Given a tensor product $A^{\otimes n}$ over a field $k$ (characteristic $\neq 2$) of $n$ copies of the $k$-algebra $A$, a premutation $\sigma \in S_n$ of order $2$ acts on the elements of $A^{\otimes n}$ by permuting each generator and then extend by linearity. It will then split $A^{\otimes n}$ into a positive and a negative eigenspace, given by the projectors $\frac{1}{2}(id-\sigma)$ and $\frac{1}{2}(id+\sigma)$, where $id$ is the identity map.

What I need to know to go on is, given an $(a_1, \ldots, a_n)\in A^{\otimes n}$ in one of the eigenspaces, will $(a_1, \ldots, a_n, a_{n+1})$ be in one of the corresponding eigenspaces of $A^{\otimes n+1}$ given $\sigma$ is extended to act on the first $n$ elements the natural way?

  • $\begingroup$ What exactly do you mean by "permuting each generator"? $\endgroup$
    – tomasz
    Sep 5, 2012 at 16:02
  • $\begingroup$ @tomasz I mean given a generator $a_1 \otimes \cdots\otimes a_n$, it permutes the indices, and since any element of the tensor product is a sum of such generators, you let the permutation act on each of the generators of a given element. $\endgroup$
    – Arthur
    Sep 5, 2012 at 17:12

1 Answer 1


The function from $A^{\otimes n}$ to $A^{\otimes n+1}$ adds the same element at the end on each of the generators, and by then it is pretty obvious by bilinearity of the tensor product $A^{\otimes n}\otimes A$ that it actually has the property I was looking for.


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