## Lectures in Representation Theory

Last update: 20 August 2013

## Lecture 13

Definition 2.24 The product $\prod _{1\le i,j\le n}\frac{1}{1-{x}_{i}{y}_{j}}$ is called the Cauchy kernel.

Theorem 2.25 $\sum _{\underset{{\lambda }_{1}\ge \cdots \ge {\lambda }_{n}\ge 0}{\lambda =\left({\lambda }_{1},\dots ,{\lambda }_{n}\right)}}{s}_{\lambda }\left(x\right){s}_{\lambda }\left(y\right)=\prod _{1\le i,j\le n}\frac{1}{1-{x}_{i}{y}_{j}}\text{.}$

 Proof. Note that $\text{det} \left(\frac{1}{1-{x}_{i}{y}_{j}}\right)$ is an alternating polynomial in the ${x}_{i}\text{'s}$ and in the ${y}_{j}\text{'s.}$ Thus, for $\lambda =\left({\lambda }_{1}\ge {\lambda }_{2}\ge \cdots \ge {\lambda }_{n}\ge 0\right),$ we wish to compute the coefficient of ${x}^{\lambda +\delta }\text{.}$ We have $11-xiyj= 1+xiyj+ (xiyj)2+ (xiyj)3+ ⋯,$ so $det (11-xiyj) = ∑w∈Wε(w) 11-x1yw(1) 11-x2yw(2)⋯ 11-xnyw(n) = ∑k1,…,kn ∑w∈Wε(w) x1k1 yw(1)k1 x2k2 yw(2)k2⋯ xnkn yw(n)kn.$ Hence the coefficient of ${x}^{\lambda +\delta }$ in $\text{det} \left(\frac{1}{1-{x}_{i}{y}_{j}}\right)$ is $∑w∈Wε(w) yw(1)λ1+n-1 yw(2)λ2+n-2⋯ yw(n)λn =aλ+δ(y).$ By the same argument, the coefficient of ${y}^{\lambda +\delta }$ is ${a}_{\lambda +\delta }\left(x\right),$ and thus $det (11-xiyj)= ∑ λ=(λ1,…,λn) λ1≥⋯≥λn≥0 aλ+δ(y) aλ+δ(x)$ Applying Cauchy’s determinant 2.23 gives us $∑ λ=(λ1,…,λn) λ1≥⋯≥λn≥0 aλ+δ(x) aλ+δ(y) aδ(x) aδ(y) =∏1≤i,j≤n 11-xiyj,$ and the result follows from the definition 2.21 of the Schur function. $\square$

Definition 2.26 We say that $\mu =\left({\mu }_{1},{\mu }_{2},\dots ,{\mu }_{n}\right)\in {ℕ}^{n}$ is a partition of $m,$ denoted $\mu ⊢m,$ if ${\mu }_{1}\ge {\mu }_{2}\ge \cdots \ge {\mu }_{n}\ge 0$ and $\mu +{\mu }_{2}+{\mu }_{n}=m\text{.}$ The length $\ell \left(\mu \right)$ of $\mu$ is the largest $j$ such that ${\mu }_{j}>0\text{.}$

If $\mu =\left({\mu }_{1},{\mu }_{2},\dots ,{\mu }_{n}\right)\in {ℕ}^{n},$ then we write $\mu =\left({1}^{{m}_{1}}{2}^{{m}_{2}}\cdots \right)$ where ${m}_{i}$ is the number of ${\mu }_{j}$ equal to ${m}_{i},$ and we let $μ?= 1m1m1! 2m2m2! 3m3m3!⋯.$

Lemma 2.27 $\prod _{i=1}^{n}\frac{1}{1-{x}_{i}}=\sum _{\mu }\frac{{p}_{\mu }\left(x\right)}{\mu ?}\text{.}$

 Proof. We have $\frac{1}{1-{x}_{i}}=1+{x}_{i}+{x}_{i}^{2}+{x}_{i}^{3}+\cdots ,$ so $∫11-xidxi= -log(1-xi)=log (11-x+i)= xi+xi22+ xi33+ xi44⋯.$ Therefore, $∏i=1n 11-xi = ∏i=1nexp (log (11-xi)) =exp (∑i=1n∑r>0xirr) = exp (∑r>0∑i=1nxirr) =exp (∑r>0pr(x)r) = ∏r>0exp (pr(x)r)= ∏r>0 (∑m>0(pr(x))mrmm!) = ∑m1,m2,… (p1(x))m1 1m1m1! (p2(x))m2 2m2m2! ⋯ = ∑μ=(1m12m2⋯) pμ(x)μ?$ Note: for the equality between the second and third line to hold, we really need to be more careful and grade things by degree. We then prove the equality first working with the monomials of degree one, then with the monomials of degree two, and so on. $\square$

Theorem 2.28 $\prod _{1\le i,j\le n}\frac{1}{1-{x}_{i}{y}_{j}}=\sum _{\mu }\frac{{p}_{\mu }\left(x\right){p}_{\nu }\left(y\right)}{\mu ?}\text{.}$

 Proof. Let $XY$ denote the set $XY= { x1y1, x1y2, …, x1yn, x2y1, x2y2, …, x2yn, ⋮ ⋮ … ⋮ xny1, xny2, …, xnyn } .$ Then, by the previous lemma, $∏1≤i,j≤n 11-xiyj= ∑μ pμ(xy)μ?.$ Note that $pr(xy)= ∑1≤i,j≤n (xiyj)r= (∑i=1nxir) (∑j=1nyjr) =pr(x)pr(y),$ so ${p}_{\mu }\left(xy\right)={p}_{\mu }\left(x\right){p}_{\mu }\left(y\right),$ and we get $∏1≤i,j≤n 11-xiyj= ∑μ pμ(x)pμ(y)μ?$ as desired. $\square$

Let $\mu ⊢m\text{.}$ Our next goal it to prove the Frobenius formula $pμ(x)= ∑λ⊢mℓ(λ)≤n χλ(μ)sλ, (2.29)$ where ${\chi }^{\lambda }\left(\lambda \right)$ is the irreducible ${S}_{m}\text{-character}$ evaluated on the conjugacy class labeled by the partition $\lambda ⊢m\text{.}$ Since $\mu ⊢m,$ we know that ${p}_{\mu }\left(x\right)$ is a polynomial of total degree $m,$ so the restriction $\ell \left(\lambda \right)\le m$ makes sense.

For $\mu ⊢m,$ let ${\eta }^{\lambda }\left(\mu \right)\in ℂ$ and ${K}_{\lambda ,\nu }^{-1}\in ℂ$ be defined by the equations $pμ = ∑λ⊢mℓ(λ)≤n ηλ(μ)sλ(x), (2.30) mλ(x) = ∑ν⊢mℓ(λ)≤n Kλ,ν-1sν(x). (2.31)$ In other words, $\left({\eta }^{\lambda }\left(\mu \right)\right)$ is the transition matrix in ${\Lambda }_{n}$ between the basis of power symmetric functions and the basis of Schur functions, and $\left({K}_{\lambda ,\nu }^{-1}\right)$ is the transition matrix between the basis of monomial symmetric functions and the basis of Schur functions.

From the previous theorem, we get $∑λ=(λ1,…,λn)λ1≥⋯≥λn sλ(x) sλ(y) = ∑μ pμ(x) pμ(y) μ? = ∑μ (∑ρηρ(μ)sρ(x)) (∑τητ(μ)sτ(y)) μ? = ∑ρ,τsρ (x)sτ(y) ∑μ ηρ(μ) ητ(μ) μ? ,$ so we have $∑μ ηρ(μ) ητ(μ) μ? =δρ,τ. (2.32)$

Property 2.33 For all $\lambda$ and $\nu ,$ we have ${K}_{\lambda ,\nu }^{-1}\in ℤ\text{.}$

(Note: This property holds for all Weyl groups.)

 Proof. For $w\in W,$ the polynomial ${m}_{\lambda }\left(x\right){x}_{\delta }\left(x\right)$ satisfies $w (mλ(x)aδ(x)) =(wmλ(x)) (waδ(x)),$ so it is an alternating symmetric function. Moreover, the coefficient of each monomial ${x}^{\nu +\delta }$ in ${m}_{\lambda }\left(x\right){a}_{\delta }\left(x\right)$ is an integer, since the coefficients of the monomials in ${m}_{\lambda }\left(x\right)$ and ${a}_{\delta }\left(x\right)$ are integers. Thus $mλ(x) aδ(x)= ∑ν mλ(x) aδ(x) |xν+δ ⏟∈ℤ aν+δ(x).$ We divide both sides of the equation by ${a}_{\delta }$ to obtain $mλ(x)=∑ν mλ(x) aδ(x) |xν+δ ⏟∈ℤ sν(x).$ The result follows by comparing with 2.31 $\square$

From 1.58 we have orthogonality of characters for the symmetric group: $1|Sm| ∑σ∈Sm χλ(σ-1) χν(σ)= δλ,ν.$ Let ${𝒞}_{\mu }$ denote the conjugacy class of ${S}_{m}$ of elements whose cycle type is $\mu ⊢m\text{.}$ If we collect elements in each conjugacy class and sum over conjugacy classes, we get $1|Sm| ∑μ⊢mχλ (μ)χν(μ) |𝒞μ|= δλ,ν. (2.34)$

## Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.