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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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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
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
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

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

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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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 2 days ago
Generally, I agree. A useful statement is that a filtered dg-algebra gives a multiplicative spectral sequence. – Sergei Ivanov 2 days ago
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 2 days ago
@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 2 days ago
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… and they don't seem to mention the Massey article at all. – Sean Tilson 2 days ago

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