I'm trying to compute $H^3(point group,\mathbb{Z})$ for all the 32 point groups in 3D which has some applications in physics. Unfortunately, I could not find literature discussing this problem. So I tried to use GAP program to compute it.

I used the following code to do the computation:

gap> GroupCohomology(PointGroup(SpaceGroup(3,x)),3);

which is basically calculating the point group corresponding to the space group No.x. I have to admit it might be a bit inefficient because I'm new to GAP program.

My problem is that when I take $x=207$, because I want to obtain $H^3(O,\mathbb{Z})$ where $O$ is the octahedral group, some error props out. And when I try $x=210$, which corresponds to the same group $O$, it took too long and did not give me any result.

My ultimate goal is to obtain the cohomology group of point group $O$. Can anyone give me the result(by some analytical method or citing from some literature) or simply help me with the GAP program to get the result?

Edit: The octahedral group $O$ (symmetry group of an octahedron with only orientation preserving elements) is isomorphic to $S_4$, therefore it is in fact very easy to deal with.

  • $\begingroup$ Here is a table of the point groups. All those groups will have documented group cohomology it should not be too hard to find. I find understanding the notation to be the most challenging part but as this is your field you should be ok. You want to look up the proper name for the abstract group for each of these groups when searching results. $\endgroup$ – Ali Caglayan Feb 9 '18 at 4:51
  • $\begingroup$ Hi Ali. Could you please send me the link to the literatures collecting cohomological group of those point groups? I tried and didn't find one. I think it might be that mathematicians are not well-motivated to study them. $\endgroup$ – Xu Yang Feb 9 '18 at 5:03
  • $\begingroup$ Ok so I have managed to find the corresponding abstract groups for the point groups here. Looking through I do not see anything too difficult. GAP should be able to do most of these but if not you can compute them by hand using the Kunneth Formula for group cohomology. (For a non-mathematician this may be slightly daunting however). I will see what I can do. $\endgroup$ – Ali Caglayan Feb 9 '18 at 5:30

The Kunneth formula for group cohomology is:

$$H^n(G_1 \times G_2; \Bbb Z) \cong \bigoplus_{i= 0}^n H^i(G_1;\Bbb Z) \otimes_{\Bbb Z} H^{n-i}(G_2;\Bbb Z) \oplus\bigoplus_{p =0}^{n+1} \text{Tor}^{\Bbb Z}(H^p(G_1;\Bbb Z),H^{n+1-p}(G_2;\Bbb Z)) $$ Most of these terms will be $0$ due to the fact that there are a lot of cyclic group cohomologies floating around. For the $n=3$ case we get an expression which relies only on $H^n$ for $n=0, \dots, 4$ of each direct product factor. So to compute $H^1(G_1 \times G_2)$ (assuming $\Bbb Z$ coefficients here) we need to compute 10 cohomology groups, 5 for each factor.

You can start to see why GAP was invented. Anyway since the entries for Tor are always abelian groups they should be pretty easy to calculate. (They will distribute over direct sums of groups etc.)

Now the idea is to keep doing this until you have cohomology groups you know. All groups in this table have factors whose cohomology groups are known.

This calculation is tedious so once your done write it down somewhere safe.

Alternatively you could get GAP to do it but I do not know how. This is sort of the brute force way of getting what you want.

One more thing, I would try and factor groups with multiple factors so that a single cyclic group is your second factor as this will naturally cancel many of your terms. Also if you do not know about tensor products of abelian groups Google is your friend.

Standard references for this kind of work would probably be Group cohomology by Brown but that is still not what you are looking for probably.

  • $\begingroup$ Thank you for your clear and detailed answer! Now that the corresponding abstract groups are known I guess calculating group cohomology of these point groups by hand should be quite straightforward. $\endgroup$ – Xu Yang Feb 9 '18 at 14:21

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.