MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(This question has been asked on Math.StackExchange where it attracted a few upvotes, but - unfortunately - no answer.)

I have been reading Chapter 2. of A. Hatcher's draft of "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 are some clever indexing changes going on between "upper" and "lower indices", namely that $X_{i} = \Sigma ^{i} X^{i}, K_{i} = \Sigma ^{i} K^{i}$.)

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.

share|cite|improve this question
I've just posted an answer on math-stackexhange where the question seems a bit more appropriate. – Christian Nassau Jul 10 '13 at 11:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.