# Multiplicative spectral sequence

I have a simple question regarding the definition of a multiplicative spectral sequence, which I couldn't answer myself by looking at the definitions in various texts:

• Is the product assumed to be associative?
• Is it assumed that there is a multiplicative unit? It should live in bidegree $(0,0)$ on each page.
• Is it usually assumed that the product is graded-commutative?

I'm basing myself on McCleary's book A user's guide to spectral sequences, sections 1.3 and 2.4. He actually calls them "spectral sequences of algebras", not "multiplicative spectral sequences". But it's probable that other sources give similar definitions (in the end, it all depends on what your applications are, I guess).

• Yes, the product is almost* always required to be associative.
• No, it's not always the case that there's a unit. (Indeed if there's one it should live in bidegree $(0,0)$, because if the unit $1$ is in bidegree $(p,q)$, then $1 \cdot 1 = 1 \implies (2p,2q) = (p,q)$.) It's extra structure.
• No, it's not usually assumed. The most common example of a multiplicative spectral sequence is the Serre one, which is graded commutative, but it need not be the case. For example, the Adams spectral sequence is multiplicative, but it's not graded commutative if I'm not mistaken.

By the way you might be interested in this article of Bauer and Scull (I haven't read it yet). They say they have identified conditions on an operad such that if some page of your spectral sequence is an algebra over the operad, then so are the next pages. In general if your operad has trivial differential I believe the structure should pass through without any problem; there is nothing special about $\mathtt{Ass}$ (or $\mathtt{Ass}_+$, or $\mathtt{Com}$) here, and I think you could very well consider spectral sequences of Lie algebras, Poisson algebras...

* I write almost because I don't want to make an absolute statement in case someone comes up with a counterexample, but I think I could drop it. Of course I'm sure you can also consider spectral sequences where some page is an $A_\infty$-algebra if you wanted to, which aren't strictly associative, but the spirit is the same (and the next page would be strictly associative anyway). For example the $E_1$ page of the Serre spectral sequence is something that looks like $C^*(B; C^*(F))$ which is only homotopy associative and commutative.

• Thank you. Can you give an example of a multiplicative spectral sequence which appears in practice and which has no unit? – Martin Brandenburg May 25 '15 at 12:59
• I can't think of a "standard" example off the top of my head (Serre's, Adams's... are all unital), but whenever you want to know the homology of a filtered nonunital algebra through spectral sequences you will have an example. – Najib Idrissi May 25 '15 at 17:35