Algebraic de Rham cohomology and the Degeneration of the Hodge spectral sequence

In this post we will discuss various properties of the algebraic de Rham cohomology of a variety $X$. We will focus, in particular, on various aspects of when the Hodge-to-de Rham spectral sequence on the first page, the most interesting case of which happens in positive characteristic.

Motivation

The definition of algebraic de Rham cohomology

As with most modern cohomology theories in algebraic geometry, algebraic de Rham cohomology intends to capture ‘topological’ aspects of a variety $X$. But, unlike $\ell$-adic cohomology which seeks to capture the topological via ‘topological like constructions’ (i.e. covering spaces), algebraic de Rham cohomology attempts to take a more indirect route. In particular, it looks to take advantage of the classic de Rham isomorphism

$H^i_\mathrm{sing}(M,\mathbb{R})\cong H^i_{\mathrm{dR}}(M)$

where $M$ is a smooth (real) manifold. This allows one to try and access the topological via the analytic.

Why is this a reasonable thing to want to do? I mean, after all, one of the main goal of algebraic geometry is to try and access the topological via the algebraic. But, thanks to Serre’s GAGA principle we know that the analytic and the algebraic are not far off- -certainly much closer than the algebraic and the topological. Thus, if we can reasonably get at a means of trying to understand cohomology via analytic-in-nature ‘forms’, then we stand to create a means of understanding the hidden ‘topology’ of varieties.

To motivate the actual definition of the algebraic de Rham cohomology of a variety, let us begin by recalling how the proof of de Rham’s theorem goes (one proof amongst many!). So, let us assume that $M$ is a smooth (real) manifold of dimension $n$. One then recalls that the de Rham complex

$0\to C^\infty\to\Omega^1\to \Omega^2\to\cdots\to\Omega^{2n}\to 0$

(where $C^\infty$ denotes the sheaf of smooth functions on $M$) forms a deleted resolution of the constant sheaf $\mathbb{R}$ on $M$ (this is essentially the content of the Poincare lemma). In other words,

$0\to \underline{\mathbb{R}}\to C^\infty\to \Omega^1\to\Omega^2\to\cdots\to\Omega^{2n}\to 0$

is exact. But, note that each $\Omega^i$ (with, definitionally, $\Omega^0:=C^\infty$) is acyclic. Indeed, they’re soft since they are $C^\infty$-modules, and $M$ possesses smooth bump functions, and soft sheaves are acyclic on $M$ since it’s paracompact Hausdorff.

Thus, by standard theory, we know that

$H^i(M,\underline{\mathbb{R}})=h^i\left(\Omega^\bullet(M)\right)$

where $h^i$ denotes the $i^\text{th}$ cohomology group (i.e. $\ker/\mathrm{im}$– -not to be confused with the hypercohomology of a complex which will come later) of the complex of global sections complex $\Omega^\bullet(M)$. But, this right hand side is precisely the definition of $H^i_{\mathrm{dR}}(M)$, and thus we have deduced an isomorphism

$H^i(M,\underline{\mathbb{R}})\cong H^i_{\mathrm{dR}}(M)$

Lastly, we note that since $M$ is locally contractible that there is an isomorphism

$H^i(M,\underline{\mathbb{R}})\cong H^i_{\mathrm{sing}}(M,\mathbb{R})$

and concatenating these two isomorphisms gives the desired isomorphism.

Now, while this is great, it is slightly misleading. How? Well, it gives one the wrong impression about how to go about creating an algebraic de Rham cohomology. Indeed, it certainly makes sense to consider the group $h^i(\Omega^\bullet_{X/k}(X))$, where $X/k$ is a variety (and the forms are algebraic forms), and this is mimicking precisely what happened in the above smooth case. Of course, as everyone knows, the smooth case is often a far cry from the rigid world of the algebraic- -and, indeed, this turns out to be the wrong definition.

To see the issue with this definition, let us take an intermediary step and try to work out this definition in the case of complex manifolds- -a case much closer (at least in the case of compact Kähler manifolds) to the algebraic setting than the smooth category. Namely, we might seek to define the ‘complex de Rham cohomology’ of a complex manifold $X$ as $h^i(\Omega^\bullet(X))$. But, what type of forms do we want here? We now have many options. Do we want real smooth forms? Do we want complex smooth forms? Or, possibly, do we want holomorphic forms?

Well, if we’re using this situation as a testing ground for the algebraic we are obliged to use holomorphic forms, since it is coherent $\mathcal{O}_X$-modules that behave ‘like algebraic ones’ (a la Serre’s GAGA principle). But, we immediately see that this poses an issue. The following isomorphism cannot hold:

$H^i_{\mathrm{sing}}(X,\mathbb{C})\cong h^i(\Omega^\bullet_{\mathrm{hol}}(X))$

Why? One very simple reason is that $\Omega^\bullet_\mathrm{hol}$ is simply just not ‘long enough’. Namely, $\Omega^i_{\mathrm{hol}}(X)=0$ for $i$ larger than the complex dimension of $X$. But, in general, $H^i_{\mathrm{sing}}(X,\mathbb{C})=0$ only for $i$ larger than the real dimension of $X$. As a very simple example of this, note that $\mathbb{CP}^1$ has complex dimension $1$, and so $h^2(\Omega^\bullet_{\mathrm{hol}}(\mathbb{CP}^1))=0$, but, of course, $H^2_\mathrm{sing}(\mathbb{CP}^1,\mathbb{C})=\mathbb{C}$.

What has gone wrong? Where does the argument which proves de Rham’s theorem in the smooth setting fail (quite spectacularly might I add) in the holomorphic? Well, a version of Poincare’s lemma still hold and shows that the sequence

$0\to \underline{\mathbb{C}}\to\mathcal{O}_X\to\Omega^1_{\mathrm{hol}}\to\cdots\to\Omega^m_\mathrm{hol}\to 0$

is exact (here $m$ is the complex dimension of $X$). But, this is where the argument breaks down. Before, the next step was to note that not only did the de Rham complex provide a resolution of the constant sheaf, but an acyclic one. The key being the existence of smooth cut-off functions on a smooth manifold. This is the point of miserable failure in the holomorphic version of the argument. There are no holomorphic bump functions, ergo the sheaves $\Omega^i$ are not soft, ergo there is no reason for them to be acyclic. There is nothing deep here. Our example of $\mathbb{CP}^1$ above gives a perfectly reasonable example of the de Rham complex not being a complex of acyclics. Indeed, $\Omega^1_{\mathrm{hol}}=\mathcal{O}(-2)$ which has non-vanishing first cohomology.

Hopefully this thoroughly convinces the reader that $h^i(\Omega^\bullet_{\mathrm{hol}}(X))$ is the wrong adaptation of de Rham cohomology to complex manifolds, and thus that $h^i(\Omega^\bullet_{X/k}(X))$ the wrong adaptation of varieties- -at least if we want to get a cohomology theory mimicking singular cohomology. But, all is not lost. While this naive definition doesn’t work, that doesn’t mean one should quite give up hope yet.

In particular, there is a nice piece of homological machinery which saves the day. The point is that we want some gadget which is able to compute the cohomology of a sheaf $\mathcal{F}$ from any resolution, not necessarily a resolution of acyclics. This is precisely the driving point to the definition of the hypercohomology of a complex. Namely, whatever the hypercohomology $\mathbb{H}^i(X,\mathcal{F}^\bullet)$ of a bounded-below complex $\mathcal{F}^\bullet$ is, the important point is that it has the following desirable properties:

1. Any quasi-isomorphism of complexes $u:\mathcal{F}^\bullet\to\mathcal{G}^\bullet$ induces an isomorphism $u_\ast:\mathbb{H}^i(X,\mathcal{F}^\bullet)\to\mathbb{H}^i(X,\mathcal{G}^\bullet)$.
2. If $\mathcal{F}$ denotes the complex concentrated in degree $0$, with the sheaf $\mathcal{F}$ there, then $\mathbb{H}^i(X,\mathcal{F})=H^i(X,\mathcal{F})$

Why is this useful to us? Well, let us suppose that we have some sheaf $\mathcal{F}_0$ and a resolution

$0\to \mathcal{F}_0\to\mathcal{F}^0\to\mathcal{F}^1\to\cdots$

Then, one can verify that this is equivalent to giving a quasi-isomorphism. $\mathcal{F}_0\to \mathcal{F}^\bullet$. So, using these two properties we deduce that

$H^i(X,\mathcal{F}_0)\cong \mathbb{H}^i(X,\mathcal{F}^\bullet)$

Thus, we can, in fact, compute the cohomology of a sheaf from any resoluton, if we are willing to use this tool of hypercohomology.

Cool. So, with this, we can unabashedly write the following holomorphic version of de Rham’s theorem:

$H^i_{\mathrm{sing}}(X,\mathbb{C})\cong H^i(X,\underline{\mathbb{C}})\cong \mathbb{H}^i(X,\Omega^\bullet_\mathrm{hol})$

(where the first isomorphism follows from the same theorem applied in the proof of the smooth de Rham’s theorem). This then gives us reasonable motivation to define the algebraic de Rham cohomology of a variety as the hypercohomology of the algebraic de Rham complex:

$H^i_{\mathrm{dR}}(X/k):=\mathbb{H}^i(X,\Omega^\bullet_{X/k})$

(Extended) Remark: A very important, and somewhat subtle point should be made. One might imagine that this definition leads to a wholly uninteresting result. In fact, I will now give a proof of this confusing fact. I claim that $H^i_{\mathrm{dR}}(X/k)=0$ for all $i>0$ assuming that, say, $X$ is irreducible. Why? Well, we still have the resolution

$0\to \underline{k}\to \mathcal{O}_X\to \Omega^1_{X/k}\to\cdots\to \Omega^n_{X/k}\to 0$

(where $n=\dim X$). So then, we may conclude from the machinery of hypercohomology that $H^i_{\mathrm{dR}}(X/k)=\mathbb{H}^r(X,\Omega^\bullet_{X/k})$ is equal to $H^i(X,\underline{k})$. But, since $X$ is irreducible the sheaf $\underline{k}$ is flasque, so $H^i(X,\underline{k})=0$.

Clearly something is wrong- -but what? The subtle inaccuracy we bought into was that the de Rham complex was still a resolution of $\underline{k}$. It’s not! Any sort of ‘Poincare type’ lemma will fail on $X$ (the Zariski site of $X$!). We no longer have neighborhoods isomorphic to polydiscs! This is something which, while perhaps obvious to most readers, confused me the first time I read about de Rham cohomology.

Of course, while this provides fairly convincing evidence that this is a good definition for the de Rham cohomology of an arbitrary variety, it doesn’t actually convince us that this is a worthwhile object of study- -why does this capture ‘topology like’ properties of $X$?

As a bare minimum we would hope/expect that if $X/\mathbb{C}$ is smooth projective that the following standard comparison theorem holds:

$H^i_{\mathrm{sing}}(X^\mathrm{an},\mathbb{C})\cong H^i_{\mathrm{dR}}(X/\mathbb{C})$

The motivation for this hope, as always is the following. If algebraic de Rham cohomology is to be capturing ‘topological information’ about a variety $X/k$, then in any setting where there is already canonical meaning of topological information (e.g. varieties over $\mathbb{C}$) our information should agree with the classical information- -this is the content of the comparison theorem. Luckily, this comparison theorem does hold true as a fairly easy consequence of the GAGA theorem (and the spectral sequence discussed below).

On the other end of the spectrum, our true dream for algebraic de Rham cohomology is that it is a so-called Weil cohomology theory– -a cohomology theory rich enough/topological enough in nature to prove the Weil conjectures. As it turns out, the world is a just and righteous place, and algebraic de Rham cohomology is indeed a Weil cohomology theory…sometimes. If $k$ is characteristic $0$ the planets align, and in positive characteristic they crash into one another- -in that case we have to deal with the wholly more complicated theory of crystalline cohomology.

That said, the behavior of de Rham cohomology in positive characteristic is an interesting topic in and of itself. In fact, a large portion of this post will be devoted to understanding when $H^i_{\mathrm{dR}}(X/k)$ behaves in expected ways when $k$ is of positive characteristic.

The degeneration of the Hodge spectral sequence: a motivation to care

To many people the title of this section might sound exceedingly scary. To quote Vakil on the subject of spectral sequences:

They have a reputation for being abstruse and difficult. It has been suggested that the name ‘spectral’ was given because, like spectres, spectral sequences are terrifying, evil, and dangerous. I have heard no one disagree with this interpretation, which is perhaps not surprising since I just made it up

All levity aside, spectral sequences are abstract, and their degeneration is a little dry if not motivated correctly. So, let us begin this discussion first with something slightly more palatable and, if you’re like me, more motivating. Later in the post we’ll see that the topic of this section is equivalent to the degeneration of the Hodge spectral sequence.

As (aspiring) mathematicians, we are doomed to endlessly compare. The most tantalizing chances for comparisons are between concepts made to measure the same thing. The measuring sticks apropos to this current post are various cohomology theories of varieties. It is an infuriating luxury of modern algebraic geometry to be practically over-encumbered with cohomology theories. This is compounded by the fact that their comparison is often times a deep and difficult task.

As an example of this difficulty, one could say, in some broad sense, that the comparison of the $p$-adic cohomology of a $p$-adic variety to various other cohomology theories comprises the geometric content of $p$-adic Hodge theory.

Today we are interested in a less lofty enterprise. In particular, we’d like to compare the de Rham cohomology of varieties (the one we just discussed) to another inostensibly ‘topological’, but patently ‘simpler’ cohomology theory. Let us define the Hodge cohomology of a variety $X/k$ as follows:

$\displaystyle H^r_{\mathrm{Hodge}}(X/k)=\bigoplus_{i+j=r}H^i(X,\Omega^j_{X/k})$

Our current goal is then ascertaining when the following equality holds true for a smooth projective variety $X/k$:

$\dim_k H^r_{\mathrm{dR}}(X/k)=\dim_k H^r_\mathrm{Hodge}(X/k)$

I avoid writing just ‘isomorphism’ to highlight that we are not looking for any sort of canonical isomorphism between the cohomology theories, but a non-canonical one (i.e. equality of dimensions).

Both sides of the expectancy divide should be discussed with regards to this equality. To be less cryptic, there are good reasons to believe/hope that this equality holds in some large portion of generality, and there are reasons to be very, very skeptical. We begin, with the latter.

There are several reasons why one would not expect such an equality to hold. Perhaps the most convincing comes by recasting the definition of the Hodge cohomology of a variety $X$. Somewhat surprisingly, one can also describe $H^r_\mathrm{Hodge}(X/k)$ as the hypercohomology of a complex of sheaves. Which complex? The complex $(\Omega^\bullet_{X/k},0)$– -the ‘de Rham complex’ but with the zero differential instead of the standard differential $d$. Thus, our desired equality says that the following holds true:

$\mathbb{H}^r\left(\left(\Omega^\bullet_{X/k},0\right)\right)\cong \mathbb{H}^r\left(\left(\Omega^\bullet_{X/k},d\right)\right)$

This is very surprising, at least to me. Usually when one wants to prove that two objects have the same hypercohomology one tries to relate the complexes themselves. That said, the complexes $\left(\Omega^\bullet_{X/k},0\right)$ and $\left(\Omega^\bullet_{X/k},d\right)$, while ostensibly similar (since they have the same terms) are incredibly different. As we shall see later phrasing it in this way, one expects in any reasonable case a strict inequality having $\mathrm{dim}_k H^r_\mathrm{Hodge}(X/k)$ the larger number.

So, on the flip side, why should any such equality between Hodge and de Rham cohomologies hold? Well, to be frank because it does. In particular, for smooth projective varieties over $\mathbb{C}$ this is essentially a consequence of the classic Hodge decomposition for compact Kahler manifolds.

Moreover, the argument against believing the equality also provides a weak reason to believe it’s not otherwordly to imagine they agree. Namely, as stated in that paragraph, phrasing Hodge cohomology in terms of the hypercohomology ‘zero de Rham complex’ provides us (by abstract nonsense involving spectral sequences) with an inequality- -some sort of comparison.

Spectral sequences and hypercohomology: a pense-bête

Before we jump into the interesting issue of when the Hodge-to-de-Rham spectral sequence degenerates, I’d like to begin by recalling the basic definitions and facts about hypercohomology and spectral sequences (especially as the latter pertains to the former). This will not only be pivotal in understanding the titular topic of this post, but will also make understanding more down-to-earth things, such as computations, wholly easier.

So as to make the material within this section as palatable as possible we focus on the spectral sequences associated with sheaves of abelian groups on a topological space $X$. Moreover, we shall focus on spectral sequences associated to filtrations.

Hypercohomology

The general case

We first begin by defining the piece of homological machinery that we trumpeted as our savior in the introduction. There are many ways to summarize what hypercohomology is, and what role it holds in our mathematical world.

1. Analogically: hypercohomology is to a complex of sheaves as normal cohomology is to a single sheaf.
2. In fancy words: the hypercohomology $\mathbb{H}^i(X,\mathcal{F}^\bullet)$ is just $h^i(Rf_\ast\mathcal{F}^\bullet)$ where $f:X\to\mathrm{Spec}(k)$ is the structure morphism, and $Rf_\ast:D^+(X)\to D^+(\mathrm{Spec}(k))$ is the induced map on the bounded-above derived category of abelian sheaves.
3. Practically: It allows us to compare cohomologies of things without having to be overly concerned with whether or not the constituents of the complexes are nice. The most reasonable example of this is that, as we discussed in the motivation, hypercohomology allows us to compute the cohomology of a sheaf from any resolution, whether the constituents are acyclic or not.

Sadly, we will not spend any serious amount of time discussing the details of hypercohomology. This is a largely formal (albeit very important!) topic, and is best relegated to a reading of Weibel’s Homological Algebra. That said, we provide here the bare-bones definition.

Let us start very abstractly with any two abelian categories $\mathcal{A}$ and $\mathcal{B}$– -we assume that $\mathcal{A}$ has enough injectives. Let’s assume moreover that $F:\mathcal{A}\to\mathcal{B}$ is a left exact functor.

We recall that the ‘bounded below derived category’ of $\mathcal{A}$, denoted $D^+(\mathcal{A})$ is essentially the bounded below complexes (i.e. complexes which are zero for sufficiently small indices) of objects of $\mathcal{A}$ where morphisms are up to homotopy, and we’ve inverted quasi-isomorphisms. Note that we have a natural inclusion

$\mathcal{A}\to D^+(\mathcal{A})$

taking any $A$ to the complex, also denoted $A$, concentrated in degree $0$.

Now, by standard homological algebra we have the derived functors $R^iF:\mathcal{A}\to\mathcal{B}$. Our goal is to create some functor $\mathbb{R}F:D^+(\mathcal{A})\to D^+(\mathcal{B})$ such that the following composition equals the functor $R^iF$:

$\mathcal{A}\to D^+(\mathcal{A})\xrightarrow{\mathbb{R}F}D^+(\mathcal{B})\xrightarrow{h^i}\mathcal{B}$

where $h^i$ denotes the standard cohomology of a complex. We will call this functor $\mathbb{R}F$ the hyperderived functor of $F$.

Describing how $\mathbb{R}F$ acts is actually surprisingly easy. Namely, let $\mathcal{F}^\bullet$ be any bounded below complex of $\mathcal{A}$. Choose a bounded complex $\mathcal{I}^\bullet$ of injectives of $\mathcal{A}$ (i.e. a complex whose terms are injectives) such that $\mathcal{F}^\bullet$ is quasi-isomorphic to $\mathcal{I}^\bullet$. Then, define the hyperderived functor $\mathbb{R}F(\mathcal{F}^\bullet)$ as follows:

$\mathbb{R}F(\mathcal{F}^\bullet)=F(\mathcal{I}^\bullet)$

where $F(\mathcal{I}^\bullet)$ is the complex given by applying $F$ to each term of $\mathcal{I}^\bullet$.

Of course, while this is very easy to state, there are some obvious questions. Does this descend to a functor on the bounded below derived category- – if I replace $\mathcal{F}^\bullet$ by a quasi-isomorphic complex, does the image under $\mathbb{R}F$ change similarly? Is it clear that, up to quasi-isomorphism, this definition is independent of $\mathcal{I}^\bullet$? Is it even clear that such an $\mathcal{I}^\bullet$ exists? All of these questions are totally valid and, as stated before, formal manipulations best left to checking on a lazy afternoon (or peeking in Weibel).

Note though that it’s clear from construction (modulo these just mentioned questions) that $\mathbb{R}F$ satisfies the two properties mentioned in the desired properties of hypercohomology:

1. By construction (since it only depends on the derived category), any two quasi-isomorphic complexes give rise to the same value under $\mathbb{R}F$.
2. For all $i$ and all $\mathcal{F}$ in $\mathcal{A}$, we have that $h^i(\mathbb{R}F(\mathcal{F}))=R^iF(\mathcal{F})$. Indeed, take any injective resolution $\mathcal{F}\to\mathcal{I}^\bullet$ of $\mathcal{F}$. Then, $\mathcal{F}$ is quasi-isomorphic to $\mathcal{I}^\bullet$. So, $\mathbb{R}F(\mathcal{F})$ equals $F(\mathcal{I}^\bullet)$. But, definitionally $h^i(F(\mathcal{I}^\bullet))=R^iF(\mathcal{F})$ as desired.

which tells us that it is a reasonably defined notion.

Also one can show with a little bit of elbow grease that, just like the in the classical case of derived functors, one can compute $\mathbb{R}F(\mathcal{F}^\bullet)$ as $F(\mathcal{K}^\bullet)$ for any bounded below complex $\mathcal{K}^\bullet$ of $F$-acyclic objects of $\mathcal{A}$.

The case of sheaves

Let us discuss what this notion looks like in the case of sheaves. Namely, let $X$ be some topological space, and consider the left exact functor

$\Gamma:\mathsf{Ab}(X)\to\mathsf{Ab}$

(the global sections functor). Then, by the above we get a map

$\mathbb{R}\Gamma:D^+(\mathsf{Ab}(X))\to D^+(\mathsf{Ab})$

let us define the $i^\text{th}$ hypercohomology of a bounded below complex of abelian sheaves $\mathcal{F}^\bullet$ on $X$ as follows:

$\mathbb{H}^i(X,\mathcal{F}^\bullet):=h^i(\mathbb{R}\Gamma(\mathcal{F}^\bullet))$

an element of $\mathsf{Ab}$. Of course, if we restrict to an abelian subcategory of $\mathsf{Ab}(X)$ with enough injectives whose image lies in an abelian subcategory of $\mathsf{Ab}$ then hypercohomology takes values in that subcategory. As an example, if we take the subcategory $\mathsf{Mod}_X(\underline{k})$ of $k$-vector space valued sheaves on $X$, then hypercohomology takes values in $\mathsf{Vect}_k$.

Now, for this specific case of hyperderived functors there is actually a very nice ‘canonical’ (i.e. functorial) bounded below complex of acyclics $\mathcal{I}^\bullet$ associated to any bounded below complex $\mathcal{F}^\bullet$. Namely, let us denote for any sheaf $\mathcal{F}$ and each $i\geqslant 0$ the $i^\text{th}$ Godemont sheaf $\mathcal{G}^i(\mathcal{F})$. Roughly, these sheaves are built as an iterative process of taking sections of the étalé space of $\mathcal{F}$ and its previous Godemont sheaves (really their cokernels).

The upshot is that these fit together to give a resolution

$0\to\mathcal{F}\to\mathcal{G}^0(\mathcal{F})\to\mathcal{G}^1(\mathcal{F})\to\cdots$

and, in fact, a flasque resolution. We call this the Godemont resolution of $\mathcal{F}$. The striking thing about the Godemont resolution is that it’s functorial. Any map of sheaves $\mathcal{F}_1\to\mathcal{F}_2$ gives rise to a map of complexes $\mathcal{G}^\bullet(\mathcal{F}_1)\to\mathcal{G}^\bullet(\mathcal{F}_2)$.

Now, with this we can describe a ‘canonical’ injective ‘resolution’ of any bounded below complex $\mathcal{F}^\bullet$. Namely, for each $i$ we can take the Godemmont resolution $\mathcal{G}^\bullet(\mathcal{F}^i)$ of $\mathcal{F}^i$. Now, by functoriality we get maps $\mathcal{G}^\bullet(\mathcal{F}^i)\to\mathcal{G}^\bullet(\mathcal{F}^{i+1})$. This gives rise to doubly bounded (i.e. essentially first quadrant) double complex $\mathcal{J}^{p,q}=\mathcal{G}^p(\mathcal{F}^q)$. One can then form the total complex $\mathcal{I}^\bullet$ of $\mathcal{J}^{\bullet,\bullet}$.

It turns out that the natural map $\mathcal{F}^\bullet\to\mathcal{I}^\bullet$ is a quasi-isomorphism, and moreover that $\mathcal{F}^\bullet$ is a complex of acyclics. Thus, we may conclude that

$\mathbb{H}^i(X,\mathcal{F}^\bullet)=h^i(\Gamma(\mathrm{Tot}^\bullet(\mathcal{G}^\bullet(\mathcal{F}^\bullet))))$

which gives a ‘canonical’ interepretation of the hypercohomology of sheaves. We will not use this fact in the rest of the post, but it gives one a more-down-to-earth, and hopefully comforting, way to think about hyperchomomology.

Remark: One can use this definition of hypercohomology to derive some of the spectral sequences below from the spectral sequence of a double complex- -if that sort of thing makes you happy.

Remark: This ‘Godemont resolution’ of $\mathcal{F}^\bullet$ is a special case of the general notion of an Cartan-Eilenberg resolution.

Spectral sequences

There is a reason that so few people have written well, if at all, about spectral sequences. They’re hard. They are honestly one of those things that one just gets used to. You see their utility in practice and start to appreciate them for what they are.

That said, I will make no attempt at trying to recall the general definition/theory of spectral sequences. Instead, I offer some references that have helped me in the past.

1. You could have invented spectral sequence – Timothy Chow
2. Spectral sequences: friend or foe? – Ravi Vakil (check the relevant part of his book on algebraic geometry which is likely to be an updated version of this article).
3. An Introduction to Spectral Sequences (2nd ed.) – John McCleary (the first chapter).

My recommendation to you would be to take them largely as a blackbox (as most people do). Your best bet is to read things using them and ‘fake it til you make it’. The more you see them used in the wild, the easier they will be to grok.  In fact, taking what’s written below for granted is a good start!

The spectral sequence associated to a filtered complex

So, let us discuss the particular general type of spectral sequence that will be of use to us. To set up, let’s assume, once again in the abstract, that we have $\mathcal{A},\mathcal{B}$, and $F$ as before. Moreover, let’s assume that we have $\mathcal{F}^\bullet$ a bounded below complex of objects of $\mathcal{A}$. But, the new thing we’ll assume is that we have a decreasing filtration:

$\mathrm{Fil}^\bullet \mathcal{F}^\bullet$

By this, we mean that for all $i\in\mathbb{Z}$ we have a subcomplex

$\mathrm{Fil}^i\mathcal{F}^\bullet\subseteq\mathcal{F}^\bullet$

such that for all $p$ and all $i we have that

$\mathrm{Fil}^i\mathcal{F}^p\supseteq \mathrm{Fil}^j\mathcal{F}^p$

Let’s assume that this complex is biregular, just meaning that for $i\gg 0$ we have that $\mathrm{Fil}^i\mathcal{F}^\bullet=\mathcal{F}^\bullet$ and for $i\ll 0$ we have that $\mathrm{Fil}^i\mathcal{F}^\bullet=0$.

Now, note that this filtration $\mathrm{Fil}^i\mathcal{F}^\bullet$ gives a filtration on

$\mathbb{R}^\ell F(\mathcal{F}^\bullet):=h^\ell(\mathbb{R}F(\mathcal{F}^\bullet))$

defined as followed:

$\mathrm{Fil}^i\mathbb{R}^\ell F(\mathcal{F}^\bullet):=\mathrm{im}\left(\mathbb{R}^\ell F(\mathrm{Fil}^i\mathcal{F}^\bullet)\to \mathbb{R}^\ell F(\mathcal{F}^\bullet)\right)$

Of course, since $\mathrm{Fil}^\bullet\mathcal{F}^\bullet$ is biregular, so is $\mathrm{Fil}^i\mathbb{R}^\ell F(\mathcal{F}^\bullet)$.

Now, there is a very natural question one might ask. To state it most succinctly we first need a definition. For any (decreasing) filtered object $\mathrm{Fil}^\bullet\mathcal{K}$ we define the associated graded object as follows:

$\displaystyle \mathrm{gr}_\bullet \mathrm{Fil}^\bullet\mathcal{K}:=\bigoplus_{i}\mathrm{gr}_i \mathrm{Fil}^\bullet\mathcal{K}$

where

$\mathrm{gr}_i\mathrm{Fil}^\bullet\mathcal{K}:=\mathrm{Fil}^i\mathcal{K}/\mathrm{Fil}^{i+1}\mathcal{K}$

is called the $i^\text{th}$ graded piece of $\mathrm{Fil}^\bullet\mathcal{K}$.

So, now we have two natural graded objects in our current situation. Namely, $\mathrm{gr}_\bullet \mathrm{Fil}^\bullet\mathcal{F}^\bullet$ and $\mathrm{gr}_\bullet \mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)$. The first is a bounded below complex of $\mathcal{A}$, and the second is an object of $\mathcal{B}$. Since the first is a complex, we can consider its $\ell^\text{th}$ hyperderived functor $\mathbb{R}^\ell F(\mathrm{gr}_\bullet\mathrm{Fil}^\bullet\mathcal{F}^\bullet)$ which is an object of $\mathcal{B}$. We then have the following obvious question:

$\mathbb{R}^\ell(\mathrm{gr}_\bullet \mathrm{Fil}^\bullet\mathcal{F}^\bullet)\overset{?}{\cong}\mathrm{gr}_\bullet \mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)\qquad (\ast)$

as graded objects (i.e. really we want isomorphisms of the individual $\mathrm{gr}_i$‘s). In other words, do $\mathbb{R}^\ell$ and $\mathrm{gr}_\bullet\mathrm{Fil}^\bullet$ commute?

This question is not only glaringly obvious, but one of great practical importance. Note that the object $\mathrm{gr}_\bullet \mathrm{Fil}^\bullet\mathbb{R}^\ell F(\mathcal{F}^\bullet)$ ‘knows’ the object $\mathbb{R}^\ell F(\mathcal{F}^\bullet)$ (the object we are interested in) ‘up to extensions’. Meaning that if we could reconstruct $\mathrm{Fil}^i \mathbb{R}^\ell F(\mathcal{F}^\bullet)$ (up to isomorphism) from $\mathrm{gr}_i \mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)$ and $\mathrm{Fil}^{i+1} \mathbb{R}^\ell F(\mathcal{F}^\bullet)$, then we could find $\mathbb{R}^\ell F(\mathcal{F}^\bullet)$ (remember that $\mathrm{Fil}^i\mathbb{R}^\ell F(\mathcal{F}^\bullet)=0$ for $i\ll 0$!).

As an example of this, let’s suppose that $\mathcal{B}$ is the category of finite dimensional $k$-vector spaces (this will be the case of interest to us). Then, this category has a trivial ‘extension problem’. In particular, to know $\mathrm{Fil}^i\mathbb{R}^\ell F(\mathcal{F}^\bullet)$ is to know its dimension. But, its dimension is just

$\dim_k \mathrm{gr}_i \mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)+\dim_k \mathrm{Fil}^{i+1}\mathbb{R}^\ell F(\mathcal{F}^\bullet)$

so we can reconstruct $\mathbb{R}^\ell F(\mathcal{F}^\bullet)$, up to isomorphism, from $\mathrm{gr}_\bullet\mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)$. More specifically we have the equality:

$\displaystyle \dim_k \mathbb{R}^\ell F(\mathcal{F}^\bullet)=\sum_i \mathrm{gr}_i\mathrm{Fil}^\bullet\mathbb{R}^\ell F(\mathcal{F}^\bullet)=\dim_k \mathrm{gr}_\bullet\mathrm{Fil}^\bullet\mathbb{R}^\ell F(\mathcal{F}^\bullet)$

which is obvious from inspection.

Thus, it’s clear that understanding $\mathrm{gr}_\bullet\mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)$ is of great importance. The practicality comes from the following simple observation. In every conceivable situation, computing $\mathbb{R}^\ell F(\mathrm{gr}_\bullet\mathrm{Fil}^\bullet\mathcal{F}^\bullet)$ is simpler than computing $\mathbb{R}^\ell F(\mathcal{F}^\bullet)$ (or morally equivalently, computing $\mathrm{gr}_\bullet\mathrm{Fil}^\bullet\mathbb{R}^\ell F(\mathcal{F}^\bullet)$). Why? Well, the hard operation is not taking graded objects, the hard operation in taking hyperderived functors. So, any way to make that operation easier makes the entire calculation much easier. But, the complex of sheaves $\mathrm{gr}_\bullet\mathrm{Fil}^\bullet \mathcal{F}^\bullet$ is practically always an ‘easier’ complex of sheaves than $\mathcal{F}^\bullet$– -and so, consequently, it’s hypercohomology is easier to compute.

Thus, we really do want to understand when $(\ast)$ holds or, more generally, what the relationship between the two sides is. As is usually the case when we have something simple and something hard (in a homological algebra setting) the relationship between them is described by  spectral sequence. The case is no different here. Indeed, we have the following spectral sequence known as the spectral sequence of a filtered complex:

$E^{ab}_1=\mathbb{R}^{a+b}F(\mathrm{gr}_a \mathrm{Fil}^\bullet\mathcal{F}^\bullet)\Rightarrow \mathbb{R}^{a+b} F(\mathcal{F}^\bullet)\qquad (\mathrm{F.C.S.S.})$

this gives us the relationship between the two sides of $(\ast)$.

For simplicitly, let’s assume, once again, that $F$ takes values in finite dimensional $k$-spaces. Then, we know from the general theory of spectral sequences that

$E^{ab}_\infty=\mathrm{gr}_a \mathrm{Fil}^\bullet\mathbb{R}^{a+b} F(\mathcal{F}^\bullet)$

and that $E^{ab}_\infty$ is a subquotient of $E^{ab}_1$. Thus, we have the following inequality

$\displaystyle \dim_k \mathbb{R}^\ell F(\mathcal{F}^\bullet) = \dim_k \mathrm{gr}_\bullet\mathrm{Fil}^\bullet \mathbb{R}^\ell F(\mathcal{F}^\bullet)=\sum_{a+b=\ell}E_\infty^{ab}\leqslant \sum_{a+b=\ell}E^{ab}_1$

with equality if and only if $\left(\text{F.C.S.S.}\right)$ degenerates on the first page.

So, we’ll now apply this super general setup to two particular filtrations that make sense on any abelian category.

The Hodge filtration/spectral sequence

Our first general filtration will be the Hodge filtration or, as its often endearingly known, the stupid filtration. And, indeed, it is a stupid filtration.

So, let’s begin with $\mathcal{F}^\bullet$ a bounded (on both sides) complex of $\mathcal{A}$. We define a filtration, called the Hodge filtration, on $\mathcal{F}^\bullet$ as follows:

$\mathrm{Fil}_h^i\mathcal{F}^j=\begin{cases} 0 & \mbox{if}\quad j

Since $\mathcal{F}^\bullet$ is bounded, it’s clear that this sequence is decreasing and biregular.

Note though that

$\mathrm{gr}_a\mathrm{Fil}_h^\bullet\mathcal{F}^\bullet=\mathcal{F}^a[-a]$

where $[-a]$ denotes shifting the complex $\mathcal{F}^a$, concentrated at $0$, $a$ units to the right. Thus, we see that the $\left(\text{F.C.S.S.}\right)$ becomes:

$E^{ab}_1=\mathbb{R}^{a+b}F(\mathcal{F}^a[-a])\Rightarrow \mathbb{R}^{a+b}F(\mathcal{F}^\bullet)$

But, as one can easily compute

$\mathbb{R}^{a+b}F(\mathcal{F}^a[-a])=\mathbb{R}^b F(\mathcal{F}^a)=R^bF(\mathcal{F}^a)$

thus we obtain the so-called Hodge spectral sequence

$E^{ab}_1=R^b F(\mathcal{F}^a)\Rightarrow \mathbb{R}^{a+b}F(\mathcal{F}^\bullet)\qquad\left(\text{H.S.S.}\right)$

This is a clear cut example of where the left hand side is much easier to compute than the right hand side.

Now if, once again, we assume that $F$ takes value in finite dimensional $k$-spaces then the degeneration on the first page of the $\left(\text{H.S.S.}\right)$ is equivalent to the following equality

$\displaystyle \dim_k \mathbb{R}^\ell F(\mathcal{F}^\bullet)=\sum_{a+b=\ell} R^b F(\mathcal{F}^a)$

The conjugate filtration/spectral sequence

We now discuss the other, less stupid, filtration which exists on a general bounded complex $\mathcal{F}^\bullet$. But, before we do, we need to define an ancillary (in this case- -it’s important in its own right!) filtration.

Let us define the truncation filtration as follows:

$\tau_{\leqslant i}\mathcal{F}^j=\begin{cases}\mathcal{F}^j & \mbox{if}\quad ji\end{cases}$

The reason for the strange inclusion of the kernel term, opposed to literal truncation, is so that $h^\ell(\tau_{\leqslant i}\mathcal{F}^\bullet)$ agrees with $h^\ell(\mathcal{F}^\bullet)$ for $\ell\leqslant i$.

Now, notice that $\tau_{\leqslant \bullet}\mathcal{F}^\bullet$ is an increasing filtration. To make it a decreasing filtration we simply reindex, and this is how we arrive at our desired filtration. Namely, we define the conjugate filtration as follows:

$\mathrm{Fil}_c^i\mathcal{F}^\bullet:= \tau_{\leqslant -i}\mathcal{F}^\bullet$

which is now a decreasing and biregular filtration.

But, we can compute that

$\mathrm{gr}_a\mathrm{Fil}^\bullet_c\mathcal{F}^\bullet=\left(h^{-a}\mathcal{F}^\bullet\right)[a]$

So, it’s the $(-a)^\text{th}$ cohomology object concentrated in degree $-a$. This gives us a weird spectral sequence if we just literally look at this instance of the $\left(\text{F.C.S.S.}\right)$. Instead, we do a reindexing of it as follows:

$E^{ab}_r\leadsto E^{-b,2b+a}_{r+1}$

We then obtain a second page spectral sequence as follows, called the conjugate spectral sequence:

$E^{ab}_2=R^a F(h^{b}\mathcal{F}^\bullet)\Rightarrow \mathbb{R}^{a+b}F(\mathcal{F}^\bullet)\qquad\left(\text{C.S.S.}\right)$

The analytic case

Our goal now is to see how these two spectral sequences look in some particular cases. We begin in the analytic world. Namely, let’s assume that $X$ is a connected complex manifold of dimension $n$.

We will take in this case the following choices of the general objects from the previous discussion. Namely, we’ll take the category $\mathcal{A}$ to be $\mathsf{Ab}(X)$, the category $\mathcal{B}$ to be $\mathsf{Ab}$, the functor $F$ to be $\Gamma$, and the complex $\mathcal{F}^\bullet$ to the holomorphic de Rham complex $\Omega^\bullet_\mathrm{hol}$.

The general discussion from before then leaves us with the following two spectral sequences:

$E^{ab}_1=H^b(X,\Omega^a_\mathrm{hol})\Rightarrow H^{a+b}_\mathrm{dR}(X/\mathbb{C})\qquad\left(\text{A.H.S.S.}\right)$

$E^{ab}_2=H^a(X,\mathcal{H}^b(\Omega^\bullet_\mathrm{hol}))\Rightarrow H^{a+b}_\mathrm{dR}(X/\mathbb{C})\qquad\left(\text{A.C.C.S}\right)$

where, as usual, we define $H^{\ell}_\mathrm{dR}(X/\mathbb{C})$ to be $\mathbb{H}^\ell(X,\Omega^\bullet_\mathrm{hol})$, and $\mathcal{H}^b$ is just the more customary notation for $h^b$ when dealing with sheaves. Also, the ‘A’ stands for ‘analytic’.

We then may begin to analyze these two sequences individually. We start with the $\left(\text{A.C.S.S.}\right)$ because it’s easier.

Analytic conjugate spectral sequence

Indeed, as mentioned before, one still has the holmorphic Poincaré lemma which says that $0\to\underline{\mathbb{C}}\to\Omega^\bullet_\mathrm{hol})$ is a resolution. So, in particular,

$\mathcal{H}^b\left(\Omega^\bullet_\mathrm{hol}\right)=\begin{cases}\underline{\mathbb{C}} & \mbox{if}\quad b=0\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Thus, we see that the $\left(\text{A.C.S.S.}\right)$ degenerates on the first page and thus we obtain the equality

$\displaystyle \dim_\mathbb{C} H^\ell_\mathrm{dR}(X/\mathbb{C})=\sum_{a+b=\ell}\dim_\mathbb{C} H^a(X,\mathcal{H}^b(\Omega^\bullet_\mathrm{hol}))=\dim_\mathbb{C} H^\ell(X,\underline{\mathbb{C}})=\dim_\mathbb{C} H^\ell_\mathrm{sing}(X,\mathbb{C})$

thus rediscovering the previously mentioned fact that the holomorphic de Rham cohomology computes singular cohomology.

Analytic Hodge spectral sequence

The case of the $\left(\text{A.H.S.S.}\right)$ is a much more nuanced affair. In particular, the Hodge spectral sequence does not always degenerate on the first page. For example, if $X$ is affine then $H^b(X,\Omega^a_\mathrm{hol})$ is, usually, infinite dimensional whereas $H^\ell_\mathrm{dR}(X/\mathbb{C})$, being singular cohomology, is finite dimensional.

So, to say something reasonable, let’s restrict ourselves to those $X$ which are compact so at least $H^b(X,\Omega^a_\mathrm{hol})$ is finite dimensional. Then, we know that the $\left(\text{A.H.S.S.}\right)$ degenerates on the first page if and only if

$\displaystyle \dim_\mathbb{C} H^\ell_\mathrm{dR}(X/\mathbb{C})=\sum_{a+b=\ell}H^b(X,\Omega^a_\mathrm{hol})=\dim_\mathbb{C} H^\ell_\mathrm{Hodge}(X/\mathbb{C})$

so that degeneration is the same thing as an isomorphism between de Rham cohomology and Hodge cohomology (as was hinted to in the previous motivation).

Now, while this is not something true in general it is something which holds true if $X$ is also assumed Kähler. If Kähler isn’t a phrase you very much like, just remember that $X$ which are both Kähler and algebraic are precisely the projective algebraic varieties.

More specifically, if $X$ is compact Kähler, then we have the following theorem:

Theorem 1(Hodge decomposition): Let $X$ be a compact Kähler manifold. Then, for all $\ell\geqslant 0$ we have a canonical isomorphism

$H^\ell_\mathrm{sing}(X,\mathbb{C})\cong H^\ell_\mathrm{Hodge}(X/\mathbb{C})$

But, by the above we know that holomorphic de Rham cohomology agrees with singular cohomology. Thus, in particular, we see that if $X$ is compact Kähler then necessarily the $\left(\text{A.H.S.S.}\right)$ degenerates on the first page.

In fact, if one is willing to work a little harder one can show that for any proper algebraic analytic variety (i.e. an analytic variety which is the analytification of a proper variety over $\mathbb{C}$) that Hodge decomposition, and thus degeneration on the first page, holds. This is essentially an application of Chow’s lemma.

Remark: One thing to remark is that another part of Hodge theory for compact Kähler manifolds is Hodge symmetry. Now, while this is clearly related to the Hodge decomposition (i.e. the degeneration of the $\left(\text{A.H.S.S.}\right)$ on the first page) they are not implied by one another. Hodge symmetry is an extra little inequivalent quirk of the characteristic 0 theory. In particular, there are cases in positive characteristic where the $\left(\mathrm{H.S.S.}\right)$ degenerates on the first page but Hodge symmetry does not hold.

Remark: A remark should be made about why the conjugate spectral sequence is called, well, the conjugate spectral sequence. We’ll try and answer this in some way below, but we want to say here why it is not called the conjugate spectral sequence. Note that once we fix an isomorphism $H^\ell_\mathrm{dR}(X/\mathbb{C})$ with $H^\ell_\mathrm{sing}(X,\mathbb{C})$ we get two filtrations on $H^\ell_\mathrm{sing}(X,\mathbb{C})$. Now, since we have a canonical isomorphism

$H^\ell_\mathrm{sing}(X,\mathbb{C})\cong H^\ell_\mathrm{sing}(X,\mathbb{Z})\otimes_\mathbb{Z}\mathbb{C}$

we have a notion of conjugation on $H^\ell_\mathrm{sing}(X,\mathbb{C})$. It is not true that the two filtrations on this spaces defined by the $\left(\text{A.C.S.S.}\right)$ and the $\left(\text{A.H.S.S}\right)$ are conjugate. This is easily seen by considering what the graded pieces are.

The algebraic case

Just as in the analytic case, for a variety $X/k$, we obtain two spectral sequences:

$E^{ab}_1=H^b(X,\Omega^a_{X/k})\Rightarrow H^{a+b}_\mathrm{dR}(X/k)\qquad\left(\text{Ag.H.S.S}\right)$

$E^{ab}_2=H^a\left(X,\mathcal{H}^b\left(\Omega^\bullet_{X/K}\right)\right)\Rightarrow H^{a+b}_\mathrm{dR}(X/k)\qquad\left(\text{Ag.C.S.S}\right)$

where, as usual, we define $H^\ell_\mathrm{dR}(X/k)$ to be the hypercohomology of the de Rham complex.

Now, these two cases act entirely differently depending on whether $k$ has positive characteristic or not. For this reason, we will split our analysis into these two cases. Moreover, we shall restrict our attention to the case when $X/k$ is smooth proper.

The characteristic zero case

The algebraic Hodge spectral sequence

It will be helpful to break this case down into two steps. First, we shall discuss what happens when $k=\mathbb{C}$. We shall then use this to analyze the situation when $k$ is an arbitrary characteristic zero field

The complex case

So, let’s assume that $X/\mathbb{C}$ is a smooth proper variety. Our goal, as one might expect, is to reduce this to the analytic case by passing to $X^\mathrm{an}$. Of course, some care must be taken since the objects we’re dealing with aren’t exactly under the purview of GAGA! Indeed, the differentials in the de Rham complex aren’t even $\mathcal{O}_X$-linear!

That said, we have the following amazing theorem of Grothendieck:

Theorem 2(de Rham-Singular Comparision Theorem): Let $X/\mathbb{C}$ be a smooth variety (no assumption on properness). Then, for all $\ell\geqslant 0$ we have the following isomorphism:

$H^\ell_\mathrm{sing}(X^\mathrm{an},\mathbb{C})\cong H^\ell_\mathrm{dR}(X^\mathrm{an}/\mathbb{C})\cong H^\ell_\mathrm{dR}(X/\mathbb{C})$

The first isomorphism in this chain was already proven, and is fairly easy. It is the second isomorphism which is the amazing one. In particular, it tells us that for an algebraic variety, we can compute its singular cohomology entirely algebraically!

Now, the proof of this result is a moderately complicated dévissage to the affine case, where one may proceed in several ways. I will not give the full proof here- -I recommend reading Grothendieck’s original paper.

That said, the proof in the case when we assume that $X/\mathbb{C}$ is proper (the case of interest to us) is considerably easier:

Proof(proper case): As previously mentioned we can’t literally apply GAGA to our problem. But, we note that the morphisms on the first page of $\left(\text{Ag.H.S.S.}\right)$ are $\mathcal{O}_X$-linear maps between coherent $\mathcal{O}_X$ modules. Thus, GAGA gives us an isomorphism between this and the first page of the $\left(\text{A.H.S.S.}\right)$ for $X^\text{an}$. The result then follows by the convergence of the two spectral sequences to the two objects we are looking to compare. $\blacksquare$

Remark: Another aspect of Grothendieck’s theorem which I think deserves discussion is the case when $X$ is affine. Note then that Grothendieck’s comparison theorem tells us that $H^\ell_\mathrm{sing}(X^\mathrm{an},\mathbb{C})=0$ for $\ell>\dim X$ (the complex dimension!). Indeed, by the comparison theorem we have that $H^\ell_\mathrm{sing}(X^\mathrm{an},\mathbb{C})$ is the hypercohomology of the de Rham complex $\left(\Omega^\bullet_{X/\mathbb{C}},d\right)$. But, since $X$ is affine, this is actually a complex of acyclics, and so the general theory of hyperchomology tells us that

$H^\ell_\mathrm{sing}(X^\mathrm{an},\mathbb{C})=H^\ell_\mathrm{dR}(X/\mathbb{C})=h^\ell\left(\Omega^\bullet_{X/k}(X)\right)$

and since $\Omega^i_{X/\mathbb{C}}=0$ for $i>\dim X$, this implies the result.

This is somewhat surprising- -affine complex varieties have no cohomology past their complex dimension, even though, in general, we only expect vanishing beyond their real dimension. That said, there is a way of seeing this result without ever having to enter the algebraic world. Namely, since $X$ is affine $X^\mathrm{an}$ is a so-called Stein manifold. But, then by Cartan’s Theorem B the complex $\left(\Omega^\bullet_{\mathrm{hol}},d\right)$ on $X^\mathrm{an}$ is a complex of acyclics. Thus, we could play the same game for the vanishing of higher singular cohomology using the holomorphic de Rham cohomology of $X^\mathrm{an}$ in place of the algebraic de Rham cohomology of $X$ (and Grothendieck’s comparison theorem).

Finally, I would feel remiss to not take an opportunity to point out one of my favorite mathematical facts. The above shows that, at least cohomologically, an affine complex variety acts like a real $n$-dimensional manifold. A spectacular application of Morse theory shows that this is not a coincidence. In particular, it shows that if $X$ is an affine complex variety, then $X$ has the homotopy type of an $n$-dimensional real manifold. See Theorem 1.2 of this excellent set of notes.

So, now, we can prove the degeneration of the $\left(\text{Ag.H.S.S.}\right)$ in the case when $k=\mathbb{C}$. Indeed, we need only show that if $X/\mathbb{C}$ is smooth proper then

$H^\ell_\mathrm{dR}(X/\mathbb{C})\cong H^\ell_\mathrm{Hodge}(X/\mathbb{C})$

But, by Grothendieck’s comparison theorem the left hand side is $H^\ell_\mathrm{dR}(X^\mathrm{an}/\mathbb{C})$. And, by a trivial application of GAGA, the right hand side is $H^\ell_\mathrm{Hodge}(X^\mathrm{an}/\mathbb{C})$. But, as we have previously discussed

$H^\ell_\mathrm{dR}(X^\mathrm{an}/\mathbb{C})\cong H^\ell_\mathrm{Hodge}(X^\mathrm{an}/\mathbb{C})$

from where the result follows.

The general case

We shall now reduce to the complex case by a pretty standard ‘pulling down/pushing up argument’ (cf. this post of mine which uses the same technique).

Namely, let $\mathrm{char}(k)=0$, and let $X/k$ be smooth proper. Now, by adjoining to $\mathbb{Q}$ the coefficients of the polynomials defining $X$, we can get a smooth proper model $X'/k'$ for $\mathbb{Q}\subseteq k'\subseteq k$ some finitely generated field extension of $\mathbb{Q}$. But, then, there exists an embedding $k'\hookrightarrow \mathbb{C}$. Consider then $X''=(X')_\mathbb{C}$.

Now, by the complex case we know that

$H^\ell_\mathrm{dR}(X''/\mathbb{C})\cong H^\ell_{\mathrm{Hodge}}(X''/\mathbb{C})$

But, since differentials respect pull-backs we easily verify that

$H^\ell_\mathrm{dR}(X/k)=H^\ell_\mathrm{dR}(X'/k')=H^\ell_\mathrm{dR}(X''/\mathbb{C})$

and

$H^\ell_\mathrm{Hodge}(X/k)=H^\ell_\mathrm{Hodge}(X'/k')=H^\ell_\mathrm{Hodge}(X''/\mathbb{C})$

from where the desired equality follows.

The conjugate spectral sequence

Somewhat interestingly, nothing too reasonable can be said about the $\left(\text{Ag.C.S.S.}\right)$ in characteristic zero without too much work. I’ve heard claims that some work of various people (including, perhaps, Ogus) says something of interest, but I don’t have any concrete sources.

Of course, the issue is that since the differentials in the de Rham complex are not $\mathcal{O}_X$-linear, the sheaves $\mathcal{H}^b(\Omega^\bullet_{X/k})$ needn’t even be $\mathcal{O}_X$-modules! So, when $k=\mathbb{C}$ we have no chance of trying to apply GAGA. Also, as remarked before, since these are sheaves in the Zariski topology, we also have no analogue of the Poincaré lemma, and so $\mathcal{H}^b(\Omega^\bullet_{X/k})$ has no right to be zero- -and, in most cases, isn’t.

The positive characteristic case

This is the case of the most interest since we cannot run to the warm embrace of the Hodge theory of compact Kähler manifolds. So, some new technique must be employed.

This technique comes from the following. What the positive characteristic case lacks in the safe haven of the complex analytic, it makes up for in the existence of a very special map. This will be the subject of our first subsection.

The relative Frobenius map

In this section we will recall what will be the key tool in the study of the spectral sequences $\left(\text{Ag.H.S.S.}\right)$ and $\left(\text{Ag.C.S.S.}\right)$ in characteristic $p$.

So, let’s begin by recalling the definition of the absolute Frobenius map. Namely, let $S$ be a scheme with $p\mathcal{O}_S=0$ (i.e. $S$ lies over $\mathrm{Spec}(\mathbb{F}_p)$). Then, $S$ possesses the so-called absolute Frobenius map $\mathrm{Fr}_S:S\to S$. It’s defined to be the identity on the underlying space of $S$, and on an affine open $\mathrm{Spec}(A)$ in $S$ it corresponds to the ring map $A\to A:a\mapsto a^p$.

So, now let $S$ be as in the last paragraph, and let $f:X\to S$ be an $S$-scheme. We can then form the Frobenius twist of $X$ relative to $S$, denoted by $X^{(p)}$ as fitting into the following Cartesian diagram:

$\begin{matrix}X^{(p)} & \to & X\\ \downarrow & & \downarrow^{f}\\ S & \xrightarrow{\mathrm{Fr}_S} & S\end{matrix}$

Concretely, if $X$ is cut out by the equations $\displaystyle f_i=\sum a_{i,j}x^j$ relative to $S$, then $X^{(p)}$ is cut out by the equations $\displaystyle f_i^{(p)}:=\sum a_{i,j}^p x^j$.

Note that the map $X^{(p)}\to X$ is not an $S$-map. To make this seem of more importance, note that if $S$ is a perfect scheme (meaning that $\mathrm{Fr}_S$ is an isomorphism) then the morphism $X^{(p)}\to X$ is an isomorphism of schemes, but not of schemes over $S$!

That said, we can create a natural map of $S$-schemes $X\to X^{(p)}$. By appealing to the cartesian nature of $X^{(p)}$, to do this, we need only give maps $X\to X$ and $X\to S$ making the obvious square commute. We take the maps $\mathrm{Fr}_X:X\to X$ and $f:X\to S$. Thus obtaining the relative Frobenius map $F=F_{X/S}:X\to X^{(p)}$. This is a map relative to $S$, but not necessarily an isomorphism.

Concretely, $F$ takes a tuple $(z_i)$ satisfying the equations $f_i$ and sends it to $(z_i^p)$ which will satisfy the equations of $f_i^{(p)}$.

So, an important property of $F$ for us will be the following:

Theorem 3: Suppose that $S$ is perfect and $X/S$ is smooth of relative dimension $n$. Then, $F$ is a locally free map of rank $p^n$.

Proof: We verify it first in the case that $f$ is étale. This means that we want to show that, in fact, $F$ is an isomorphism. To see this, note that the structure map $X^{(p)}\to S$ is also étale, being the pullback of étale. But, the composition

$X\to X^{(p)}\to S$

is also étale, since it’s just $f$. So, by the cancellation theorem, $X\to X^{(p)}$ is étale. But, we claim that it’s also radiciel. Indeed, a morphism $g:Z_1\to Z_2$ is radiciel if and only if for all $z\in Z_1$ with $z_2=g(z_1)$ we have that the extension $k(z_2)\hookrightarrow k(z_1)$ is totally inseparable. But, in our case the field extensions are obviously totally inseparable. Thus, we see that $X\to X^{(p)}$ is an open embedding. But, note that the composition $X\to X^{(p)}\to X$ is $\mathrm{Fr}_X$ which is surjective, so $X\to X^{(p)}$ is surjective, and thus an isomorphism.

Secondly, we verify the result when $X=\mathbb{A}^n_S$. This is trivial and can be done by hand.

Finally, locally the map $X\to S$ looks like $U\xrightarrow{\mathrm{\acute{e}tale}}\mathbb{A}^n_V\to V$ from where the two base cases imply the result. $\blacksquare$

The last thing we want to note is the following fact which will be crucial later:

Theorem 4: Let $S=\mathrm{Spec}(k)$ be a perfect field, and let $X/S$ be smooth projective. Then,

$H^\ell_\mathrm{Hodge}(X/k)\cong H^\ell_\mathrm{Hodge}(X^{(p)}/k)$

Proof: This is just a simple verification. Indeed, for each $i\geqslant 0$ note that the morphism $g:X^{(p)}\to X$ gives us an exact sequence

$0\to g^\ast\Omega^1_{X/k}\to \Omega^1_{X^{(p)}/k}\to \Omega^1_{X^{(p)}/X}\to 0$

where we have exactness on the left since $X^{(p)}\to X$ is an isomorphism. Moreover, for the same reason we have that $\Omega^1_{X^{(p)}/X}=0$. Thus, we have an isomorphism of $\mathcal{O}_{X^{(p)}}$-modules

$g^\ast\Omega^1_{X/k}\cong \Omega^1_{X^{(p)}/k}$

Then, by taking wedge powers we obtain isomorphisms:

$g^\ast\Omega^i_{X/k}\cong \Omega^i_{X^{(p)}/k}$

But, note that since $g$ is an isomorphism we deduce that

$H^j(X^{(p)},\Omega^i_{X^{(p)}/k})\cong H^j(X^{(p)},g^\ast\Omega^i_{X/k})\cong H^j(X,\Omega^i_{X/k})$

which gives the desired result. $\blacksquare$

The Hodge spectral sequence

For the entirety of this and the next section, we assume that $k$ is a perfect field.

Before we slip too far into the usual routine, let us consider the following example due to William Lang, and proved in his thesis. Namely, consider the surface

$S:y^2z=x^3-tz^2\subseteq\mathbb{P}^3_{\mathbb{F}_3}$

Then,

$b_{\mathrm{dR}}^1:=\dim_{\mathbb{F}_3}H^1_\mathrm{dR}(S/\mathbb{F}_3)=3$

$h^{1,0}=h^{0,1}=2$

(here $h^{i,j}=\dim_k H^i(S,\Omega^j_{S/\mathbb{F}_3})$). Thus,

$\displaystyle b_{\mathrm{dR}}^1<\sum_{a+b=1}h^{a,b}$

which implies that the $\left(\text{Ag.H.S.S.}\right)$ does not degenerate for $S$ on the first page.

Now that we have disabused ourselves of the notion that the analysis of the $\left(\text{Ag.H.S.S.}\right)$ will be business as usual (it doesn’t even always degenerate!) we can look for hypotheses on $X/k$ which will guarantee that it acts as expected.

The key, as mentioned before, will be to leverage the relative Frobenius map $F$ for $X/k$. So, we begin by noting some miracles surrounding the relationship between $F$ and de Rham cohomology.

First, we note that the maps

$\mathrm{Fr}_X^\ast\Omega^1_{X/k}\to\Omega^1_{X/k}$

and

$F^\ast\Omega^1_{X^{(p)}/k}\to\Omega^1_{X/k}$

are the zero maps. Both of these, in coordinates, are just flavors of the statement in that in characteristic $p$ the relationship $d(x^p)=0$ holds.

Slightly more astonishing is the following fact. Now, just as always, the differentials on the de Rham complex $(\Omega^\bullet_{X/k},d)$ are not $\mathcal{O}_X$-linear. That said, if we consider the push-forward of this complex to $X^{(p)}$, $(F_\ast\Omega^\bullet_{X/k},d)$ (it’s customary to write $d$ instead of $F_\ast d$) then the differentials of this complex are $\mathcal{O}_{X^{(p)}}$-linear! This comes from a calculation approximating the following:

$d(X^ps)=pX^{p-1}ds+X^p ds=X^p ds$

which one can make precise in coordinates.

While these two observations are nice, the first truly amazing property involving Frobenius and differentials if the following theorem of Cartier:

Theorem 5: Let $X/k$ be a smooth variety. Then, there is a unique isomorphism of graded $\mathcal{O}_{X^{(p)}}$-algebras

$\displaystyle \gamma:\bigoplus_i \Omega^i_{X^{(p)}/k}\to \bigoplus_i \mathcal{H}^i\left((F_\ast\Omega^\bullet_{X/k}),d)\right)$

such that the map $\gamma^1:\Omega^1{X^{(p)}/k}\to\mathcal{H}^1\left((F_\ast\Omega^\bullet_{X/k},d)\right)$ is given in coordinates by sending $d\left(\mathrm{Fr}_X^\ast(s)\right)=s^{p-1}ds$.

Note that the product structure on both sides is coming from the wedge product.

This theorem is actually surprisingly easy to prove. One verifies it by hand for $\mathbb{A}^1$, deduces the result for $\mathbb{A}^n$ by Künneth formula, and then show that it is insensitive to changing by an étale map.

Now, while this result is sort of surprising in its own right it begs an even more tantalizing question. Namely, note that as a corollary of Cartier’s theorem one obtains that the complexes $(\Omega^\bullet_{X^{(p)}/k},0)$ and $(F_\ast\Omega^\bullet_{X/k},d)$ have isomorphic cohomology. Thus, a natural question is whether the following assertion holds true:

$\left(\mathrm{QI}\right)$: there exists a quasi-isomorphism $(\Omega^\bullet_{X^{(p)}/k},0)\to (F_\ast\Omega^\bullet_{X/k},d)$ inducing the Cartier isomorphisms on cohomology.

Now, up until this point, our discussion seemed relatively far afield from the discussion of the degeneration of the $\left(\text{Ag.H.S.S.}\right)$. But, in fact we have the following claim:

$\left(\mathrm{QI}\right)\implies\text{degeneration on first page of Ag.H.S.S}$

In fact, this is relatively easy to deduce using our knowledge of the relative Frobenius map obtained in the last section. Assume that $\left(\mathrm{QI}\right)$ holds. Then:

\begin{aligned}H^\ell_\mathrm{Hodge}(X/k) &\overset{(1)}{\cong} H^\ell_{\mathrm{Hodge}}(X^{(p)}/k)\\ &\cong \mathbb{H}^\ell\left((\Omega^\bullet_{X^{(p)}/k},0)\right)\\ &\overset{(2)}{\cong} \mathbb{H}^\ell\left((F_\ast\Omega^\bullet_{X/k},d)\right)\\ &\overset{(3)}{\cong} \mathbb{H}^\ell\left((\Omega^\bullet_{X/k},d)\right)\\ &\cong H^\ell_{\mathrm{dR}}(X/k)\end{aligned}

where $(1)$ is our previous computation that Frobenius twists don’t change Hodge cohomology, $(2)$ follows by $\left(\mathrm{QI}\right)$, and $(3)$ follows from the fact that $F$ is a finite map.

Thus, it seems of immediate interest to determine when $\left(\mathrm{QI}\right)$ holds. There is only two reasonable cases (that I know of) in which $\left(\mathrm{QI}\right)$ holds:

Theorem 6: Let $X/k$ be a smooth proper variety. Then, $\left(\mathrm{QI}\right)$ holds in either of the following two situations:

1. $X/k$ has a (smooth) lift to $W_2(k)$ and $\dim X<\mathrm{char}(k)$.
2. $X/k$ has a (smooth) lift to $W_2(k)$ and $F:X\to X^{(p)}$ lifts to a map

$F':X'\to (X')^{(p)}:= X'\times_{W_2(k)}W_2(k)$

where $W_2(k)$ acts on itself by the natural lift of Frobenius.

Here $W_2(k)$ denotes the length two Witt vectors of $k$. Explicitly, $W_2(k)$ is the ring $k^2$ with the addition

$(a_1,a_2)+(b_1,b_2)=(a_1+b_1,a_2+b_2+p^{-1}\left(a_1^{p-1}+b_1^{p-1}-(a_1+b_1)^p\right))$

and with multiplication

$(a_1,a_2)(b_1,b_2)=(a_1b_1,b_1^p a_2+b_2a_1^p)$

For example, if $k=\mathbb{F}_p$, then $W_2(k)=\mathbb{Z}/p^2\mathbb{Z}$.

Now, both of these results are proved in this seminal paper of Deligne-Illusie. In fact, the sufficiency of 2. for $\left(\mathrm{QI}\right)$ is proved as a lemma in proving the following result which implies sufficiency of 1.:

Theorem 7(Deligne-Illusie): Let $k$ be a perfect field and $X/k$ a smooth scheme. Suppose that $X$ admits a smooth lift $X'/W_2(k)$. Then, there is a quasi-isomorphism

$\tau_{

inducing the Cartier isomorphism on cohomology.

Here, in the language of the truncation filtration $\tau_{\leqslant i}$ defined above, $\tau_{ is just $\tau_{\leqslant p-1}$.

In particular, note that if $\dim X then $\tau_{ does nothing (since the chains are already length $p$), and thus we obtain sufficiency for 1. in this case.

I will say nothing of the proofs of these results, deferring them to the extremely well written paper of Deligne-Illusie. The only thing I’d like to remark on is the obvious question: why $? What is causing a necessity of this inequality?

The key step in the proof is the following. Suppose that we could find a morphism:

$\varphi^1:\Omega^1_{X^{(p)}/k}[-1]\to F_\ast\Omega^\bullet_{X/k}$

which induces the Cartier isomorphism $\gamma^1$ on cohomology. Then, I claim that we can build the necessary maps

$\varphi^i:\Omega^i_{X^{(p)}/k}[-i]\to F_\ast\Omega^\bullet_{X/k}$

for $i\geqslant 1$ for free. Indeed, one can define it as the composition:

$\Omega^i_{X^{(p)}/k}\xrightarrow{a}\left(\Omega^1_{X^{(p)}/k}\right)^{\otimes i}\xrightarrow{\left(\varphi^1\right)^{\otimes i}}\left(F_\ast\Omega^\bullet_{X/k}\right)^{\otimes i}\xrightarrow{\text{mult.}}F_\ast\Omega^\bullet_{X/k}$

here $\mathrm{mult.}$ is the multiplication, and $a$ is the anti-symmetrization map:

$\displaystyle a(\omega_1\wedge\cdots\wedge\omega_i)=\frac{1}{i!}\sum_\sigma (-1)^{\mathrm{sgn}(\sigma)}\omega_{\sigma(1)}\otimes\cdots\otimes\omega_{\sigma(i)}$

and this map only makes sense for $i. This is the place where we require the inequality involving $p$!

As an immediate application, we obtain the following:

Theorem 8: Let $k$ be a perfect field, and $X/k$ a smooth proper curve. Then, the $\left(\mathrm{Ag.H.S.S.}\right)$ degenerates on the first page.

Proof: Since $\dim X=1<\mathrm{char}(p)$ it suffices to know, by Deligne-Illusie, that $X$ lifts to a curve $X'/W_2(k)$. But, in general, the obstruction space for smooth lifts of a variety $Z/k$ is $H^2(Z,T_Z)$, where $T_Z$ is the tangent bundle. In particular. since $H^2(X,T_X)=0$ (since $X$ is a curve!) we have that $X$ automatically lifts to $W_2(k)$ (and, in fact, $W(k)$). $\blacksquare$

Note, incidentally, that Hodge symmetry also holds for curves- -it’s just Serre duality!

The conjugate spectral sequence

We will now discuss what happens in the case of the conjugate spectral sequence in positive characteristic. It will turn out that there is not much more than can be said.

The key calculation is as follows. Let’s suppose, as per usual, that $k$ is perfect and $X/k$ is smooth proper. Then, we have, as discussed in the previous section, the Cartier isomorphism

$\Omega^i_{X^{(p)}/k}\xrightarrow{\approx}\mathcal{H}^i\left(F_\ast\Omega^\bullet_{X/k}\right)$

Thus, of course, we obtain an isomorphism of cohomology groups

$H^{n-i}(X^{(p)},\Omega^i_{X^{(p)}/k})\xrightarrow{\approx} H^{n-i}(X^{(p)},\mathcal{H}^i(F_\ast\Omega^\bullet_{X/k}))$

But, since $F$ is finite we have that

$H^{n-i}(X^{(p)},\mathcal{H}^i(F_\ast\Omega^\bullet_{X/k}))\cong H^{n-1}(X^{(p)},F_\ast\mathcal{H}^i(\Omega^\bullet_{X/k}))\cong H^{n-i}(X,\mathcal{H}^i(\Omega^\bullet_{X/k}))$

Finally, since, as mentioned before

$H^i(X^{(p)},\Omega^j_{X^{(p)}/k})\cong H^i(X,\Omega^j_{X/k})$

we may conclude that

$E^{i,n-i}_1=H^{n-i}(X,\Omega^i_{X/k})\cong H^{n-i}(X,\mathcal{H}^i(\Omega^\bullet_{X/k}))=E^{n-i,i}_2$

thus, we have obtained an ‘isomorphism’ between the first page of the $\left(\mathrm{Ag.H.S.S.}\right)$ and the $\left(\mathrm{Ag.C.S.S.}\right)$. I put isomorphism in quotes since it doesn’t take $E^{p,q}_1$ to $E^{p,q}_2$ but instead switches $p$ and $q$– –$E^{p,q}_1\leadsto E^{q,p}_2$. This is one reason why it makes sense to call the $\left(\mathrm{Ag.C.S.S.}\right)$ the conjugate spectral sequence- -it ‘conjugates’ the indices.

All terminological philosophizing aside, we may conclude from the above that

Theorem 9: Let $k$ be a perfect field, and $X/k$ a smooth proper variety. Then, the $\left(\mathrm{Ag.H.S.S.}\right)$ degenerates on the first page for $X$ if and only if the $\left(\mathrm{Ag.C.S.S.}\right)$ degenerates on the second page for $X$.

Indeed, degeneration is a condition on the sums of the dimensions of the first (resp. second) page of the spectral sequence. Since the terms of these pages are permuted as proven above, the result is obvious.

Applications of Deligne-Illusie

I’d like to end this post by just pointing out two amazing facts which follow from Deligne-Illusie.

Theorem 10(Hodge decomposition): Let $k$ be a characteristic $0$ field, and $X/k$ a smooth proper variety. Then, the $\left(\mathrm{Ag.H.S.S.}\right)$ degenerates on the first page for $X$.

Theorem 11:(Kodaira-Nakano-Akizuki vanishing): Let $k$ be a characteristic $0$ field, and $X/k$ a smooth proper variety. Then, for any ample line bundle $\mathscr{L}$, and $i+j>\dim X$ the following holds:

$H^j(X,\mathscr{L}\otimes\Omega^i_{X/k})=0$

Now, neither of these results are ‘new’. In fact, the analogous statements for projective complex varieties are standard results proved in any complex geometry course (we’ve used Hodge decomposition for such manifolds in this post!). The results of the proof are entirely analytic. For example, Hodge decomposition holds in a much more general form- -it’s really a statement about harmonic forms representing de Rham classes on a Riemannian manifold. One then deduces both results by reducing to the analytic case (we did this for Hodge decomposition above).

No, the amazing thing is not that these theorems are true. It’s that one can give a purely algebraic proof of these results using Deligne-Illusie. While the details are a little cumbersome to check (cf. Deligne-Illusie’s article), the main idea is pretty simple. It suffices to prove it in the case when $k=\mathbb{C}$. But, by standard techniques one can find a smooth proper model of $X$ over some finitely generated $\mathbb{Z}$-algebra. One then shows that there exists a prime $p$ larger than $\dim X$ for which the reduction of this model modulo $p$ has a lift to $W_2$ (of the relevant field). Thus, $\left(\mathrm{QI}\right)$ holds for the reduction of the model. One then shows that $\left(\mathrm{QI}\right)$ implies the two above theorems or, rather, their positive characteristic analogies (this was done for Hodge decomposition above). Finally, one checks that the numerology works out and that this actually gives the desired results.

So, if one loves Hodge decomposition (for smooth proper complex algebraic varieties), but doesn’t feel like studying elliptic operators, one can just follow their algebraic nose to the results of Deligne-Illusie!