If $V$ is the standard representation of $S_n$, how to show that $\bigwedge^k V$ is irreducible $(0 \leq k \leq n-1)$?

Question

Edit: As Nate pointed out, the hint was wrong. I added the correct answer (from Fulton & Harris) below.

The hint I get: We have a decomposition of regular representation $R$=$U \oplus V$ where $U$ is the trivial representation. Since we have the expansion

$$ \bigwedge^k(U \oplus V) \cong \bigoplus_{a+b=k}\bigwedge^a U \otimes \bigwedge^b V $$

It suffices to compute $(\chi_{\wedge^k R},\chi_{\wedge^k R})$, and it should be $2$. Then we can deduce that $\bigwedge^k V$ is irreducible whenever $0 \leq k \leq n-1$.

I managed to compute $(\chi_{\wedge^2 R},\chi_{\wedge^2 R})$ when $n=3$. And it is indeed $2$. But I don't know how to step on. Is there any way to expand $(\chi_{\wedge^k R},\chi_{\wedge^k R})$ with respect to $U$ and $V$ (some massive computation involved I guess?). I know how to compute $\chi_{\wedge^2 R}(g)$ but how to compute $\chi_{\wedge^k R}(g)$? My guess is there is some much simpler approach for regular representation.

Any hint or solution will be sincerely appreciated!

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In case people are confused with terminologies:

1. Standard representation.
2. Regular representation.

Answer 1

The inner product $(χ_{∧^kR},χ_{∧^kR})$ can be interpreted as the expectation of $(\text{Tr} \phi(g))^2$, where $\phi$ is the representation of $S_n$ on the exterior product space ${∧^k\mathbb{C}^n}$, and $g$ is a random element of $S_n$ chosen uniformly.

If we define random variables $r_a$ ($a$ runs over the $k$-element subsets of $[1, 2, ... ,n]$) as

$$r_a=\left\{ \begin{array} & 1 & \text{if the action of $g$ maps the set $a$ to $a$ and preserves the orientation} \\-1 & \text{if the action of $g$ maps the set $a$ to $a$ and reverses the orientation} \\ 0 & \text{otherwise}\end{array} \right .$$

Then $\text{Tr} \phi (g)$ is just the sum of all the $r_a$s, by the definition of exterior product space.

To compute the expectation of $(\text{Tr} \phi(g))^2$, we expand it as $$\mathbb E(\text{Tr} \phi(g) )^2= \sum_{a\in {n\choose k}} \mathbb Er_a^2+\sum_{a, b\in {n\choose k}\\{|a\cap b|=k-1} } \mathbb Er_ar_b+\sum_{a, b\in {n\choose k}\\{|a\cap b| \leq k-2} } \mathbb Er_ar_b$$

where $n \choose k $ denotes all the $k$-element subsets of $[1,2,...,n]$, and the last two sums are over ordered pairs.

It is straightforward that $\mathbb Er_a^2=\frac{k!(n-k)!}{n!}$, so the first sum evaluates to $1$.

To study the second sum, we consider the action of $g$ on $a \cap b$. For the summand to be nonzero, $g$ must map this set to itself and map the elements $a \text{\\} (a \cap b) $ and $b \text{\\} (a \cap b) $ to themselves respectively.

This happens with probability $\frac{(n-k-1)!(k-1)!}{n!}$; and whenever this happens, the value of $r_ar_b$ is $1$, whether the orientation of $a \cap b$ is reversed by $g$ or not. By counting the number of such ordered pairs $(a,b)$, i.e. $\frac{n!}{(n-k-1)!(k-1)!}$, we find out that the second term also evaluates to $1$.

Every summand in the third term equals $0$. To see this, we may assume that $b_1$ and $b_2$ are two elements in $B\text{\\}A$, and let $\tau$ be the element in $S_n$ that exchanges $b_1$ and $b_2$ and fix everything else.

The expectation $\mathbb E r_ar_b$ can be written as $\frac{1}{n!}\sum _{g\in S_n} r_a(g)r_b(g)$, just to emphasize that $r_a$ and $r_b$ depends on $g$.

But $\sum _{g\in S_n} r_a(g)r_b(g)=\sum _{g\in S_n} r_a(\tau g)r_b(\tau g)=-\sum _{g\in S_n} r_a(g)r_b(g)$, as $\tau$ reverses the orientation on $b$ but has no effect on $a$ if either of the summands are nonzero.

So the third sum evaluates to $0$ and we are done.

Answer 2

Here I provide a (relatively) more expanded version of Fulton & Harris' proof.

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From the decomposition $\mathbb{C}^n = V \oplus U$, we see since

$$ (\chi_{\mathbb{C}^n},\chi_{\mathbb{C}^n}) = (\chi_U,\chi_U)+(\chi_V,\chi_V), $$

$(\chi_V,\chi_V)=1$ if and only if $(\chi_{\mathbb{C}^n},\chi_{\mathbb{C}^n})=2$. Similarly, since

$$ \wedge^k\mathbb{C}^n = (\wedge^k V \otimes \wedge^0 U) \oplus (\wedge^{k-1}V \otimes \wedge^1 U) = \wedge^k V \oplus \wedge^{k-1}V, $$

we have

$$ (\chi_{\wedge^k \mathbb{C}^n},\chi_{\wedge^k\mathbb{C}^n})=(\chi_{\wedge^k V},\chi_{\wedge^kV})+(\chi_{\wedge^{k-1} V},\chi_{\wedge^{k-1}V}). $$

It suffices to show that $(\chi_{\wedge^k \mathbb{C}^n},\chi_{\wedge^k\mathbb{C}^n})=2$, which will force the two on the right hand side to be $1$. Let $A= \{1,2,\cdots,n\}$. For a subset $B$ of $A$ with $k$ elements, and $g \in S_n$, let

$$ \{g\}_B = \begin{cases} 0 \quad &\text{if $g(B) \neq B$} \\ 1 \quad &\text{if $g(B) = B$ and $g\vert_B$ is an even permutation} \\ -1 \quad & \text{if $g(B)=B$ and $g\vert_B$ is odd} \end{cases} $$ Here, if $g(B)=B$, $g\vert_B$ denotes the permutation of the set $B$ determined by $g$. Then $\chi_{\wedge^k\mathbb{C}^n}(g)=\sum\{g\}_B$, and $$ \begin{aligned} (\chi_{\wedge^k \mathbb{C}^n},\chi_{\wedge^k\mathbb{C}^n}) &= \frac{1}{n!}\sum_{g \in G}\left(\sum_B \{g\}_B\right)^2 \\ &= \frac{1}{n!}\sum_{g \in G}\sum_B \sum_C \{g\}_B\{g\}_C \\ &= \frac{1}{n!}\sum_B\sum_C\sum_g(\operatorname{sgn} g \vert_B) \cdot (\operatorname{sgn} g\vert_C) \end{aligned} $$

where the sum are over subsets $B$ and $C$ of $A$ with $k$ elements, and in the last equation, the sum is over those $g$ with $g(B)=B$ and $g(C)=C$ (those $\{g\}_B=\{g\}_C=0$ are eliminated). Such $g$ is given by four permutations:

1. of $B \cap C$;
2. of $B \setminus B \cap C$;
3. of $C \setminus B \cap C$;
4. of $A \setminus B \cup C$.

Put $l=|B \cap C|$, the last sum can be rewritten as $$ \frac{1}{n!} \sum_B \sum_C \sum_{a \in S_l} \sum_{b \in S_{k-l}} \sum_{c \in S_{k-l}} \sum_{h \in S_{n-2k+l}}(\operatorname{sgn}a)^2(\operatorname{sgn}b)(\operatorname{sgn}c) \\ = \frac{1}{n!} \sum_B \sum_C l!(d-2k+l)! \left(\sum_{b \in S_{k-l}} \operatorname{sgn}b\right)\left(\sum_{c \in S_{k-l}} \operatorname{sgn}c\right) $$

These last sums are $0$ unless $k=l$ or $k=l+1$. The summand when $k=l$ is

$$ \frac{1}{n!}\sum_B k!(n-k)! = \frac{1}{n!} {n \choose k}k!(n-k)!=1. $$

The summand when $k=l+1$ is $1$ as well, hence $(\chi_{\wedge^k \mathbb{C}^n},\chi_{\wedge^k\mathbb{C}^n})=1+1=2$ as we wanted.