4
$\begingroup$

Let $A$ be a finite $\mathbb{Z}$-module (i.e., a finite abelian group). My question is: for what $n\in \mathbb{Z}^{n\geq 2}$ the map \begin{align} \alpha_{n}:\bigwedge^nA&\to A^{\otimes n}\\ a_1\wedge \cdots \wedge a_n&\mapsto \sum_{\pi\in \mathbb{S}_n}(sig(\pi))a_{\pi(1)}\otimes \cdots\otimes a_{\pi(n)} \end{align} is injective? For $n=2$, $\alpha_2$ is injective. It follows, for example, just by induction on $r$, if $A=\mathbb{Z}/\mathbb{Z}m_1\oplus \cdots \oplus\mathbb{Z}/\mathbb{Z}_{m_r}$.

Thanks!

$\endgroup$

1 Answer 1

3
$\begingroup$

I think that this is always true. We can write $A$ as $A_1\oplus\dotsb\oplus A_r$, where each $A_i$ is cyclic. Let $I(n)$ denote the set of sequences $(i_1,\dotsc,i_n)$ with $1\leq i_1<i_2<\dotsb <i_n\leq r$. For $i\in I^n$ put $A(i)=A_{i_1}\otimes\dotsb\otimes A_{i_n}$, so $A^{\otimes n}=\bigoplus_{i\in I^n}A(i)$. Put $A[n]=\bigoplus_{i\in I(n)}A(i)$, so there is an evident inclusion $\iota\colon A[n]\to A^{\otimes n}$ and projection $\pi\colon A^{\otimes n}\to A[n]$. There is also a map $\mu\colon A^{\otimes n}\to\Lambda^nA$ sending $a_1\otimes\dotsb\otimes a_n$ to $a_1\wedge\dotsb\wedge a_n$. Because $A_i$ is cyclic we have $\Lambda^2A_i=0$, so $\Lambda^*A_i=\mathbb{Z}\oplus A_i$. Moreover, for any $B$ and $C$, the multiplication map $\Lambda^*(B)\otimes\Lambda^*(C)\to\Lambda^*(B\oplus C)$ is an isomorphism. Using this, we see that the map $\mu\iota\colon A[n]\to\bigwedge^nA$ is an isomorphism. On the other hand, it is also easy to see that the composite $\pi\alpha\mu\iota$ is the identity. It follows that $\alpha$ is injective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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