# Hodge and de Rham Cohomology Revisited

I was going to talk about how the moduli of K3 surfaces is stratified by height in positive characteristic and some of the cool properties of this (for instance, “most” K3 surfaces have height 1). Instead I’m going to shift gears a little. We’ve talked about ${\ell}$-adic étale cohomology, Witt cohomology, cohomology on any site you want to put on ${X}$, de Rham cohomology, and we’ve implicitly used Hodge theory in places. Secretly we’ve been heading straight towards cyrstalline cohomology. I think it might be neat to start a series of posts on how each of these relate to eachother and then really motivate the need for crystalline stuff.

Away from this blog I’ve been thinking about degeneration of the Hodge-de Rham Spectral Sequence a lot. Suppose for a minute we’re in the nicest situation possible. We have a smooth variety ${X}$ over ${\mathbb{C}}$. This means we can look at the ${\mathbb{C}}$-points and get an actual complex manifold. We defined the algebraic de Rham cohomology awhile ago to be ${H^i_{dR}(X/\mathbb{C}):=\mathbf{H}^i(\Omega_{X/\mathbb{C}}^\cdot)}$ the hypercohomology of the complex ${0\rightarrow \mathcal{O}_X\rightarrow \Omega^1\rightarrow \Omega^2\rightarrow \cdots}$. Since we’re in this nice case, this actually agrees perfectly with the standard singular cohomology on the manifold with coefficients in ${\mathbb{C}}$ (and by the de Rham theorem, the standard de Rham cohomology).

On a complex manifold we also have a nice working notion of Hodge theory. The Hodge numbers are $h^{ij}=\dim_{\mathbb{C}}H^j(X, \Omega^i)$ which we would normally derive through the Dolbeault resolution. We also have a Hodge decomposition ${H_{dR}^j(X/\mathbb{C})=\bigoplus_{p+q=j} H^q(X, \Omega^p)}$.

How do we see this using fancy language? Well, merely from the fact that de Rham cohomology is defined as the hypercohomology of a complex, we get the spectral sequence arising from hypercohomology. Without doing any work we can just check what this spectral sequence is and we find ${E_1^{ij}=H^j(X, \Omega^i)\Rightarrow H_{dR}^{i+j}(X/\mathbb{C})}$. This is because the ${E_1^{ij}}$ terms come from resolving each individual part of the complex which by definition just gives sheaf cohomology of the ${\Omega^i}$.

Of course, there was nothing special about ${X}$ being over ${\mathbb{C}}$, we could just as easily be over an arbitrary field and all of this still works. There is a great theorem that says that this spectral sequence degenerates at ${E_1}$ if ${X}$ is smooth over a characteristic ${0}$ field. There are several known proofs, some more analytic and some more algebraic. The coolest one is certainly by Deligne and Illusie.

They prove this preliminary result that if ${X}$ is smooth over a field ${k}$ of characteristic ${p}$ where ${p>\dim(X)}$ and ${X}$ has a lift to ${W_2(k)}$, then the Hodge-de Rham spectral sequence degenerates at ${E_1}$. Maybe we’ll talk about how this is done some other day, but if you know about the Cartier isomorphism then it is related to that. Using this side result that seems to be about as unrelated to the characteristic ${0}$ case as possible they then amazingly prove the characteristic ${0}$ case by reducing to positive characteristic and using a Lefschetz principle type argument.

Now despite the fact that H-dR degenerating being the norm in characteristic ${0}$, it turns out to be not so much the case in positive characteristic, so it is really shocking that to prove the characteristic ${0}$ case they moved themselves to this situation where it was likely not to degenerate. But we’ll get a better intuition later for why this wasn’t as risky as it sounds. Namely that since it came from characteristic ${0}$, there wasn’t going to be a problem lifting it back so the lifting to ${W_2(k)}$ was not a problem. It seems that the obstruction to being able to do this is almost exactly the failure of degeneracy. Recall that every K3 surface lifts to characteristic ${0}$, so (if you don’t know the proof of this) you’d expect the H-dR SS to degenerate at ${E_1}$. It might be a fun exercise for you to try to figure out why this is (very important hint: there are no global vector fields on a K3 so ${h^{1,0}=0}$).

Before ending this post it should be pointed out that all of this can be done in the relative setting as well. We actually originally defined de Rham cohomology purely in the relative setting without thinking about it over a field like we did today. Suppose ${\pi: X\rightarrow S}$ is a smooth scheme. The relative H-dR SS is given by ${E_1^{ij}=H^j(X, \Omega^i_{X/S})\Rightarrow \mathbf{R}^{i+j}\pi_*(\Omega_{X/S}^\cdot)=H_{dR}^{i+j}(X/S)}$.

We’ll continue with this next time, but I’ll just leave you with the thought that you can basically formulate for any class of schemes you want a large open problem by asking yourself whether or not the HdR SS degenerates at ${E_1}$ or at all.