1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What exactly do you mean by "permuting each generator"? $\endgroup$ – tomasz Sep 5 '12 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 '12 at 17:12
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.