Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
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

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.