Irreducibility of the standard representation of $S_n$.

The permutation representation of $S_n$ is $\mathbb C^n$ with elements of $S_n$ permuting the basis vectors $\{e_1, e_2, \ldots, e_n\}$. It has a trivial subrepresentation spanned by the vector $v = \sum_i e_i$. By Maschke's theorem there is a complement subrepresentation (given by the condition $\sum_i x_i = 0$, where $x_i$ is the i'th coordinate). This last representation (the standard representation of $S_n$) is irreducible. Why is this?

If we try characters, we can get the following: denote the standard representation by $V$ (dimension n-1), and the full permutation representation by $Perm$. Then $$Perm = V\oplus Triv,$$ where the latter is the trivial representation. This means that $\chi_V(g) = |X^g| - 1$, and $$\langle\chi_V, \chi_V\rangle = \frac{1}{|G|}\sum(|X^g|-1)^2|C(g)| = 1+\frac{1}{|G|}\sum (|X^g|^2 - 2|X^g|)|C(g)|,$$ where $X = \{1,2,\ldots, n\}$ and $X^g$ is the set of fixed points of permutation $g\in S_n$ and $C(g)$ is the conjugacy class of $g$. The sum is taken over all different conjugacy classes of $S_n$.

Now, since I know that $V$ is irreducible, the last sum must be zero. I don't see, however, why that should be.

Perhaps there is a direct proof or irreducibility?

Thank you.

-
There is a direct proof (it is worth trying to figure out yourself; take a vector and explicitly show the subrepresentation it generates is the whole thing). The character-theoretic proof is explained here: mathoverflow.net/questions/17230/… –  Qiaochu Yuan Apr 26 '12 at 16:44
@Qiaochu: What do you mean by the subrepresentation a vector generates? –  joriki Apr 26 '12 at 17:27
@joriki: I mean the intersection of all subrepresentations containing a vector. Concretely, I mean $\text{span}\{ gv : g \in G \}$. –  Qiaochu Yuan Apr 26 '12 at 18:14
@Qiaochu: At first I thought that's what you meant, but if so, I don't understand your argument that "the subrepresentation it generates is the whole thing". In the above sense, each $e_i$ generates the entire permutation representation, but that doesn't show that the permutation representation is irreducible. I don't see what's different in the case of the complement of the trivial subrepresentation. –  joriki Apr 26 '12 at 18:35
possible duplicate of Permutation module of $S\_n$ (and I needed to upvote some of the answers there to be able to mark this as duplicate! please consider doing some more upvoting) –  Marc van Leeuwen Apr 17 '13 at 7:47

Coming late to this, but it's worth pointing out that there is an interesting proof by induction that avoids characters and whatnot, "from scratch", in Hernández-Lamoneda, L.; Juárez, R.; Sánchez-Sánchez, F. Dissection of solutions in cooperative game theory using representation techniques. Internat. J. Game Theory 35 (2007), no. 3, 395–426.

Let $\mathbb{R}^n$ be generated by $\{e_i\}_{1\leq i\leq n}$. Basically, when $n=2$ it's clear that $V$ is irreducible, so let $n>2$ and use induction. Note that the subgroup that fixes $e_n$ is (isomorphic to) $S_{n-1}$, so the permutation representation of of that subgroup (by induction) decomposes as desired. Then you look back at the standard representation under $S_n$ and note if it were reducible, its invariant subspaces would also be invariant under $S_{n-1}$ and hence decompose into the specific subspaces given by induction. But those subspaces aren't $S_n$-modules, so it's not reducible.

I have no idea whether this is original to this paper, nor whether there is something a little incomplete about that last step, which somehow seems fishy (but I prefer using characters so I don't often make such arguments). I do like that it doesn't use a lot of heavy tools, and that it uses induction.

-

Another nice proof comes the following fact:

if a finite group $G$ acts on a set $X$ two-transitively, then the obvious representation of $G$ on the vector space with $X$ has a basis is irreducible,

whose proof, in turn, I will leave as an exercise.

-