2
$\begingroup$

Let $S_n$ be the symmetric group on $n$ symbols and let $\lambda = (\lambda_1,\dots,\lambda_r)$ be a partition of $n$. Considering the Young diagram associated to this partition we can define the subgroups $P$ and $Q$ of $S_n$ whose elements respectively preserve each row and each column of the diagram.

In that case if $\mathbb{C}S_n$ is the group algebra we define

$$a_\lambda = \sum_{g\in P}e_g,\quad b_\lambda = \sum_{g\in Q}\operatorname{sgn}(g) e_g,$$

we also define $c_\lambda = a_\lambda b_\lambda$ the associated Young symmetrizer.

Now let $V$ be a vector space and suppose $S_n$ acts on $V^{\otimes n}$ by permutating the factors. We can extend this action to allow $c_\lambda$ act on $V^{\otimes n}$ and thus we get one representation of $\mathbb{C}S_n$ on $V^{\otimes n}$.

Thus if we take the partition $\lambda = (n)$ we have $c_\lambda = \sum_{g\in S_n} e_g$ and hence

$$c_\lambda(V^{\otimes n})=\operatorname{Sym}^n V.$$

I have read that if $[\lambda_1,\dots,\lambda_r]$ denotes the irreducible of the symmetric group representation associated to the partition $(\lambda_1,\dots,\lambda_r)$, we then have the decomposition into irreducible parts

$$[n]\otimes [1]=[n,1]\oplus [n+1],$$

which in index notation decomposes one $F^{a_1\dots a_n a_{n+1}}$ symmetric on the first $n$ indices into the parts $F_1^{a_1\dots a_n a_{n+1}}$ and $F_2^{a_1\dots a_{n} a_{n+1}}$ where $F_1$ is totally symmetric and $F_2$ is antisymmetric on the last two indices.

Now I want to understand this and understand how to find this kind of decomposition in general.

I believe the point is: $[n]$ corresponds to the representation associated with the partition $(n)$ and $[1]$ that with partition $(1)$. These give symmetrizers $c_{(n)},c_{(1)}$ acting respectively on $V^{\otimes n}$ and $V$. Thus we get one symmetrizer $c_{(n)}\otimes \mathbf{1}+\mathbf{1}\otimes c_{(1)}$ acting on $V^{\otimes {n+1}}$ such that

$$c_{(n)}\otimes \mathbf{1}+\mathbf{1}\otimes c_{(1)}(V^{\otimes n}\otimes V)=\operatorname{Sym}^n V\otimes V$$

and this is decomposed.

Now my doubts are:

  1. Which representations are being dealt with here? Representations of $S_n$ or $\mathbb{C}S_n$? I believe it is of the later.

  2. How one decomposes $[n]\otimes [1]=[n,1]\otimes [n+1]$ in practice? I believe this has something to do with the so-called Weyl construction, but I'm totally unsure.

$\endgroup$
0
$\begingroup$

For question 1: there is no difference between representations of $S_n$ and representations of $\mathbb{C}S_n$.

For question 2: If I am translating your notation correctly, then $[n]\otimes[1]$ refers to induction from $S_n$ to $S_{n+1}$. For an arbitrary partition $\lambda\vdash n$ (which I will identify with its Young diagram), we have $$ [\lambda]\otimes[1]=\bigoplus_{\lambda'}[\lambda'] $$ where the sum is over all partitions $\lambda'\vdash n+1$ which can be obtained from $\lambda$ by adding a box. For example, $$ [2,1]\otimes[1]=[3,1]\oplus[2,2]\oplus[2,1,1]. $$ In terms of Ferrers diagrams (which is the best I know how to do on this site), we get $$ \begin{bmatrix}\bullet&\bullet\\\bullet&\end{bmatrix}\otimes\begin{bmatrix}\bullet\end{bmatrix}=\begin{bmatrix}\bullet&\bullet&\bullet\\\bullet&&\end{bmatrix}\oplus \begin{bmatrix}\bullet&\bullet\\\bullet&\bullet\end{bmatrix}\oplus\begin{bmatrix}\bullet&\bullet\\\bullet&\\\bullet&\end{bmatrix} $$

$\endgroup$
  • $\begingroup$ BTW the proof of this branching rule can be found in Sagan's book on Symmetric groups. $\endgroup$ – David Hill Feb 16 '18 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.