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

Sn-Harmonic Polynomials

Solutions of the system (∂1 + . . . + ∂n)f(x1, . . . , xn) = (∂2

1 + . . . + ∂2 n)f(x1, . . . , xn)

= . . . (∂n

1 + . . . + ∂n n)f(x1, . . . , xn)

= are said to be Sn-harmonic polynomials. Here ∂i :=

∂ ∂xi .

They can be characterized as all the linear combinations of ∂k1

1 ∂k2 2 · · · ∂k2 n

• i<j

(xi − xj)

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Graded Sn-Module

The space HSn of harmonic polynomials for the symmetric group is a graded Sn-module, i.e.: HSn ≃

• d≥0

πd(HSn), where πd is the linear projection sending polynomials to there degree d homogeneous component. Recall that the group Sn acts on polynomials in n variables x = x1, x2, . . . , xn by permuting variables: σ · xi = xσ(i).

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Graded Irreducible Decomposition

From the now classical decomposition Q[x] ≃ Q[x]Sn ⊗ HSn, as graded Sn-modules, we get Q[x] ≃

• ℓ(µ)≤n
• λ⊢n
• sh(τ)=λ

eµVτ, where: (1) Vτ is some copy of an irreducible representation of Sn, of Frobenius characteristic sλ, in the homogeneous component, in HSn, 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, eµ denotes the elementary polynomials in the variables x, considered as linear operator on Q[x].

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Hilbert Series

Recall that we have

• d≥0

dim(πd(Q[x]))td =

• 1

1 − t n and the previous decomposition gives

• 1

1 − t n =

n

• i=1

1 1 − ti

• λ⊢n
• sh(τ)=λ

nλtco(τ) with nλ equal to the number of standard tableaux of shape λ. Recall that n(λ) :=

• i

(i − 1) λi. is the smallest possible value for the cocharge of a standard tableaux of shape λ.

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Twisted Steenrod Operators and Associated Harmonics

Dk;q :=

n

• i=1

q xi∂k+1

i

+ ∂k

i

Hn;q := {f | Dk;qf = 0, ∀k ≥ 1}

• Dk :=

n

• i=1

xi∂k+1

i

• Hx := {f |

Dkf = 0, ∀k ≥ 1}

• Dk :=

n

• i=1

xi∂k+1

i

+ (k + 1)∂k

i

• Hx := {f |

Dkf = 0, ∀k ≥ 1} Dk :=

n

• i=1

ai ∂k

i

Hn := {f | Dkf = 0, ∀k ≥ 1} All of these spaces are homogeneous, and we write Hn;q(t),

• Hn(t),
• Hn(t),

and Hn(t), for the respective Hilbert series.

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Conjecture of Hivert-Thi´ ery

Conjecture (HT). The space Hn;q is isomorphic, as a graded

Sn-module, to the space of Sn-harmonic polynomials, for

“generic” values of q. In fact Hn;q = {f | D1;q f = 0, and D2;q f = 0} since [Dk;q, Dj;q] = q(k − j)Dk+j;q

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General Conjecture

For the general operators D1 :=

n

• i=1

bi xi∂2

i + ai ∂i,

and D2 :=

n

• i=1

di xi∂3

i + ci ∂2 i ,

set Hn = {f | D1 f = 0, and D2 f = 0}. Then Conjecture (B). There is a graded space isomorphism between the space Hn and the space of Sn-harmonic polynomials, for “generic” values of ai, bi, ci, and di.

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Dual Point of View

For the scalar product on Q[x] defined by xa, xb :=      a! if, xa = xb,

• therwise,

for two monomials xa and xb (in vector notation), with a! standing for a1!a2! · · · an!, we easily check that xk

i ∂j i f, g = f, xj i∂k i g

thus we get the dual operators D∗

k;q = n

• i=1

q xk+1

i

∂i + xk

i

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Hit Polynomials

Following R. Wood we say that a polynomial is hit, for the

• perators D∗

k;q, if it can be expressed in the form

f(x) =

• k≥1

D∗

k;q gk(x),

for some polynomials gk. We write Cn;q for the graded quotient of the space of polynomial by the subspace of hit-polynomials for the

• perators D∗

k;q. Likewise, we write

• Cn,

and

• Cn

for the spaces respectively associated to the operators

• D∗

k := n

• i=1

xk+1

i

∂i, and

• D∗

k := n

• i=1

xk+1

i

∂i + (k + 1)xk

i .

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Wood’s Conjecture

Conjecture (W). The space Cn contains a copy of the regular representation spanned by the monomials ex xa, with a = (a1, . . . , an), 0 ≤ ai < i, with ex = x1x2 · · · xn. In fact, we will see that the entire space can be described as follows. Conjecture (BGW). The space Cn affords the basis ey ya, with a = (a1, . . . , ak), 0 ≤ ai < i, with ey = y1y2 · · · yk, k varying from 0 to n, and y varying in all k-subsets of x. Clearly the spaces Cn;q, Cn, Cnare respectively isomorphic, as graded Sn-modules, to the spaces Hn;q, Hn, Hn.

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Example

For the space C3 we have the basis 1, x1, x2, x3, x1x2, x1x2

2,

x1x3, x1x2

3,

x2x3, x2x2

3,

x1x2x3, x1x2

2x3, x1x2x2 3, x1x2 2x2 3, x1x2x3 3, x1x2 2x3 3.

Modulo the conjecture, the associated Hilbert series is

• Hn(t) =

n

• k=0

n k

• tk[k]t!.

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Graded Frobenius Characteristic

Recall that the graded Frobenius characteristic of an invariant homogeneous subspace V =

• d≥0

Vd,

• f Q[x] is

FV(t) :=

• d≥0

td 1 n!

• σ∈Sn

χVd(σ)pλ(σ) Since the associated operators are symmetric, the spaces Hn;q, Hn,

• Hn are invariant homogeneous spaces, we have corresponding

Fn;q(t),

• Fn(t),

and

• Fn(t),

graded Frobenius characteristics.

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First Results

We have the following Theorem (1). If q is considered as a formal parameter, then the space Hx;q is isomorphic, as a graded Sn-module, to a submodule

• f the Sn-harmonics.

Theorem (2). Let the Hilbert series of Hx;q be Hx;q(t) =

• d≥0

cd,n td, then cd,n = [n]t!

• td,

∀d ≤ n. Theorem (3). The space of tilde-harmonics has the direct sum decomposition

• Hx =
• y⊆x

ey Hy.

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Proof of Theorem 3

Decompose f in Q[x] in the form f =

• y⊆x

eyfy with fy in Q[y]. Then one checks that f is in Hx if and only if all fy are chosen to lie in Hy, using the operator identity

• Dk ex = ex

Dk. In other words, we get

• y⊆x

ey Hy = Hx, thus finishing the proof.

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Implication for the Frobenius

It follows from this proof that the graded Frobenius characteristic

• f

Ha

x is given by the symmetric function

• Fa(t) =

n

• k=0

tk Fa(t) hn−k. Here a stands for the characteristic function for selection of some subset of indices for which we set

• Ha

x

:= {f | Dkf = 0, if a(k) = 0}, and

• Ha

x

:= {f | Dkf = 0, if a(k) = 0}.

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Kernel of Dk

The Hilbert series of the kernel of Dk :=

n

• i=1

bi xi∂k+1

i

+ ai ∂k

I

is (1 + t + . . . + tk−1)

• 1

1 − t n−1 , and we have an explicit description of it. For k = 1, the elements of the kernel take the form f =

• r

cr(yr + Ψ1(yr)), where, setting x = xn, a=an, b=bn and y = x1, . . . , xn−1; we have Ψ1(g) :=

• m≥1

(−1)m Dm

1 (g)

[a; b]m xm m! .

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One Generic Case

We have the following Theorem (4). For all choices of ai’s such that

• k∈K

ak = 0, ∀K ⊆ {1, . . . , n}, K = ∅, the Hilbert series of the space {f |

n

• i=1

ai ∂m+j

i

f = 0, for 1 ≤ j ≤ n}, is m + n n

• t

[n]!t for all m ≥ 0.

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Proof of Theorem 4

To prove the theorem, we use the fact that

• Proposition. Polynomials θ1(x), θ2(x), . . . , θn(x) form a regular

sequence in Q[x] if and only if the system of equations θ1(x) = 0 , θ2(x) = 0 , . . . , θn(x) = 0 has, for x ∈ Qn, the unique solution x1 = 0 , x2 = 0 , . . . , xn = 0. For our case, we formulate this in the format         xm+1

1

xm+1

2

. . . xm+1

n

xm+2

1

xm+2

2

. . . xm+2

n

. . . . . . ... . . . xm+n

1

xm+n

2

. . . xm+n

n

                a1 a2 . . . an         =         . . .        

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An Hyperplane Arrangement

For the hyperplane arrangement

• K⊆{1,...,n}

K=∅

k∈K

ak

• = 0,

the number of chambers are 1, 2, 6, 32, 370, 11292, . . . For n = 3 we get

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Diagonal version

Conjecture (B). The space corresponding to the set of common zeros of the operators

n

• i=1

ai ∂k

xi∂j yi,

for all k, j ∈ N such that k + j > 0, is of dimension (n + 1)n−1, whenever we have

• k∈K

ak = 0, 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.

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References/see arXiv:0812.3566

[1] F. Bergeron, Algebraic Combinatorics and Coinvariant Spaces, CMS Treatise in Mathematics, CMS and A.K.Peters, 2009. [2] F. Hivert and N. Thi´ ery, Non-commutative deformation of symmetric functions and the integral Steenrod algebra, in: Invariant theory in all characteristics, volume 35 of CRM Proc. Lecture Notes, pages 91-125, Amer. Math. Soc., Providence, RI, 2004. [3] R.M.W. Wood, Problems in the Steenrod algebra, Bull. London Math. Soc. 30 (1998) 449–517.

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