6 , 32 , 370 , 11292 , . . . For n = 3 we get 20 FPSAC, August 2010

  21. Diagonal version Conjecture (B) . The space corresponding to the set of common zeros of the operators n � a i ∂ k x i ∂ j y i , i =1 for all k, j ∈ N such that k + j > 0 , is of dimension ( n + 1) n − 1 , whenever we have � a k � = 0 , k ∈ K for all nonempty subsets K of { 1 , . . . , n } . A stronger statement can be made in term of bigraded Hilbert series, and several sets of variables. 21 FPSAC, August 2010

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