# An invitation to p-adic Hodge theory, or: How i learned to stop worrying and love fontaine

This is the rough outline of a talk I recently gave at the Berkeley Student Algebraic Geometry Seminar on the progression of ideas that might lead one to define the Hodge-Tate decomposition.

# Motivation/Outline

In this post we will do a few things. First, we will discuss Hodge decomposition in its classical form: the Hodge decomposition theorem for compact Kahler manifold. We will focus on what such a decomposition does for–why it is important, both in a practical, and in a philosophical sense. We then move on to discuss how one might develop such a decomposition theorem over a local $p$-adic field, stating the Hodge-Tate decomposition theorem.

This is meant to be a very gentle (read ‘no gory details’) discussion. Everything here should be viewed only as a means to build intuition, and to get a broad idea of the topics involved. Consequently, a lot of the below is not fully explained, nor is it formal.

# Act I: classical Hodge decomposition

To understand where we are going, we must first understand where we have been. In our case, this means understanding a result about complex manifold theory, even though our ultimate goal is arithmetic geometry.

We start by assuming $M$ is a compact Kahler manifold. The definition of this object is somewhat incidental. Technically, it’s a compact complex manifold $M$ equipped with a Hermitian metric $h=g-i\omega$, such that, as a $2$-form, $\omega$ is closed. Intuitively, Kahler manifolds are complex manifolds $M$ which have a metric $h=g-i\omega$ such that the three resulting structures, the complex structure of $M$, the Riemannian manifold structure $(M,h)$, and the symplectic structure $(M,\omega)$ all work in concert.

For us, the important thing is the following: if $X/\mathbf{C}$ is a smooth, projective, integral variety, then $X^\text{an}$ is a Kahler manifold. Recall that $X^\text{an}$ can be defined as follows: choose an embedding $X\xrightarrow{\approx} V(f_1,\ldots,f_\ell)\subseteq\mathbf{P}^m_\mathbf{C}$. Take then the $\mathbf{C}$-points $X(\mathbf{C})$ to obtain the literal common zeros of $f_1,\ldots,f_\ell$ inside of $\mathbf{CP}^m$. From there, since the Jacobian of the equations $f_1,\ldots,f_\ell$ is invertible, we can use the holomorphic version of the implicit function theorem to define holomorphic charts on $X(\mathbf{C})$. The resulting complex manifold is what we denote $X^\text{an}$.

Remark: For those curious, the Kahler metric on $X^\text{an}$ can just be taken to be the pull-back of the Fubini-Study [pronounced ‘Foo-bean-ee’ and ‘Stew-dee’ ‘Shtoo-di’, see Georges’s comment below! :)] from $\mathbf{CP}^m$ from the embedding $X^\text{an}\hookrightarrow \mathbf{CP}^m$. It is a deep, very difficult, and astounding (but entirely unrelated to our current post) fact that all Kahler metrics on projective Kahler manifolds are of this form.

Now, in the land of Kahler manifolds, there is one theorem to rule them all:

Theorem(Hodge decomposition theorem): Let $M$ be a compact Kahler manifold. Then, for all $k$ we have a decomposition of $\mathbf{C}$-spaces

$\displaystyle H^k_\text{sing}(M,\mathbf{C})\cong\bigoplus_{i+j=k}H^i(M,\Omega^j_\text{hol})$

Moreover, $h_M^{i,j}=h_M^{j,i}$ as $\mathbf{C}$-spaces.

Here $\Omega^j_\text{hol}$ denotes the sheaf of holomorphic $j$-forms, and $H^k_\text{sing}$ denotes the singular cohomology. The numbers $h_M^{i,j}$ denote $\dim_\mathbf{C} H^i(M,\Omega^j_\text{hol})$ are called the Hodge numbers of $M$, and the result that $h_M^{i,j}=h_M^{j,i}$ is called Hodge symmetry.

Remark: For those that are familiar with complex geometry, and who are sticklers for notational tradition, the above is the equivalent of a living hell–I know since I am one such person. Namely, the usage of $i$ and $j$ as above is highly non-standard–traditionally, one uses $p$ and $q$ instead. Saying ‘$j$-form’ sounds dirty to me. But, alas, we are fighting two conflicting terminologies since below, we shall be discussing primes. And, as everyone knows, the only letters one can use for primes are $p$, $q$, and $\ell$. And, as everyone equally knows, $\ell$ is for primes not equal to $p$ (see below for what this strange statement means)! Thus, trapped between a rock and a hard place, I opted for the less painful ‘$j$-forms’. Please forgive me.

Even though the goal of this post is not to understand the complex geometry behind the Hodge decomposition (for this, I highly recommend Claire Voisin’s book Hodge Theory and Complex Algebraic Geometry I), I’d like to sketch the proof of this theorem. This is to culture a greater appreciation for what comes after. Namely, you’ll see in the proof below that the non-formal steps are entirely analytical (very hard analysis involving elliptic partial differential operators!). This makes the ability to transfer such results to the $p$-adic case even more astounding!

Remark: The word ‘sketch’ in what follows cannot be stressed enough. Namely, not only am I not filling in details, I am not even revealing the true character of the proof. Namely, in this proof sketch you won’t get any sense (at least not explicitly) as to why we require $M$ to be Kahler (or, more to the point, why the metric which shows up in the proof needs to be Kahler). But, as noted above, since our goal is not to really understand the proof, but to get an idea of what broad ideas go into it, this is ok.

Proof of Hodge decomposition(sketch): Since singular and complex de Rham cohomology agrees on $M$, it suffices to prove the following:

$\displaystyle H^k_\text{dR}(M/\mathbf{C})\cong\bigoplus_{i+j=k}H^i(M,\Omega^j_\text{hol})$

Now, here’s the hard part of the proof. For any Riemannian manifold, there is a theorem (also called Hodge decomposition!) which essentially says that every complex de Rham class contains a unique harmonic representative. So, what we certainly can say is the following:

$\displaystyle H^k_\text{dR}(M/\mathbf{C})=\bigoplus_{i+j=k}\mathcal{H}^{i,j}(M)$

where $\mathcal{H}^{i,j}(M)$ denotes the de Rham classes of harmonic $(i,j)$-forms. Now, one can prove that for our choice of metric, the de Rham classes of harmonic $(i,j)$-forms can be naturally identified with the Dolbeault cohomology $H^{i,j}(M)$. Or, in symbols, we have an isomorphism $\mathcal{H}^{i,j}(M)\cong H^{i,j}(M)$. But, by the Dolbeault isomorphisms (for a brief introduction to Dolbeault cohomology and the Dolbeault theorem, you can see my post here) we have that $H^{i,j}(M)\cong H^j(M,\Omega^i_\text{hol})$. So combining these results, we have:

$\displaystyle H^k_\text{dR}(M/\mathbf{C})=\bigoplus_{i+j=k}\mathcal{H}^{i,j}(M)\cong\bigoplus_{i+j=k} H^{i,j}(M)\cong\bigoplus_{i+j=k}H^i(M,\Omega^j_\text{hol})$

which proves the decomposition.

To prove Hodge symmetry, we note that implicit in the definition of $\mathcal{H}^{i,j}(M)$ we have a literal equality

$\mathcal{H}^{i,j}(M)=\overline{\mathcal{H}^{j,i}(M)}$

Considering the isomorphisms $\mathcal{H}^{i,j}(M)\cong H^j(M,\Omega^i_\text{hol})$ as discussed above, we obtain Hodge symmetry. $\blacksquare$

Now, while it may be obvious why this theorem is incredible, let me point out why it is so astounding for an algebraic geometer.

Let’s fix a smooth, projective, integral variety $X/\mathbf{C}$. Like we said above, the compact complex manifold $X^\text{an}$ is Kahler. Thus, we can apply Hodge decomposition to it to find

$\displaystyle H^k_\text{sing}(X^\text{an},\mathbf{C})\cong\bigoplus_{i+j=k}H^i(X^\text{an},\Omega^j_\text{hol})$

Ok, great, this is nothing new. But, we can use the following basic, but incredibly important fact (a special instance of the GAGA principle) to make things even more astounding:

$H^i(X,\Omega^j_{X/\mathbf{C}})\cong H^i(X^\text{an},\Omega^j_\text{hol})$

where $\Omega^j_{X/\mathbf{C}}$ are the algebraic differentials on the variety $X$. Thus, combining this with Hodge decomposition we find that

$\displaystyle H^k(X^\text{an},\mathbf{C})=\bigoplus_{i+j=k}H^i(X,\Omega^j_{X/\mathbf{C}})$

and, by Hodge symmetry, $h^{i,j}_X=h^{j,i}_X$ (where these algebraic Hodge numbers have obvious meaning).

Before we get into applications of this result, let’s point out it’s incredible ‘philosophical’ meaning. When doing algebraic geometry over $\mathbf{C}$ (or, more generally, varieties over any field), one is often times accosted by various cohomology theories: $\ell$-adic cohomology, singular cohomology of the analytification, de Rham cohomology of the analytification, algebraic de Rham cohomology/crystalline cohomology, sheaf cohomology, etc. One gets somewhat overwhelmed. How, if at all, do these cohomology theories, varying both in difficulty of definition and computation, relate?

Well, the Hodge decomposition can be read as precisely the unification of two of these (perhaps furthest from each other!) cohomologies (often called ‘comparison theorems’). Namely, it tells us that while the singular cohomology of the complex manifold $X^\text{an}$ should (by all rights!) should have nothing to do with sheaf cohomology on the scheme $X$ it does. Moreover, the relationship is one of utter simplicity.

Remark: In fact, there are many other ‘comparison theorems’ between the above cohomologies. One of them fits very nicely into the above discussion. Namely, one can define the algebraic de Rham cohomology $H^k_\text{dR}(X/\mathbf{C})$ of $X/\mathbf{C}$, as mentioned above. It’s defined to be the hypercohomology $\mathbb{H}^k(\Omega^\bullet_{X/\mathbf{C}})$ of the de Rham sequence. A powerful theorem of Grothendieck then states that for any smooth integral variety $X/\mathbf{C}$ one has that $H^k_\text{dR}(X/\mathbf{C})=H^k_\text{dR}(X^\text{an}/\mathbf{C})$ where, on the right, this is the usual complex de Rham cohomology of the complex manifold $X^\text{an}$.

There are comparisons between $\ell$-adic cohomology and singular cohomology, but not with $\mathbf{C}$-coefficients. Namely, another theorem of Grothendieck says that $H^i(X,\mathbf{Z}_\ell)\cong H^i_\text{sing}(X^\text{an},\mathbf{Z}_\ell)$, where the left is, as usual, $\varprojlim H^i(X_\text{et},\underline{\mathbf{Z}/\ell^n\mathbf{Z}})$ and the right hand side is singular cohomology with coefficients in the $\ell$-adic integers.

Moreover, not only does it relate two cohomology theories on the scheme $X$, but it relates two entire fields: algebraic topology and algebraic geometry. This go-between intermediating topological and algebraic information is astounding.

Let’s see an example of an explicit waysin which this connection between topology and algebraic geometry aids our intuition.

Example: Let’s see what the above tells us when $X$ is some smooth, projective, integral curve over $\mathbf{C}$. Well, the Hodge decomposition for $k=1$ becomes

$H^1(X^\text{an},\mathbf{C})=H^1(X,\mathcal{O}_X)\oplus H^0(X,\Omega_{X/\mathbf{C}})$

Now, by Serre duality (or Hodge symmetry!) $H^1(X,\mathcal{O}_X)\cong H^0(X,\Omega_{X/\mathbf{C}})$, and so the right hand side has dimension $2g(X)$ where $g(X):=\dim_\mathbf{C} H^0(X,\Omega_{X/\mathbf{C}})$ is the algebraic genus of $X$. Now, since $X^\text{an}$ is a Riemann surface, and so in particular an orientable surface, it has a genus $g_\text{top}(X^\text{an})$. Moreover, we know then how to compute $H^1(X^\text{an},\mathbf{C})$–it’s just $\mathbf{C}^{2g_\text{top}(X^\text{an})}$. So, comparing dimensions on both sides we get $2g_\text{top}(X^\text{an})=2g(X)$, and so $g_\text{top}(X^\text{an})=g(X)$.

This is great! The relationship between topology and algebraic geometry tells us that our definition of algebraic genus (as $\dim H^0(X,\Omega_X)$) agrees with our topological intuition, at least in the case where it makes sense to ask such a question (over $\mathbf{C}$).

Not only is this good at proving intuitive things, it’s also helpful in proving useful theorems for any characteristic $0$ field.

Example: Let $K$ be a characteristic $0$ field, and $X/K$ a smooth, integral, projective variety. Then, $X/K$ satisfies Hodge symmetry. This means that $h^{p,q}_X=h^{q,p}_X$. To see how one can use the Hodge decomposition theorem to prove this, see my post here.

# Act II: Hodge decomposition p-adic fields

Now that we have seen what the classical case held for us, we now begin the discussion of Hodge decomposition over $p$-adic fields.

The first issue we have to contend with is a simple one: what should Hodge decomposition even mean over, say, $\mathbf{Q}_p$? Namely, it’s not just the proof of the Hodge decomposition theorem that seems $\mathbf{C}$-centric (with all of its analysis!) but even the statement.

To start to get an idea of how one might concoct a version of Hodge decomposition over $\mathbf{Q}_p$, let’s look back at the classical case again. Namely, let’s now start with a projective, smooth, integral variety $X/\mathbf{Q}$. Then, as silly as it seems let’s write the Hodge decomposition for $X^\text{an}:=(X_\mathbf{C})^\text{an}$ as follows:

$\displaystyle H^k_\text{sing}(X^\text{an},\mathbf{R})\otimes_\mathbf{R}\mathbf{C}\cong\bigoplus_{i+j=k}H^i(X,\Omega^j_{X/\mathbf{Q}})\otimes_\mathbf{Q}\mathbf{C}$

But, to be quite zen here, what is $\mathbf{C}$ but $\overline{\mathbf{Q}_\infty}$, where $\infty$ is the infinite place of $\mathbf{Q}$? What happens if we try and replace $\infty$ with the other finite places of $\mathbf{Q}$. Namely, what happens if we try and replace $\mathbf{C}$ with $\overline{\mathbf{Q}_p}$?

So, if we’re going to try and replace $\mathbf{C}$ with $\overline{\mathbf{Q}_p}$, what are we going to do with $X^\text{an}$ and $H^k_\text{sing}$? One possible approach would be to try and develop some form of rigid analytic analytification, but let’s be more simple minded. Namely, what if we just replace with $X^\text{an}$ with $X_{\overline{\mathbf{Q}_p}}$, and $H^k_\text{sing}$ with some appropriate form of cohomology? Well, appropriate should mean ‘acts like singular cohomology’ which, in turn means ‘is a Weil cohomology theory‘. And, the only Weil cohomology theory I know with values in $\mathbf{Q}_p$ is $p$-adic cohomology.

So, making the formal replacements

$\mathbf{R}\leadsto\mathbf{Q}_p\qquad \mathbf{C}\leadsto\overline{\mathbf{Q}_p} \qquad X^\text{an}\leadsto X_{\overline{\mathbf{Q}_p}}$

the Hodge decomposition theorem becomes

$\displaystyle H^k(X_{\overline{\mathbf{Q}_p}},\mathbf{Q}_p)\otimes_{\mathbf{Q}_p}\overline{\mathbf{Q}_p}\cong\bigoplus_{i+j=k}H^i(X,\Omega_{X/\mathbf{Q}})\otimes_{\mathbf{Q}}\overline{\mathbf{Q}_p}$

Great! We now at least have a candidate for what Hodge decomposition might mean over $\mathbf{Q}_p$!

That said, now that we are in Arithmetic Land we expect more out of our isomorphisms. Namely, we don’t expect them to just be isomorphisms of vector spaces, we expect our isomorphisms to respect their god-given extra structure. In this case, this extra structure manifests as natural $G_{\mathbf{Q}_p}:=\text{Gal}(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)$ actions. Namely, $G_{\mathbf{Q}_p}$ acts on the left hand side of the above diagonally $g(x\otimes y)=g(x)\otimes g(y)$, and on the right hand side it acts only on the factor $\overline{\mathbf{Q}_p}$. So, if this isomorphism is to be at all up to snuff, it better be $G_{\mathbf{Q}_p}$-equivariant. That said, we have been raised so high only to be let fall–there is no such $G_{\mathbf{Q}_p}$-equivariant isomorphism.

To explain why there can be no such isomorphism of $G_{\mathbf{Q}_p}$-modules, let us first examine a simpler situation. Namely, let’s forget about tensoring with $\overline{\mathbf{Q}_p}$. Is it possible that there is a $G_{\mathbf{Q}_p}$-equivariant isomorphism of the following form:

$\displaystyle H^k(X_{\overline{\mathbf{Q}_p}},\mathbf{Q}_p)\cong\bigoplus_{i+j=k}H^i(X,\Omega^j_{X/\mathbf{Q}})\otimes_{\mathbf{Q}}\mathbf{Q}_p$

This is easy to answer: no. Why? Simply because the left-hand side contains too much information, and the right-hand side does not. Namely, the right-hand side is ‘constant’ as $G_{\mathbf{Q}_p}$-module–it’s not changing with respect to $X$. If this isomorphism were true, then the same would have the same constancy for the left-hand side. But, this cannot be the case. For example, if $X$ is an abelian variety over $\mathbf{Q}_p$, then the left-hand side, as $G_{\mathbf{Q}_p}$-module contains the information of whether or not $X$ has good reduction at $p$ (if it is a so-called ‘crystalline’ representation).

So, the obvious thing does not work–que sera sera. That said, it doesn’t mean we shouldn’t ask if it can be fixed. The above tells us that for a hope for some sort of $G_{\mathbf{Q}_p}$-module isomorphism, we have to ‘lose information’. It turns out that there is a natural way to do this due to, amongst others, Tate, Raynaud, and Faltings. It is the big result of this post:

Theorem(Hodge-Tate decomposition): Let $K$ be a finite extension of $\mathbf{Q}_p$, and $X/K$ a smooth, integral, proper variety. Then, for all $k\geqslant 0$, we have a decomposition of $G_K$-modules

$\displaystyle H^k(X_{\overline{K}},\mathbf{Q}_p)\otimes_{\mathbf{Q}_p}\mathbf{C}_p\cong\bigoplus_{i+j=k}H^{i}(X,\Omega^j_{X/\mathbf{Q}})\otimes_\mathbf{Q}\mathbf{C}_p(-j)$

Here $G_K$ is still acting diagonally on the left, and just on the $\mathbf{C}_p(-j)$ factor on the right.

There a few pieces of notation here that I would like to recall.

Recall first that $\mathbf{C}_p$ is $\widehat{\mathbf{Q}_p}$, and that it is algebraically closed. Now, a priori, there is no reason for $G_K\subseteq G_{\mathbf{Q}_p}$ to act on $\mathbf{C}_p$. But, note that $G_{\mathbf{Q}_p}$ acts on  $\mathbf{Q}_p$ by isometries. Thus, the elements extend by continuity (and the definition of completion!) to an action on $\mathbf{C}_p$. These extended maps are still field automorphisms since commuting with addition/multiplication can be checked on the dense subset $\overline{\mathbf{Q}_p}$. In fact, you can identify $G_{\mathbf{Q}_p}$ with the subset of isometric automorphisms in $\text{Gal}(\mathbf{C}_p/\mathbf{Q}_p$. Thus, in this way is how we get a $G_K$ action on $\mathbf{C}_p$.

Let us now remind ourselves of the notation $\mathbf{C}_p(r)$ for an integer $r$. Recall that we have the $p$-adic cyclotomic character $\chi_p:G_K\to \mathbf{Z}_p^\times$ defined as the following composition:

$G_K\twoheadrightarrow\text{Gal}(K(\mu_{p^\infty})/K)\cong\mathbf{Z}_p^\times$

We then denote by $K(1)$ the one-dimensional $G_{K}$-representation associated to $\chi_p$. Similarly, we denote by $K^(r)$, for $r\in\mathbb{N}$ the representation $K(1)^{\otimes r}$. Finally, we define $K(-r)$ for $r\in\mathbb{N}$ $K(r)^\vee$, with the obvious $G_K$-action. Then, for any $K$-linear $G_K$-module $V$ and any $r\in\mathbb{Z}$ we denote by $V(r)$ the $G_K$-module $V\otimes_K K(r)$. Note that as a $K$-space we can identify $V\otimes_K K(r)$ with $V$, in which case the action becomes $g\cdot x=\chi_p(g)^r gx$, where $gx$ is the $G_K$-action on $V$ we started with.

This theorem is astounding for three reasons:

The first is, it, as in the case of classical Hodge decomposition, unites two different cohomology theories. Again, on one-side, we have coherent cohomology on the scheme $X$, but now on the other we have $\ell$-adic cohomology which, a priori, is much more complicated. That said, the above tells us, in particular the following:

$\displaystyle \dim_{\mathbf{Q}_p} H^k(X_{\overline{K}},\mathbf{Q}_p)=\sum_{i+j=k}h^{i,j}$

This is amazing. The definition, and the computation of $H^k(X_{\overline{K}},\mathbf{Q}_p)$ are incredibly difficult at first glance. But, this tells us that, at least as a vector space, we can recover this object just by computing the Hodge numbers (which is just ‘easy’ coherent cohomology) of $X$!

The second thing it tells us is the amount of information that we lose by tensoring to $\mathbf{C}_p$. To emhasize this, we rewrite the Hodge-Tate decomposition in another form

$\displaystyle H^k(X,\mathbf{Q}_p)\otimes_{\mathbf{Q}_p}\mathbf{C}_p\cong\bigoplus_{i+j=k}\mathbf{C}_p(-j)^{h^{i,j}}$

Indeed, since the action on the $H^i(X,\Omega_{X/\mathbb{Q}}^j)\otimes_\mathbf{Q}\mathbf{C}_p(-j)$ is just on the $\mathbf{C}_p(-j)$ factor, the right-hand side is just a bunch of copies of the pieces $\mathbf{C}_p(-j)$. Now, this shows that while the $G_K$-module $H^i(X_{\overline{K}},\mathbf{Q}_p)$ contained so much information about $X$, tensoring up to $\mathbf{C}_p$ somehow equalizes the whole spectrum of possible $X$.

The third fact is not nearly as obvious, and is based off a somewhat obvious question. Like we said in the first implication, from the Hodge numbers, we can recover the dimension of $\mathbf{Q}_p$-space $H^i(X_{\overline{K}},\mathbf{Q}_p)$. Can we reverse this process? Namely, does the $G_K$-module $H^i(X_{\overline{K}},\mathbf{Q}_p)$ determine the Hodge number?

Somewhat surprisingly, the answer is yes. The result follows from the following theorem:

Theorem(Tate, Sen): Let $K/\mathbb{Q}_p$ be a finite extension of $\mathbf{Q}_p$. Then,

$H^0(G_K,\mathbf{C}_p(r))=\begin{cases}K & \mbox{if}\quad r=0\\ 0 & \mbox{if}\quad r\ne 0\end{cases}$

Remark: The proof of this theorem is not difficult, assuming the following technical result of Tate:

Theorem(Tate): Let $K/\mathbb{Q}_p$ be finite, and let $K_\infty$ be a totally ramified $\mathbb{Z}_p$-extension. Then, for all finite extensions $M/K_\infty$ $\text{Tr}_{M/K_\infty}(\mathcal{O}_M)\supseteq \mathfrak{m}_{K_\infty}$.

Once one has this theorem, which, in particular tells you that $\text{Tr}_{M/K_\infty}(\mathcal{O}_M)$ contains elements of arbitrarily small valuation, one can prove the theorem of Tate and Sen by well-approximating any fixed point in $\mathbf{C}_p$, say, by elements of $\overline{K}^G_K=K$.

This is an instance of what Faltings called ‘almost mathematics’. In particular, note that if $\mathcal{O}_L/\mathcal{O}_K$ is a finite etale extension of complete DVRs (i.e. integer rings of local fields), then

$\text{Tr}_{\mathcal{O}_L/\mathcal{O}_K}:\mathcal{O}_L\to\mathcal{O}_K$

is surjective. The above theorem of Tate says that while finite extensions of totally ramified $\mathbb{Z}_p$-extensions might not be etale, they almost have surjectivity of trace, and so are ‘almost etale’.

So, how is this result useful for recovering the Hodge numbers from the $G_K$-module $H^k(X_{\overline{K}},\mathbf{Q}_p)$? Simple! If we know $H^k(X_{\overline{K}},\mathbf{Q}_p)$ as a $G_K$-module, then we also know $H^k(X_{\overline{K}},\mathbf{Q}_p)\otimes_{\mathbf{Q}_p}\mathbf{C}_p(r)$ for each $r$. But, note then that by the theorem of Sen and Tate that for each $r=0,\ldots,k$

\begin{aligned}\left(H^k(X_{\overline{K}},\mathbf{Q}_p)\otimes_{\mathbf{Q}_p}\mathbf{C}_p(r)\right)^G_K &= \left(\bigoplus_{i+j=k}\mathbf{C}_p(r-j)^{h^{i,j}}\right)\\ &=\bigoplus_{i+j=k}\left(\mathbf{C}_p(r-j)^{G_K}\right)^{h^{i,j}}\\ &= K^{h^{i,j}}\end{aligned}

Thus, we recover $h^{i,j}$ as $\dim_K \left(H^{i+j}(X_{\overline{K}},\mathbf{Q}_p)\otimes_{\mathbf{Q}_p}\mathbf{C}_p(j)\right)^{G_K}$. Which, while complicated, still tells us that we can recover the Hodge numbers of $X$ purely from the $G_K$-modules $H^k(X_{\overline{K}},\mathbf{Q}_p)$.

1. Georges Elencwajg says:

Dear Alex, congratulations on your fine blog, certainly one of the finest on the Internet. Just a nitpick: Eduard Study’s name is pronounced [Shtoo-Di], with the stress on the first syllable. Who suggested otherwise ?

1. Hey Georges! Thanks very much for the kind words!

Also, thanks for the correction! As you may already be well-aware, there is a tendency in American English to drop the ‘sh’ sound from Germanic names–even though we often times retain other aspects of the correct German pronunciation. So, for example, Wittgenstein becomes ‘Vit-gen-styn’ instead of ‘Vit-gen-shtein’. I assume it’s the same for Study–I’ve only ever heard it pronounced ‘Stew-dee’, and never ‘Shtoo-di’.

Once again, this mispronounciation is not shocking to me. I’m in the department that Hartshorne has spent the majority of his career, and who still frequents the department, and yet we all still knowingly say ‘Heart-shorne’. You’d also be mortified by our pronunciation of Polish names, ‘Loss’s’ theorem being one of my favorites.

I will modify the above comment!

1. Georges Elencwajg says:

Dear Alex, rereading our exchange I realize that my comment was rather pedantic: your original suggestion was an excellent compromise, warning against the too anglicized pronunciation of Study as in the verb “to study”.
In general I find the question of pronunciation of foreign nouns a fascinating subject at the boundary of linguistics and “étiquette” [slipping in some French, just to be a bit self-referential :-)].
For example how should one pronounce “Einstein” in French?
The French pronunciation would be “instin”, where “in” is the nasal French phoneme in “vin”. But that sounds completely ridiculous, although I think the French pronounced it that way in the 1920’s.
A completely German pronunciation in a French conversation sounds pedantic to my ears and personally I go for something like “inesh’tine” (wit the “i” of “fine”) , but with the stress on the second syllable according to French usage rather than more correctly on the first.

But enough of these digressions: the main motivation for my comment was to congratulate you on your blog in general and this entry in particular.
They attest to a maturity and depth of understanding that many professional mathematicians of double your age could envy you !
I’m looking forward to many, many more entries…