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.
 A: 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.
