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 12

Continued proof.

Step 3. The polynomial i<j(xi-xj) is homogeneous of degree (n2).

Proof.

Each monomial in the expansion of i<j(xi-xj) is obtained by choosing a factor of either xi or xj from each factor (xi-xj) for 1i<jn. Therefore, this polynomial is homogeneous of total degree |{(i,j)|1i<jn}|=(n2).

Step 4. The coefficient of xλ+δ in aλ+δ(x) is one.

Proof.

Observe that λ+δ has distinct parts, so wxλ+δ=xλ+δ implies w=1. Then aλ+δ(x) |xλ+δ= wW (ε(w)wxλ+δ) |xλ+δ=1.

Step 5. (i<j(xi-xj))|xδ=1.

Proof.

To obtain the monomial xδ=x1n-1+x2n-2++xn-1 as a term in the expansion of i<j(xi-xj), it is necessary to choose a factor of x1 from all factors of the form (x1-xj), j>1. The only source of factors x2 then are the factors of the form (x2-xj), j>2, from each of which we must choose x2. We proceed in this manner and are forced to choose all factors of xi from the terms (xi-xj) where i<j for 1in-1. Hence, there is only one monomial equal to xδ in the expansion of this product.

We now conclude the proof of the Weyl Denominator Formula for Type A.

Proof [WDF].

The polynomials aλ+δ(x) are a basis for the space of alternating symmetric functions, hence i<j(xi-xj)=λncλ+δaλ+δ(x) for some cα. The left hand side is homogeneous of degree (n2) by step 3; however, only aδ has this degree among the aλ+δ by step one. Thus, i<j(xi-xj)=cδaδ. Comparing coefficients of xδ using steps 4 (with λ=0) and 5 yields that cδ=1.

Definition 2.21 The Schur function, denoted sλ(x) associated to the partition λn is the symmetric function defined by sλ(x)= aλ+δ(x)aδ(x)

Note that the Schur function is a symmetric function. If f(x)=αfαxα and g(x)=βgβxβ are arbitrary polynomials in [x1,x2,,xn], then for all wW w·(fg) = α,β fαgβw· xα+β = α,β fαgβ xwα+wβ = αfαxwα βgβxwβ = (w·f) (w·g) In particular, w·(aδ(x)sλ(x)) =ε(w)aδ(x)(w·sλ(x)). However w·(aδ(x)sλ(x)) =w·aλ+δ(x)= ε(w)aλ+δ(x) from which it follows that w·sλ(x)=sλ(x) for all wW.

Moreover, we may define a linear map ΛnAn by sending a symmetric function f(x) to aδf(x)An. The inverse map AnΛn defined by g(x)g(x)aδ(x) is well defined, since the set of aλ+δ(x) forms a basis for An and these polynomials are divisible by aδ(x). Hence this map is a vector space isomorphism of ΛnAn. Furthermore, since the Schur functions map onto a basis of An, we have that

Proposition 2.22 The Schur functions {sλ(x)|λn} form a basis for the vector space Λn.

Remark. This works for any finite Weyl group W.

We will next establish a very interesting relationship between the Schur functions and the Power symmetric functions. First, we will need the following formula due to Cauchy.

Lemma 2.23 (Cauchy‚Äôs Determinant) |11-xiyj|1i,jn =i<j 11-xiyj aδ(x) aδ(y)

Proof.

We roll up our sleeves and calculate. Let Δ=|11-xiyj|. Subtract the first row from all other rows; the (i,j) entry of rows two through n becomes 11-xiyj- 11-x1yj= (xi-x1)yj (1-xiyj) (1-x1yj) . Hence, we may pull out a common factor of (xi-x1) from row i for i2. Note that the product of these common factors may be written (-1)n-1i=2n(x1-xi). The determinant then becomes Δ=(-1)n-1 i=2n (x1-xi) 11-x1y1 11-x1y2 11-x1yn y1(1-x2y1)(1-x1-y1) y2(1-x2y2)(1-x1-y2) yn(1-x2yn)(1-x2-yn) y1(1-xny1)(1-x1-y1) y2(1-xny2)(1-x1-y2) yn(1-xnyn)(1-x1-yn) . Extracting a common factor of (1-x1yj)-1 from the jth column for 1jn, we obtain Δ=(-1)n-1 i=2n (x1-xi) j=1n 11-x1yj 111 y11-x2y1 y21-x2y2 yn1-x2yn y11-xny1 y21-xny2 yn1-xnyn . Next subtract column one from each of the remaining columns. For rows 2in, the (i,j) entry is given by yj1-xiyj- y11-xiy1= yj-y1 (1-xiyj) (1-xiy1) . Hence we may extract a factor of (-1)(y1-yj) from column j for 2jn. Combining factors of (-1)n-1, we obtain Δ = i=2n (x1-xi) (y1-yi) j=1n 11-x1yj 100 y11-x2y1 1(1-x2y2)(1-x2y1) y1(1-x2yn)(1-x2y1) y11-xny1 1(1-xny2)(1-xny1) 1(1-xnyn)(1-xny1) = i=2n (x1-xi) (y1-yi) j=1n 11-x1yj |1(1-xiyj)(1-xiy1)| 2i,jn by expanding along the first row. We may pull out a factor of (1-xiy1)-1 from each column (2jn); it follows from the inductive hypothesis for the variables x2,x3,xn,y2,y3,,yn that Δ = i=1n (x1-xi) (y1-yi) j=1n 11-x1yj 11-xjy1 |1(1-xiyj)| 2i,jn = i=1n (x1-xi) (y1-yi) j=1n 11-x1yj 11-xjy1 { 2i<jn (xi-xj) (yi-yj) 11-xiyj } = aδ(x) aδ(y) 1i<jn 11-xiyj

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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