1
$\begingroup$

I need some help comparing two 'different' constructions of group homology.

Let $G$ be a finite group and let $A$ be a $G$-module.

I understand the definition of group homology as it is usually done: $H_{q}(G,A)=\operatorname{Tor}_{q}^{\mathbb{Z}[G]}(\mathbb{Z},A)$.

Namely, we take a projective resolution (in the category of $\mathbb{Z}[G]$-modules) of $\mathbb{Z}$ (trivial action):

\begin{equation} \dots\to P_{2}\to P_{1}\to P_{0}\to \mathbb{Z}\to 0, \end{equation} then we tensor this with $A$, obtaining an exact sequence. Then we chop off the term involving $\mathbb{Z}$:

\begin{equation} \dots\to A\:\otimes P_{2}\to A\:\otimes P_{1}\to A\:\otimes P_{0}\to 0, \end{equation} and the homology of this complex gives group homology. Okay. (This is correct, right?)

The other construction is found in the book "Cohomology of Number Fields", by J. Neukirch.

He does pretty much the same, but after tensoring with $A$, he takes "coinvariants" of the tensored exact sequence

(recall that the coinvariants of a $G$-module $M$ is defined as $M_{G}=M/DM$ where $DM$ is the submodule generated by elements of the form $gm-m$)

So, after taking these coinvariants, he obtains a complex and then takes homology of this complex:

enter image description here

The author does not make explicit whether the tensoring is happening in the category of abelian groups or $G$-modules and I think this may be the difference. Also, in the second version, there is no mention to "chopping off" the $\mathbb{Z}$ term.

How are they related?

Help is appreciated. Thanks

$\endgroup$
3
$\begingroup$

I think what is happening here is that one can compute the Tor-functors by resolving the first argument or the second argument. Here one is computing $\text{Tor}_*^{\Bbb ZG}(\Bbb Z,A)$. One can do this by taking a projective resolution $(P_*)$ of $\Bbb Z$ and then take the homology of $(P_*\otimes_{\Bbb ZG}A)$. Alternatively one could take a projective resolution $(Q_*)$ of $A$, and then take the homology of $(\Bbb Z\otimes_{\Bbb ZG}Q_*)$. This will be the complex of co-invariants of $(Q_*)$ as $\Bbb Z_{\otimes \Bbb ZG}M\cong M/I_GM$.

We get two different complexes, but their homology is the same.


[added in edit]

Now you have posted Neukirch's text I see what is happening. He has $(P_n\otimes A)_G$. Here $\otimes $ refers to the tensor product over $\Bbb Z$ and $M_G$ is the co-invariant group $M/I_GM$. In general for $\Bbb ZG$-modules $A$ and $B$, $A\otimes_{\Bbb Z G}B=(A\otimes_{\Bbb{Z}} B)_G$. In your original description, $A\otimes P_n$ is really $A\otimes_{\Bbb ZG}P_n$ so the complexes really are identical.

$\endgroup$
  • $\begingroup$ I think they both resolve $\mathbb{Z}$ :(. I'll edit the post with a picture of Neukirch's construction. And The pdf is this one: math.ucla.edu/~sharifi/groupcoh.pdf $\endgroup$ – Shoutre Jan 27 '18 at 13:55
  • $\begingroup$ Thank you so much! I was suspecting this, but he didn't make it explicit $\endgroup$ – Shoutre Jan 29 '18 at 21:05

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.