harmonics for twisted steenrod operators
play

Harmonics for Twisted Steenrod Operators. Fran cois Bergeron - PowerPoint PPT Presentation

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

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend