Let $V$ be a $k$-vector space and $V^{\otimes n}$ the $n$-fold tensor power of $V$ and let $\mathbb{S}_n$ be the symmetric group of an n-element set, with its signum representation denoted by $(-1)^\sigma$ for $\sigma\in \mathbb{S}_n$.

Now on one side we have the "usual" symmetric tensor power $V^{\odot n}$ of $V$ defined by quotienting the ordinary tensor power by the maximal ideal generated by "sums over antisymmetriezed permutations", i.e.:

$V^{\odot n}:= V^{\otimes n} / \langle\{\sum_{\sigma\in \mathbb{S}_n} (-1)^\sigma v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}\;|\; v_1\otimes...\otimes v_n \in V^{\otimes n}\}\rangle$

On the other side it is said, that the same symmetric tensor power can be obtained as the following tensor product:

$V^{\otimes n} \otimes_{k[\mathbb{S}_n]} k$

where $k[\mathbb{S}_n]$ is the group algebra over $\mathbb{S}_n$, with $k$ a left $\mathbb{S}_n$-module, induced from the trivial representation and with $V^{\otimes n}$ a right $\mathbb{S}_n$ module induced from the right representation of $\mathbb{S}_n$ which permutes the indices.

Can someone explain, how $V^{\otimes n} \otimes_{k[\mathbb{S}_n]} k$ is equivalent to the symmetric tensor product $V^{\odot n}$? I'm not sure, how we should think about $V^{\otimes n} \otimes_{k[\mathbb{S}_n]} k$ or tensor products $\bullet \otimes_{k[\mathbb{S}_n]} \bullet$ in general.

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    $\begingroup$ Might belong on stackexchange, but alas. Let us note that we should assume $n\geq 2$. You can think of the right factor of $V^{\otimes n} \otimes_{k[S_n]} k$ as swallowing all of the $S_n$-action on the left factor. In particular, for an element of the subspace (not ideal) of antisymmetrized tensors, we get $\sum_{\sigma\in S_n} (-1)^{\sigma}v_{\sigma(1)}\otimes\ldots\otimes v_{\sigma(n)}\otimes_{k[S_n]} 1 = \sum_{\sigma\in S_n} (-1)^{\sigma}v_1\otimes\ldots\otimes v_n\otimes_{k[S_n]} 1 = 0$. Hence the obvious map is well-defined. For the other direction, similarly the obvious map is bilinear. $\endgroup$ Aug 10, 2015 at 7:58
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    $\begingroup$ I believe your presentation of the symmetric tensor power is not correct when $n\geq 3$. When $n=3$ and $\dim V = 2$ with basis $x,y$, $V^{\odot 3}$ has basis $x^3, x^2y,xy^2,y^3$ and so is 4-dimensional. The RHS is an 8-dimensional vector space quotiented by the 0-dimensional vector space, as there are no totally-antisymmetric tensors in two variables. $\endgroup$ Aug 10, 2015 at 19:53
  • $\begingroup$ I see. Yes the common def of $V^{\odot n}$ is quotienting by the ideal generated just by $x\otimes y - y\otimes x$. But this makes the equivalence even more mysterious and then I don't understand Englouties comment. $\endgroup$ Aug 10, 2015 at 23:21
  • $\begingroup$ Well, I must apologize. I just assumed that the presentation was correct, in which case my comment hopefully makes sense. Basically, there is indeed a well-defined map, but its "inverse" is only well-defined if we map to the actual symmetric power. $\endgroup$ Aug 11, 2015 at 7:12
  • $\begingroup$ Why is everyone voting for close, if on the other side no one can really explain it in detail? I already tried at math.stackexchange but got nothing in a week. So I deleted and tried here. Does't seem to be THAT basic after all. $\endgroup$ Aug 11, 2015 at 9:09

1 Answer 1


For any $k[\mathbb{S}_n]$-module $M$, $M\otimes_{k[\mathbb{S}_n]}k$ is the largest quotient of $M$ on which $\mathbb{S_n}$ acts trivially.

Factoring $V^{\otimes n}$ by the ideal generated by differences of elements that differ just by swapping two tensor factors gives the largest quotient on which transpositions act trivially.

So it comes down to the fact that $\mathbb{S}_n$ is generated by transpositions.


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