# Some examples of Geometric Galois representations

In this post we discuss the Galois representation associated to a projective scheme $X/K$, where $K$ is a number field. We also discuss how this representation can be computed in several simple cases.

# Motivation

Galois representations are a central focus of much of modern number theory. The desire for Galois representations is simple. As aspiring number theorists, we’d very much like to understand number fields (finite extensions of $\mathbb{Q}$). In the abstract though, this is a hugely formidable problem. Namely, how could we go about describing such complicated objects? Well, one possible technique of attack is to describe the Galois group $G_\mathbb{Q}:=\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Indeed, after all, every number field $K/\mathbb{Q}$ is a subextension of a Galois extension $L/\mathbb{Q}$, the arithmetic of which is largely informed by its Galois group $\text{Gal}(L/K)$. But, $\text{Gal}(L/K)$ is a quotient of $G_\mathbb{Q}$. Thus, understanding $G_\mathbb{Q}$ is largely equivalent to understanding certain aspects of number fields.

Now, while ‘understanding $G_\mathbb{Q}$‘ certainly seems like a less vague goal than ‘understanding number fields’, it is no less formidable. The absolute Galois group conjecturally has every finite group as a quotient. This gives a qualitative way in which understanding $G_\mathbb{Q}$ is hard. Thus, instead of trying to study $G_\mathbb{Q}$ directly, we instead try to understand a simplified problem. While there are many ways of understanding a group, perhaps the most historically successful approach is to understand its representation theory. Namely, to understand homomorphisms $G_\mathbb{Q}\to\text{GL}_n(F)$ for various fields $F$. That said, studying all representations are somewhat too hard for our purposes. Since we only care about continuous surjections $G_\mathbb{Q}\to H$ (say, $H$ is $\text{Gal}(L/\mathbb{Q})$ for some subextension $\overline{\mathbb{Q}}\supseteq L\supseteq\mathbb{Q}$), we might as well restrict ourselves to continuous homomorphisms $G_\mathbb{Q}\to \text{GL}_n(F)$, where $F$ is a topological field. If $F$ happens to be $\mathbb{Q}_\ell$, with its standard topology, we call such representations $\ell$-adic.

Considering the representation theory of $G_\mathbb{Q}$ has another appealing feature besides being a more tenable foothold to start attacking ‘understanding $G_\mathbb{Q}$‘. Namely, $G_\mathbb{Q}$ comes with a lot of local information bundled inside of it. Namely, for each $p$ prime, we can choose an embedding $\overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}_p}$, which gives us an embedding $G_{\mathbb{Q}_p}\hookrightarrow G_\mathbb{Q}$. That said, both $G_{\mathbb{Q}}$, and these local pieces are only determined up to conjugacy, much like the fundamental group of a path connected space is determined ‘up to conjugacy’ by the choice of base point (this is literally what happens if one thinks of $G_{\mathbb{Q}}=\pi_1^{et}(\text{Spec}(\mathbb{Q}),\overline{x})$, where $\overline{x}:\text{Spec}(\overline{\mathbb{Q}})\to\text{Spec}(\mathbb{Q})$ is a chosen geometric base point).

Tangential Remark: The map $G_{\mathbb{Q}_p}\to G_{\mathbb{Q}}$ is injective by Krasner’s lemma. Namely, Krasner’s lemma tells us that the embedding $\overline{\mathbb{Q}}\hookrightarrow\overline{\mathbb{Q}_p}$ is dense (since every finite extension of $\mathbb{Q}_p$ is of the form $L_\mathfrak{p}$ for some finite extensions $L/\mathbb{Q}_p$, and some prime $\mathfrak{p}$ of $L$ lying over $p$). But, since $G_{\mathbb{Q}_p}$ consists of continuous (even isometric!) automorphisms $\overline{\mathbb{Q}_p}$ over $\mathbb{Q}_p$, this implies the restriction map $G_{\mathbb{Q}_p}\to G_{\mathbb{Q}}$ is injective. In fact, choosing a prime $\mathfrak{p}$ of $\overline{\mathbb{Z}}$ (the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$) lying over $p$, we may identify $G_{\mathbb{Q}_p}$ with the decomposition group $D(\mathfrak{p}\mid p)\subseteq G_{\mathbb{Q}}$.

This non-canonicalness, which should sit badly with us, sort of disappears when we move to the representation theoretic point of view. Not only does conjugation not much matter in $\text{GL}_n(F)$ (inner morphisms define equivalent representations!) but we have a veritable cornucopia of class functions on $\text{GL}_n(F)$: $\text{tr}$, $\text{det}$, all of which are bundled together by the conjugation invariant characteristic polynomial.

Ok, now that I’ve hopefully convinced you that Galois representations are a desirable thing to study, a question begs to be asked: are there a lot of them? In fact, we’d like to not only know that there are numerous examples of Galois representations, but these examples are, in some sense, ‘natural’. Objects of study which don’t seem too contrived.

That said, not only do we have a huge source of naturally occurring, hugely impactful Galois representations, we even have them coming in two flavors. Not cherry and grape, but analytic and geometric. More specifically, we have automorphic representations which come, roughly, from the analytic objects of automorphic forms (e.g. modular forms). We also have geometric representations which show up in the $\ell$-adic cohomology of varieties. Both of these are extremely important, not only in modern number theory, but in many neighboring fields (most immediately, algebraic geometry).

In this post, we’d like to discuss the latter type of representations, in particular taking the time to write down some concrete examples. But, as often happens in mathematics (almost necessarily from a philosophical point of view!), the more powerful an invariant, the more difficult it is to compute. This is no more true in the computation of geometric Galois representations, which seem to require incredible effort to compute the simplest of examples. But, the restriction to $K$ a number field, opens up for us whole new avenues of attack which, in some cases (such as projective space), offer simpler alternatives to ‘brute calculation’.

# Geometric representations and basic properties

Let’s begin by precisely what we mean when we speak of geometric representations (at least in the context of this post). Thanks to the work of Grothendieck, Artin, and many others, we now have for a field $K$ (assumed perfect purely for conveinience), whose characteristic differs from $\ell$, a contravariant functor

$H^i(-,\mathbb{Q}_\ell):\left\{\begin{array}{c}\text{projective, smooth varieties}\\ \text{over }K\end{array}\right\}\longrightarrow \left\{\begin{array}{c}\text{finite dimensional }\ell\text{-adic}\\ \text{Galois representations}\end{array}\right\}$

This is relatively simple to define, assuming you take the entirety of étale cohomology as a blackbox, which, honestly, is not a terrible idea. Namely:

$H^i(X,\mathbb{Q}_\ell):=\left( \varprojlim H^i_{\text{et}}\left(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z}\right)\right)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$

This is a finite-dimensional $\mathbb{Q}_\ell$-space by general theory. The Galois action is defined as follows. For each $g\in G_K$ we have a morphism $\widetilde{g}:X_{\overline{K}}\to X_{\overline{K}}$. We then get from standard étale cohomology theory a map

$H^i_\text{et}(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})\to H^i\left(X_{\overline{K}},\widetilde{g}^\ast\left(\mathbb{Z}/\ell^n\mathbb{Z}\right)\right)$

But, the pull-back of a constant sheaf is constant, and so we get that the right hand side is naturally $H^i_\text{et}(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})$. This map is $\mathbb{Z}/\ell^n\mathbb{Z}$-linear, and respects composition. Thus, we obtain a continuous map

$G_K\to \text{Aut}_{\mathbb{Z}}\left(H^i_\text{et}(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})\right)$

Passing to the limit in $n$ gives us a continuous map $G_K\to\text{Aut}_{\mathbb{Z}_\ell}\left(\varprojlim H^i_\text{et}(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})\right)$. Then, finally, tensoring with $\mathbb{Q}_\ell$ gives us a continuous homomorphism $G_K\to \text{Aut}_{\mathbb{Q}_\ell}\left(H^i(X,\mathbb{Q}_\ell)\right)$.

Remark: A more fancy way of phrasing the above is as follows. Consider the structure morphism $f:X\to\text{Spec}(K)$. Then, since $f$ is smooth and proper, we have a nice theory of the pushforward $R^if_\ast\left(\mathbb{Z}/\ell^n\mathbb{Z}\right)$ (e.g. it’s constructible). By general theory, we know that sheaves on $\text{Spec}(K)$ correspond to $G_K$-modules by moving to the stalk at a geometric point $\overline{x}:\text{Spec}(\overline{K})\to \text{Spec}(K)$. But, as a group, proper base change (the ‘real’ proper base change) tells us that $\left(R^if_\ast\left(\mathbb{Z}/\ell^n\mathbb{Z}\right)\right)_{\overline{x}}=H^i(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})$. Thus, we obtain a $G_K$-modules structure on $H^i(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})$. Since, in general, cohomology plays well with limits, we can take the limit of this to obtain a $G_K$-module $\displaystyle \varprojlim H^i(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z})$ which is a $\mathbb{Z}_\ell$-module, and for which $G_K$ acts $\mathbb{Z}_\ell$-linearly. Finally, tensoring with $\mathbb{Q}_\ell$ gives the $\ell$-adic representation described above: $G_K\to\text{Aut}_{\mathbb{Q}_\ell}\left(H^i(X,\mathbb{Q}_\ell\right)$. This is the correct way to think if one is thinking of all of the above as some sort of ‘monodromy’.

After seeing the definition (which even takes all of étale cohomology for granted), one can see why the computation of the Galois representation coming from $H^i(X,\mathbb{Q}_\ell)$ can be quite difficult. Lucky for us, the functor $H^i(-,\mathbb{Q}_\ell)$ is a so-called Weil cohomology theory and so enjoys, amongst others, the following helpful properties:

1. As we’ve already mentioned, $H^i(X,\mathbb{Q}_\ell)$ is finite-dimensional.
2. We have a pairing of Galois modules $H^i(X,\mathbb{Q}_\ell)\times H^j(X,\mathbb{Q}_\ell)\to H^{i+j}(X,\mathbb{Q}_\ell)$ called the cup product.
3. For $i>2\dim X$, we have $H^i(X,\mathbb{Q}_\ell)=0$.
4. We have the so-called orientation isomorphism $H^{2\dim X}(X,\mathbb{Q}_\ell)\xrightarrow{\approx}\mathbb{Q}_\ell(-\dim X)$.
5. We have a version of Pontryagin duality, in particular, that for $i+j=2\dim X$, the pairing $H^i(X,\mathbb{Q}_\ell)\times H^j(X,\mathbb{Q}_\ell)\to H^{i+j}(X,\mathbb{Q}_\ell)\cong \mathbb{Q}_\ell(-\dim X)$ is non-degenerate. Thus, in particular, we have that $H^i(X,\mathbb{Q}_\ell)\cong H^j(X,\mathbb{Q}_\ell)^\vee\otimes\mathbb{Q}(-\dim X)$.
6. We have a version of the Kunneth formula, so that as graded rings (with the cup product as above, and with the obvious grading), we have an isomorphism of graded $\mathbb{Q}_\ell[G_K]$-algebras $H^\ast(X,\mathbb{Q}_\ell)\otimes_{\mathbb{Q}_\ell}H^*(Y,\mathbb{Q}_\ell)\cong H^*(X\times_\text{Spec}(K)Y,\mathbb{Q}_\ell)$.

Moreover, while not strictly one of the axioms of a Weil cohomology theory, $H^i(X,\mathbb{Q}_\ell)$ also benefits from the following extremely useful fact (an extremely specific application of smooth base change!):

(Smooth proper base change): Let$A$ is a complete DVR with function field $K$ and residue field $k$, neither of which have characteristic $\ell$, and suppose that $\mathfrak{X}/\text{Spec}(A)$ is a smooth projective scheme. Then, there is an isomorphism $\varphi:H^i(\mathfrak{X}_K,\mathbb{Q}_\ell)\xrightarrow{\approx} H^i(\mathfrak{X}_k,\mathbb{Q}_\ell)$ such that for $g\in G_K$ we have

$\varphi(gx)=\overline{g}\varphi(x)$

where $\overline{g}$ is the quotient map $G_K\to G_k$ (whose kernel is the inertia group $I_K$).

This would be a good time to define the notation $\mathbb{Q}_\ell(n)$, for $n$ an integer, as defined above. For any field $K$, note that $\text{Gal}(K(\mu_{\ell^n})/K)\hookrightarrow (\mathbb{Z}/\ell^n\mathbb{Z})^\times$, in the obvious way, for all $n$. Thus, by passing to the limit, we obtain an injection of $\text{Gal}(K(\mu_{\ell^\infty})/K)\to \mathbb{Z}_\ell^\times\subseteq\mathbb{Q}_\ell^\times=\text{GL}_1(\mathbb{Q}_\ell)$. The composition of this injection with the quotient map $G_K\to \text{Gal}(K(\mu_{\ell^\infty})/K)$ is called the cyclotomic character of $K$, and is denoted $\chi_{\ell,K}$ (or just $\chi_\ell$ when $K$ is clear from context). It’s a one-dimensional Galois representation (continuity comes from being built by finite quotients), and perhaps the simplest of all such representations.

We denote the space $\mathbb{Z}_\ell$ with the $G_K$-action $g\cdot x:\chi_\ell(g)x$ by $\mathbb{Z}_\ell(1)$, and the space $\mathbb{Q}_\ell$ with the similar action $\mathbb{Q}_\ell(1)$. For $n>0$, we denote $\mathbb{Q}_\ell(n)$ the representation $\mathbb{Q}_\ell(1)^{\otimes n}$. Choosing a basis, this is just the space $\mathbb{Q}_\ell$ with the action $g\cdot x=\chi_\ell(g)^n x$. For $n<0$ we define $\mathbb{Q}_\ell(-n)$ to be $\mathbb{Q}_\ell$ with the action $g\cdot x=\chi_\ell(g)^{-n}x$, this is the same as the dual representation of $\mathbb{Q}_\ell(n)$. For a $\mathbb{Q}_\ell[G_K]$-module $M$, we define its dual $M^\vee$, to be the $\mathbb{Q}_\ell$-space $\text{Hom}_{\mathbb{Q}_\ell}(M,\mathbb{Q}_\ell)$ together with the action $(g\varphi)(x)=\varphi(g^{-1}x)$. One can check that this is, indeed, a $G_K$-representation, if $M$ was (in particular, we still have continuity). This dual is the same dual showing up in 5. above.

Using just these properties, we can prove amazing things about these Galois representations. For example:

Theorem 1: Let $K$ be a number field. Then, if $X/K$ is a smooth projective variety, then for almost all primes $\mathfrak{p}$ of $K$, the representation $H^i(X,\mathbb{Q}_\ell)$ is unramified.

What it means for a representation of $G_K$ to be unramified at $\mathfrak{p}$ is that for a choice of an embedding $G_{K_\mathfrak{p}}\hookrightarrow G_K$, we have the composition

$I_{K_\mathfrak{p}}\hookrightarrow G_{K_\mathfrak{p}}\hookrightarrow G_{\mathbb{Q}}\to \text{Aut}_{\mathbb{Q}_\ell}(H^i(X,\mathbb{Q}_\ell))$

is trivial. Or, said in words, the local inertia acts trivially on $H^i(X,\mathbb{Q}_\ell)$. While the embeddings are only defined up to conjugation, acting trivially is insensitive to this ambiguity.

So, how might we go about proving Theorem 1? Well, let’s begin by noticing that $X/K$ actually has a smooth model $\mathfrak{X}/U$, for some open subscheme $U\subseteq\mathrm{Spec}(\mathcal{O}_K)$. This follows from the general principle of ‘spreading out’ (cf. Bjorn Poonen’s excellent notes Rational Points on Varieties). That said, there is a simple way of describing this model.

Namely, take the equations $\{f_i\}$ cutting out $X$ in $\mathbb{P}^n_K$, then since projective space is insensitive to clearing of denominators, we may assume without loss of generality that these equations lie in $\mathcal{O}_K[T_0,\ldots,T_n]$. Then, we just take the projective variety $\mathfrak{X}'\subseteq\mathbb{P}^n_{\mathcal{O}_K}$ cut out by these same equations. Now, the morphism $\mathfrak{X}'\to\text{Spec}(\mathcal{O}_K)$ might not be smooth. But, smoothness is open on the target for a finite type map, and is non-empty since it’s smooth at the generic point. Thus, for some open subscheme $U\subseteq\text{Spec}(\mathcal{O}_K)$ we have that the pull-back $\mathfrak{X}/U$ is smooth and projective, which gives us our desired model.

Now, since $\text{Spec}(\mathcal{O}_K)$ is a curve, the open subscheme $U$ must be just $\text{Spec}(\mathcal{O}_K)-S$, for some finite set of primes $S$. Also, since it won’t affect finiteness, we assume without loss of generality that $S$ contains all primes of $K$ lying over $\ell$. So, now, take any prime $\mathfrak{p}\notin S$. Then, we have a morphism $\text{Spec}(\mathcal{O}_{K_\mathfrak{p}})\to U$. We can then consider $\mathfrak{X}_{\text{Spec}(\mathcal{O}_{K_\mathfrak{p}})}$. Since $\mathcal{O}_{K_\mathfrak{p}}$ is a complete DVR, smooth proper base-change tells us that we have an isomorphism $H^i(\mathfrak{X}_{K_{\mathfrak{p}}},\mathbb{Q}_\ell)\cong H^i(\mathfrak{X}_{k_{\mathfrak{p}}},\mathbb{Q}_\ell)$ in such a way which intertwines the map $G_{K_\mathfrak{p}}\to G_{k_\mathfrak{p}}$. In particular, note that this implies that $I_{K_\mathfrak{p}}$ acts trivially on $H^i(\mathfrak{X}_{K_\mathfrak{p}},\mathbb{Q}_\ell)$.

That said, one easily sees that $\mathfrak{X}_{K_\mathfrak{p}}=X_{K_\mathfrak{p}}$ and, in particular, the action of $G_{K_\mathfrak{p}}$ on $H^i(X_{K_\mathfrak{p}},\mathbb{Q}_\ell)$ formed by the composition $G_{K_\mathfrak{p}}\hookrightarrow G_K$ is the same as the action of $G_{K_\mathfrak{p}}$ on $H^i(\mathfrak{X}_{K_\mathfrak{p}},\mathbb{Q}_\ell)$ described above.

Since $S$ is a finite set, putting all of this together gives the desired result stated in Theorem 1.

# The ostensibly unreasonable power of the Frobenius

Before we begin actually calculating specific examples of geometric Galois representations, as above, we want to develop one of the most powerful ways of getting at Galois representations, at least over number fields. Let us begin by recalling one of the most famous theorems of basic representation theory. One may not have heard this theorem, in this context or generality, but likely one has run into under various guises before:

Theorem(Brauer-Nesbitt Theorem): Let $k$ be a characteristic zero field, and $A$ a (not necessarily commutative!) $k$-algebra. Then, two semi-simple $A$-modules $E$ and $E'$, which are finite-dimensional as $k$-spaces, are isomorphic, if and only if $\text{tr}(m_a)=\text{tr}(m'_a)$ for for all $a\in A$ (here $m_a$ is multiplication by $a$ as a map $E\to E$, and similarly for $E'$.

As a corollary, we derive the following

Corollary 2: Let $L$ be a characteristic $0$ field, and $K$ an arbitrary field. Then, two Galois representations $\rho_1:G\to\text{GL}_n(L)$ and $\rho_2:G\to\text{GL}_m(L)$ are equivalent if and only if $\text{tr}(\rho_1(g))=\text{tr}(\rho_2(g))$ for all $g\in G$.

If we define the character of a Galois representation $\rho:G\to\text{GL}_n(L)$ (poor notation, since it’s NOT a ‘character’ in the sense of a one-dimensional representation) to be $g\mapsto \text{tr}(\rho(g))$, then the corollary above says that $\rho$ is entirely determined by its character.

Now, while this theorem is clearly nice on first glance, it’s true power comes when considered in the context of Galois representations where the field $K$ is a number field, and the representation is unramified almost everywhere (as in the case of geometric representations). The true all-star in this set-up is the famous Chebotarev density theorem:

Theorem(Chebotarev density):Let $L/K$ be a Galois extension of number fields. Then, for any conjugacy class $\mathcal{C}$ in $\text{Gal}(L/K)$, the Dirichlet density of the set

$\left\{\mathfrak{p}\in\text{Spec}(\mathcal{O}_K):\{\text{Frob}_{\mathfrak{q}}\}_{\mathfrak{q}\mid\mathfrak{p}}=\mathcal{C}\right\}$

is $\displaystyle \frac{\#\mathcal{C}}{|G|}$.

Recall that for any unramified prime $\mathfrak{p}$ of $\mathcal{O}_K$, we can define, for each $\mathfrak{q}\mid\mathfrak{p}$, a Frobenius element $\text{Frob}_{\mathfrak{q}}\in\text{Gal}(L/K)$. This is the unique $\sigma\in\text{Gal}(L/K)$ such that $\displaystyle \sigma(x)\equiv x^{p^r}\mod\mathfrak{q}$, where $p^r=\#(\mathcal{O}_K/\mathfrak{p})$. Now, by the transitivity of $\text{Gal}(L/K)$‘s action on primes of $\mathcal{O}_L$ dividing $\mathfrak{p}$, we see that the set $\left\{\text{Frob}_\mathfrak{q}\right\}_{\mathfrak{q}\mid\mathfrak{p}}$, is, in fact, a full conjugacy class in $\text{Gal}(L/K)$. Thus, for almost all $\mathfrak{p}$ (the unramified ones!) the question of whether $\{\text{Frob}_\mathfrak{q}\}_{\mathfrak{q}\mid\mathfrak{p}}=\mathcal{C}$ is a sensical one. Since Dirichlet density (a natural extension of the usual notion of ‘natural density’) is insensitive to the finitely many primes where this statement doesn’t make sense, we see that the Dirichlet density of the set in Chebotarev density is a sensical thing.

Now, Dirichlet density seems like it has nothing to do with the Brauer-Nesbitt theorem. To make the connection, let us first begin with some notation. Let $K$ be a number field, and let $S$ be a finite set of places. We define $K^S$ to be the maximal extension of $K$ in $\overline{K}$ which is unramified outside of $S$–so, the only ramification that can occur is at the finitely many places of $S$. We shall denote $\text{Gal}(K^S/K)$ by $G_{K,S}$.

For example, if $K=\mathbb{Q}$, and $S=\{p,\infty\}$, then $K^S\supseteq \mathbb{Q}(\mu_{p^\infty})$, and, in fact, $\mathbb{Q}(\mu_{p^\infty})$ is the maximal abelian subextension of $K^S$ (this follows from class field theory). Thus, we see that $G_{K,S}^\text{ab}$ (recall that for a profinite group $G$, $G^\text{ab}$ is the maximal Hausdorff abelian quotient of $G$–i.e. $G/\overline{[G,G]}$) is just $\mathbb{Z}_p^\times$.

Now, the Chebotarev density theorem has this following surprising consequence:

Corollary 3:Let $K$ be a number field, and $S$ a finite set of primes. For each finite prime $\mathfrak{p}\notin S$, choose an embedding $\iota:\overline{K}\hookrightarrow\overline{K_\mathfrak{p}}$. The conjugacy classes of the maps $\text{Frob}_{\mathfrak{p},\iota}\in G_{K,S}$ are dense.

Before we begin the proof, let us clarify precisely what we mean by the above. The embedding $\iota$ gives rise to a map $G_{K_\mathfrak{p}}\hookrightarrow G_K$, and so on composition with the quotient map $G_K\to G_{K,S}$, a map $G_{K_\mathfrak{p}}\to G_{K,S}$. Since $K^S$ is unramified outside of $S$, we see that $I_{K_\mathfrak{p}}\subseteq G_{K_\mathfrak{p}}$ maps to zero in $G_{K,S}$. Thus, we get a well-defined map $G_{k_\mathfrak{p}}\to G_{K,S}$ where, as usual $G_{k_\mathfrak{p}}=G_{K_\mathfrak{p}}/I_{K_\mathfrak{p}}$ is the absolute Galois group of the residue field of $\mathcal{O}_{K_\mathfrak{p}}$. But, $G_{k_\mathfrak{p}}$ has a canonical Frobenius element (the one defined by $x\mapsto x^{\#(k_{\mathfrak{p}})}$) and we denote the image of this Frobenius element in $G_{K,S}$ by $\text{Frob}_{\iota,\mathfrak{p}}$. Clearly, as the notation suggests, this Frobenius element not only depends on $\mathfrak{p}$, but the embedding $\iota:\overline{K}\hookrightarrow\overline{K_\mathfrak{p}}$.

As mentioned in the motivation, here is another way to view this is as follows. For our given prime $\mathfrak{p}$, let us choose a prime $\mathfrak{P}$ of $\overline{\mathbb{Z}}$ (the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}=\overline{K}$) lying over $\mathfrak{p}$. The Galois group $G_K$ acts on $\overline{\mathbb{Z}}$ by ring automorphisms, and since it fixes $\mathcal{O}_K$, permutes the primes of $\overline{\mathbb{Z}}$ lying over $\mathfrak{p}$. We can then define, in analogy with the finite case, the decomposition group $D(\mathfrak{P}\mid\mathfrak{p})$ to be those elements of $G_K$ which send $\mathfrak{P}$ to itself. Since every element of $D(\mathfrak{P}\mid\mathfrak{p})$ fixes $\mathfrak{p}$ (not pointwise, just as a set!) we get a well-defined map $D(\mathfrak{P}\mid\mathfrak{p})\to G_{k_\mathfrak{p}}$, where $k_\mathfrak{p}=\mathcal{O}_K/\mathfrak{p}$ (note that, as the notation suggests, this is the same field $k_\mathfrak{p}$ as above!). The kernel of this maps is denoted $I(\mathfrak{P}\mid\mathfrak{p})$. Now, consider the composition $D(\mathfrak{P}\mid\mathfrak{p})\to G_K\to G_{K,S}$. Since $K^S/K$ is unramified, one can show that $I(\mathfrak{P}\mid\mathfrak{p})$ maps to zero in $G_{K,S}$, and thus, once again, we get a well-defined map $G_{k_\mathfrak{p}}\to G_{K,S}$. We denote the image of the Frobenius element in $G_{K,S}$ as $\text{Frob}_\mathfrak{P}$.

It’s a good test of understanding to show that any two $\text{Frob}_{\iota,\mathfrak{p}}$ and $\text{Frob}_{\tau,\mathfrak{p}}$, for different embeddings $\overline{K}\hookrightarrow\overline{K_{\mathfrak{p}}}$ are conjugate. Moreover, a similar conjugacy statement holds for $\text{Frob}_{\mathfrak{P}}$ and $\text{Frob}_{\mathfrak{Q}}$ for any two primes $\mathfrak{P}$ and $\mathfrak{Q}$ of $\overline{\mathbb{Z}}$ lying over $\mathfrak{p}$. In fact, one can show that the two notions of $\text{Frob}_{\iota,\mathfrak{p}}$ and $\text{Frob}_\mathfrak{P}$, as $\iota$ and $\mathfrak{P}$ range over their respective indexing sets, describe the exact same set of elements!

Proof(of Chebotarev density): Since $G_{K,S}$ is a profinite group, a basis of open sets is given by sets of the form $gU$, for $g\in G_{K,S}$, and $U$ an open subgroup. Consider the quotient $G_{K,S}/U$ which, by Galois theory, is equal to $\text{Gal}(K_U/K)$ for some finite subextension of $K^S/K$. Now, by Chebotarev density, there are infinitely many primes $\mathfrak{p}$ of $\mathcal{O}_K$ which are unramified, and for which $\text{Frob}_{\mathfrak{q}}$ is conjugate to $gU$, for some prime $\mathfrak{q}$ of $\mathcal{O}_{K^S}$ dividing $\mathfrak{p}$. Since $S$ is finite, this implies that there is some $\mathfrak{p}\notin S$. Then, by construction, there exists $\overline{h}\in G_{K,S}/U$ (here $h\in G_{K,S}$) such that $\overline{h}\text{Frob}_\mathfrak{q}\overline{h}^{-1}=gU$. So, choose a prime $\mathfrak{P}$ of $\overline{\mathbb{Z}}$ lying over this $\mathfrak{q}$. We claim then that $h\text{Frob}_\mathfrak{P}h^{-1}\in gU$. Indeed, this is the same thing as checking that $\overline{h}\overline{\text{Frob}_\mathfrak{P}}\overline{h}^{-1}=gU$. But, by functoriality, in $G_{K,S}/U$, we have that $\overline{\text{Frob}_\mathfrak{P}}=\text{Frob}_\mathfrak{q}$, and so the conclusion follows. $\blacksquare$

Remark: At first glance, it looks like finding a $\text{Frob}_\mathfrak{q}$ conjugate to $gU$ in $\text{Gal}(K_U/K)$ didn’t need that $S$ was finite–there are always such $\mathfrak{q}$ by Chebotarev density! The fear is that if $S$ were infinite, then it may be equal to the set of $\mathfrak{p}$ for which $\text{Frob}_\mathfrak{q}$, for $\mathfrak{q}\mid\mathfrak{p}$, is conjugate to $gU$. This is no impediment to finding things conjugate to $gU$ in $\text{Gal}(K_U/K)$, but then we can’t lift those Frobenii up to $G_{K,S}$, since they make no sense there. In short, we need to be cognizant that we need to find $\mathfrak{q}$ for which $\text{Frob}_\mathfrak{q}$ is conjugate to $gU$, and $\mathfrak{q}\cap\mathcal{O}_K$ is not ramified in $K^S$!

So, now armed with this corollary of Chebotarev density theorem, we can make the following claim.

Theorem 4: Let $K$ be a number field, and let $\rho_1:G_K\to\text{GL}_n(L)$ and $\rho_2:G_K\to\text{GL}_m(L)$ be two semi-simple Galois representations, where $L$ is a Hausdorff field of characteristic $0$, and which are unramified outside of finite sets $S_1$ and $S_2$ respectively. Then, $\rho_1$ is equivalent to $\rho_2$ if and only if $\text{tr}(\rho_1(\text{Frob}_\mathfrak{P}))=\text{tr}(\rho_2(\text{Frob}_\mathfrak{P}))$ for all $\mathfrak{P}$ primes in $\overline{\mathbb{Z}}$ lying above a prime not in $S_1\cup S_2$.

Note that $\text{tr}(\text{Frob}_\mathfrak{P})$ is also independent of which $\mathfrak{P}$ dividing $\mathfrak{p}$ we choose!

Proof: Note that since $\rho_1$ and $\rho_2$ are unramified outside of $S:=S_1\cup S_2$, we have that both factor through $G_{K,S}$, call these factorizations $\widehat{\rho_1}$ and $\widehat{\rho_2}$. Then, by assumption, the maps $g\mapsto \text{tr}(\rho_1(g))$ and $g\mapsto \text{tr}(\rho_2(g))$ agree on the dense set $\{\text{Frob}_\mathfrak{P}\}_{\mathfrak{P}\cap\mathcal{O}_K\notin S}$. Since $L$ is Hausdorff, this implies that these two functions must be equal. So, the rest follows by Brauer-Nesbitt. $\blacksquare$

Thus, understanding Galois representations of the above form (or at least their semi-simplifications!) is equivalent to understanding just the character’s value on the set of Frobenii. This shows why Frobenii are so important in the field of algebraic number theory/geometry–they are a small package, which carries an unreasonable amount information.

# Finally a computation; the case of curves

So, if the $\ell$-adic Galois representations $H^i(X,\mathbb{Q}_\ell)$ are to be of any use, we’d at least hope that we can have a reasonable description of them in the case when $X$ is the simplest non-trivial object–a curve over $K$ (NB: for the purposes of this post, a curve will mean smooth, projective, and geometrically integral). Indeed, while it must be admitted that any invariants worth studying are, in general, hard to compute, if we can’t do the simplest case, what’s the point?

Tangential anecdote: I’m reminded of something a friend once told me, the details of which I’ll almost certainly bungle. He told me of a wonderful, famous algebraic invariant associated to CW-complexes $X$, call it $\mathrm{WK}(X)$ (this is because I vaguely remember it being ‘Waldhausen K-theory’, but I am nowhere near certain). He said that when it was created the topological world was abuzz since it carried such sophisticated information, that it should turn the topological world on its head! Interested, I asked him what $\mathrm{WK}(S^1)$ was. He said, “well, that’s hard”. To this, I responded “Well, OK. But, certainly $\mathrm{WK}(\text{point})$ is trivial, right?” Looking a bit ill he dropped this bomb: “well, computing the value of $\mathrm{WK}(\text{point})$ is equivalent to computing the homotopy groups of spheres…”

Now, while I said that we can compute Galois representations $H^i(C,\mathbb{Q}_\ell)$, for a curve $C$ over the field $K$ (with characteristic not $\ell$), I never said it would be easy. By far, the most involved part of the computation is the following, deceivingly benign, calculation:

Lemma 5: Let $C/K$ be a curve. Then, the following holds:

$H^i_\text{et}(C_{\overline{K}},\mathbf{G}_m)=\begin{cases}\overline{K}^\times & \mbox{if}\quad i=0\\ \mathrm{Pic}(C_{\overline{K}}) & \mbox{if}\quad i=1\\ 0 & \mbox{if}\quad i>1\end{cases}$

Moreover, the Galois action on $H^1_\text{et}(C_{\overline{K}},\mathbf{G}_m)$ agrees with the obvious Galois action on $\mathrm{Pic}(C_{\overline{K}})$ (i.e. given $\sigma\in G_K$, the action $\sigma^\ast:\mathscr{L}\to\sigma^\ast\mathscr{L}$).

The proof is incredibly difficult, and relies on a lot of very complicated Galois cohomology computations. For a proof, you can either see Milne’s relevant book, or see Daniel Litt’s approach using twisted sheaves (which is the modern way of thinking about the Brauer group). That said, while the proof is very difficult in its technical nuances, it’s proof follows precisely the idea you’d want to adapt from the analogous statement for the Zariski site.

Namely, if $X/K$ is any factorial, integral variety, then one can show that

$H^i(X,\mathbf{G}_m)=\begin{cases}\mathcal{O}_X(X)^\times & \mbox{if}\quad i=0\\ \mathrm{Pic}(X) & \mbox{if}\quad i=1\\ 0 & \mbox{if}\quad i>1\end{cases}$

The $i=0$ and $i=1$ cases are trivial and famous respectively. The $i>1$ case is the slightly tricky fact. It follows because one has a canonical flasque resolution of $\mathbf{G}_m$:

$0\to\mathbf{G}_m\to \mathcal{K}_X^\times\to \mathcal{K}_X^\times/\mathbf{G}_m\to 0\to 0\to\cdots$

Here $\mathcal{K}_X^\times$ is the constant sheaf with values in $K(X)^\times$ (here $K(X)$ is the function field of $X$). Now, it’s clear that $\mathcal{K}_X^\times$ is flasque, but the quotient $\mathcal{K}_X^\times/\mathbf{G}_m$ is much less obviously so. The trick is to use the fact that $X$ is factorial and integral to identify this sheaf of Cartier divisors with the sheaf $\mathrm{Div}_X$, defined by $U\mapsto \mathrm{Div}_X(U)$, the group of Weil divisors on $U$, with the obvious restriction maps. This is obviously flasque. The cohomology computation then immediately follows.

In the étale cohomology computation, the proof goes much the same. The first two cases are trivial, and famous respectively (the proof of the latter is hard, but somewhat ‘natural’). The huge amount of annoying Galois cohomology goes into showing that the analogous sheaf $\mathrm{Div}_{C_{\overline{K}}}$ on $(C_{\overline{K}})_{\text{et}}$ is acyclic. This is the part which you should reference the two aforementioned sources for.

So, with that lemma in hand, we can prove the lemma which will imply the main result:

Lemma 6: Let $K$ be a field with characteristic coprime to $\ell$, and let $C/K$ be a curve. Then, for all $n\geqslant 0$

$H^i_{\text{et}}\left(C_{\overline{K}},\mu_{\ell^n}\right)=\begin{cases}\mu_{\ell^n}\left(\overline{K}^\times\right) & \mbox{if}\quad i=0\\ \mathrm{Pic}^0(C_{\overline{K}})[\ell^n] & \mbox{if}\quad i=1\\ \mathbb{Z}/\ell^n\mathbb{Z} & \mbox{if}\quad i=2\\ 0 & \mbox{if}\quad i>2\end{cases}$

where the identification is of $G_K$-modules.

Before we begin the proof, let us say that the action of $G_K$-modules is similar to the action described above. Namely, it’s the $G_K$-module associated to the stalk of, say, $R^1f_\ast\left(\mu_{\ell^n}\right)$ at some geometric point.

Proof:  So, we’re trying to compute $H^i_\text{et}(C_{\overline{K}},\mu_{\ell^n})$. But, we then consider the obvious complex of abelian sheaves:

$0\to\mu_{\ell^n}\to\mathbf{G}_m\xrightarrow{x\mapsto x^{\ell^n}}\mathbf{G}_m\to 0$

the so-called Kummer sequence (when applied to the spectrum of a field, the long exact sequence of Galois cohomology from this sequence gives the famed Kummer’s theorem). To see that this sequence is exact, it suffices to check at geometric points $\overline{x}:\text{Spec}(\overline{\Omega})\to C_{\overline{K}}$. But, taking stalks leaves us to show that we have a short exact sequence

$0\to\mu_{\ell^n}(A)\to A^\times\xrightarrow{x\mapsto x^{\ell^n}}A^\times\to 0$

where $A$ is the strictly Henselian ring $A:=\mathcal{O}_{C_{\overline{K}},x}^\text{sh}$ (the $\mathrm{sh}$ meaning strict Henselization and $x$ is the image of $\overline{x}$). Exactness on the left, and middle are clear. To see exactness at the right, we need to show that any element $u$ of $A^\times$ has a $\ell^n$-th root. So, consider the polynomial $X^{\ell^n}-u\in A[X]$. Moving to the residue $k$ field we have the separable polynomial (because $(\ell,\text{char}(K))$ again!) $X^{\ell^n}-\overline{u}\in k[X]$. This has a root since $k$ is algebraically closed, and so by Henselianess, $X^{\ell^n}-u$ has a root in $A[X]$. The exactness at the right follows.

So, now, taking the long exact sequence in étale cohomology gives

\begin{aligned}0\to &\mu_{\ell^n}(\overline{K}^\times)\to \overline{K}^\times\xrightarrow{x\mapsto x^{\ell^n}}\overline{K}^\times\to H^1_\text{et}(C_{\overline{K}},\mu_{\ell^n})\\ &\to H^1_\text{et}(C_{\overline{K}},\mathbf{G}_m)\to H^1_\text{et}(C_{\overline{K}},\mathbf{G}_m)\to H^2(C_{\overline{K}},\mu_{\ell^n})\to 0\end{aligned}

which is a sequence of $G_K$-modules, since the Kummer sequence itself was a sequence of $G_K$-sheaves. Now, we’ve already taken the computation of $H^0(C_{\overline{K}},\mu_{\ell^n})$ for granted. Now, note that $\overline{K}^\times\xrightarrow{x\mapsto x^{\ell^n}}\overline{K}^\times$ is surjective, and thus the map $H^1_\text{et}(C_{\overline{K}},\mu_{\ell^n})\to H^1(C_{\overline{K}},\mathbf{G}_m)$ is injective. Thus, as a $G_K$-module, we can identify $H^1_\text{et}(C_{\overline{K}},\mu_{\ell^n})$ with $\ker\left(H^1(C_{\overline{K}},\mathbf{G}_m)\to H^1(C_{\overline{K}},\mathbf{G}_m)\right)$ which, by additivity, is just ‘multiplication by $\ell^n$‘. But, identifying $H^1(C_{\overline{K}},\mu_{\ell^n})$ with $\mathrm{Pic}^0(C_{\overline{K}})$, we see that we have an identification of $G_K$-modules

$H^1_\text{et}(C_{\overline{K}},\mu_{\ell^n})\xrightarrow{\approx} \ker\left(\mathrm{Pic}(C_{\overline{K}})\xrightarrow{\mathscr{L}\mapsto \mathscr{L}^{\otimes n}}\mathrm{Pic}(C_{\overline{K}})\right)=\mathrm{Pic}^0(C_{\overline{K}})[\ell^n]$

which gives the desired result for $H^1(C_{\overline{K}},\mu_{\ell^n})$.

Finally, to compute $H^2(C_{\overline{K}},\mu_{\ell^n})$ the above shows us that we have and identification

$H^2(C_{\overline{K}},\mu_{\ell^n})\xrightarrow{\approx}\text{coker}\left(\mathrm{Pic}(C_{\overline{K}})\xrightarrow{\mathscr{L}\mapsto \mathscr{L}^{\otimes n}}\mathrm{Pic}(C_{\overline{K}})\right)=\mathbb{Z}/\ell^n\mathbb{Z}$

where the last step uses the fact that $\mathrm{Pic}(C_{\overline{K}})$ decomposes as $\mathrm{Pic}^0(C_{\overline{K}})\times\mathbb{Z}$, and uses the fact that $\mathrm{Pic}^0(C_{\overline{K}})$ is $\ell^n$-divisible. $\blacksquare$

So, now we can finally prove the big result of this section:

Theorem 7: Let $K$ be a field with $(\ell,\text{char}(K))=1$, and $C/K$ a curve. Then,

$H^i(C,\mathbb{Q}_\ell)\cong\begin{cases}\mathbb{Q}_\ell &\mbox{if}\quad i=0\\ V_\ell(\mathrm{Jac}(C))^\vee & \mbox{if}\quad i=1\\ \mathbb{Q}_\ell(-1) & \mbox{if}\quad i=2\\ 0 & \mbox{if}\quad i>2\end{cases}$

Before we give the proof, let us quickly remind ourselves what $V_\ell(\mathrm{Jac}(C))$ is. Recall that for a curve $C/K$ of genus $g$, there is an abelian variety, the Jacobian of $C$, denoted $\mathrm{Jac}(C)$, representing the relative degree $0$ Picard functor $\mathrm{Pic}^0_{C/k}$ (for more information, see Kleiman’s famous notes). Then, for any abelian variety $A$ we have the associated Tate module $T_\ell A$ defined as $\displaystyle \varprojlim A(\overline{K})[\ell^n]$, with the obvious Galois action (the Galois group acts additively!). This is a free $\mathbb{Z}_\ell$-module of rank $2g$. Thus, upon tensoring with $\mathbb{Q}_\ell$ we obtain a $2g$-dimensional $\ell$-adic Galois representation $V_\ell A$ of $G_K$. This is an object which, ostensibly complicated, is fairly well-understood.

Proof: Let us begin with the most important case of $i=1$. Now, above we computed that $H^i_\text{et}(C_{\overline{K}},\mu_{\ell^n})$ as a $G_K$-module was $\mathrm{Pic}^0(C_{\overline{K})}[\ell^n]$ which, by the precious paragraph, is just $\text{Jac}(C)(\overline{K})[\ell^n]$. Thus, upon passing to the limit, we obtain that, as a $G_K$-module we have that $\varprojlim H^i_\text{et}(C_{\overline{K}},\mu_{\ell^n})$ is just $T_\ell \text{Jac}(C)$, and so tensoring with $\mathbb{Q}_\ell$ gives

$H^i\left(C_{\overline{K}},\mu_{\ell^n}\right)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell\cong V_\ell \text{Jac}(C)$

Then, finally, we appeal to one last abstract nonsense fact from $\ell$-adic cohomology. Namely, for any smooth projective variety $X/K$, the $G_K$-module

$\varprojlim H^i\left(X_{\overline{K}},\mu_{\ell^n}\right)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$

is known as $H^i(X,\mathbb{Q}_\ell(1))$. It then follows (essentially from a version of the projection formula) that $H^i(X,\mathbb{Q}_\ell(1))=H^i(X,\mathbb{Q}_\ell)\otimes_{\mathbb{ Q}_\ell}\mathbb{Q}_\ell(1)$. So, finally we see that

\begin{aligned}H^i(C,\mathbb{Q}_\ell) &\cong H^i(C,\mathbb{Q}_\ell(1))\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell(-1)\\ & \cong V_\ell\text{Jac}(C)\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell(-1)\\ &\cong \left(V_\ell \text{Jac}(C)\right)^\vee\end{aligned}

where we’ve used that for any abelian variety $A$, one has

$V_\ell A^t\cong (V_\ell A)^\vee\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell(1)$

and that $\text{Jac}(C)^t=\text{Jac}(C)$ (i.e. Jacobians are self-dual). $\blacksquare$

Remark: Martin Olsson pointed out to me that there is an easier way to prove that $H^i(X,\mathbb{Q}_\ell(1))\cong H^i(X,\mathbb{Q}_\ell)\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell(1)$. There is always a map of $G_K$-modules

$H^i_\text{et}(X,\mathscr{F})\otimes H^j(X,\mathscr{G})\to H^{i+j}(X,\mathscr{F}\otimes\mathscr{G})$

Taking $j=0$, $\mathscr{F}=\mathbb{Q}_\ell(1)$, and $\mathscr{G}=\mathbb{Q}_\ell$ gives us an injective map of $\mathbb{Q}_\ell[G_K]$-modules

$H^i(X,\mathbb{Q}_\ell)\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell(1)\to H^i(X,\mathbb{Q}_\ell(1))$

But, these two $\mathbb{Q}_\ell$-spaces have the same dimension since, if one forgets about the action of $G_K$, $\mathbb{Q}_\ell=\mathbb{Q}_\ell(1)$ (i.e. as sheaves on $(C_{\overline{K}})_\text{et}$ one has a NON $G_K$-equivariant isomorphism $\mu_{\ell^n}\cong \mathbb{Z}/\ell^n\mathbb{Z}$). Thus, it is actually an isomorphism of $\mathbb{Q}_\ell[G_K]$-modules.

Tangential remark: One of the very confusing things about the above proof is the following. When one first learns about étale cohomology, to makes things easier, one usually ignores the $G_K$-action. In fact, the above Theorem 7 is usually stated, but one forgets the $G_K$-action, essentially just stating that $\dim_{\mathbb{Q}_\ell} H^1(C,\mathbb{Q}_\ell)=2g$. This is proven by actually computing $H^1(C,\mathbb{Q}_\ell(1))$ as above, and doing the identification $\mathbb{Q}_\ell=\mathbb{Q}_\ell(1)$ (i.e. $\mu_{\ell^n}\cong \mathbb{Z}/\ell^n\mathbb{Z}$). This, as remarked above, is totally fine if all you care about is the structure as a $\mathbb{Q}_\ell$-space. In fact, if you only care about these Betti numbers, then you can just reduce the computation to the case over $\mathbb{C}$ (by using base change properties, and smooth proper base-change), where the calculation is trivial (using Grothendieck’s comparison theorem).

But, these methods completely disfigures the Galois action. This identification, without further statement, is something that messed me when I was first trying to understand the $G_K$-module $H^i(C,\mathbb{Q}_\ell)$ since you, having done it before, just willy-nilly make this identification. I wish that this comment, and more generally a focus on the $G_K$-module structure, would be paid more attention in books on étale cohomology.

# Projective space; Frobenius to the rescue

I now would like to explain how one would calculate the cohomology of another ‘obvious’ smooth projective variety: $\mathbb{P}^m_K$. Once again, a lot of pain can be ignored if we are willing to forget about the Galois action, and just find the Betti numbers (i.e. $\dim_{\mathbb{Q}_\ell}H^i(\mathbb{P}_K^m,\mathbb{Q}_\ell$). For lack of motivation, I will not record here the more ‘standard way’ of computing the $G_K$-module $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$. The key is to use the fact that the $\ell$-adic cohomology is generated by cycles, and so one can use a souped version of the cycle class map to find the answer.

No, instead, I’m going to compute the cohomology of $\mathbb{P}^m_K$, where $K$ is a number field, using a somewhat different technique. Namely, we’ll figure out what the action of Frobenii are on $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$ which, once we see it’s necessarily semi-simple (which is easier than it sounds!), we’ll have our answer by the Brauer-Nesbitt theorem.

That said, before we proceed with our computations, we’ll need another key tool from the theory of $\ell$-adic cohomology. In fact, it’s the reason that Weil wanted an alternate cohomology theory on varieties, and how he envisioned a proof of the conjectures which bear his name. I speak, of course, of the Grothendieck trace formula. The trace formula is fairly general, but we state the most relevant version here.

So, let us begin with a quick recollection of the various types of Frobenii on $X_{\overline{\mathbb{F}_q}}$.

Recall first that if $S$ is any scheme over $\mathbb{F}_p$, then there is a morphism $F_S:S\to S$ called the absolute Frobenius of $S$. This is the map which is the identity on the underlying topological space of $S$, and which on an affine open $\mathrm{Spec}(A)$ corresponds to the ring map $a\mapsto a^p$. Note that if $S$ is a scheme over $\mathbb{F}_q$, with $q=p^m$, then $F_S$ is not a morphism of $\mathbb{F}_q$-schemes. Indeed, the composition $S\xrightarrow{F_S}S\to \mathrm{Spec}(\mathbb{F}_q)$ corresponds precisely to the usual Frobenius map on $\mathbb{F}_q$ which is not the identity. That said, one can easily see that $\mathrm{Fr}_S^m$ is a morphism of $\mathbb{F}_q$-schemes.

This absolute Frobenius then allows us to define four different types of Frobenius on $X_{\overline{\mathbb{F}_q}}$ where $X/\mathbb{F}_q$ is some variety. Indeed, we can take $F_{X_{\overline{F}_q}}^m$, we can take $\mathrm{Fr}_r^m:=\mathrm{id}\times \mathrm{Fr}_X^m$, we can take $\mathrm{Fr}_a^m:=\mathrm{id}\times \mathrm{Fr}_{\overline{k}}$, or we can take $\mathrm{Fr}_g^m:=\mathrm{id}\times \mathrm{Fr}_{\overline{k}}^{-m}$. These are called the absolute, relative, arithmetic, and geometric Frobenius respectively.

As a clarifying example of their differences, let’s consider what happens in the case when $X=\mathbb{F}_q[T_1,\ldots,T_n]$, so $X_{\overline{\mathbb{F}_q}}=\overline{\mathbb{F}_q}[T_1,\ldots,T_n]$. Then, $\mathrm{Fr}_{X_{\overline{\mathbb{F}_q}}}^m$ takes $f$ to $f^m$. The relative Frobenius takes $T_i$ to the $T_i^m$, but leaves the coefficients alone. The arithmetic Frobenius acts by taking the coefficients to the $q^\text{th}$-power, and the geometric Frobenius acts by taking $q^\text{th}$-roots of the coefficients.

Now, the important thing to note is that all the Frobenii are (iso)morphisms of schemes $X_{\overline{\mathbb{F}_q}}\to X_{\overline{\mathbb{F}_q}}$, and so induce (iso)morphism of $H^i_{\mathrm{\acute{e}t}}(X_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)$. The only two important facts we’ll need are:

1. The (iso)morphisms $F_r$ and $F_g$ induce the same (iso)morphisms of $H^i_\mathrm{\acute{e}t}(X_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)$.
2. The (iso)morphism induced by $F_g$ is the one corresponding to the action of the geomtric Frobenius in $G_{\mathbb{F}_q}$ on $H^i_{\mathrm{\acute{e}t}}(X_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)$ (definitionally).

We then have the following famed theorem of Grothendieck:

Theorem(Grothendieck-Lefschetz formula): Let $X/\mathbb{F}_q$ be a smooth projective scheme. Denote by $F$ the morphism $\mathrm{Fr}_r^m$. Then, for all $r\geqslant 1$, we have the following equality:

$\displaystyle \# X(\mathbb{F}_{q^r})=\sum_i (-1)^i \mathrm{tr}\left((F^r)^\ast\mid H^i_\mathrm{\acute{e}t}(X_{\overline{\mathbb{F}_q}},\mathbb{Q}_\ell)\right)$

Remark: Note that $X(\mathbb{F}_{q^n})$ is the set of fixed points of the map $F^n$ on $X\left(\overline{\mathbb{F}_q}\right)$, and so this can be thought of as a generalization of Lefschetz fixed point theorem. In fact, there is a generalization of the above formula to much more general situations, which directly correlate to the Lefschetz fixed point theorem! Roughly it says that if you have an algebraically closed field $\overline{K}$, a variety $X/K$, and any map $F:X\to X$, then the number of non-degenerate fixed points is counted by a similar formula.

So, let’s begin our computation of $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$ by computing what the Betti numbers of $\mathbb{P}^m_K$ are. This is easy. One can either use standard Mayer-Vietoris arguments from $\ell$-adic cohomology. Or, after choosing an embedding $K\hookrightarrow\mathbb{C}$, use the fact that

$\dim_{\mathbb{Q}_\ell} H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)=\dim_{\mathbb{Q}_\ell} H^i\left(\mathbb{P}^m_\mathbb{C},\mathbb{Q}_\ell\right)=\dim_{\mathbb{Q}_\ell} H^i_\text{sing}\left(\mathbb{CP}^m,\mathbb{Q}_\ell\right)$

where we have used the so-called Grothendieck comparsion theorem, telling us how $\ell$-adic cohomology for varieties over $\mathbb{C}$ relates to the singular cohomology (with coefficients in $\mathbb{Q}_\ell$) of their analytification–they’re equal. In particular, we see that

$\dim_{\mathbb{Q}_\ell} H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)=\begin{cases}1 & \mbox{if}\quad i=0,2,\ldots,2m\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Now, what this automatically tells us is that $H^i(\mathbb{P}^m_K,\mathbb{Q}_\ell)$ is semi-simple! Indeed, any one (or zero) dimensional space is. Thus, we can entirely get our hands on $H^i(\mathbb{P}^m_K,\mathbb{Q}_\ell)$ if we can understand how $\text{Frob}_{\iota,\mathfrak{p}}$ acts for all $\iota$, and for all $\mathfrak{p}$.

So, now, let us begin by reducing the task of understanding $\text{Frob}_{\iota,\mathfrak{p}}$ on $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$, to a scenario for which Grothendieck’s trace formula comes into play. Namely, begin by considering $\mathbb{P}^m_{\mathcal{O}_{K_\mathfrak{p}}}/\text{Spec}(\mathcal{O}_{K_\mathfrak{p}})$. This is clearly smooth and projective. And so, assuming $\mathfrak{p}\nmid\ell$, we see by smooth proper base change, we have an isomorphism $H^i\left(\mathbb{P}^m_{K_\mathfrak{p}},\mathbb{Q}_\ell\right)\cong H^i\left(\mathbb{P}^m_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)$ in such a way which intertwines the map $G_{K_\mathfrak{p}}\to G_{k_\mathfrak{p}}$. So, we see that $\text{Frob}_{\mathfrak{p}}$ acts on $H^i\left(\mathbb{P}^m_{K_\mathfrak{p}},\mathbb{Q}_\ell\right)$ in the same way that $\text{Frob}_q$ (here $q=\# k_\mathfrak{p}$, and this is the standard map $x\mapsto x^q$) acts on $H^i\left(\mathbb{P}^m_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)$–in particular, they have the same trace. But, $H^i(\mathbb{P}^m_{K_\mathfrak{p}},\mathbb{Q}_\ell)$ with the obvious $G_{K_\mathfrak{p}}$-action coincides, as a $G_{K_\mathfrak{p}}$-representation, with $H^i\left(\mathbb{P}^m_k,\mathbb{Q}_\ell\right)$ with the action coming from an embedding $G_{K_\mathfrak{p}}\hookrightarrow G_K$.

Thus, we see that we’ve reduced our computation to understanding how $\text{Frob}_q$ works on $H^i\left(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell\right)$, where we assume that $\ell\ne p$ (where $q=p^m$) which is what we now do.

So, begin by noting that the Betti numbers $\dim_{\mathbb{Q}_\ell}H^i\left(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell\right)$ are the same as the ones computed above (remember our smooth proper base change computation!). Thus, if we denote $\text{tr}\left(F^\ast:H^i(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell)\to H^i(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell)\right)$ by $\tau_i$, for $i=0,\ldots,2m$, then necessarily we have that $\tau_i=0$ for $i=1,\ldots,2m-1$. Moreover, if we compute the traces of $(F^n)^\ast$, these are clearly just $\tau_i^n$.

Now, note that with this notation, the Grothendieck trace formula just says:

$\displaystyle \#\mathbb{P}^m_{\mathbb{F}_q}\left(\mathbb{F}_{q^n}\right)=\sum_{i=0}^{2m}(-1)^i \tau_i=1+\tau_2^n+\cdots+\tau_{2m}^n$

for all $n$. That said, we can compute the left hand side by purely combinatorial reasoning. Namely, we’re trying to count the number of distinct lines in $(\mathbb{F}_{q^n})^{m+1}$. Now, we can uniquely identify a line by choosing a non-zero point on it, of which there are $q^{n(m+1)}-1$ such points. But, we’ve greatly over-counted since choosing two different points on the same line will yield the same line. Thus, we need to divide by the number of non-zero points on a given line, which is just $\#(\mathbb{F}_{q^n})-1=q^n-1$. Thus, the left hand side is nothing but

$\displaystyle \frac{q^{n(m+1)}-1}{q^n-1}=1+q^n+\cdots+\left(q^n\right)^m$

So, now, let’s prove a slightly tricky, albeit elementary lemma:

Lemma 8: Suppose that $F$ is a field, and we have elements $\alpha,\beta_1,\ldots,\beta_m$ such that for all $n\geqslant 1$ the following holds

$1+\alpha^n+\left(\alpha^2\right)^n+\cdots+\left(\alpha^m\right)^n=1+\beta_1^n+\cdots+\beta_m^n$

and $\alpha$ is not a root of unity. Then, $\beta_i=\alpha^{a_i}$ for $a_i\in\{1,\ldots,m\}$.

Proof: Using Newton’s identities, since $p_i(1,\alpha,\alpha^2,\ldots,\alpha^m)=p_i(1,\beta_1,\ldots,\beta_m)$ you know that $e_i(1,\alpha,\ldots,\alpha^m)=e_i(1,\beta_1,\ldots,\beta_m)$. Thus, each $\beta_i$ is a root of

$X^{m} - e_{m-1}(1,\alpha,\ldots,\alpha^m)X^{m-1} + \dotsb + (-1)^{m-1}e_1(1,\alpha,\ldots,\alpha^m)x + (-1)^{m}e_0(1,\alpha,\ldots,\alpha^m)$

whose roots are precisely $\{1,\alpha,\ldots,\alpha^m\}$. Thus, we have that $\{1,\alpha,\ldots,\alpha^m\}=\{1,\beta_1,\ldots,\beta_m\}$. Now, since $\alpha$ is not a root of unity, the left hand side has exactly $m+1$ elements. So, now, if $\beta_i=1$, this would be a contradiction since the right hand side would the have at most $m$ elements. Thus, $\beta_i=\alpha^{a_i}$ for some $a_i=1,\ldots,m$. $\blacksquare$

Thus, the Grothendieck trace formula and this lemma allow us to deduce that $\tau_{2i}=q^{a_{2i}}$ for each $i=1,\ldots,m$. That said, from the Weil cohomology axioms, we actually can deduce much more. For example, using the orientation isomorphism, one can see that $\tau_{2m}=q^m$. But, the real important thing we can deduce is that $\tau_{2i}\tau_{2j}=\tau_{2(i+j)}$. Indeed, identify each $H^{2i}(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell)$ with $\mathbb{Q}_\ell$. Then, the cup-product gives us a $G_{\mathbb{F}_q}$-morphism $\mathbb{Q}_\ell\otimes_{\mathbb{Q}_\ell}\mathbb{Q}_\ell\to\mathbb{Q}_\ell$. One can check that this map is non-zero (just check it on one cover!). Thus, the $G_{\mathbb{F}_q}$-equivariance tells us precisely that $q^{a_{2i}}q^{a_{2j}}=q^{a_{2(i+j)}}$.

So, now we claim that this implies that $\tau_{2i}=q^i$ for all $i$. Indeed, note that $\tau_{2i}=\alpha_2^i$ by the above, and so it suffices to prove that $\alpha_2=q$. But, by the lemma we know that $\tau_2=q^i$ for some $i$. Consider $f(x)=1+\cdots+x^m$. This is an everywhere strictly increasing function on $x>0$. So, the fact that $f(\tau_2)=f(q)$ (by checking $n=1$ of the Grothendieck trace formula!) we conclude that necessarily $\tau_2=q$.

Thus, with the above, we can finally compute $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$:

Theorem 9: Let $K$ be a number field. Then, as $G_K$-modules

$\displaystyle H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)=\begin{cases}\mathbb{Q}_\ell\left(\frac{-i}{2}\right) & \mbox{if}\quad i=0,2,\ldots,2m\\ 0 & \mbox{if}\quad \text{otherwise}\end{cases}$

Proof: Our calculation of Betti numbers, as above, shows that for $i\ne 0,\ldots,2m$ the claim is correct. Thus, we’re left with the case $i=0,2,\ldots,2m$.

Now, since $\mathbb{P}^m_K$ admits a smooth projective model over all of $\text{Spec}(\mathcal{O}_K)$ (just $\mathbb{P}^m_{\mathcal{O}_K}$!), our discussion in the first section (the proof of Theorem 1), shows that $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$ is unramified outside of primes of $K$ lying over $\ell$ (remember, we’re ignoring the inifnite places here). Thus, by Chebotarev density, it suffices to understand $\text{tr}\left(\rho\left(\text{Frob}_{\iota,\mathfrak{p}}\right)\right)$, where $\rho:G_K\to \text{Aut}_{\mathbb{Q}_\ell}\left(H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)\right)$, $\mathfrak{p}\nmid \ell$, and $\iota:\overline{\mathbb{Q}}\hookrightarrow \overline{K_\mathfrak{p}}$ is some embedding.

But, as we mentioned before, the trace of $\text{Frob}_{\iota,\mathfrak{p}}$ acting on $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$ is the same as the trace of $\text{Frob}_q$ (once again, $q:= \# k_{\mathfrak{p}}$ acting on $H^i\left(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell\right)$. But, since $H^i\left(\mathbb{P}^m_{\mathbb{F}_q},\mathbb{Q}_\ell\right)$ is either $0$ or $1$-dimensional, one can quickly check that the trace of the action of $\text{Frob}_q$ is the reciprocal of the trace of the action of $F^\ast$. But, we calculated this to be $\displaystyle q^{\frac{i}{2}}$. Thus, we see that $\text{tr}\left(\rho\left(\text{Frob}_{\iota,\mathfrak{p}}\right)\right)=(\# k_\mathfrak{p})^{\frac{-i}{2}}$.

But, note that $\mathbb{Q}_\ell\left(\frac{-i}{2}\right)$ also has $\left(\# k_\mathfrak{p}\right)^{\frac{-i}{2}}$ as the trace of $\text{Frob}_{\iota,\mathfrak{p}}$ for all $\mathfrak{p}\nmid\ell$. Thus, by Chebotarev density, and Brauer-Nesbitt, it follows that $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)=\mathbb{Q}_\ell\left(\frac{-i}{2}\right)$.

Putting this all together, we get obtain the desired result. $\blacksquare$

Thus, we see that by applying the power of Brauer-Nesbitt/Chebotarev density, and just the basic properties of $\ell$-adic cohomology (the ones one would ‘want’ it to satisfy: Weil cohomology axioms, and trace formula), we can tease out the $G_K$-representation $H^i\left(\mathbb{P}^m_K,\mathbb{Q}_\ell\right)$.

# Could we have done the case for curves in the same way?

Now, while the computation of $H^i(C,\mathbb{Q}_\ell)$ above is undoubtedly the ‘correct’ way of computing the Galois representations of curves, and in particular, elliptic curves, the last section may put doubt in one’s mind. Namely, you may wonder whether or not we can check the guessed identity of $H^i(C,\mathbb{Q}_\ell)$ by computing traces of Frobenii.

Even if we don’t expect to be able to do this for all curves $C$, one case seems like a particularly good candidate for such a technique. Namely, if $C$ happens to be an elliptic curve $(E,O)$, then $H^1(E,\mathbb{Q}_\ell)$ (the non-trivial cohomology group) has something internal to do with $E$. Namely, since $\text{Jac}(E)\cong E$ in this case, we’d hope that we could do some internal-to-$E$ computation which would give the result. So, let’s run with this idea and see what happens.

Now, as discussed above, our elliptic curve $E/K$ has a smooth model $\mathcal{E}/U$, for some dense open subscheme $U\subseteq\text{Spec}(\mathcal{O}_K)$, which is necessarily of the form $\text{Spec}(\mathcal{O}_K)-S$, for some finite set $S$ of primes, which, without loss of generality, we assume contain the primes dividing $\ell$. So, for each $\mathfrak{p}\notin S$, and for each embedding $\iota:\overline{K}\hookrightarrow\overline{K_\mathfrak{p}}$, let’s compute $\text{tr}\left(\rho\left(\text{Frob}_{\iota,\mathfrak{p}}\right)\right)$, where $\rho:G_K\to\text{Aut}_{\mathbb{Q}_\ell}\left(H^1(E,\mathbb{Q}_\ell)\right)$ is our Galois representation.

To do this, we begin by consider $\mathcal{E}_{\mathcal{O}_{K_\mathfrak{p}}}/\text{Spec}(\mathcal{O}_{K_\mathfrak{p}})$, which is smooth and projective. Now, by smooth proper base change, we know that we have an isomorphism of $\mathbb{Q}_\ell$-spaces

$H^1(\mathcal{E}_{K_\mathfrak{p}},\mathbb{Q}_\ell)\cong H^1(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell)$

which intertwines the map $G_{K_\mathfrak{p}}\to G_{k_\mathfrak{p}}$. Moreover, the $G_{K_\mathfrak{p}}$-representation $H^1(\mathcal{E}_{K_\mathfrak{p}},\mathbb{Q}_\ell)$ is the same as the $G_{K_\mathfrak{p}}$-representation $H^i(E,\mathbb{Q}_\ell)$ given by the composition $G_{K_\mathfrak{p}}\to G_K$ induced by the inclusion $\iota:\overline{K}\hookrightarrow \overline{K_\mathfrak{p}}$. Thus, we see that $\text{tr}\left(\rho\left(\text{Frob}_{\iota,\mathfrak{p}}\right)\right)$ is the same as $\text{tr}\left(\text{Frob}_q\right)$ (as usual, $q:=\# k_\mathfrak{p}$) where we denote by $\text{Frob}_q$ the action of $\text{Frob}_q\in G_{k_\mathfrak{p}}$ on $H^1\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)$.

That said, a quick calculation, by the definition of $\text{Frob}_q$, and the geometric Frobenius map one checks that $\text{Frob}_q$‘s action on $H^1(\mathcal{E},\mathbb{Q}_\ell)^\vee$ is the same as $F$‘s action on $H^1(\mathcal{E},\mathbb{Q}_\ell)$. But, by the Grothendieck trace formula, we have that

$\displaystyle \# E\left(k_{\mathfrak{p}}\right)=\sum_{i=0}^{2}\text{tr}\left(F^\ast:H^i\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\to H^i\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\right)$

Then, just by the Weil cohomology theory axioms, we know that $H^0(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell)\cong\mathbb{Q}_\ell$ (as $G_{k_\mathfrak{p}}$-modules–i.e. the trivial action), and $H^2\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\cong \mathbb{Q}_\ell(-1)$ as $G_{k_\mathfrak{p}}$-module. So, since $F^\ast$‘s action on $\mathbb{Q}_\ell$ is the same as $\text{Frob}_q$‘s action on $\mathbb{Q}_\ell^\vee=\mathbb{Q}_\ell$, which is trivial, we see that

$\displaystyle \text{tr}\left(F^\ast:H^0\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\to H^0\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\right)=1$

Similarly, $F^\ast$‘s action on $H^2\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)=\mathbb{Q}_\ell(-1)$ is the same as $\text{Frob}_q's$ action on $\mathbb{Q}_\ell(-1)^\vee=\mathbb{Q}_\ell(1)$ which, by the very definition is $p$, so

$\text{tr}\left(F^\ast:H^2\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\to H^2\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\right)=\# k_{\mathfrak{p}}$

Thus, solving, we find

\begin{aligned}\text{tr}\left(\text{Frob}:H^1(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell)^\vee\to H^1(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell)^\vee\right) &=\text{tr}\left(F^\ast:H^1(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell)\to H^1\left(\mathcal{E}_{k_\mathfrak{p}},\mathbb{Q}_\ell\right)\right)\\ &=1+\# k_{\mathfrak{p}}-\# E(k_{\mathfrak{p}})\end{aligned}

On the other hand, it’s a super common fact (see, for example, Silverman’s text) that the action of $\text{Frob}_{\iota,\mathfrak{p}}$ on $V_\ell E$ is intertwined by the action of $\text{Frob}_q$ on $V_\ell \mathcal{E}_{k_{\mathfrak{p}}}$, which has trace $1+\# k_{\mathfrak{p}}-\# E(k_{\mathfrak{p}})$.

Thus, we have shown that for all $\mathfrak{p}\notin S$:

\begin{aligned}\text{tr}\left(\text{Frob}_{\iota,\mathfrak{p}}:H^1(E,\mathbb{Q}_\ell)^\vee\to H^1(E,\mathbb{Q}_\ell)^\vee\right) &= 1+\# k_{\mathfrak{p}}-\# E(k_{\mathfrak{p}})\\ &= \text{tr}\left(\text{Frob}:V_\ell \mathcal{E}_{k_{\mathfrak{p}}}\to V_\ell \mathcal{E}_{k_{\mathfrak{p}}}\right)\\ &= \text{tr}\left(\text{Frob}_{\iota,\mathfrak{p}}:V_\ell E\to V_\ell E\right)\end{aligned}

This is great! It seems like that, at least for elliptic curves, we actually can bypass much of the messiness in the previous computation of $H^1(E,\mathbb{Q}_\ell)$ in lieu of a nicer ‘Frobenius check’. Well, as usual, life is not so generous. Namely, we’re overlooking a simple, but important thing. Namely, we don’t a priori know that $H^1(E,\mathbb{Q}_\ell)$ is semi-simple! A posteriori we do, since it’s equal to $(V_\ell E)^\vee$ (which is semi-simple), but, as far as I can tell, there is no way of non-circularly deducing this. In fact, while it’s likely that $H^i(X,\mathbb{Q}_\ell)$ should be semi-simple in general, this is still an open conjecture (it’s implied by Tate’s conjecture, in fact).

Thus, all we’ve really shown is that the semi-simplification of $\rho$, usually denoted $\rho^{\text{ss}}$, is isomorphic to $(V_\ell E)^\vee$. But, of course, this is not satisfying. Thus, we see that while the method of checking a representation’s Frobenii action is unreasonably poweful, it requires at least some (often times difficult to deduce!) information about the representation to begin with.

1. As an aspiring arithmetic geometer, I’ve really enjoyed your blog! It’s filled a lot of conceptual gaps I had, so I’d like to thank you. I figure I’d just point out a few tiny pedantic typos here—it’s easy to infer what you actually mean, but since I noticed them anyway I thought I might as well let you know.

• “$\text{Gal}(L/\mathbb{Q})$ for some subextension $\mathbb{Q}\supseteq L\supseteq\mathbb{Q}$” should read “$\text{Gal}(L/\mathbb{Q})$ for some subextension $\overline{\mathbb{Q}}\supseteq L\supseteq\mathbb{Q}$

• “Thus, we obtain a continuous map $G_K\to\text{Aut}_{\mathbb{Z}}(H^i_{\text{et}}(X_{\overline{k}},\mathbb{Z}/\ell^n\mathbb{Z}))$” should read “Thus, we obtain a continuous map $G_K\to\text{Aut}_\mathbb{Z}(H^i_\text{et}(X_{\overline{K}},\mathbb{Z}/\ell^n\mathbb{Z}))$

• “note that $\text{Gal}(K(\mu_{\ell^n})/K)\hookrightarrow(\mathbb{Z}/\ell^n\mathbb{Z})$” should read “note that $\text{Gal}(K(\mu_{\ell^n})/K)\hookrightarrow(\mathbb{Z}/\ell^n\mathbb{Z})^\times$

• “an injection of $\text{Gal}(K(\mu_{\ell^\infty}))\rightarrow\mathbb{Z}_\ell^\times\subseteq\mathbb{Q}_\ell^\times=\text{GL}_1(\mathbb{Q}_\ell)$” should read “an injection of $\text{Gal}(K(\mu_{\ell^\infty}/K))\rightarrow\mathbb{Z}_\ell^\times\subseteq\mathbb{Q}_\ell^\times=\text{GL}_1(\mathbb{Q}_\ell)$

• “together with the action $(g\circ\varphi)(x) = \varphi(g^{-1}x)$” should probably read “together with the action $(g\varphi)(x) = \varphi(g^{-1}x)$,” since we’re forming the dual representation, which is actually a space of functions, so using $\circ$ for the $G_K$-action might be confusing

• “for some open subscheme $U\subseteq\mathfrak{X}$” should read “for some open subscheme $U\subseteq\text{Spec}\mathcal{O}_K$

• “the maximal Hausdorff quotient of $G$” should read “the maximal Hausdorff abelian quotient of $G$

• “We claim then that $h\text{Frob}_\mathfrak{P}h^{-1}\in g_U$” should read “We claim then that $h\text{Frob}_\mathfrak{P}h^{-1}\in gU$

• “we can identify $H^1_{\text{et}}(C_{\overline{K}},\mu_{\ell^n})$ with $\text{ker}\left(H^1_{\text{et}}(C_{\overline{K}},\mathbf{G}_m)\to H^1_{\text{et}}(C_{\overline{K}},\mu_{\ell^n})\right)$” should read “we can identify $H^1_{\text{et}}(C_{\overline{K}},\mu_{\ell^n})$ with $\text{ker}\left(H^1_{\text{et}}(C_{\overline{K}},\mathbf{G}_m)\to H^1_{\text{et}}(C_{\overline{K}},\mathbf{G}_m)\right)$

• “This morphism is the identity on $X_{\overline{\mathbb{F}_q}}$” should be “This morphism is the identity on $X_{\mathbb{F}_q}$

• When acting on an affine open subset, I think the arithmetic Frobenius $\text{Frob}_q$ only takes $q$-th powers of the coefficients, not the functions—in other words, if $\varphi\in\text{Gal}_{\mathbb{F}_q}$ is the $q$-th power map, then $\text{Frob}_q$ is the morphism $\text{id}_X\otimes_{\text{Spec}\mathbb{F}_q} vp:X_{\overline{\mathbb{F}_q}}\to X_{\overline{\mathbb{F}_q}}$.

• “unramified outside of primes of $K$ lying over $p$” should read “unramified outside of primes of $K$ lying over $\ell$

• There’s slight confusion in how you denote the traces of Frobenius. When you introduce them (“Thus, if we denote $\text{tr}$…”), you don’t explicitly assign symbols to them, though you seem to want to call them $a_i$. Then you write “$\beta_i = \alpha^{a_i}$” in Lemma 8—with the intent of letting $\alpha=q$, letting $\{\beta_1,\ldots,\beta_m\}=\{a_1,\ldots,a_m\}$, and letting $a_i\in\{1,\ldots,m\}$. As this conflicts with your earlier choice of notation for the traces of Frobenius (and this conflict spills over a little into the following paragraphs), might I recommend that you replace $\alpha$ with $r$, $\beta_i$ with $a_i$, and $a_i$ with $c_i$ in Lemma 8?

Hopefully these trivial details aren’t a total waste of your time—again, thanks for the great post!

1. Whoops, I seem to have used the wrong brackets for LaTeX—I’m still getting used to LaTeXing in WordPress!

1. PS Don’t worry about the LaTeX—I’ve fixed it. The general formula is $latex(SPACE)(Math)$ where (SPACE) is a literal space.

2. Hey!

Thank you very much for your kind comments, as well as your helpful suggestions.

I fixed all the typos that you mentioned.

With regards to the geometric Frobenius question. I think I must have been very sleepy when I wrote that portion since, at least to me, it was largely not understandable. I have rewritten it—hopefully now it is both clearer, and more instructive!

I have fixed the issue with the traces. I decided to denote the traces by $\tau_i$ (for the obvious reason), which I believe fixes everything.

Lastly, you should feel free to email me if you like—I always enjoy chatting with people interested in the same field.

Best,
Alex