# Completed Cohomology

Let $F$ be a number field, and let $G_{F,S}$ be the Galois group over $F$ of the maximal extension of $F$ unramified outside a finite set of primes $S$. It should follow from the Langlands correspondence that $n$-dimensional continuous (we shall only be talking about continuous Galois representations in this post, so we omit the word “continuous” from here on) representations $\rho:G_{F,S}\to \mathrm{GL_{n}}(\overline{\mathbb{Q}}_{p})$ should correspond to certain automorphic representations $\pi$ of $\mathrm{GL}_{n}$ unramified outside $S$ (see also Automorphic Forms).

The Fontaine-Mazur-Langlands conjecture further states that such Galois representations $\rho$ that are irreducible and “geometric” (i.e. its restrictions to the primes above $p$ satisfy some conditions related to p-adic Hodge theory, see also p-adic Hodge Theory: An Overview) should match up with “algebraic” (we shall explain this shortly) cuspidal $\pi$. Furthermore this conjecture expects that certain “Hodge numbers” associated to the Galois representation $\rho$ via p-adic Hodge theory should match up to “Hodge numbers” defined for the automorphic representation $\pi$ via its “infinitesimal character” at the archimedean primes (note that they are defined differently, since they are associated to different kinds of representations; they only share the same name because they are expected to coincide).

Generally, whether $\rho$ is “geometric” or not, its Hodge numbers going to be $p$-adic numbers, and whether $\pi$ is “algebraic” or not, its Hodge numbers are complex numbers. However, if $\rho$ is geometric, then its Hodge numbers are integers, and if $\pi$ is algebraic, its Hodge numbers are also integers (in fact the definition of “algebraic” here just means that its Hodge numbers are integers), and this allows us to match them up.

To see things in a little more detail, let us consider the case of a $1$-dimensional representation $\rho:G_{F,S}\to \overline{\mathbb{Q}}_{p}$. We have seen in Galois Representations that an example of this is given by the p-adic cyclotomic character which we can also view as follows. Let $S=\lbrace p,\infty\rbrace$. Let $G_{F,S}^{\mathrm{ab}}$ be the abelianization of $G_{F,S}$. It follows from the Kronecker-Weber theorem that $G_{F,S}^{\mathrm{ab}}$ is isomorphic to $\mathbb{Z}_{p}^{\times}$, and it is precisely the p-adic cyclotomic character that gives this isomorphism. Since $\mathbb{Z}_{p}^{\times}$ embeds into $\overline{\mathbb{Q}}_{p}^{\times}$, which is also $\mathrm{GL}_{1}(\overline{\mathbb{Q}_{p}})$, we have our $1$-dimensional Galois representation. We can also take a power of the p-adic cyclotomic character to get another $1$-dimensional Galois representation.

But the p-adic cyclotomic character and its powers are not the only $1$-dimensional Galois representations. For instance, we have a map from $\mathbb{Z}_{p}^{\times}\to \mathbb{Q}_{p}^{\times}$ given by reducing $\mathbb{Z}_{p}$ mod $p^{r}$ and then composing it with the map $\chi$ that sends this element of $(\mathbb{Z}/p^{r})^{\times}$ to the corresponding $p^{r}$-th root of unity in $\overline{\mathbb{Q}}_{p}^{\times}$. This is a finite-order character. We also have another map from $\mathbb{Z}_{p}^{\times}\to \overline{\mathbb{Q}}_{p}^{\times}$ which sends $x$ to $x^{s}$, for some $s$ in $\overline{\mathbb{Q}}_{p}$ such that $\vert s\vert<\frac{p}{p-1}$. If we compose the p-adic cyclotomic character with either of these maps, we get another $1$-dimensional Galois representation. It turns out the Hodge number of the latter representation is given by $s$.

The $1$-dimensional Galois representations form a rigid analytic space (see also Rigid Analytic Spaces), and their Hodge numbers form p-adic analytic functions on this space. The geometric representations are the ones that are from a power of the p-adic cyclotomic character composed with a finite-order character, and these form a countable dense subset of this rigid analytic space.

Some form of this phenomena happens more generally for higher dimensional Galois representations – they form a rigid analytic space and the geometric ones are a subset of these.

It is convenient that our Galois representations form a rigid analytic space, and suppose we want to do something similar for our automorphic representations. The problem is that the automorphic representations aren’t really “p-adic”, as we may see from the fact that their Hodge numbers are complex instead of p-adic. This is the problem that p-adically completed cohomology, also simply known as completed cohomology, aims to solve.

Let us look at how we want to find automorphic representations in cohomology. Let $G_{\infty}=\mathrm{GL_{n}}(F\otimes_{\mathbb{Q}}\mathbb{R})$. If $F$ has $r_{1}$ real embeddings and $r_{2}$ complex embeddings, then $G_{\infty}$ will be isomorphic to $\mathrm{GL}_{n}(\mathbb{R})^{r_{1}}\times\mathrm{GL}_{n}(\mathbb{C})^{r_{2}}$. Let $K_{\infty}^{\circ}$ be a maximal connected compact subgroup of $G_{\infty}$. With $r_{1}$ and $r_{2}$ as earlier, $K_{\infty}^{\circ}$ will be isomorphic to $\mathrm{SO}(n)^{r_{1}}\times \mathrm{U}(n)^{r_{2}}$.

Let $X$ be the quotient $G_{\infty}/\mathbb{R}_{>0}^{\times}K_{\infty}^{\circ}$. This is an example of a symmetric space – for example, if $F=\mathbb{Q}$ and $n=2$, $X$ is going to be $\mathbb{C}\setminus \mathbb{R}$.

The space $X$ has an action of $G_{\infty}$, and its subgroup $\mathrm{GL}_{n}(\mathcal{O}_{F})$. Letting $N\geq 1$, we may therefore take the quotient

$\displaystyle Y(N)=\mathrm{GL}_{n}(\mathcal{O}_{F})\backslash (X\times \mathrm{GL}_{n}(\mathcal{O}_{F}/N\mathcal{O}_{F}))$

For example, if $F=\mathbb{Q}$ and $n=2$, then $Y(N)$ consists of copies of the (uncompactified) modular curve of level $N$ (the number of copies is equal to the number of primes less than $N$).

It is this space $Y(N)$ whose cohomology we are interested in. For instance $H^{i}(Y(N),\mathbb{C})$ is related to automorphic forms by a theorem of Jens Franke. However, it is complex, and not the p-adically varying one that we want. There is an isomorphism between $\mathbb{C}$ and $\overline{\mathbb{Q}}_{p}$, but the important part of this cohomology comes from the cohomology with $\mathbb{Q}$ coefficients, which is unchanged when we do this isomorphism, and therefore does not really add anything.

This is now where we introduce completed cohomology. Let us require that $N$ and $p$ be mutually prime. We define the completed cohomology $\widetilde{H}^{i}$ as follows:

$\displaystyle \widetilde{H}^{i}:=\varprojlim_{s\geq 1}\varinjlim_{r\geq 0}H^{i}(Y(Np^{r}),\mathbb{Z}/p^{s}\mathbb{Z})$

The order of the limits here is important (we will see shortly what happens when they are interchanged). By first taking the direct limit we are essentially considering the union of $H^{i}(Y(Np^{r}),\mathbb{Z}/p^{s}\mathbb{Z})$ for all $r$ with $\mathbb{Z}/p^{s}\mathbb{Z}$ coefficients. This is a very big abelian group that might not even be finitely generated. Then the inverse limit means we are taking the $p$-adic completion – having this as the last step guarantees that the result is something that is p-adically complete (hence the name p-adically completed cohomology). So the completed cohomology $\widetilde{H}^{i}$ is a p-adically complete module over $\mathbb{Z}_{p}$, which again may not be finitely generated. Taking the tensor product of $\widetilde{H}^{i}$ with $\mathbb{Q}_{p}$ over $\mathbb{Z}_{p}$ gives us a vector space $\widetilde{H}_{\mathbb{Q}_{p}}^{i}$ which moreover is a Banach space.

Let us consider now what happens if the order of the limits were interchanged. Let us denote the result by $H^{i}$:

$\displaystyle H^{i}:=\varinjlim_{r\geq 0}\varprojlim_{s\geq 1}H^{i}(Y(Np^{r}),\mathbb{Z}/p^{s}\mathbb{Z})$

By taking the inverse limit first we are simply considering $H^{i}(Y(Np^{r}),\mathbb{Z}_p$, and taking the direct limit means we are taking the union of $H^{i}(Y(Np^{r}),\mathbb{Z}_p)$ for all $r$. If we take the tensor product of $H^{i}$ with $\mathbb{Q}_{p}$ over $\mathbb{Z}_{p}$, then what we get is $H_{\mathbb{Q}_{p}}^{i}$, the union of $H^{i}(Y(Np^{r}),\mathbb{Q}_p$ for all $r$. Being the cohomology with characteristic zero coefficients, this may once again be related to the automorphic forms, as earlier.

Therefore, $H_{\mathbb{Q}_{p}}^{i}$, via the Fontaine-Mazur-Langlands conjecture, should be related to the geometric Galois representations. Now it happens that we can actually embed $H_{\mathbb{Q}_{p}}^{i}$ into the completed cohomology $\widetilde{H}_{\mathbb{Q}_{p}}^{i}$, because there is a map from $H^{i}(Y(Np^{r}),\mathbb{Z}_p)$ to $H^{i}(Y(Np^{r}),\mathbb{Z}/p^{s}\mathbb{Z})$, and then we can take the direct limit over $r$ followed by the inverse limit over $r$ and then tensor over $\mathbb{Q}_{p}$ as previously.

This embedding of $H_{\mathbb{Q}_{p}}^{i}$ into $\widetilde{H}_{\mathbb{Q}_{p}}^{i}$ should now bring to mind the picture with the geometric Galois representations which sit inside the rigid analytic space of Galois representations which may not necessarily be geometric, as discussed earlier. It is in fact a conjecture that $\widetilde{H}_{\mathbb{Q}_{p}}^{i}$ should know about the rigid analytic space of Galois representations.

In the case $F=\mathbb{Q}$ and $n=2$, the completed cohomology is some space of p-adic modular forms, and there is much that is known via the work of Matthew Emerton, who also showed that the p-adic local Langlands correspondence appears inside the completed cohomology. This has led to a proof of many cases of the Fontaine-Mazur conjecture for $2$-dimensional odd Galois representations.

We have only provided a rough survey of the motivations behind the theory of completed cohomology in this post. We will discuss further deeper aspects of it, and its relations to the p-adic local Langlands correspondence and the Fontaine-Mazur conjecture in future posts.

References:

Completed cohomology and the p-adic Langlands correspondence by Matthew Emerton on YouTube

Completed cohomology and the p-adic Langlands program by Matthew Emerton

Completed cohomology – a survey by Frank Calegari and Matthew Emerton