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I have been reading Chapter 2. of A. Hatcher's "Spectral Sequences in Algebraic Topology", which is freely available at the author's website.

I have trouble understanding the Adams exact couple, which is pictured at the top of page 17. It's obtained by applying the homotopy group functor to the Adams tower, ie. a sequence of spectra

$X \leftarrow X^{1} \leftarrow X^{2} \leftarrow \ldots$

such that the cofibre of any consecutive morphisms forms a distinguished triangle

$X^{i+1} \rightarrow X^{i} \rightarrow K^{i}$,

where $K_{i}$ is a generalized Eilenberg-Maclane spectrum and the map $X_{i} \rightarrow K_{i}$ is surjective on mod $p$ cohomology. (There is some clever indexing changes going on between "upper" and "lower indices", that I will not describe here since my question is only about this specific reference.)

The picture of the exact couple at page 17 seems to show the nice "middle part" and I have trouble understanding at what happens if we go downstairs. According to the illustration, we should have a downstairs going map

$\pi_{t}(X_{0}) \rightarrow \pi_{t-1}(X_{-1})$,

but it's not clear to me how is $X_{-1}$, or - more generally - any of the negative indexed $X_{i}$, defined in the text. At first I thought that maybe we should extend the exact couple by putting $0$ groups outside of the defined range, but it seems to me that this would not preserve exactness of the couple.

I thought a little bit more and now it seems to me that we should follow the pattern by defining $X^{i}, X_{i}$ by imposing the requirement that $X^{i+1} \rightarrow X^{i} \rightarrow K^{i}$ is an exact triangle and $K^{i}$ are zero for negative $i$. (So that the associated spectral sequence is of the required form.) In practice, it seems to mean that $X^{i} = X$ or equivalently $X_{i} = \Sigma ^{i} X$ for negative $i$.

I believe this way one can work out through the convergence of the spectral sequence as it is presented and still prove the needed theorem. However, it seems that this would introduce two, three additional, small arguments in the given proof.

Is that the correct way to proceed or am I missing something obvious here? Note that I am leaning to the latter, since I learnt most of my algebraic topology from A. Hatcher's excellent handbook.

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I think you are on the right track: setting $K_i=0$ and $X^i=X^0=X$ for $i<0$ should indeed give you a consistent picture.

I would recommend trying to reconcile/check your understanding of the ASS with its usual pictorial representation (i.e., the picture on page 22 of Hatcher's draft). The homotopy of the $X^k$ is what you get when you truncate the spectral sequence to the region $s\ge k$, while the individual rows $s=k$ correspond to the Eilenberg Mac Lane spectra $K_k$.

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  • $\begingroup$ Indeed, I feel I have worked that out. :) Thank you! Would you mind pointing out which exactly picture on p. 22 do you mean? $\endgroup$ – Piotr Pstrągowski Jul 12 '13 at 16:00
  • $\begingroup$ Oops, I got that wrong: the picture I meant is on page 21, not 22. It's just the usual Adams SS picture that you can, e.g. also find in Ravenel's green book. $\endgroup$ – Christian Nassau Jul 12 '13 at 16:07

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