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.


1 Answer 1


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 $$ \begin{bmatrix}\bullet&\bullet\\\bullet&\end{bmatrix}\otimes\begin{bmatrix}\bullet\end{bmatrix}=\begin{bmatrix}\bullet&\bullet&\bullet\\\bullet&&\end{bmatrix}\oplus \begin{bmatrix}\bullet&\bullet\\\bullet&\bullet\end{bmatrix}\oplus\begin{bmatrix}\bullet&\bullet\\\bullet&\\\bullet&\end{bmatrix} $$

  • 1
    $\begingroup$ BTW the proof of this branching rule can be found in Sagan's book on Symmetric groups. $\endgroup$
    – David Hill
    Feb 16, 2018 at 18:22

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