I would like to read Atiyah's paper Characters and cohomology of finite groups; but when I started reading the introduction, Atiyah mentions that he will prove that there is a "spectral sequence $\{E^p_r\}$ with $E^p_2 = H^p(G,\mathbb{Z})$ and $E^p_\infty = R_p(G)/R_{p+1}(G)$"

I can't understand what this means, because a spectral sequence is supposed to have $2$ indices, like $E^{p,q}_2$.

My best guess is that it would be an old notation, for instance that what he denotes $E^p_2$ would the graded module $\displaystyle\bigoplus_{q}E_2^{p,q}$; but I'm not sure, and I would like not to have to guess to understand the paper.

Is this a common notation, an old notation ? What does it mean ?

  • $\begingroup$ A few pages later he uses the same notation for the cohomology-to-K-theory AHSS, so I think that you're right (to a degree). He defines "spectral sequence" later to just mean a sequence of complexes (with $E_r$'s differential of degree $r$) with fixed isomorphisms $H(E_r, d_r) \cong E_{r+1}$. This includes the case we're used to (bigraded things). But remember that $R(G)$ is not a graded object! His subscript refers to a filtration as opposed to a grading. So it makes sense to me that a bigraded SS is not the right thing to converge to this. $\endgroup$ – user98602 Feb 17 at 15:21
  • $\begingroup$ More importantly: march on! It's common to be briefly confused at something that the author will clarify later, even from the best of authors; the introduction especially sometimes has varying levels of detail so as to appeal to many different people. It's valuable to be able to skip ahead for a while and only come back to something later when it's necessary, or the author has given you the appropriate gear. $\endgroup$ – user98602 Feb 17 at 15:23
  • $\begingroup$ @MikeMiller Oh thank you for having looked that up ! Thank you for your advice as well, I usually do that, it's just that being confused by that I tried to look around in the article for a definition of this sort of spectral sequence, but didn't seem to find it and he seemed to just use the same terminology for a different object so I was really confused - so I asked the question. Perhaps you can write a short answer (essentially your first comment) to remove this from the unanswered queue ? $\endgroup$ – Max Feb 17 at 15:27
  • $\begingroup$ That's fair, I should have written this in the answer box. $\endgroup$ – user98602 Feb 17 at 15:28
  • 1
    $\begingroup$ Well, there exists single graded spectral sequence, the Bockstein spectral sequence (en.wikipedia.org/wiki/Bockstein_spectral_sequence) is such an example. The construction of spectral sequence from an exact couple does not require a bigraduation. (Though of course you can add a bigraduation if you prefer, as does the wikipedia article). $\endgroup$ – Roland Feb 17 at 15:44

He writes what he means in a section titled 'spectral sequences' at the start of page 34 (end of section 3). He means a sequence of graded complexes $(E_r, d_r)$, with $d_r$ of degree $r$, with specified isomorphisms $H(E_r, d_r) \cong E_{r+1}$.

He writes this with the meaning you expect (collapsing a bigraded spectral sequence to a single grading) for the usual cohomology-to-K-theory AHSS in propositions 2.4 and 2.6.

However, the target for the spectral sequence you're talking about is the representation ring $R(G)$, which in particular is not a graded object; Atiyah introduces a filtration (which I think is from the dimension of an irrep) and then argues that a spectral sequence converges to this target.

I am not familiar enough with the paper to tell you how that spectral sequence is constructed. However, I will point out that because $R(G)$ is not naturally graded, it would be odd to have a bigraded spectral sequence converging to it; this would naturally construct a grading (at least on each filtered piece), which seems like it would be odd; I have no idea where it would come from!

So it seems likely that the SS he constructs does not naturally have a second grading, and that it only fits into the formalism I mentioned at the start of this post.


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.