# Schur -Weyl duality for $sl_2$ and $S_n$

$V$ is an $m$ dimensional vector space having a structure of $sl_2(\mathbb{C})$-module, where $sl_2(\mathbb{C})$ is the Lie algebra of the Lie group $SL_2(\mathbb{C})$. The symmetric group $S_n$ acts on the tensor product $V^{\otimes n}$.

What does Schur-Weyl duality say in this case?

What is the irreducible decomposition of $V^{\otimes n}$?

If we have $S_n$ irreducible decomposition can we get $sl_2(\mathbb{C})$ decomposition and vice versa?

I would be very grateful if someone could give a detailed answer. Thanking you in advance.

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– Marc van Leeuwen Jun 8 '12 at 13:46

## 1 Answer

Here is an answer to your question about the decomposition of $V^{\otimes n}$ as an $\mathfrak{sl}_2(\mathbb{C})$-module (I assume $V$ is the standard $2$-dimensional irreducible module).

The irreducible $\mathfrak{sl}_2(\mathbb{C})$-modules are indexed by non-negative integers and the one corresponding to the integer $m$ will be denoted $L(m)$ (it has dimension $m+1$ and we know exactly what it looks like, see for example Humphrey's Introduction to Lie Algebras and Representation Theory chapter 7).

So to see how to decompose $V^{\otimes n}$ we need to know what $V\otimes L(m)$ is for some integer $m$. This is easiest to see if we look at these as $\mathfrak{gl}_2(\mathbb{C})$-modules where the decomposition of tensor products is given by the Littlewood-Richardson rule. In this case, since $V = L(1)$, we simply get that $V\otimes L(m) = L(m+1)\oplus L(m-1)$.

Now we want to apply this to see how many times $L(m)$ appears as a summand in $V^{\otimes n}$. Let us denote this multiplicity by $a_{m,n}$.

We see from the above that $a_{m,n} = a_{m-1,n-1} + a_{m+1,n-1}$.

One can check that $$a_{m,n} = \binom{n}{\frac{m+n}{2}} - \binom{n}{\frac{n - m - 2}{2}}$$ satisfies the above recursive formula when $n$ and $m$ have the same parity (and when they don't, $a_{m,n} = 0$). Note that $a_{0,2k} = a_{1,2k-1}$ is the $k$'th Catalan number.

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Thank you for your answer, i was trying to get some information about $V$ not necessarily an irreducible module, having dimension m... – Jacob Feb 20 at 6:42
If $V$ is not irreducible, it will be the direct sum of some irreducible modules. It is possible to calculate the tensor powers of an arbitrary module in the same way as above, but the combinatorics become very complicated in the general case. – Tobias Kildetoft Feb 20 at 13:48
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