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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.

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    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: http://mathoverflow.net/questions/17230/permutation-representation-inner-product2012-04-26
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    @Qiaochu: What do you mean by the subrepresentation a vector generates?2012-04-26
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    @joriki: I mean the intersection of all subrepresentations containing a vector. Concretely, I mean $\text{span}\{ gv : g \in G \}$.2012-04-26
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    @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.2012-04-26
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    @joriki: "a vector" should be interpreted as meaning "any nonzero vector," not "a particular vector."2012-04-26
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    I think that submodule is a more standard name than subrepresentation. But Qiaocho did not give an argument that the subrepresentation it generates is the whole thing. He said it was worth trying to prove it yourself, and I agree with that! It's not too hard.2012-04-26
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    possible duplicate of [Permutation module of $S\_n$](http://math.stackexchange.com/questions/212025/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)2013-04-17

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