Question on the proof that exterior powers of standard representations are irreducible

Question

I am reading Fulton and Harris representation theory Chapter 3.2 right now and I got a question on the proof of the fact that exterior powers of standard representations are irreducible.

Let $\mathbb{C}^n$ be the permutation representation of symmetric group $S_n$. Then we can decompose this representation into two irreducible represetations: trivial representation $U$ and standard representation $V$. The authors state that if $(\chi_{\mathbb{C}^n},\chi_{\mathbb{C}^n})=2$, then $V$ is irreducible, and it was shown in this question.

Then authors claim that "similarly", since $\bigwedge^k(\mathbb{C}^n)=(\bigwedge^k(V)\otimes \bigwedge^0(U))\oplus(\bigwedge^{k-1}(V)\otimes\bigwedge^1(U))$ $=\bigwedge^k(V)\oplus\bigwedge^{k-1}(V)$, then in order to show the irreducibility of each $\bigwedge^k(V)$ with $1 \leq k \leq n-1$, it is enough to show that $(\chi_{\bigwedge^k(\mathbb{C}^n)},\chi_{\bigwedge^k(\mathbb{C}^n)})=2$. But unfortunately I cannot see similarity to previous case, because I see no connection between $\bigwedge^k(V)$ and $\bigwedge^{k-1}(V)$, which I saw in previous case.

I know that $(\chi_{\bigwedge^k(\mathbb{C}^n)},\chi_{\bigwedge^k(\mathbb{C}^n)})=(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^k(V)})+2(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^{k-1}(V)})+(\chi_{\bigwedge^{k-1}(V)},\chi_{\bigwedge^{k-1}(V)})$.

The natural idea would be trying to use induction, then we would have $(\chi_{\bigwedge^{k-1}(V)},\chi_{\bigwedge^{k-1}(V)})=1$, and if we would know that $(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^{k-1}(V)})=0$, then that would be enough. But I'm not sure if this works.

Thank you in advance!

EDIT: Wouldn't it be enough if I show in inductive basis that $(\chi_{\bigwedge^1(V)},\chi_{\bigwedge^{2}(V)})=0$?

Answer 1

Ok, actually I got a solution, although it wasn't quite obvious without following considerations:

We can decompose arbitrary character $\chi_V$ into the sum of irreducible characters $\chi_{\lambda_1\oplus ... \oplus\lambda_n}$ If we look at the scalar product of an arbitrary character with itself, then

$$(\chi_V,\chi_V)=(\chi_{\lambda_1\oplus ... \oplus\lambda_n},\chi_{\lambda_1\oplus ... \oplus\lambda_n})=\sum_{i,j=1}^n(\chi_{\lambda_i},\chi_{\lambda_j})$$

and this is the sum of $1$'s and $0$'s, because character of irreducible representations is orthonormal. Assume we have $V\neq0$, so there exists $\lambda_k\neq 0$ and $\sum_{i,j=1}^n(\chi_{\lambda_i},\chi_{\lambda_j})\geq(\chi_{\lambda_k},\chi_{\lambda_k})=1$, so $(\chi_{V},\chi_{V})\in \mathbb{N}_+$

So if we are given $(\chi_{\bigwedge^k(\mathbb{C}^n)},\chi_{\bigwedge^k(\mathbb{C}^n)})=2$ and $(\chi_{\bigwedge^k(\mathbb{C}^n)},\chi_{\bigwedge^k(\mathbb{C}^n)})=(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^k(V)})+2(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^{k-1}(V)})+(\chi_{\bigwedge^{k-1}(V)},\chi_{\bigwedge^{k-1}(V)})$,

then, since we know that $(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^k(V)})\in\mathbb{N}_+$ and $ (\chi_{\bigwedge^{k-1}(V)},\chi_{\bigwedge^{k-1}(V)})\in\mathbb{N}_+$ and scalar product is positive definite, then the only way to get equality is to have $(\chi_{\bigwedge^k(V)},\chi_{\bigwedge^k(V)})=(\chi_{\bigwedge^{k-1}(V)},\chi_{\bigwedge^{k-1}(V)})=1$.