Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 20 August 2013

Lecture 13

Definition 2.24 The product 1i,jn11-xiyj is called the Cauchy kernel.

Theorem 2.25 λ=(λ1,,λn)λ1λn0 sλ(x)sλ(y) = 1i,jn 11-xiyj.

Proof.

Note that det(11-xiyj) is an alternating polynomial in the xi's and in the yj's. Thus, for λ=(λ1λ2λn0), we wish to compute the coefficient of xλ+δ. We have 11-xiyj= 1+xiyj+ (xiyj)2+ (xiyj)3+ , so det(11-xiyj) = wWε(w) 11-x1yw(1) 11-x2yw(2) 11-xnyw(n) = k1,,kn wWε(w) x1k1 yw(1)k1 x2k2 yw(2)k2 xnkn yw(n)kn. Hence the coefficient of xλ+δ in det(11-xiyj) is wWε(w) yw(1)λ1+n-1 yw(2)λ2+n-2 yw(n)λn =aλ+δ(y). By the same argument, the coefficient of yλ+δ is aλ+δ(x), and thus det(11-xiyj)= λ=(λ1,,λn) λ1λn0 aλ+δ(y) aλ+δ(x) Applying Cauchy‚Äôs determinant 2.23 gives us λ=(λ1,,λn) λ1λn0 aλ+δ(x) aλ+δ(y) aδ(x) aδ(y) =1i,jn 11-xiyj, and the result follows from the definition 2.21 of the Schur function.

Definition 2.26 We say that μ=(μ1,μ2,,μn)n is a partition of m, denoted μm, if μ1μ2μn0 and μ+μ2+μn=m. The length (μ) of μ is the largest j such that μj>0.

If μ=(μ1,μ2,,μn)n, then we write μ=(1m12m2) where mi is the number of μj equal to mi, and we let μ?= 1m1m1! 2m2m2! 3m3m3!.

Lemma 2.27 i=1n 11-xi = μ pμ(x)μ?.

Proof.

We have 11-xi=1+xi+xi2+xi3+, 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=1nr>0xirr) = exp(r>0i=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.

Theorem 2.28 1i,jn 11-xiyj = μ pμ(x)pν(y) μ? .

Proof.

Let XY denote the set XY= { x1y1, x1y2, , x1yn, x2y1, x2y2, , x2yn, xny1, xny2, , xnyn } . Then, by the previous lemma, 1i,jn 11-xiyj= μ pμ(xy)μ?. Note that pr(xy)= 1i,jn (xiyj)r= (i=1nxir) (j=1nyjr) =pr(x)pr(y), so pμ(xy)=pμ(x)pμ(y), and we get 1i,jn 11-xiyj= μ pμ(x)pμ(y)μ? as desired.

Let μm. Our next goal it to prove the Frobenius formula pμ(x)= λm(λ)n χλ(μ)sλ, (2.29) where χλ(λ) is the irreducible Sm-character evaluated on the conjugacy class labeled by the partition λm. Since μm, we know that pμ(x) is a polynomial of total degree m, so the restriction (λ)m makes sense.

For μm, let ηλ(μ) and Kλ,ν-1 be defined by the equations pμ = λm(λ)n ηλ(μ)sλ(x), (2.30) mλ(x) = νm(λ)n Kλ,ν-1sν(x). (2.31) In other words, (ηλ(μ)) is the transition matrix in Λn between the basis of power symmetric functions and the basis of Schur functions, and (Kλ,ν-1) 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)λ1λn 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 λ and ν, we have Kλ,ν-1.

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

Proof.

For wW, the polynomial mλ(x)xδ(x) satisfies w (mλ(x)aδ(x)) =(wmλ(x)) (waδ(x)), so it is an alternating symmetric function. Moreover, the coefficient of each monomial xν+δ in mλ(x)aδ(x) is an integer, since the coefficients of the monomials in mλ(x) and aδ(x) are integers. Thus mλ(x) aδ(x)= ν mλ(x) aδ(x) |xν+δ aν+δ(x). We divide both sides of the equation by aδ to obtain mλ(x)=ν mλ(x) aδ(x) |xν+δ sν(x). The result follows by comparing with 2.31

From 1.58 we have orthogonality of characters for the symmetric group: 1|Sm| σSm χλ(σ-1) χν(σ)= δλ,ν. Let 𝒞μ denote the conjugacy class of Sm of elements whose cycle type is μm. If we collect elements in each conjugacy class and sum over conjugacy classes, we get 1|Sm| μmχλ (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.

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