A representation-theoretical solution to MathOverflow question #88399

A representation-theoretical solution to MathOverflow question

#88399 Darij Grinberg

version 2 (4 December 2013)

Sorry for hasty writing. Please let me know about any mistakes or unclar- ities (A@B.com with A=darijgrinberg and B=gmail).

§1. Statement of the problem Letn ∈N.

For every w ∈ S_n, let σ(w) denote the number of cycles in the cycle decomposition of the permutation w (this includes cycles consisting of one element).

We can consider the matrix

x^σ(^gh−1)

g,h∈Sn

; this is a matrix over the polynomial ring Q[x], whose rows and whose columns are indexed by the elements of S_n. (So this is a matrix with n! rows and n! columns, although there is no explicit ordering on the set of rows/columns given.)

The claim of MathOverflow question #88399 is:

Theorem 1. The polynomial det

x^σ(^gh−1)

g,h∈S_n

∈Q[x]

factors into linear factors of the formx−`with` ∈ {−n+ 1,−n+ 2, ..., n−1}.

Before we head to the proof of this theorem, let us show some examples:

Example. If n = 1, then the matrix

x^σ(^gh−1)

g,h∈Sn

has only one row and one column, and its only entry isx. Its determinant thus is x, which is in agreement with Theorem 1.

Ifn = 2, then the matrix

x^σ(^gh−1)

g,h∈Sn

has two rows and two columns.

Picking a reasonable ordering on Sn, we can represent it as the 2×2-matrix x² x

x x²

, which has determinant x²(x−1) (x+ 1).

If n = 3, then the matrix

x^σ(^gh−1)

g,h∈Sn

can be represented (by picking an ordering on Sn) by the 6×6-matrix

















x³ x² x² x x x² x² x³ x x² x² x x² x x³ x² x² x x x² x² x³ x x² x x² x² x x³ x² x² x x x² x² x³















 ,

and thus has determinant x⁶(x−2) (x+ 2) (x−1)⁵(x+ 1)⁵. This, again, matches the claim of Theorem 1.

Forn = 4, we have det

x^σ(^gh−1)

g,h∈Sn

= (x−3) (x+ 3) (x−2)¹⁰(x+ 2)¹⁰(x−1)²³(x+ 1)²³x²⁸.

Exercise 1. Prove that the polynomial det

x^σ(^gh−1)

g,h∈Sn

is even (that is, a polynomial in x²) for everyn ≥2. (See the end of this note for a hint.)

§2. Reduction to representation theory

Let us first reduce Theorem 1 to a representation-theoretical statement:

For any finite group G, let IrrepG denote a set of representatives of all irreducible representations ofG overC modulo isomorphism.¹

From the theory of group determinants (more precisely, the results of [1], or the proof of Theorem 4.7 in [2]), we know that ifGis a finite group, andX_g is an indeterminate² for every g ∈G, then the matrix (X_gh⁻¹)_g,h∈G (both rows and columns of this matrix are indexed by elements of G) has determinant

det

(X_gh⁻¹)_g,h∈G

= Y

ρ∈IrrepG



det X

g∈G

ρ(g)X_g

!dimρ

.

Applying this toG =S_n and evaluating this polynomial identity at X_g = x^σ(g), we obtain

det

x^σ(^gh−1)

g,h∈Sn

= Y

ρ∈IrrepSn



det X

g∈Sn

ρ(g)x^σ(g)

!dimρ

. (1)

Hence, in order to show that the polynomial det

x^σ(^gh−1)

g,h∈Sn

∈Q[x]

factors into linear factors of the formx−`with`∈ {−n+ 1,−n+ 2, ..., n−1}, it is enough to prove that, for every irreducible representationρ of S_n overC,

1Remark. We are considering irreducible representations overChere for simplicity, but actually the argument works more generally: We can replaceCby any fieldKof character- istic 0 such that the group algebraK[G] factors into a direct product of matrix rings overK. In particular, the algebraic closure ofQdoes the trick. In the caseG=S_n (this is the case we are going to consider!), it is known thatanyfield of characteristic 0 can be taken asK, because the Specht modules are defined overQand thus provide a factorization of the group algebraK[G] into a direct product of matrix rings over Kfor any field Kof characteristic 0. See any good text on representation theory ofSn for details (the main reason for this to work is Corollary 4.38 of [2]).

2Distinct indeterminates are presumed to commute.

the polynomial det P

g∈Sn

ρ(g)x^σ(g)

!

factors into linear factors of the formx−`

with` ∈ {−n+ 1,−n+ 2, ..., n−1}.

We are going to show something better:

Theorem 2. Let λ = (λ₁, λ₂, ..., λ_n) be a partition of n. Let m_λ be the number of nonzero parts of the partition λ. Let ρ_λ be the irreducible representation of S_n over C corresponding to the partition λ. Then,

X

g∈S_n

ρ_λ(g)x^σ(g)= n!

dimρ Y

1≤i<j≤m_λ

λi−λj+j−i j−i ·

mλ

Y

i=1

x+λi−i λ_i

λ_i+m_λ−i

m_λ−i

·id_ρλ. (2)

Let us first see how Theorem 1 follows from Theorem 2:

Proof of Theorem 1. For every partition λ of n, let us denote by ρ_λ the irreducible representation of Sn overC corresponding to λ, and let us denote by m_λ the number of nonzero parts of the partition λ. It is known that the isomorphism classes of irreducible representations of S_n over C are in 1-to-1 correspondence with the partitions of n, and this correspondence sends every partition λ to the representationρ_λ. Thus,

Y

ρ∈IrrepSn



det X

g∈Sn

ρ(g)x^σ(g)

!dimρ



= Y

λpartition ofn



det X

g∈S_n

ρ_λ(g)x^σ(g)

!dimρ_λ



= Y

λpartition ofn

















 n!

dimρ Y

1≤i<j≤m_λ

λ_i−λ_j +j −i j −i ·

mλ

Y

i=1

x+λ_i−i λi

λ_i+m_λ −i

m_λ−i

·id_ρλ









dimρλ







 (by (2)).

Combined with (1), this yields det

x^σ(^gh−1)

g,h∈Sn

= Y

λpartition ofn

















 n!

dimρ Y

1≤i<j≤m_λ

λ_i−λ_j +j −i j −i ·

mλ

Y

i=1

x+λ_i−i λ_i

λ_i+m_λ −i

m_λ−i

·id_ρλ









dimρ_λ







 .

Now, the right hand side of this equation is clearly a polynomial in x which factors into a product of a constant and linear factors. All of the linear factors have the form x+λ_i −i−α for α ∈ {0,1, ..., λ_i−1} for various partitions λ of n and various i ∈ {1,2, ..., m_λ}. ³ By very simple combinatorics, it is easy to see that each of these factors has the form x −` for some ` ∈ {−n+ 1,−n+ 2, ..., n−1}. Thus, the polynomial det

x^σ(^gh−1)

g,h∈Sn

∈ Q[x] factors into a product of a constant and linear factors of the formx−`with

`∈ {−n+ 1,−n+ 2, ..., n−1}. Moreover, the constant is 1 because the poly- nomial det

x^σ(^gh−1)

g,h∈Sn

is monic⁴. Hence, the polynomial det

x^σ(^gh−1)

g,h∈Sn

∈ Q[x] factors into linear factors of the formx−`with`∈ {−n+ 1,−n+ 2, ..., n−1}.

Thus, Theorem 1 is proven (using Theorem 2).

§3. Proof of Theorem 2

Proof of Theorem 2. First of all, (2) is a polynomial identity inx. Hence, we can WLOG assume thatx is not a polynomial indeterminate in Q[x], but an integer greater than n (because if a polynomial identity over Q holds for infinitely many integers, then it must always hold). Assume this.

Sincex is an integer greater thann, we havex∈N. This allows us to find aQ-vector space of dimension x. Let V be such a vector space.

For everyS_n-moduleP, letχ_P denote the character of this moduleP. Note that everyh∈S_n satisfies

χ_V^⊗n(h) =x^σ(h). (3)

5

Let Lλ be the representation of GL (V) corresponding to the partition λ of n. In other words, let L_λ be the image of V under the λ-th Schur functor.

3In fact, the only place where x occurs on the right hand side of this equation is

x+λ_i−i λi

, and this factors as

x+λ_i−i λi

= (x+λ_i−i) (x+λ_i−i−1)...(x+λ_i−i−(λ_i−1))

λi! .

4Proof. In order to see this, it is enough to show that when the determinant det

x^σ(^gh−1)

g,h∈S_n

is written as a sum over all permutations of the setS_n (nota bene:

permutations ofS_n, not permutations in S_n), the highest degree ofxis contributed by the product of the main diagonal. But this is clear, because the main diagonal of the matrix

x^σ(^gh−1)

g,h∈Sn

is filled with x^σ(id) =xⁿ terms, while all other entries of the matrix are lower powers ofx.

5Proof. Leth∈S_n. Denote the action of honV^⊗n byh|_V⊗n. Then, by the definition of a character,χ_V⊗n(h) = Tr (h|_V⊗n).

Pick a basis (e1, e2, ..., ex) of V. This basis induces a basis (ei₁⊗ei₂⊗...⊗ei_n)_(i

1,i2,...,in)∈{1,2,...,x}ⁿ of V^⊗n. By the definition of the action of

Then, L_λ = Hom_Q[Sn](ρ_λ, V^⊗n) (by one of the definitions of Schur functors), so that

dimL_λ = dim Hom_Q[Sn] ρ_λ, V^⊗n

=hχ_V^⊗n, χ_ρλi

by Theorem 3.8 of [2], applied to V =V^⊗n and W =ρ_λ

= 1

|S_n|

|{z}

=1 n!

X

g∈Sn

χρ_λ(g)

| {z }

=Tr(ρλ(g))

χ_V^⊗n g⁻¹

| {z }

=x^σ(^g−1)

(by (3))

(by one of the definitions of the inner product of characters)

= 1 n!

X

g∈Sn

Tr (ρ_λ(g))x^σ(^g−1) = 1

n!Tr X

g∈Sn

ρ_λ(g)x^σ(^g−1)

!

= 1

n!Tr X

g∈Sn

ρ_λ(g)x^σ(g)

!

since everyg ∈S_n satisfies σ g⁻¹

=σ(g) . (4)

Sn onV^⊗n, every (i1, i2, ..., in)∈ {1,2, ..., x}ⁿ satisfies

h(ei₁⊗ei₂⊗...⊗ei_n) =e_h⁻¹_(i1)⊗e_h⁻¹_(i2)⊗...⊗e_h⁻¹_(in).

Thus, if h^(×n) denotes the permutation of the set {1,2, ..., x}ⁿ which sends every (i1, i2, ..., in)∈ {1,2, ..., x}ⁿ to h⁻¹(i1), h⁻¹(i2), ..., h⁻¹(in)

, then the linear maph|_V^⊗n is represented by the permutation matrix of the permutationh^(×n)with respect to the basis (ei₁⊗ei₂⊗...⊗ei_n)_(i

1,i₂,...,i_n)∈{1,2,...,x}ⁿ ofV^⊗n. Hence, Tr (h|V^⊗n) = Tr

permutation matrix of the permutation h^(×n)

=

number of fixed points ofh^(×n) (because the trace of a permutation matrix always equals the number of fixed points of the

corresponding permutation). Now, let us count the fixed points ofh^(×n).

Clearly, ann-tuple (i1, i2, ..., in)∈ {1,2, ..., x}ⁿis a fixed point ofh^(×n)if and only if every j∈ {1,2, ..., n}satisfiesij=i_h⁻¹_(j). In other words, ann-tuple (i1, i2, ..., in)∈ {1,2, ..., x}ⁿ is a fixed point ofh^(×n)if and only if each pair of elementsj andkof{1,2, ..., n}which lie in the same cycle ofhsatisfies ij =ik. Hence, if we want to choose a fixed point ofh^(×n), we need only to specify, for every cyclecofh, the value ofijfor some elementjof this cycle c(which elementj we choose doesn’t matter). Thus, we have to choose one element of the set{1,2, ..., x}for each cycle ofh; these choices are arbitrary and independent, but beside them we have no more freedom. Thus, there is a total ofx^σ(h)ways to choose a fixed point ofh^(×n)(because there areσ(h) cycles ofh, and there arexelements of the set{1,2, ..., x}).

In other words, x^σ(h)=

number of fixed points ofh^(×n)

= Tr (h|_V^⊗n) =χ_V^⊗n(h). This proves (3).

On the other hand, Theorem 4.63 of [2] (the Weyl character formula) yields dimL_λ = Y

1≤i<j≤x

λ_i−λ_j +j−i

j −i (whereλ<sub>`</sub> denotes 0 for all ` > m_λ)

= Y

1≤i<j≤m_λ

λ_i−λ_j +j−i

j−i · Y

1≤i≤m_λ<j≤x

| {z }

=^mλQ

i=1 x

Q

j=mλ+1

λ_i−λ_j +j −i j−i

| {z }

=λ_i+j−i j−i

(sincem_λ<jyieldsλj=0)

· Y

m_λ≤i<j≤x

λ_i−λ_j +j −i j−i

| {z }

=1 (sincem_λ≤i<j yields that bothλiandλjare 0)

= Y

1≤i<j≤m_λ

λ_i−λ_j +j−i j−i ·

mλ

Y

i=1

x

Y

j=m_λ+1

λ_i+j−i j −i

| {z }

=

x+λ_i−i λ_i

λ_i+m_λ−i

m_λ−i

(this is straightforward to check)

· Y

m_λ≤i<j≤x

1

| {z }

=1

= Y

1≤i<j≤m_λ

λ_i−λ_j +j−i j−i ·

mλ

Y

i=1

x+λ_i−i λ_i

λ_i+m_λ−i

m_λ−i .

Combined with (4), this yields

1

n!Tr X

g∈S_n

ρ_λ(g)x^σ(g)

!

= Y

1≤i<j≤m_λ

λ_i−λ_j +j−i j −i ·

mλ

Y

i=1

x+λi−i λ_i

λ_i+m_λ−i

m_λ−i ,

so that

Tr X

g∈Sn

ρ_λ(g)x^σ(g)

!

=n! Y

1≤i<j≤mλ

λ_i−λ_j +j−i j −i ·

m_λ

Y

i=1

x+λ_i−i λ_i

λ_i+m_λ−i

mλ−i

. (5)

But P

g∈S_n

gx^σ(g) is a central element of Q[S_n] (since the map S_n → Q, g 7→ σ(g) is a class function), so that P

g∈Sn

gx^σ(g) acts on any irreducible rep- resentation ofSn as a scalar multiple of id (by Schur’s lemma). In particular, this yields that ρ_λ P

g∈Sn

gx^σ(g)

!

= κ· id_ρλ for some κ ∈ C (since ρ_λ is an

irreducible representation ofS_n). Consider this κ. Then, X

g∈Sn

ρ_λ(g)x^σ(g) =ρ_λ X

g∈Sn

gx^σ(g)

!

=κ·id_ρλ, (6) so that

Tr X

g∈Sn

ρ_λ(g)x^σ(g)

!

= Tr (κ·id_ρλ) =κ·dimρ_λ. Combined with (5), this yields

κ·dimρ_λ =n! Y

1≤i<j≤m_λ

λ_i−λ_j+j−i j−i ·

mλ

Y

i=1

x+λ_i−i λ_i

λ_i+m_λ−i

m_λ−i , so that

κ= n!

dimρ Y

1≤i<j≤m_λ

λ_i−λ_j +j−i j−i ·

mλ

Y

i=1

x+λ_i−i λ_i

λ_i+m_λ−i

m_λ−i . Thus, (6) becomes

X

g∈S_n

ρ_λ(g)x^σ(g) = n!

dimρ Y

1≤i<j≤m_λ

λ_i−λ_j+j−i j−i ·

mλ

Y

i=1

x+λi−i λ_i

λ_i+m_λ−i

m_λ−i

·id_ρλ. This proves Theorem 2.

Hints to exercises Hint to exercise 1: Letn≥2. Expand det

x^σ(^gh−1)

g,h∈Sn

as a product over all permutations ofS_n(a total of (n!)! permutations, but you don’t have to actually do the computations...). It is clearly enough to show that every such permutation gives rise to a product which simplifies to x^m for some even m.

To prove this, show that any permutationα∈S_n satisfies signα= (−1)^n−σ(α). References

[1] Keith Conrad,The Origin of Representation Theory.

http://www.math.uconn.edu/~kconrad/articles/groupdet.pdf

[2] Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwend- ner, Dmitry Vaintrob, Elena Yudovina,Introduction to representation theory, arXiv:0901.0827v5.

http://arxiv.org/abs/0901.0827v5