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Let $V$ be a vector space and $\dim V=n$. Under a basis, the vector $\mathbf v$ is represented by the coordinate $(a_1,a_2,\ldots, a_n)$. Let $S_n$ be the group of all permutations on the set $\{1,2\ldots, n\}$ represented by matrices. $S_n\subseteq M_n(\mathbb R)$.

Let's form the set $$ P_\mathbf v=S_n\mathbf v=\{M\mathbf v:M\in S_n\}. $$

I try to investigate the dimension of $\text{span }P_\mathbf v $. It appears to me that a maximal subset of independent vectors in $P_\mathbf v$ can either be very large ($n$ vectors) or very small (one vector), but not something in between.

What are the possible dimensions of $\text{span }P_\mathbf v $?

Ma Joad
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1 Answers1

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The symmetric group of order $n!$ acting as permutation matrices on an n-dimensional vector space (non-modular characteristic) is well-known to have irreducible decomposition given by the trivial representation $1 = \mathrm{span}\{(1, \ldots, 1)\}$ and the $(n-1)$-dimensional "standard representation" $S$. In any case, your $P_v$ is a subrepresentation, so the only possibilities are $P_v \in \{0, 1, S, 1 \oplus S\}$. Correspondingly, the possible dimensions are $0, 1, n-1, n$.

The first two possibilities are realized by $v=0, v=(1, \ldots, 1)$. You can project onto $1$ by applying the averaging ("Reynolds") operator $\frac{1}{n!} \sum_{\sigma \in S_n} \sigma$, which says you get $1$ as a component of $P_v$ if and only if the average of your entries is non-zero. Hence $v=(1, 2, \ldots, n)$ gives you $1 \oplus S$, while $v=(1, -1, 0, \ldots, 0)$ gives you $S$.

Joshua P. Swanson
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  • Very interesting. Can I ask you an embarrassingly naive question? I know that a "representation" is a homomorphism of a group into $GL(V)$. When you say, for example, "the trivial representation $1=\text{span} (1, \ldots, 1)$", where is the homomorphism? – Giuseppe Negro Oct 15 '19 at 08:35
  • What is "Standard representation" $S$? – Ma Joad Oct 15 '19 at 11:47
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    @GiuseppeNegro: There are two perspectives on representations: (1) a homomorphism $G \to \mathrm{GL}(V)$, and (2) $V$ can be thought of as a $kG$-module, where $kG$ is the group algebra of $G$ and $k$ is $V$'s field of scalars. Here I was using $V = \mathrm{span}{(1, \ldots, 1)}$ where $\sigma \in S_n$ acts trivially, $\sigma \cdot v = v$ for all $v \in V$. Thus, each $\sigma$ acts on $V$ as the identity, so the homomorphism is literally trivial. – Joshua P. Swanson Oct 15 '19 at 22:25
  • @Jethro: the standard representation is by definition this one, i.e. permutation matrices acting on the quotient $k^n/\mathrm{span}{(1, \ldots, 1)}$. The "hard part" of my answer would be verifying the standard representation is irreducible. Reversing this answer is one elementary approach. – Joshua P. Swanson Oct 15 '19 at 22:33
  • @JSwanson: Oh, this is very interesting for me, I had never grasped this alternative point of view. I now see that "an irreducible representation" is exactly an irreducible submodule of $V$. Thank you very much. – Giuseppe Negro Oct 16 '19 at 11:23
  • @JSwanson Thanks a lot. By the time I see this I have already proven it myself :) – Ma Joad Oct 17 '19 at 08:04