Let $n \geq 7$ and $V = \mathbb{C}^n$ be the standard representation for $S\_{n+1}$, the symmetric group of cardinal $(n+1)!$ 


Let $k$ be an integer such that $2 \leq k \leq n$. Is it true or false that $\bigwedge^k V$ does not appear as an irreducible sub-representation of $\mathrm{Sym}^k V$? I am looking for a reference which would either prove or give counter-examples to this.


Thanks in advance!


**EDIT** : Is there an online software (analogous to [LIE](http://wwwmathlabo.univ-poitiers.fr/~maavl/LiE/form.html)) that would compute the decompositions into irreducible representations of the tensor product of two ireps for the symmetric group?



alternating and symmetric powers of the standard representation of the symmetric group
Let $W = \mathbb C^n$, with the obvious representation of $S\_n$. 


**Lemma** Let $\lambda$ be a partition of $n$. The multiplicity of the irreducible representation $V\_\lambda$ inside $W^{\otimes k}$ is given by the number of ways of obtaining the partition $\lambda$ by starting with the partition $[n]$, and then alternating between removing a box from the Ferrers diagram and adding a box to the Ferrers diagram $k$ times.


**Proof** Induction on $k$, the base case $k=0$ being trivial. We have
$$ W^{\otimes (k+1)} = W^{\otimes k} \otimes \mathrm{Ind}\_{S\_{n-1}}^{S\_n} \mathbb C = \mathrm{Ind}\_{S\_{n-1}}^{S\_n} \mathrm{Res}^{S\_n}\_{S\_{n-1}} W^{\otimes k},$$
see e.g. Exercise 3.16 of Fulton-Harris. Thus the result follows from Pieri's rule. **QED**


Let $V = W/\mathbb C$ be the irreducible representation corresponding to the partition $[n-1,1]$. Then $\wedge^k V$ is also irreducible, with corresponding partition $[n-k,1,1,\ldots,1]$, for $k\leq n-1$. By the preceding lemma, it is easy to see that the multiplicity of $\wedge^k V$ inside $W^{\otimes k}$ is exactly $1$ for $k \leq n-1$. 


It follows in particular that $\wedge^k V$ occurs with multiplicity $0$ inside $\mathrm{Sym}^k(W)$ for $k=2,\ldots,n-1$. Indeed $\wedge^k(V) \subset \wedge^k(W)$ and $\wedge^k(W) \cap \mathrm{Sym}^k(W) = \{0\}$ for $k \geq 2$, where the intersection is taken inside of $W^{\otimes k}$. In particular this representation doesn't occur inside $\mathrm{Sym}^k(V)$, either.




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Also, here is some SAGE code that will compute things for you. 


`s = SymmetricFunctions(QQ).schur()
V = s([n-1,1])
ext = s([1]*k)
sym = s([k])
print ext.inner_plethysm(V) ## decomposition of \wedge^k(V)
print sym.inner_plethysm(V) ## decomposition of Sym^k(V)`


