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I have a question about the interpretation of multiplicities and dimensions using Schur-Weyl duality.

$V$ is an n-dimensional complex vector space. Then $V$ $\otimes$ $V$ $\otimes$ $V$ decomposes as:

$V$ $\otimes$ $V$ $\otimes$ $V$ = $Sym^3$ $V$ $\oplus$ $\wedge^3$ $V$ $\oplus$ $S_{(2,1)}V$ $\oplus$ $S_{(2,1)}V$

where $S_{(2,1)}V$ is a Schur module for the partition (2,1). (Fulton-Harris, Chapter 6).

Then Schur-Weyl duality says that the multiplicity of $S_{(2,1)}V$ in this decomposition (it is 2) should be the dimension of the irrep of $S_3$ labelled by the partition (2,1) - which is correct.

My question is about the other half of this duality: the dimension of $S_{(2,1)}V$ in this decomposition (it is 8) should correspond to some sort of multiplicity for the irrep of $S_3$ labelled by the partition (1,2) - but I am unable to see exactly what...

Any help is most welcome.

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I suppose you know that $S_\lambda(V)=F_\lambda\otimes_{S_k} V^{\otimes k}$, where $F_\lambda$ is the $\lambda$-irrep of the symmetric group $S_k$. –  Jim Conant Sep 27 '11 at 11:54
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3 Answers 3

The dimension of $S_{21}(V)$ is $(n+1)n(n-1)/3$; it is only $8$ if $n=3$. The other half of Schur Weyl duality says that $S_{21}(V)$ is a $GL_n$ (not $S_3$) irrep; namely, the one which is indexed by the partition $(2,1)$. Similarly, $\mathrm{Sym}^3(V)$ and $\bigwedge^3 V$ have dimensions $(n+2)(n+1)n/6$ and $n(n-1)(n-2)/6$ and are $GL_n$, not $S_3$, irreps.

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the dim of $S_\lambda(V)$ is approximately $c_\lambda dim^n V / n!$ where $c_\lambda$ is the dimension of the corresponding module over the symmetric group. –  kassabov Sep 27 '11 at 14:10
    
@kassabov: I think you meant to leave your comment for the OP. In any case, David's answer addresses the OP's concerns perfectly. –  Faisal Sep 27 '11 at 17:07
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The multiplicity of a Specht module $Specht(\lambda)$ (Fulton-Harris Ch.4) in $V^{\otimes n}$ is the number semi-standard Young tableaux of shape $\lambda$ and entries in $\{1,2,\dots,\dim V\}$. In your example (where I've guessed you're assuming $\dim V = 3$) the multiplicity of the Spetch module $Specht((2,1))$ is the number of SSYT of this shape with entries in $\{1,2,3\}$. This is computed using the hook-content formula, and is 8 (or just write down the 8 tableuax).

It is instructive to find a basis of the Schur module of shape $\lambda$ indexed by SSYT of shape $\lambda$. See Fulton's book on Young tableuax for this, or the book Constructive invariant theory by Sturmfels.

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Thank you! That helped. Ravi –  Ravi Sep 28 '11 at 10:23
    
I could only find lecture notes called "Constructive Invariant Theory" (Sturmfels & Tsai, 28 pages). Is there a book too? Ravi –  Ravi Sep 28 '11 at 10:40
    
I was wrong. It is called Algorithms in invariant theory. –  Andy B Sep 28 '11 at 16:34
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Let $V$ be the vector representation of $GL_n(\mathbb{C})$, and let $d\leq n$.

You want to see the multiplicities of a given irreducible $S_d$ module in $V^{\otimes d}$ in terms of the dimension of an associated irreducible $GL_n(\mathbb{C})$-module. To do this, observe that $S_d$ acts on $V^{\otimes d}$ by permuting the tensor factors and this action commutes with the diagonal action of $GL_n(\mathbb{C})$ on $V^{\otimes d}$. Therefore, for a partition $\lambda=(\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_n\geq 0)$ (equivalently, a polynomial weight for $GL_n(\mathbb{C})$) we can consider the space $W_\lambda$ of highest weight vectors in $V^{\otimes d}$ of highest weight $\lambda$. Since the action of $S_d$ commutes with the action of $GL_n(\mathbb{C})$ (and, in particular, the action of the standard maximal torus, and standard Borel), $W_\lambda$ is an $S_d$-module. By character considerations, one can show that this space is the irreducible Specht module, $W_\lambda\cong S^{\lambda}$.

More generally, if $X\in GL_n(\mathbb{C})$, then $X\cdot W_\lambda$ is another $S_d$-module since, again, the action of $S_d$ commutes with the action of $GL_n(\mathbb{C})$. Moreover, this module is obviously isomorphic to $S^\lambda$. This means that there are $\dim L(\lambda)$ copies of $S^\lambda$ in $V^{\otimes d}$, where $L(\lambda)$ is the irreducible $GL_n(\mathbb{C})$-module of highest weight $\lambda$.

This proves half of the statement that, as a $(GL_n(\mathbb{C}),S_d)$-bimodule, $V^{\otimes d}\cong \bigoplus_{\lambda\vdash d}L(\lambda)\otimes S^\lambda.$ From your question, I assume you understand the other half.

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Thank you! That helped. Ravi –  Ravi Sep 28 '11 at 10:23
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