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EDIT: I've decided to rephrase my question in order for it to be more concise and to the point.

Let $G$ be a group, and let $F_\bullet\rightarrow\mathbb{Z}$ be a free $\mathbb{Z}[G]$-resolution of the trivial $G$-module. In Ken Browns book "Cohomology of groups", given a $G$-module $M$, he defines the homology of $M$ to be the homology of the complex $F_\bullet\otimes_G M$, and defines functoriality as follows:

If $\alpha : G\rightarrow G'$ is a homomorphism of groups, let $F_\bullet'$ be a free $\mathbb{Z}G'$-resolution of $\mathbb{Z}$, and let $\tau : F_\bullet\rightarrow \alpha^\# F_\bullet'$ be a morphism of complexes of $G$-modules, extending the identity on $\mathbb{Z}$.

For $M\in Mod_G, M'\in Mod_{G'}$ and $f : M\rightarrow \alpha^\#M'$ a homomorphism of $G$-modules, then the the map $H_*(G,M)\rightarrow H_*(G',M')$ induced by $(\alpha,f)$ is itself by the map $\tau\otimes f : F_\bullet\otimes_G M\rightarrow \alpha^\#F'_\bullet\otimes_G \alpha^\#M' = F'_\bullet\otimes_{G'}M'$ on chains.

My confusion stems from the following. I am used to the definition of group homology as the derived functors of the $G$-coinvariants functor : $Mod_G\rightarrow Ab$ sending $M\mapsto M_G = M\otimes_{\mathbb{Z}G}\mathbb{Z}$, which identifies $H_*(G,M) = Tor_*^{\mathbb{Z}G}(\mathbb{Z},M)$, which is why the complex $F_\bullet\otimes_G M$ computes the homology of $M$. What's unclear is the functorial dependence of the homology on the complex $F_\bullet\otimes_G M$.

From the derived functor definition of group homology, one has the following notion of functoriality in $G$ and $M$:

Given $\alpha : G\rightarrow G'$ as above, $M\in Mod_G, M'\in Mod_{G'}$, and $f : M\rightarrow\alpha^\# M'$, the usual derived functor functoriality gives us a map $$f_* : H_*(G,M)\rightarrow H_*(G,\alpha^\#M')$$ Next, the fact that $H_*(G',-)$ is a universal $\delta$-functor implies that the natural map $(\alpha^\#(-))_G\rightarrow (-)_G$ extends functorially to a morphism of $\delta$-functors: $H_*(G,\alpha^\#(-))\rightarrow H_*(G',-)$, whence a map: $$cor : H_*(G,\alpha^\#M')\rightarrow H_*(G',M')$$

My question can be phrased as follows:

  1. Why is $f_*$ computable using $id\otimes f : F_\bullet\otimes_G M\rightarrow F_\bullet\otimes_G \alpha^\#M'$? (Normally one would take a projective resolution of $M$, then tensor by $\mathbb{Z}$)

  2. Why is the corestriction map $cor$ computable using $\tau\otimes id : F_\bullet\otimes_G \alpha^\# M'\rightarrow \alpha^\#F_\bullet'\otimes_G \alpha^\#M'$?

I've tried to convince myself of the equivalence, to which I've had a little success, though I'd appreciate it if experts could sketch how they would see that these two definitions are equivalent (different perspectives are always valuable), or perhaps explain why it's obvious from some standard facts which I'm missing. References which explain this would also be appreciated.

BEGIN OLD QUESTION

I'm trying to understand Weibel's description of the action of conjugation on group homology (this is p190 in his "Introduction to Homological Algebra"). I'm having trouble understanding exactly what makes his description valid.

In general, for a morphism of groups $\rho : H\rightarrow G$ and a $G$-module $A$, we may form the $H$-module $\rho^\#A$ with $H$-module structure given via $\rho$, and this functor $\rho^\#$ is exact.

In our case, let $H\le G$ be a subgroup, and let $c_g : H\rightarrow gHg^{-1}$ be the conjugation isomorphism $h\mapsto ghg^{-1}$.

If $A$ is a $G$-module, the abelian group map $\mu_g : A\rightarrow A$ given by $a\mapsto ga$ is actually an $H$-module map from $A$ to $c_g^\#A$.

Thus, $\mu_g$ determines a natural map $(\mu_g)_* : H_*(H,A)\rightarrow H_*(H,c_g^\#A)$.

Moreover, the functors $Mod_H\rightarrow Ab$ given by $M\mapsto H_*(H,c_g^\#A)$ is a homological $\delta$-functor (since $c_g^\#$ is exact), and hence the natural abelian group homomorphism $$H_0(H,c_g^\#A) = (c_g^\#A)_H\rightarrow A_{gHg^{-1}} = H_0(gHg^{-1},A)$$ extends to a morphism of $\delta$-functors $$\zeta_* : H_*(H,c_g^\#A)\rightarrow H_*(gHg^{-1},A)$$ (This $\zeta_*$ is normally called the corestriction map)

I'd like to understand the composition: $$\zeta_*\circ(\mu_g)_* : H_*(H,A)\rightarrow H_*(gHg^{-1},A)$$

Weibel gives the following way of understanding this composition. He picks a projective $\mathbb{Z}G$-resolution $P_\bullet\rightarrow\mathbb{Z}$ (with $\mathbb{Z}$-the trivial $G$-module), and notes that it is simultaneously projective for $\mathbb{Z}G,\mathbb{Z}H$, and $\mathbb{Z}[gHg^{-1}]$. Then he notes that $\mu_g : P_i\rightarrow P_i$ sending $p\mapsto gp$ is an $H$-module chain map $P_i\rightarrow c_g^\# P_i$ lying over the identity map on $\mathbb{Z}$. He deduces that $\zeta_*\circ(\mu_g)_*$ is induced from $$P\otimes_{\mathbb{Z}H}A\rightarrow P\otimes_{\mathbb{Z}[gHg^{-1}]}A\qquad x\otimes a\mapsto gx\otimes ga$$

Why does this calculate the morphism $\zeta_*\circ(\mu_g)_*$?

My model of what's going on is the following. Suppose you have a right-exact functor $F : \mathcal{A}\rightarrow\mathcal{B}$ where $\mathcal{A}$ has enough projectives. Then for a morphism $f : X\rightarrow Y$ in $\mathcal{A}$, one can calculate $L_iF(f)$ by picking projective resolutions $P_\bullet\rightarrow X$ and $Q_\bullet\rightarrow Y$, extending $f$ to a morphism of complexes $f_\bullet : P_\bullet\rightarrow Q_\bullet$ (unique up to homotopy), applying $F$ to $f_\bullet$, and taking the induced maps on homologies. (Roughly speaking this works because in the derived category one computes everything by replacing what we care about by projective resolutions.)

However, this doesn't seem to directly apply to Weibel's situation, since we have multiple $\delta$-functors at work. I would very much appreciate it if an expert could describe the situation more clearly for me.

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    $\begingroup$ Tor is balanced. $\endgroup$ – Wilberd van der Kallen Nov 2 '18 at 8:16
  • $\begingroup$ Tor being balanced is precisely correct. The new question 1 precisely asks why we can compute $Tor^{\Bbb Z[G]}(\Bbb Z, M)$ by resolving either $\Bbb Z$ or $M$ $\endgroup$ – Tyler Lawson Nov 2 '18 at 9:30
  • $\begingroup$ The new question 2 is about being balanced and functorial. It asks about $Tor^{\Bbb Z[G]}(\Bbb Z, \alpha^\# M) = Tor^{\Bbb Z[G']} (\Bbb Z[G \backslash G'], M)$, and the corestriction map can be interpreted as the map on $Tor$ induced by the map $\Bbb Z[G \backslash G'] \to \Bbb Z$. The complex $F_\bullet \otimes_{\Bbb Z[G]} \Bbb Z[G']$ is a projective resolution of $\Bbb Z[G \backslash G']$, and so if you know Tor is balanced and functorial in both variables then you answer the second question. $\endgroup$ – Tyler Lawson Nov 2 '18 at 9:35
  • $\begingroup$ @TylerLawson what is $\mathbb{Z}[G\backslash G’]$? $\endgroup$ – stupid_question_bot Nov 2 '18 at 16:32
  • $\begingroup$ @stupid_question_bot The free abelian group on the right cosets, or equivalently $\Bbb Z \otimes_{\Bbb Z[G]} \Bbb Z[G']$. $\endgroup$ – Tyler Lawson Nov 2 '18 at 18:38
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We are basically spelling out that group (co)homology is a functor in both variables. For conjugation by $g\in G$, we have two maps $c_g:H\to gHg^{-1}$ and $\mu_g:A\to A$ which can be composed on the homology level, but it is better to think of the pair $(c_g,\mu_g)$ as a morphism in the category of pairs {(group, coefficient module)}. To compute the induced map on (co)homology, you need a chain map $\tau$ between two projective resolutions (over $H$ and $gHg^{-1}$) which are compatible with $c_g$ (here you take the same resolution $P$ as in your post), and then $\tau\otimes \mu_g$ is a chain map (it's the diagonal one you wrote in your post) that defines the induced map!

By the way, not bashing on Weibel's book at all, but Ken Brown's bible "Cohomology of groups" is the clearest description in my opinion (and this discussion about conjugation is explained in Chapter III.8 in lieu of II.6).

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  • $\begingroup$ Thanks for your response and your reference to Brown's book! However, I'm still unclear why this works. For example, as you describe, $\tau$ is indeed a chain map, but it isn't a chain map between projective resolutions of the objects for which we're trying to compute the induced map on homology! That's my main issue. If we were using projective resolutions of $A$, that would be one thing, but instead we're starting with projective resolutions of $\mathbb{Z}$, whose relationship to (a proj resolution) of $A$ is unclear to me $\endgroup$ – stupid_question_bot Nov 1 '18 at 17:14
  • $\begingroup$ $\tau$ is not inducing the desired map itself, you use it to define the actual chain map: $\tau$ tensored by $\mu_g$. That's the map you wrote down immediately above your question. $\endgroup$ – Chris Gerig Nov 1 '18 at 17:20
  • $\begingroup$ But $\tau\otimes\mu_g$ is also not a map between projective resolutions of $A$. Usually to compute the map on derived functors induced by a morphism of objects $f : X\rightarrow Y$, you pick projective resolutions of $X$ and $Y$, at which point the chain maps are automatic (determined by $f$ up to homotopy) by projectivity, and you apply the functor to this chain map. How does $\tau\otimes\mu_g$ fit into this picture? $\endgroup$ – stupid_question_bot Nov 1 '18 at 17:34
  • $\begingroup$ I really recommend the relevant chapters of Brown (III.1, III.8, II.6): The definition of $H_\ast(G,A)$ (see Chapter III.1) is the homology of the complex $P\otimes_GA$, viewed as a tensor product of chain complexes (where $P$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}[G]$). Because $\tau$ is compatible with $c_g$ (see Chapter II.6), $\tau\otimes \mu_g$ is a chain map between the desired tensor-product-chain-complexes (see Chapter III.8). Note that in our scenario, $G$ changes to $H$ and $gHg^{-1}$ while $P$ remains the same. $\endgroup$ – Chris Gerig Nov 1 '18 at 17:42
  • $\begingroup$ I've edited my question to be more to the point. I've looked at Brown's book (including the chapters you mentioned), but he doesn't seem to suggest an answer to my question. $\endgroup$ – stupid_question_bot Nov 1 '18 at 21:30

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