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.

`$E^{r+1}=H(E^r)$`

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