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Let $S_n$ be the permutation group and $V = \operatorname{Fun}(X,\mathbb{k})$ functions from $X=\{1,\dotsc,n\}$ to some field $\mathbb{k}$. How can I prove that every irreducible representation of $S_n$ occurs in $V^{\otimes m}$ for integer $m$ large enough?

This is a standard fact when $n!\neq 0$ in $\mathbb{k}$.

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    $\begingroup$ This is a standard fact that holds for any faithful representation of any finite group. $\endgroup$
    – lambda
    Jul 31, 2020 at 13:27
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    $\begingroup$ @lambda, is it true for any faithful representation in any characteristic? I don't know that it isn't, but I thought that there might be some problems in case the characteristic of $\mathbb k$ divides $n!$. $\endgroup$
    – LSpice
    Jul 31, 2020 at 13:33
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    $\begingroup$ Not to keep beating the same (possibly wrong, to mix metaphors) horse, but, @DianbinBao, isn't that a characteristic-$0$ argument? $\endgroup$
    – LSpice
    Jul 31, 2020 at 13:40
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    $\begingroup$ See ams.org/journals/proc/1962-013-05/S0002-9939-1962-0141710-X/… $\endgroup$ Jul 31, 2020 at 14:01
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    $\begingroup$ Does "occurs" mean as a subfactor, as a subrepresentation, as a summand, ... ? $\endgroup$ Jul 31, 2020 at 14:08

2 Answers 2

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Turning my comment into an answer, Robert Steinberg proved in Complete sets of representations of algebras that if you have a faithful representation of a finite semigroup $S$, then every irreducible representation of $S$ appears as a composition factor of a tensor power of that representation. In fact, he shows that the semigroup algebra $KS$ acts faithfully on the direct sum of the tensor powers, which is a stronger statement since finite semigroups are not usually completely reducible. My favorite proof of this is Passman’s variant of a proof of Rieffel in Elementary bialgebra properties of group rings and enveloping rings: An introduction to Hopf algebras.

If $G$ is a finite group, then $KG$ is a Frobenius algebra and so every irreducible representation is a subrepresentation of the regular representation. Thus in the case of a finite group, every irreducible representation will be a subobject of a tensor power.

For semigroups, this is not true. There are many examples. The easiest is the monoid $T_n$ of all maps on $n$ letters. For $n\geq 2$, the natural representation on $\mathbb C^n$ is faithful and has the trivial representation as a quotient but not as a subrepresentation. The same remains true after taking tensor powers.

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    $\begingroup$ It seems to me that the regular representation of $S_n$ occurs, specifically, in $V^{\otimes n}$ where $V$ is functions on an $n$-element set, as the subspace of functions on the $n$th power of an $n$ element set which vanish on all $n$-tuples that are not permutations. So we can take $m=n$ here. $\endgroup$
    – Will Sawin
    Jul 31, 2020 at 18:10
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    $\begingroup$ That is true in this case since that module is just the action of the group on functions n to n. The general case can require n the size of S. $\endgroup$ Jul 31, 2020 at 18:46
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Benjamin Steinberg answered the question, but I wanted to unwind his idea into an explicit formula.

Let $V$ be the representation with basis $e_1,\dotsc, e_n$, where a permutation $\sigma$ acts by sending $e_i$ to $e_{\sigma(i)}$.

Let $W$ be an irreducible representation of $S_n$.

Fix a linear form $l$ on $W$. We can map $W$ to $V^{\otimes n}$ by sending $w \in W$ to $$f(w) = \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) }.$$

Then we have \begin{align*} \sigma' (f(w)) & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot \sigma'( e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm e_{ \sigma(n) } ) \\ & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (w))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\ & {}= \sum_{ \sigma \in S_n} l ( (\sigma' \circ \sigma)^{-1}( \sigma'(w)))\cdot e_{ \sigma'(\sigma(1))} \otimes e_{ \sigma'(\sigma(2))} \otimes \dotsm \otimes e_{ \sigma'(\sigma(n))} \\ & {}= \sum_{ \sigma \in S_n} l ( \sigma^{-1} (\sigma' (w) ))\cdot e_{\sigma (1) }\otimes e_{\sigma(2)} \otimes \dotsm\otimes e_{ \sigma(n) } = f(\sigma' (w)) \end{align*} using the change of variables $\sigma \mapsto \sigma' \circ \sigma$ in the last line. So $f$ is a homomorphism, and because $W$ is irreducible, and $f$ is nontrivial (as long as $l$ is a nontrivial linear form, since all the terms in the sum give different basis vectors in the tensor product), this map is the inclusion of a subrepresentation, as desired.

One can get embeddings into $V^{\otimes m}$ for higher values of $m$ by just putting repetitions in the sequence of basis vectors being tensored, or for $n-1$ by removing the last term, but one can't go lower than $n-1$, because of the sign representation (except maybe in characteristic $2$, I guess).

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  • $\begingroup$ Sorry for the silly question—why does the sign representation prevent going lower than $n - 1$? $\endgroup$
    – LSpice
    Jul 31, 2020 at 19:25
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    $\begingroup$ @LSpice Suppose there is a copy of the sign representation in the tensor product. Take a generator (or the generator, since it's unique up to scaling). Write it in a basis of tensors of basis elements. Find one tensor of basis elements which has a nonzero coefficient. Since only $n-2$ basis elements can appear, two don't. Swap those and derive a contradiction. $\endgroup$
    – Will Sawin
    Jul 31, 2020 at 19:34

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