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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 ?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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