# Higher cohomology operations and spectral sequences

I was wondering how higher cohomology operations interact with spectral sequences. For example if $X$ is a simplicial set. Then $C^*(X,\mathbb{F}_2)$ has an algebra structure over the Surjection Operad. Suppose we are given a filtration of $C^*(X,\mathbb{F}_2)$ which behaves well with respect to the upper algebra structure (whatever this means). Then one can ask how higher cohomology operations interact with the induced filtration on $H^*(X)$ (whose subquotients is the $E_\infty$-page).

I was wondering whether this has already been studied, googling only gave me results for primary cohomology operations.

EDIT: Let us first look at the secondary cohomology operation $Ker(Sq^1)\subset H^1\rightarrow Cok(Sq^2)\subset H^2$ corresponding to the Adem-Relation $Sq^1Sq^1=0$. Note that source and target inherit a filtration from the filtration on $H^*$ and so it makes sense to ask how those interact.

Let me elaborate a bit on that particular one (aiming more at the earlier pages).

For example we could obtain a cochain formula for that operation in the following way. Let us start with choices $x,y\in C^1$ with $dy=(x\cup x)=\langle 12\rangle (x\otimes x)$. The right hand side is just the expression for the $cup$-product in the Surjection Operad and $Sq^1$ is just the cup square.

Now we need two different reasons why $(x \cup x) \cup_1 (x\cup x)=(\langle 13412\rangle+\langle 12342\rangle)(x^{\otimes 4})$ is boundary. The first comes from the Adem-relation. One could choose for example $\langle 123412\rangle(x^{\otimes 4})$ (its boundary is exactly the upper expression ) and the second one comes from the well-definedness-proof of $Sq^1$. There it is shown that $\langle 121\rangle(dy\otimes dy)$ is also boundary of something.

This gives one explicit cochain formula for that secondary cohomology operation, but this one might not behave well with respect to filtrations and it might make more sense to choose a better one. If we plug in only an $r$-almost cycle we might not get zero, but only something that lives hopefully further down in the filtration. For example for Steenrod-Squares it turns out that the definition $[x]\mapsto [x\cup_{n-k}x + x\cup_{n-k+1}dx]$ behaves sometimes better than the definition $[x]\mapsto [x\cup_{n-k}x]$, the boundary of the first term is $dx \cup_{n-k}x + x\cup_{n-k}dx$ and the boundary of the latter is $dx\cup_{n-k-+1}dx$ and depending on what "interacts nicely with the filtration" means, it could happen that the latter lies way further down in the filtration then the first expression.

Now we could ask similar questions for the cochain formula for that higher operation. Can we add some extra stuff that vanishes for cocycles, but that makes it behave better with respect to that filtration?

• The naive answer is that, for sufficiently natural spectral sequences, the higher differentials are higher cohomology operations, and identifying them is pretty much one of the central goals of studying the spectral sequence. – Denis Nardin Jan 17 '18 at 13:07
• This is an interesting question. In the interaction between filtrations and power operations that I am familiar with, the filtration is usually some kind of canonical thing. Do you have a specific filtration in mind? What might be easier is to work with a filtration of the cochain complex rather than one coming from the space. I would guess it more likely to look at what happens in the specific instance when the cochain complex IS an algebra over the surjection operad and then you can maybe say something. ... – Sean Tilson Jan 19 '18 at 10:23
• ... The stuff I am familiar with all goes back to Bruner. I hate doing this, but I trained to explain the idea in section 4 of arxiv.org/pdf/1702.04632.pdf. – Sean Tilson Jan 19 '18 at 10:24
• @SeanTilson I was just thinking of the Serre spectral sequence. Somewhere I read that the $cup_i$ products interact with the filtration in some way and my hope was that one could also examine how all those operations from that operad interact with the filtration and then use the formulas for higher cohomology operations to say something. – HenrikRüping Jan 29 '18 at 12:56
• Jeez, I clearly meant tried and not trained in the above. Have you looked at the work of Bill Singer? He wrote a book that you might find helpful. – Sean Tilson Jan 30 '18 at 11:01