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Let $G=S_n$ and let $V$ be the permutation module of $G$ with basis $\{x_1,\cdots,x_n\}.$

Let $\lambda, \mu \in \mathbb{C}$ to allow one to define a $\mathbb{C}G$-homomorphism $\rho:V \to V$ by $$\rho(x_j):=\lambda x_j+\mu\sum_{i \neq j}x_i.$$

By using the above fact or otherwise, how can we prove that $V$ is the direct sum of two non-isomorphic irreducible $\mathbb{C}G$ -submodules?

I tried to prove this by construction. A familiar irreducible submodule in this case is the $1$-dimensional space $U:=span\{x_1+\cdots+x_n\}$. I intend to find another $(n-1)$-dimensional submodule $W$ which makes $V=U\oplus W$ hold, but it's hard to do so. Is there a way to use the fact instead of a random construction?

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Find an inner product in your space which is preserved by the action of the group. Then the orthogonal complement of every submodule is a submodule. (Of course, you need to prove this at some point!) –  Mariano Suárez-Alvarez Oct 13 '12 at 6:45
    
Thanks! But probably that is not the expected way as I have not been introduced to the inner product and orthogonal complement yet. –  user31899 Oct 13 '12 at 7:16
    
Now that is very, very weird! One more dent off my faith in the way we teach math nowadays :-/ –  Mariano Suárez-Alvarez Oct 13 '12 at 7:59
    
I am curious. How does one come across representation theory without a basic knowledge of linear algebra? I thought the later was virtually everywhere a prerequisite for the former. –  Sebastian Schoennenbeck Dec 6 '12 at 8:36
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3 Answers 3

I have self learnt some knowledge on the inner product of characters. First of all, by Maschke's theorem there exists an $(n-1)$-dimensional submodule $W$ of $V$ such that $V =U \oplus W$.

It's clear that $U$ is the trivial module, so it remains to show $W$ is irreducible:

Let $\nu$ be the character of $W$, denote the character of $V$ and $U$ be $\pi$ and $1_G$ respectively, then $\nu=\pi-1_G$. Now it's sufficient to prove that $\langle \nu ,\nu \rangle=1$. We need $$\langle \pi ,\pi \rangle-2\langle \pi ,1_G \rangle +\langle 1_G, 1_G \rangle =0$$ Since $1_G$ is irreducible, so $\langle 1_G,1_G \rangle =1$, by $\mathbb{Hom}_{\mathbb{C}G}(V,U)$ is one dimensional to get $\langle \pi ,1_G \rangle=1$. So it left to show $\langle \pi,\pi \rangle=2.$.

It's clear that $\pi (g)=\mathbb{number~ of~ points~ fixed by~ g}:=|\mathbb{fix}(g)|$, but how to deduce $\langle \pi,\pi \rangle=2$?

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Maybe like here. –  Martin Sleziak Feb 14 '13 at 13:20
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I'll try to give a simple solution (not using characters); but my solution is not using the homomorphism $\rho$, which was suggested in your post as a hint.

This solution is based on a hint given by Qiaochu Yuan in this comment.

We work with the permutation FG-module for $S_n$, i.e. we choose a basis $v_1,\dots,v_n$ for $U$ and the action of $S_n$ is given by $$\left(\sum x_iv_i\right)g = \sum x_iv_{ig}.$$ We denote this FG-module as $U$.

The vector $v=v_1+\dots+v_n$ generates a one-dimensional FG-submodule $U_1$.

It is relatively easy to find FG-submodule $U_2$ such that $U=U_1\oplus U_2$. (From Maschke's theorem we know that such a submodule exists.) This sumbodule is precisely $$U_2=\{\sum x_iv_i; \sum x_i=0\},$$ i.e. it contains precisely the vectors, for which the sum of coordinates is zero; $x_1+\dots+x_n=0$. (It is easy to see, that it is indeed an FG-submodule, its dimension is $n-1$ and $U_1\cap U_2=\{0\}$.)

As a basis for $U_2$ we can choose, for example, $v_1-v_2,v_2-v_3,\dots,v_{n-1}-v_n$.

$U_2$ is irreducible

If $v=x_1v_1+\dots+x_nv_n$ is a non-zero vector from $U_2$, then $x_i\ne x_j$ for some $i$, $j$. (Since $v\notin U_1$.)

We can choose a permutation $g$ in a such way, that for $w=vg=y_1v_1+\dots+y_nv_n$ we have $y_1\ne y_2$. Of course, $w\in U_2$.

The submodule $U_2$ contains also the vector $w(12)$, which is the same as $w$, only the first two coordinates are swapped.

Thus $$u=w_1-w_2=(y_1-y_2)(v_1-v_2),$$ and $y_1-y_2\ne 0$. We can multiply this vector and get $\underline{v_1-v_2\in U_2}$.

By applying the permutation $(12\dots n)$ to the vector $v_1-v_2$ we get all basic vectors $\underline{v_i-v_{i+1}\in U_2}$.

So we have in fact shown that if we have some non-zero submodule $V$ of $U_2$ (i.e., if $V$ contains at least one non-zero vector), then this submodule contains the whole basis of $U_2$, an thus $V=U_2$. This means that $U_2$ is irreducible.

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Cool, thank you! This is much easier to understand than those answers on MO thread. –  haemhweg Jun 23 '13 at 13:57
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Well if you don't need to find the two submodules there is an easy way:

If for all $\lambda, \mu \in \mathbb{C}$ $\rho$ is an endomorphism of $V$ the ring of endomorphisms has at least $\mathbb{C}$-Dimension $2$. Hence $V$ is not simple since $\mathbb{C}$ is obviously a splitting field for the group algebra. By Maschke's theorem $V$ is the direct sum of two non-trivial submodules.

Now all you have to show is that every endomorphism of $V$ is of the given form.

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