Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 27 August 2013

Lecture 15

Theorem 2.45 pr(x)sλ (x)=μ (-1)k-1 sμ(x), where the sum is over partitions μλ such that μ/λ is a border strip of length r, and k is the number of rows in μ/λ.


Since the function pr(x) is symmetric, we have pr(x) aλ+δ(x) = wW ε(w)w pr(x) xλ+δ = wW ε(w)w i=1n xir x1λ1+n-1 x2λ2+n-2 xnλn = i=1n wW ε(w)w x1λ1+n-1 x2λ2+n-2 xiλi+n-i+r xnλn = i=1n aλ+δ+rεi (x), where εi=(0,0,,1,0,,0) has a 1 in the ith slot and 0 everywhere else.

The function aλ+δ+rεi(x) is alternating, so aλ+δ+rεi (x)=0, if there is a repeated entry in the exponents. That is, it is zero if (n-i)+λi +r=(n-j)+λj for some 1ji. If this is not the case, then there exists some j with 1j<i so that w=sjsj+1si-1 satisfies wxλ+δrεi =xμ+δ for some partition μ. In other words, w rearranges the exponents so that they are in the form μ+δ. Then we have aλ+δ+rεi (x)=(-1)(w) aμ+δ(x) To see the relation to border strips, consider the case where λ=(3,3,2,1,1,1,0), r=6, and i=4. Then λ+δ+rεi= | | | | | | | . Since aλ+δ+rεi(x) is alternating, we can apply s3, as long as we multiply by -1. Multiplying by s3 switches row three and row four. If we do this and then write the result with δ left of the wall, we get | | | | | | | . Continuing in this way, i.e., multiplying by s2 and then s1, gives the diagram | | | | | | | , which is a border strip of length 6 that has 4 rows. For each i, we add a border strip of length r at row i, and then sum over i. If adding such a strip does not produce a partition, then at some point in the process, we have violated the condition that all of the exponents (thus the length of each row) must be distinct, and we get 0. Thus, we have shown pr(x) aλ+δ(x)= μ(-1)k-1 aμ+δ(x). Dividing both sides by aδ(x) proves the theorem.

Corollary 2.46 p1(x) sλ(x) = μλμ/λ= sμ(x).

Corollary 2.47 p(1m)(x) = λ fλsλ (x) for some positive integer fλ.


Use induction and the previous corollary.

Since fλ>0, we have finished the proof of the Frobenius formula 2.29.

Theorem 2.48 [Murnaghan-Nakayama Rule] Let μ=(μ1,,μk) and μˆ=(μ1,,μk-1). Then χλ(μ)= ν(-1)s-1 (μˆ), where the sum is over νλ such that λ/ν is a border strip of length μk, and s is the number of rows in this border strip.


From the Frobenius formula, we have λχλ (μ)sλ (x) = pμ(x) = pμ1(x) pμ2(x) pμk(x) = pμˆ(x) pμk(x) = ( νχν (μˆ) sν(x) ) pμk(x) = νχν (μˆ) λ(-1)s-1 sλ(x) = λsλ(x) ν(-1)s-1 χν(μˆ) where λν and λ/ν is a border strip of length μk with s rows. Comparing coefficients of sλ gives the theorem.

Definition 2.49 The weight wt of a border strip λ/μ is given by wt(λ/μ)=(-1)s-1, where s is the number of rows in λ/μ.

Definition 2.50 Let μ=(μ1,,μk) be a partition. A μ-border strip tableau of shape λ is a sequence T of partitions T= ( =λ(0) λ(1) λ(2) λ(k)=λ ) such that λ(j)/λ(j+1) is a border strip of length μj. The weight of T is given by wt(T)= j=1kwt (λ(j)/λ(j+1)) .

With these definitions, the Murnaghan-Nakayama rule can be written in the following form

Theorem 2.51 [Murnaghan-Nakayama Rule II] χλ(μ)= Twt(T), where the sum is over all μ-border strip tableaux T.

Now let μ=(1m). We can compute all of the μ-border strip tableaux as follows. This graph is called the Young lattice. It is characterized by the following properties:

(1) On the mth level, the nodes are labeled by partitions of m.
(2) A node on the mth level is connected to a node on the (m+1)st level if the partitions differ by a single box.

Fix a partition λ. The paths from to λ that are always increasing in level are the (1m)-border strip tableaux of shape λ. Moreover, from the Murnaghan-Nakayama rule, we see that the number of these paths is equal to χλ(1m) which is the dimension of the irreducible Sm-module labeled by λ.

Definition 2.52 A standard tableau of shape λm is a filling in of the boxes in the Ferrers diagram of λ with the integers {1,2,,m} such that numbers increase in both the rows (moving left to right) and the columns (moving top to bottom).

The increasing paths in the Young diagram correspond exactly to standard tableaux. For example, the path 1 1 2 1 2 3 1 2 3 4 , is completely determined by the standard tableau 1 2 3 4 , and vice-versa.

Recall that from the semisimplicity of [Sm] we have |Sm|= λmdim (Vλ)2= λm (fλ)2, where fλ denotes the number of standard tableaux of shape λ. For example, in the 4th row of the Young diagram, we have 4!=1+32+22 +32+1.

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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