5
$\begingroup$

This is an exercise from Mcleary's book on Spectral sequence which I have been stuck with for some time. Let us recall what a Cartan-Eilenberg system is: IT consists of a module $H(p,q)$ for each pair of integers, $-\infty \leq p \leq q \leq \infty$ along with:

1) Homomorphisms $\eta:H(p',q') \rightarrow H(p,q)$ whenever $p \leq p'$, $q \leq q'$.

2) For $-\infty \leq p \leq q \leq r \leq \infty$ we have a connecting homomorphism $\delta:H(p,q) \rightarrow H(q,r)$.

3) $H(p,q) \rightarrow H(p,q)$ is the identity.

4) If $p \leq p' \leq p''$ and $q \leq q' \leq q''$ then the following diagram commutes $$\require{AMScd} \begin{CD} H(p'', q'') @>>> H(p,q) \\ @VVV @AAA \\ H(p',q') @= H(p',q') \end{CD}$$ 5) If $p \leq p'$, $q\leq q'$ and $r \leq r'$ then the following diagram commutes $$\require{AMScd} \begin{CD} H(p', q') @>>> H(q',r') \\ @VVV @AAA \\ H(p,q) @>>> H(q,r) \end{CD}$$ 6) For $-\infty \leq p \leq q \leq r \leq \infty$ , the following sequence is exact $$\cdots \rightarrow H(q,r) \rightarrow H(p,r) \rightarrow H(p,q) \xrightarrow{\delta} H(q,r) \rightarrow \cdots$$ 7) $H(-\infty,q)$ is the direct limit of the system $$H(q,q) \rightarrow H(q-1,q) \rightarrow H(q-2,q) \rightarrow \cdots$$

Then, Mcleary claims that we get a spectral sequence by letting $Z^p_r = im(H(p,p+r) \rightarrow H(p,p+1)$ , $B^p_r = im(H(p-r+1,p) \rightarrow H(p,p+1)$ and $E^p_r = Z^p_r/B^p_r$ . However, how does this give a spectral sequence? Mcleary defines a spectral sequence to be bigraded, first of all. Then I don't see any nice method to show that this gives a spectral sequence, so any help would be appreciated.

$\endgroup$
3
$\begingroup$

This procedure is described (without many details) in the book homological algebra, chapter XV, section 7. To obtain a bigrading, assume the $H(p,q)$ are graded, the morphisms in (1) are degree preserving, and the connecting homomorphisms in (2) have degree 1. There are several grading conventions for spectral sequences, so you have to decide for yourself if you would like to put $E_1^{p,q}=Z_1^{p,q}=H^{p+q}(p,p+1)$ as in the Leray-Serre spectral sequence, or choose some other convention.

The long exact sequences (6) allow one to construct the spectral sequence from this. This is indeed an exercise, though not a pleasant one. You find at least some of the details explained in Switzer's book, chapter 15, where $h_\bullet(X^p,X^q)$ corresponds to $H(p,q)$.

$\endgroup$

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.