Group cohomology

Shadow · **Posted:** Fri, 23 Mar 2012 02:44:57 UTC

I'm interested in group cohomology calculations, specific ones in particular, so I can manage a better understanding of the theory.

The basic setup is one is given a group,

, and

-modules (i.e. modules over $\mathbb{Z}[G]$ ), A,B

. And the starting point is calculating $\hom_{\mathbf{Ab}}(A,B)$ , (this is before I look at

-equivariance), and I"m trying to figure it out for say a free $\mathbb{Z}/n\mathbb{Z}$ -module (from there I should be able to get all the finitely generated abelian groups by breaking the homs across direct sums), however the only thing I know for sure from the start is that $$\mathbb{Z}[\mathbb{Z}/n\mathbb{Z}]=\mathbb{Z}[x]/(x^n-1)\cong\bigoplus_{d|n}\mathbb{Z}[x]/\Phi_d(x)$ , which looks like $\mathbb{Z}^{n}$ as an abelian group if I'm not mistaken. And if my group is free, I'm pretty sure the group ring is just

itself (as a free, abelian group) right? And if so, do I just calculate hom via the isomorphism type, or is there merit in explicitly understanding the generators? From there, how can one proceed to figure out the G-equivariant homomorphisms?

outermeasure · **Posted:** Fri, 23 Mar 2012 05:13:49 UTC

Shadow wrote:

I'm interested in group cohomology calculations, specific ones in particular, so I can manage a better understanding of the theory.

The basic setup is one is given a group,

, and

-modules (i.e. modules over $\mathbb{Z}[G]$ ), A,B

. And the starting point is calculating $\hom_{\mathbf{Ab}}(A,B)$ , (this is before I look at

-equivariance), and I"m trying to figure it out for say a free $\mathbb{Z}/n\mathbb{Z}$ -module (from there I should be able to get all the finitely generated abelian groups by breaking the homs across direct sums), however the only thing I know for sure from the start is that $$\mathbb{Z}[\mathbb{Z}/n\mathbb{Z}]=\mathbb{Z}[x]/(x^n-1)\cong\bigoplus_{d|n}\mathbb{Z}[x]/\Phi_d(x)$ , which looks like $\mathbb{Z}^{n}$ as an abelian group if I'm not mistaken. And if my group is free, I'm pretty sure the group ring is just

itself (as a free, abelian group) right? And if so, do I just calculate hom via the isomorphism type, or is there merit in explicitly understanding the generators? From there, how can one proceed to figure out the G-equivariant homomorphisms?

For any group

, the (integral) group ring $\mathbb{Z}G$ as an abelian group is the direct sum of $\# G$ copies of your ring, not

itself.

To check equivariance, you only need to check the action of generators of

. If

is cyclic, that is easy enough (and very concrete). I'm not sure what you mean by isomorphism type here.

Shadow · **Posted:** Fri, 23 Mar 2012 16:43:50 UTC

outermeasure wrote:

Shadow wrote:

I'm interested in group cohomology calculations, specific ones in particular, so I can manage a better understanding of the theory.

The basic setup is one is given a group,

, and

-modules (i.e. modules over $\mathbb{Z}[G]$ ), A,B

. And the starting point is calculating $\hom_{\mathbf{Ab}}(A,B)$ , (this is before I look at

-equivariance), and I"m trying to figure it out for say a free $\mathbb{Z}/n\mathbb{Z}$ -module (from there I should be able to get all the finitely generated abelian groups by breaking the homs across direct sums), however the only thing I know for sure from the start is that $$\mathbb{Z}[\mathbb{Z}/n\mathbb{Z}]=\mathbb{Z}[x]/(x^n-1)\cong\bigoplus_{d|n}\mathbb{Z}[x]/\Phi_d(x)$ , which looks like $\mathbb{Z}^{n}$ as an abelian group if I'm not mistaken. And if my group is free, I'm pretty sure the group ring is just

itself (as a free, abelian group) right? And if so, do I just calculate hom via the isomorphism type, or is there merit in explicitly understanding the generators? From there, how can one proceed to figure out the G-equivariant homomorphisms?

For any group

, the (integral) group ring $\mathbb{Z}G$ as an abelian group is the direct sum of $\# G$ copies of your ring, not

itself.

To check equivariance, you only need to check the action of generators of

. If

is cyclic, that is easy enough (and very concrete). I'm not sure what you mean by isomorphism type here.

I mean if say I don't have a free $\mathbb{Z}[G]$ module, it might help to know an explicit description of the modules

or

as quotients of some polynomial algebra and knowing a description of exacly what maps generate the abelian group hom-set. Or do I not need to be so specific in general? (i.e. if I know that say $A\cong \mathbb{Z}^n$ and

is whatever then from abstract theory I just have that the abelian group homomorpisms look like n copies of

, but I'm wondering if--given a presentation of each module as, say, polynomial algebras knowing which are the generators--would that be helpful in understanding the cohomology, or generally do people only care about the isomorphism type in these kinds of calculations?) My backgroung with (co)homology is from topology where sometimes maps (like the Bott map) are useful to know explicitly is a reason I ask. In particular I know that the general theory can easily give me the isomorphism type of $\hom_{\mathbf{Ab}}(A,B)$ , but I'm wondering if I need to know a better description than just that to get at the

-equivariant parts, i.e. in order to check the action (I know that checking on generators is enough).

outermeasure · **Posted:** Sat, 24 Mar 2012 02:08:11 UTC

Shadow wrote:

outermeasure wrote:

Shadow wrote:

I'm interested in group cohomology calculations, specific ones in particular, so I can manage a better understanding of the theory.

The basic setup is one is given a group,

, and

-modules (i.e. modules over $\mathbb{Z}[G]$ ), A,B

. And the starting point is calculating $\hom_{\mathbf{Ab}}(A,B)$ , (this is before I look at

-equivariance), and I"m trying to figure it out for say a free $\mathbb{Z}/n\mathbb{Z}$ -module (from there I should be able to get all the finitely generated abelian groups by breaking the homs across direct sums), however the only thing I know for sure from the start is that $$\mathbb{Z}[\mathbb{Z}/n\mathbb{Z}]=\mathbb{Z}[x]/(x^n-1)\cong\bigoplus_{d|n}\mathbb{Z}[x]/\Phi_d(x)$ , which looks like $\mathbb{Z}^{n}$ as an abelian group if I'm not mistaken. And if my group is free, I'm pretty sure the group ring is just

itself (as a free, abelian group) right? And if so, do I just calculate hom via the isomorphism type, or is there merit in explicitly understanding the generators? From there, how can one proceed to figure out the G-equivariant homomorphisms?

For any group

, the (integral) group ring $\mathbb{Z}G$ as an abelian group is the direct sum of $\# G$ copies of your ring, not

itself.

To check equivariance, you only need to check the action of generators of

. If

is cyclic, that is easy enough (and very concrete). I'm not sure what you mean by isomorphism type here.

I mean if say I don't have a free $\mathbb{Z}[G]$ module, it might help to know an explicit description of the modules

or

as quotients of some polynomial algebra and knowing a description of exacly what maps generate the abelian group hom-set. Or do I not need to be so specific in general? (i.e. if I know that say $A\cong \mathbb{Z}^n$ and

is whatever then from abstract theory I just have that the abelian group homomorpisms look like n copies of

, but I'm wondering if--given a presentation of each module as, say, polynomial algebras knowing which are the generators--would that be helpful in understanding the cohomology, or generally do people only care about the isomorphism type in these kinds of calculations?) My backgroung with (co)homology is from topology where sometimes maps (like the Bott map) are useful to know explicitly is a reason I ask. In particular I know that the general theory can easily give me the isomorphism type of $\hom_{\mathbf{Ab}}(A,B)$ , but I'm wondering if I need to know a better description than just that to get at the

-equivariant parts, i.e. in order to check the action (I know that checking on generators is enough).

OK, I see what you are trying to do.

Yes, you need to know more to get the

-equivariant part --- consider $A=\mathbb{Z}[G]$ with the usual

-action ( $\# G=n$ ), versus $A=\mathbb{Z}^n$ with the trivial

-action. These two give the same $\mathop{\mathrm{Hom}}_{\mathbb{Z}}(A,B)$ (namely, B^n

), but usually different $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(A,B)$ --- $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}[G],B)\cong B$ but $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}^n,B)\cong (B^G)^n$ . The important

-action which was thrown out in calculating $\mathop{\mathrm{Hom}}_{\mathbb{Z}}(A,B)$ is definitely needed.

Of course, if you are calculating cohomology by taking free resolution of $\mathbb{Z}$ (such as the bar resolution, which nobody use in practice to compute cohomology) then you just need $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}[G],B)\cong B$ and $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z},B)\cong B^G$ (together with the induced maps).

Shadow · **Posted:** Sat, 24 Mar 2012 07:47:27 UTC

outermeasure wrote:

OK, I see what you are trying to do.

Yes, you need to know more to get the

-equivariant part --- consider $A=\mathbb{Z}[G]$ with the usual

-action ( $\# G=n$ ), versus $A=\mathbb{Z}^n$ with the trivial

-action. These two give the same $\mathop{\mathrm{Hom}}_{\mathbb{Z}}(A,B)$ (namely, B^n

), but usually different $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(A,B)$ --- $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}[G],B)\cong B$ but $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}^n,B)\cong (B^G)^n$ . The important

-action which was thrown out in calculating $\mathop{\mathrm{Hom}}_{\mathbb{Z}}(A,B)$ is definitely needed.

Of course, if you are calculating cohomology by taking free resolution of $\mathbb{Z}$ (such as the bar resolution, which nobody use in practice to compute cohomology) then you just need $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}[G],B)\cong B$ and $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z},B)\cong B^G$ (together with the induced maps).

Oh, so you're saying

-equivariant abelian group homomorphisms are just homomorphisms in $\mathbb{Z}[G]-\text{Mod}$ ? If so I think that makes a lot of sense, and definitely will make my life easier in terms of computations. How does one actually end up doing computations in this subject? I'm not in a class for it, so I don't have to be pedantic and take resolutions if it's not really necessary.

Thanks a lot for the clarification so far, it's been quite useful!

outermeasure · **Posted:** Sat, 24 Mar 2012 09:40:17 UTC

Shadow wrote:

outermeasure wrote:

OK, I see what you are trying to do.

Yes, you need to know more to get the

-equivariant part --- consider $A=\mathbb{Z}[G]$ with the usual

-action ( $\# G=n$ ), versus $A=\mathbb{Z}^n$ with the trivial

-action. These two give the same $\mathop{\mathrm{Hom}}_{\mathbb{Z}}(A,B)$ (namely, B^n

), but usually different $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(A,B)$ --- $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}[G],B)\cong B$ but $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}^n,B)\cong (B^G)^n$ . The important

-action which was thrown out in calculating $\mathop{\mathrm{Hom}}_{\mathbb{Z}}(A,B)$ is definitely needed.

Of course, if you are calculating cohomology by taking free resolution of $\mathbb{Z}$ (such as the bar resolution, which nobody use in practice to compute cohomology) then you just need $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z}[G],B)\cong B$ and $\mathop{\mathrm{Hom}}_{\mathbb{Z}[G]}(\mathbb{Z},B)\cong B^G$ (together with the induced maps).

Oh, so you're saying

-equivariant abelian group homomorphisms are just homomorphisms in $\mathbb{Z}[G]-\text{Mod}$ ? If so I think that makes a lot of sense, and definitely will make my life easier in terms of computations. How does one actually end up doing computations in this subject? I'm not in a class for it, so I don't have to be pedantic and take resolutions if it's not really necessary.

Thanks a lot for the clarification so far, it's been quite useful!

Yes,

-equivariant here just means you are working in $\mathbb{Z}[G]\mathrm{-Mod}$ .

Usually one avoids doing any computation

. The point of the bar resolution (and variants) is to demonstrate you have enough in/pro-jectives so that you can sleep at night, and I think they are the ones used by GAP and Magma to calculate H^1 and H^2. Of course, there are some tools like the LES and various spectral sequences which, if you are lucky enough, will be usable, but even for finite group and mod-p coefficient the task is quite daunting (the mod-p cohomology ring can be reconstructed by gluing that of the element p-subgroups and possible nilradical, but if you want to know about the nilradical you need to work a lot harder).

Shadow · **Posted:** Sat, 24 Mar 2012 18:08:50 UTC

I see, so the real reality is I DO need to do the resolution for the first couple (like for free modules and the basic torsion modules), then once I know it for some projectives/injectives I can setup the same s.e.s.s as usual and do things like the snake lemma or a spectral sequence. So it's just like in topology really, just with more algebra involved for the baseline.

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Group cohomology

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