MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Fix some $1 \leq k \leq n$. I'm looking for finite-dimensional vector spaces $M_{n,k}$ over $\mathbb{Q}$ on which $\mathbb{Z}^n$ acts such that the natural map $H_k(\ell \mathbb{Z}^n,M_{n,k}) \rightarrow H_k(\mathbb{Z}^k,M_{n,k})$ is not an isomorphism for some $\ell \geq 2$. Here $\ell \mathbb{Z}^n$ is the subgroup of $\mathbb{Z}^n$ consisting of vectors each of whose entries is divisible by $\ell$ and the map on group homology is induced by the inclusion $\ell \mathbb{Z}^n \hookrightarrow \mathbb{Z}^n$.

share|cite|improve this question
In your target homology group, I think $\mathbb{Z}^k$ should be $\mathbb{Z}^n$. – Mark Grant Feb 29 '12 at 13:50
Thanks Mark!!!! – Ron Mar 2 '12 at 6:21
up vote 1 down vote accepted

$k=n=1$, $M_{n,k}=M=\mathbb{Q}$. Let $\mathbb{Z}$ act on $\mathbb{Q}$ by $n \cdot q = (-1)^n q$.

The homology is $H_i(\mathbb{Z};M)=\mathbb{Q}$ for $i=1$ and $0$ otherwise. The subgroup $2 \mathbb{Z}$ acts trivially on $M$ and so $H_i (2 \mathbb{Z};M)=\mathbb{Q}$ for $i=0$ and $0$ otherwise.

share|cite|improve this answer
I think you can generalize your example to all $n,l$ by letting $\mathbb{Z}$ act on $\mathbb{C}$ by $n \cdot z = \zeta_l^nz$. Then extend the action to an action of $\mathbb{Z}^n$ on $\mathbb{C} \otimes \cdots \otimes \mathbb{C}$. The case for a fixed $k$ should then be manageable by dimension-shifting. – Ralph Feb 29 '12 at 10:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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