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Let $E$ be a spectral sequence and assume that there is a product

$E^{r}_{p_1,q_1} \times E^r_{p_2,q_2} \to E^r_{p_1+p_2,q_1+q_2}$

which satisfies the Leibniz rule (for all $p_i,q_i$, but $r$ fixed). Then it extends to a product

$E^{r+1}_{p_1,q_1} \times E^{r+1}_{p_2,q_2} \to E^{r+1}_{p_1+p_2,q_1+q_2}$.

In the errata for Weibel 5.2.13 it is suggested that the Leibniz rule does not hold automatically in $E^{r+1}$. I've convinced myself of this through lengthy calculations with the conclusion: Ok, nothing is compatible, and there is no reason why the product has something to do with the isomorphisms $H(E^r) \cong E^{r+1}$.

  • Is there an easy insightful example where the Leibniz rule fails?
  • Is there a simple-to-check criterion which makes the Leibniz rule valid?

Just to clarify: Of course there are lots of special cases where you can write down a product on $E^r$ for each $r$. For example if $E$ is the spectral sequence coming from a filtered differential graded algebra. But I'm rather interested in the general case. I think often you only know $E^2$ well and it would be interesting if you can lift the product to $E^{\infty}$ without actually make your hands dirty.

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3  
I don´t think there is any non-tautological way to answer your two questions. If you allow yourself to define each multiplication independently of the others, then it is clear that there is no reason to expect the multiplication induced on $E^{r+1}=H(E^r)$ by the one you defined on $E^r$ will relate in any sensible way to the differential of $E^{r+1}$! So you can construct examples by just picking a product on $E^3$, say, which does not do what you want it not to do... – Mariano Suárez-Alvarez May 9 '10 at 15:16
4  
Right, e.g. if the $d_2$ differential is zero then it satisfies the Leibniz rule for any random multiplication you might pick. – Tyler Lawson May 9 '10 at 16:26
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is there a reason you want to look at a spectral sequence that is somehow multiplicative but is not coming from some filtered DGA? In order for the product on $E^2$ or $E^1$ to induce anything meaningful at $E^\infty$ doesn't it need to come from somewhere? – Sean Tilson May 19 '10 at 17:42

As far as I know that 1954 paper of Massey is faulty, and you cannot get multiplicative spectral sequences just from such stucture on an exact couple. The best I know that you can do is to use Cartan-Eilenberg systems. Pairings of Cartan-Eilenberg systems are discussed in Douady's Cartan seminar paper from 1958/59 on products in the Adams spectral sequence. You can find it at numdam.org.

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1  
Do you know the details about the flaw in that Massey paper, or do you know somebody who would? And is there no other paper on these multiplicative structures than a seminar note that is not even indexed in mathscinet (at least I did not find it there)? – Sebastian Goette Feb 13 at 10:12
    
Joseph Neisendorfer's AMS Memoir "Primary Homotopy Theory" (#232) uses exact couples in section 4, and writes: "But exact couples are not sufficient for the introduction of products into the spectral sequence." He therefore introduces Cartan-Eilenberg systems in section 5 and pairings of Cartan-Eilenberg systems in section 12, referring to Douady's note. I suspect that Neisendorfer (and I) received this wisdom from John Moore, who was active in the Cartan seminar around that time. – John Rognes Feb 16 at 9:46
    
In John McCleary's book "A User's Guide ..." (2. ed.), Cartan-Eilenberg systems are mentioned in Exercise 2.2, and pairing of Cartan-Eilenberg systems are mentioned in Exercise 2.3. He refers to Neisendorfer's Memoir (see my comment above) and David Anick's book "Differential Algebras in Topology" on page 462 for their discussions of pairings in spectral sequences. – John Rognes Feb 16 at 9:53
    
Tsuyoshi Ogawara, a master's student of Lars Hesselholt, wrote out details of the passage from a multiplicative filtration of spectra to a multiplicative Cartan-Eilenberg system, and from a multiplicative Cartan-Eilenberg system to a multiplicative spectral sequence. The master's thesis is unpublished. – John Rognes Feb 16 at 9:57
    
Dear John, thanks for all your hints, I will check them in the library this afternoon. It is nice to know that all this is not just folklore. – Sebastian Goette Feb 16 at 11:01

To my knowledge, all useful examples of spectral sequences come from exact couples. Here

http://www.jstor.org/stable/1969719?seq=1#page_scan_tab_contents

you can find a condition on exact couples which implies that the corresponding spectral sequence is a spectral sequence of algebras. In particular, it is proved that a filtered dg-algebra gives a spectral sequence of algebras.

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1  
This paper, as I recall, essentially boils down to "a spectral sequence is multiplicative iff it is multiplicative". I don't think that the conditions here are helpful. – Sean Tilson Feb 9 at 13:11
    
Generally, I agree. A useful statement is that a filtered dg-algebra gives a multiplicative spectral sequence. – Sergei Ivanov Feb 9 at 13:19
    
A similarly useful statement is the following. Given a symmetric monoidal model category M (with probably lots of other adjectives like stable and suitably enriched) filtered objects inherit a monoidal structure such that monoids with respect to that structure have associated spectral sequences that are multiplicative. – Sean Tilson Feb 9 at 13:23
    
@SeanTilson You are a bit too pessimistic about Massey's paper. Please have a look at section 7, where he discusses filtered rings. The conditions there look as if they were easy enough to verify in special cases. What worries me however, is the age of this paper compared to problems in more recent books (Ravenel, and Hatcher's tentative fifth chapter) that deal with multiplicative spectral sequences. Did anybody discover a mistake in Massey's article? – Sebastian Goette Feb 9 at 14:59
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I don't think that there is a mistake. Usually one wants to not have to check conditions on the level of the exact couple. Also, most spectral sequences that topologists care about don't come from filtered DGAs but from filtered spaces or filtered spectra. It might then take a great deal of care to show that one can obtain a filtered DGA from that set up , or a filtered $E_{\infty}$-DGA. I looked up ams.org/journals/tran/1973-177-00/S0002-9947-1973-0372860-2/… and they don't seem to mention the Massey article at all. – Sean Tilson Feb 9 at 15:08

This is an expansion of John Rognes' answer. I have filled in a few details in Douady's seminare notes and noticed that one gets away with slightly weaker axioms. If there is already a reliable reference for all this, please let me know.

Recall that a Cartan-Eilenberg system $(H,\eta,\partial)$ (see here, chapter XV.7) consists of modules $H(p,q)$ for each $p\le q$, morphisms $\eta\colon H(p',q')\to H(p,q)$ for all $p\le p'$, $q\le q'$, and boundary morphisms $\partial\colon H(p,q) \to H(q,r)$ for all $p\le q\le r$, such that

  1. $\eta=\mathrm{id}\colon H(p,q)\to H(p,q)$,

  2. $\eta=\eta\circ\eta\colon H(p'',q'')\to H(p',q')\to H(p,q)$,

  3. $\eta$ and $\partial$ commute,

  4. there are long exact sequences $\cdots\to H(q,r)\stackrel\eta\to H(p,r)\stackrel\eta\to H(p,q)\stackrel\partial\to H(q,r)\to\cdots$.

The conditions needed for convergence have been omitted. A typical example is $H(p,q)=\tilde h_\bullet(X_p/X_q)$, where $\cdots\supset X_{-1}\supset X_0\supset X_1\supset\cdots$ is a decreasing sequence of cofibrations and $\tilde h_\bullet$ is some generalised homology theory. The grading is suppressed in the following, but you can easily fill it in.

To set up a spectral sequence from $(H,\eta,\partial)$, one defines $$Z^r_p=\mathrm{im}\bigl(H(p,p+r)\stackrel\eta\to H(p,p+1)\bigr)\;,$$ $$B^r_p=\mathrm{im}\bigl(H(p-r+1,p)\stackrel\partial\to H(p,p+1)\bigr)\;,$$ $$E^r_p=Z^r_p/B^r_p\;,$$ $$d^r_p\colon Z^r_p/B^r_p\twoheadrightarrow Z^r_p/Z^{r+1}_p\cong B^{r+1}_{p+r}/B^r_{p+r}\hookrightarrow Z^r_{p+r}/B^r_{p+r}\;.$$ Details are in Switzer's book, chapter 15. In particular $$\ker(d^r_p)=Z^{r+1}_p/B^r_p\qquad\text{and}\qquad \mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p\;.$$ For $a=\eta(a_0)\in H(p,p+1)$, $a_0\in H(p,p+r)$, one has $$d^r_p([a])=[\partial a_0]\in E^r_p\qquad\text{with}\qquad\partial a_0\in H(p+r,p+r+1)\;.$$

Definition (Douady, II.A) Let $(H,\eta,\partial)$, $(H',\eta',\partial')$ und $(H'',\eta'',\partial'')$ be Cartan-Eilenberg systems. A spectral product $\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial)$ is a sequence of maps $$\mu_r\colon H'(m,m+r)\otimes H''(n,n+r)\to H(m+n,m+n+r)$$ such that for all $m$, $n$, $r\ge 1$, the following two diagrams commute. $\require{AMScd}$ \begin{CD} H'(m,m+r)\otimes H''(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\ @V\eta'\oplus V\eta''V@VV\eta V\\ H'(m,m+1)\otimes H''(n,n+1)@>\mu_1>>H(m+n,m+n+1)\rlap{\;,} \end{CD} \begin{CD} H'(m,m+r)\otimes H''(n,n+r)@>\mu_r>>H(m+n,m+n+r)\\ @V\partial'\otimes\eta''\oplus V\eta'\otimes\partial''V@VV\partial V\\ {\begin{matrix}H'(m+r,m+r+1)\otimes H''(n,n+1)\\\oplus\\H'(m,m+1)\otimes H''(n+r,n+r+1)\end{matrix}}@>\mu_1+\mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)\rlap{\;.} \end{CD}

The first diagram is weaker than in Douady's notes. The second can be read as a Leibniz rule.

Theorem (Douady, Thm II) A spectral product $\mu\colon(H',\partial')\times(H'',\partial'')\to(H,\partial)$ induces products $$\mu^r\colon E^{\prime r}_m\otimes E^{\prime\prime r}_n\to E^r_{m+n}\;,$$ such that

  1. $\mu^1=\mu_1$

  2. $d^r_{m+n}\circ\mu^r=\mu^r\circ(d^{\prime r}_m\otimes\mathrm{id})\pm\mu^r\circ(\mathrm{id}\circ d^{\prime\prime r}_n)$,

  3. $\mu^{r+1}$ is induced by $\mu^r$.

Proof. Assume by induction that $\mu^r$ is induced by $\mu_1$. In particular, $$Z^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to Z^r_{m+n}\;,$$ $$B^{\prime r}_m\otimes Z^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;,$$ $$Z^{\prime r}_m\otimes B^{\prime\prime r}_n\stackrel{\mu_1}\to B^r_{m+n}\;.$$ This is clear for $r=1$ if we put $\mu^1=\mu_1$ because $E^1_p=Z^1_p=H(p,p+1)$ and $B^1_p=0$.

Let $[a]\in Z^{\prime r}_m$, $[b]\in Z^{\prime\prime r}_n$ be represented by $a=\eta'(a_0)\in H'(m,m+1)$, $b=\eta''(b_0)\in H''(n,n+1)$ with $a_0\in H'(m,m+r)$, $b_0\in H''(n,n+r)$. Using the first diagram and the construction of $d^r_{m+n}$, we conclude that $$(d^r_{m+n}\circ\mu^r)([a]\otimes[b])=d^r_{m+n}[\mu_1(a\otimes b)]=d^r_{m+n}[\eta(\mu_r(a_0\otimes b_0))]=(\partial\circ\mu_r)(a_0\otimes b_0)\;.$$ From the second diagram, we get $$(\partial\circ\mu_r)(a_0\otimes b_0)=\mu_1(\partial'a_0\otimes\eta''b_0)\pm\mu_1(\eta'a_0\otimes\partial''b_0)=\mu^r(d^{\prime r}_m[a]\otimes[b])\pm\mu^r([a]\otimes d^{\prime\prime r}_n[b])\;.$$ This proves the Leibniz rule (2).

From the Leipniz rule and the facts that $\ker(d^r_p)=Z^{r+1}_p/B^r_p$ and $\mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p$, we conclude that $\mu^r$ induces a product on $E^{r+1}_p\cong\ker(d^r_p)/\mathrm{im}(d^r_p)$, which proves (3). Because $\mu^r$ is induced by $\mu_1$, so is $\mu^{r+1}$, and we can continue the induction.

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As far as I understand, your weak definition of spectral product does not imply that the pairing $H'(-\infty,+\infty) \otimes H''(-\infty,+\infty)\to H(-\infty,+\infty)$ respects the filtration. So the pairing on the spectral sequences does not converge to the pairing on limits in this case. – Sergei Ivanov Apr 1 at 19:09
    
@SergeiIvanov. You are right. I did not care about convergence here. If we assume in addition that the first diagram commutes when $+1$ is replaced by $+s$ with $s\le r$, and that $H(p,\infty)=\lim H(p,p+r)$, then the product first of all extends to $H(p,\infty)$. Cartan-Eilenberg also want $H(-\infty,p)=\operatorname{colim}H(p-r,p)$, and now you would need Douady's full first axiom to prove that $\mu$ extends to $H(-\infty,\infty)$ in a compatible way. Of course, even without these assumptions the product structure is already very helpful in understanding the differentials. – Sebastian Goette Apr 4 at 19:42

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