Let $A$ be an associative algebra with a unit over a field $k$. fix $n > 1$. Define a $k$-algebra structure on the vector space $A^{\otimes n} = A \otimes_k \dots \otimes_k A$ (where there are $n$ copies of $A$) by$$(a_1 \otimes \dots \otimes a_n) \cdot (a_1' \otimes \dots \otimes a_n') := (a_1a_1') \otimes \dots \otimes (a_na_n').$$I know that provided $A$ is commutative, that $S^nA$ and $\bigwedge^n A$ are closed under multiplication in $A^{\otimes n}$. My question is, can we drop the commutativity assumption?
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9$\begingroup$ How are you regarding $S^nA$ and $\bigwedge^nA$ as subspaces of $A^{\otimes n}$ (they're more naturally quotients)? And however you do it, is $\bigwedge^nA$ really closed under multiplication, even in the commutative case? $\endgroup$– Jeremy RickardAug 30, 2015 at 14:06
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4$\begingroup$ By construction, the multiplication map $A^{\otimes n}\otimes A^{\otimes n}\to A^{\otimes n}$ is an intertwiner for the permutation representation of the symmetric group $S_n$. Since $S^nA$ is the invariant subspace of this representation, this intertwiner restricts to $S^n A\otimes S^n A\to S^n A$, independently of whether the original multiplication is commutative or not. On the other hand, the resulting map $\bigwedge^n A\otimes\bigwedge^n A\to A^{\otimes n}$ also lands in $S^n A$, so that $\bigwedge^n A$ is closed under multiplication only in the degenerate cases $char(k)=2$ or $dim(A)<n$. $\endgroup$– Tobias FritzAug 30, 2015 at 14:24
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