Some examples of Geometric Galois representations

In this post we discuss the Galois representation associated to a projective scheme , where 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 ). 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 . Indeed, after all, every number field is a subextension of a Galois extension , the arithmetic of which is largely informed by its Galois group . But, is a quotient of . Thus, understanding is largely equivalent to understanding certain aspects of number fields.

Now, while ‘understanding ‘ 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 is hard. Thus, instead of trying to study 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 for various fields . That said, studying all representations are somewhat too hard for our purposes. Since we only care about continuous surjections (say, is for some subextension ), we might as well restrict ourselves to continuous homomorphisms , where is a topological field. If happens to be , with its standard topology, we call such representations -adic.

Considering the representation theory of has another appealing feature besides being a more tenable foothold to start attacking ‘understanding ‘. Namely, comes with a lot of local information bundled inside of it. Namely, for each prime, we can choose an embedding , which gives us an embedding . That said, both , 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 , where is a chosen geometric base point).

Tangential Remark: The map is injective by Krasner’s lemma. Namely, Krasner’s lemma tells us that the embedding is dense (since every finite extension of is of the form for some finite extensions , and some prime of lying over ). But, since consists of continuous (even isometric!) automorphisms over , this implies the restriction map is injective. In fact, choosing a prime of (the integral closure of in ) lying over , we may identify with the decomposition group .

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 (inner morphisms define equivalent representations!) but we have a veritable cornucopia of class functions on : , , 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 -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 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 (assumed perfect purely for conveinience), whose characteristic differs from , a contravariant functor

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:

This is a finite-dimensional -space by general theory. The Galois action is defined as follows. For each we have a morphism . We then get from standard étale cohomology theory a map

But, the pull-back of a constant sheaf is constant, and so we get that the right hand side is naturally . This map is -linear, and respects composition. Thus, we obtain a continuous map

Passing to the limit in gives us a continuous map . Then, finally, tensoring with gives us a continuous homomorphism .

Remark: A more fancy way of phrasing the above is as follows. Consider the structure morphism . Then, since is smooth and proper, we have a nice theory of the pushforward (e.g. it’s constructible). By general theory, we know that sheaves on correspond to -modules by moving to the stalk at a geometric point . But, as a group, proper base change (the ‘real’ proper base change) tells us that . Thus, we obtain a -modules structure on . Since, in general, cohomology plays well with limits, we can take the limit of this to obtain a -module which is a -module, and for which acts -linearly. Finally, tensoring with gives the -adic representation described above: . 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 can be quite difficult. Lucky for us, the functor is a so-called Weil cohomology theory and so enjoys, amongst others, the following helpful properties:

1. As we’ve already mentioned, is finite-dimensional.
2. We have a pairing of Galois modules called the cup product.
3. For , we have .
4. We have the so-called orientation isomorphism .
5. We have a version of Pontryagin duality, in particular, that for , the pairing is non-degenerate. Thus, in particular, we have that .
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 -algebras .

Moreover, while not strictly one of the axioms of a Weil cohomology theory, also benefits from the following extremely useful fact (an extremely specific application of smooth base change!):

(Smooth proper base change): Let is a complete DVR with function field and residue field , neither of which have characteristic , and suppose that is a smooth projective scheme. Then, there is an isomorphism such that for we have

where is the quotient map (whose kernel is the inertia group ).

This would be a good time to define the notation , for an integer, as defined above. For any field , note that , in the obvious way, for all . Thus, by passing to the limit, we obtain an injection of . The composition of this injection with the quotient map is called the cyclotomic character of , and is denoted (or just when 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 with the -action by , and the space with the similar action . For , we denote the representation . Choosing a basis, this is just the space with the action . For we define to be with the action , this is the same as the dual representation of . For a -module , we define its dual , to be the -space together with the action . One can check that this is, indeed, a -representation, if 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 be a number field. Then, if is a smooth projective variety, then for almost all primes of , the representation is unramified.

What it means for a representation of to be unramified at  is that for a choice of an embedding , we have the composition

is trivial. Or, said in words, the local inertia acts trivially on . 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 actually has a smooth model , for some open subscheme . 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 cutting out in , then since projective space is insensitive to clearing of denominators, we may assume without loss of generality that these equations lie in . Then, we just take the projective variety cut out by these same equations. Now, the morphism 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 we have that the pull-back is smooth and projective, which gives us our desired model.

Now, since is a curve, the open subscheme must be just , for some finite set of primes . Also, since it won’t affect finiteness, we assume without loss of generality that contains all primes of lying over . So, now, take any prime . Then, we have a morphism . We can then consider . Since is a complete DVR, smooth proper base-change tells us that we have an isomorphism in such a way which intertwines the map . In particular, note that this implies that acts trivially on .

That said, one easily sees that and, in particular, the action of on formed by the composition is the same as the action of on described above.

Since 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 be a characteristic zero field, and a (not necessarily commutative!) -algebra. Then, two semi-simple -modules and , which are finite-dimensional as -spaces, are isomorphic, if and only if for for all (here is multiplication by as a map , and similarly for .

As a corollary, we derive the following

Corollary 2: Let be a characteristic field, and an arbitrary field. Then, two Galois representations and are equivalent if and only if for all .

If we define the character of a Galois representation (poor notation, since it’s NOT a ‘character’ in the sense of a one-dimensional representation) to be , then the corollary above says that 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 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 be a Galois extension of number fields. Then, for any conjugacy class in , the Dirichlet density of the set

is .

Recall that for any unramified prime of , we can define, for each , a Frobenius element . This is the unique such that , where . Now, by the transitivity of ‘s action on primes of dividing , we see that the set , is, in fact, a full conjugacy class in . Thus, for almost all (the unramified ones!) the question of whether 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 be a number field, and let be a finite set of places. We define to be the maximal extension of in which is unramified outside of –so, the only ramification that can occur is at the finitely many places of . We shall denote by .

For example, if , and , then , and, in fact, is the maximal abelian subextension of (this follows from class field theory). Thus, we see that (recall that for a profinite group , is the maximal Hausdorff abelian quotient of –i.e. ) is just .

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

Corollary 3:Let be a number field, and a finite set of primes. For each finite prime , choose an embedding . The conjugacy classes of the maps are dense.

Before we begin the proof, let us clarify precisely what we mean by the above. The embedding gives rise to a map , and so on composition with the quotient map , a map . Since is unramified outside of , we see that maps to zero in . Thus, we get a well-defined map where, as usual is the absolute Galois group of the residue field of . But, has a canonical Frobenius element (the one defined by ) and we denote the image of this Frobenius element in by . Clearly, as the notation suggests, this Frobenius element not only depends on , but the embedding .

As mentioned in the motivation, here is another way to view this is as follows. For our given prime , let us choose a prime of (the integral closure of in ) lying over . The Galois group acts on by ring automorphisms, and since it fixes , permutes the primes of lying over . We can then define, in analogy with the finite case, the decomposition group  to be those elements of which send to itself. Since every element of fixes (not pointwise, just as a set!) we get a well-defined map , where (note that, as the notation suggests, this is the same field as above!). The kernel of this maps is denoted . Now, consider the composition . Since is unramified, one can show that maps to zero in , and thus, once again, we get a well-defined map . We denote the image of the Frobenius element in as .

It’s a good test of understanding to show that any two and , for different embeddings are conjugate. Moreover, a similar conjugacy statement holds for and for any two primes and of lying over . In fact, one can show that the two notions of and , as and range over their respective indexing sets, describe the exact same set of elements!

Proof(of Chebotarev density): Since is a profinite group, a basis of open sets is given by sets of the form , for , and an open subgroup. Consider the quotient which, by Galois theory, is equal to for some finite subextension of . Now, by Chebotarev density, there are infinitely many primes of which are unramified, and for which is conjugate to , for some prime of dividing . Since is finite, this implies that there is some . Then, by construction, there exists (here ) such that . So, choose a prime of lying over this . We claim then that . Indeed, this is the same thing as checking that . But, by functoriality, in , we have that , and so the conclusion follows.

Remark: At first glance, it looks like finding a conjugate to in didn’t need that was finite–there are always such by Chebotarev density! The fear is that if were infinite, then it may be equal to the set of for which , for , is conjugate to . This is no impediment to finding things conjugate to in , but then we can’t lift those Frobenii up to , since they make no sense there. In short, we need to be cognizant that we need to find for which is conjugate to , and is not ramified in !

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

Theorem 4: Let be a number field, and let and be two semi-simple Galois representations, where is a Hausdorff field of characteristic , and which are unramified outside of finite sets and respectively. Then, is equivalent to if and only if for all primes in lying above a prime not in .

Note that is also independent of which dividing we choose!

Proof: Note that since and are unramified outside of , we have that both factor through , call these factorizations and . Then, by assumption, the maps and agree on the dense set . Since is Hausdorff, this implies that these two functions must be equal. So, the rest follows by Brauer-Nesbitt.

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 -adic Galois representations are to be of any use, we’d at least hope that we can have a reasonable description of them in the case when is the simplest non-trivial object–a curve over (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 , call it (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 was. He said, “well, that’s hard”. To this, I responded “Well, OK. But, certainly is trivial, right?” Looking a bit ill he dropped this bomb: “well, computing the value of is equivalent to computing the homotopy groups of spheres…”

Now, while I said that we can compute Galois representations , for a curve over the field (with characteristic not ), 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 be a curve. Then, the following holds:

Moreover, the Galois action on agrees with the obvious Galois action on (i.e. given , the action ).

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 is any factorial, integral variety, then one can show that

The and cases are trivial and famous respectively. The case is the slightly tricky fact. It follows because one has a canonical flasque resolution of :

Here is the constant sheaf with values in (here is the function field of ). Now, it’s clear that is flasque, but the quotient is much less obviously so. The trick is to use the fact that is factorial and integral to identify this sheaf of Cartier divisors with the sheaf , defined by , the group of Weil divisors on , 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 on 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 be a field with characteristic coprime to , and let be a curve. Then, for all

where the identification is of -modules.

Before we begin the proof, let us say that the action of -modules is similar to the action described above. Namely, it’s the -module associated to the stalk of, say, at some geometric point.

Proof:  So, we’re trying to compute . But, we then consider the obvious complex of abelian sheaves:

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 . But, taking stalks leaves us to show that we have a short exact sequence

where is the strictly Henselian ring (the meaning strict Henselization and is the image of ). Exactness on the left, and middle are clear. To see exactness at the right, we need to show that any element of has a -th root. So, consider the polynomial . Moving to the residue field we have the separable polynomial (because again!) . This has a root since is algebraically closed, and so by Henselianess, has a root in . The exactness at the right follows.

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

which is a sequence of -modules, since the Kummer sequence itself was a sequence of -sheaves. Now, we’ve already taken the computation of for granted. Now, note that is surjective, and thus the map is injective. Thus, as a -module, we can identify with which, by additivity, is just ‘multiplication by ‘. But, identifying with , we see that we have an identification of -modules

which gives the desired result for .

Finally, to compute the above shows us that we have and identification

where the last step uses the fact that decomposes as , and uses the fact that is -divisible.

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

Theorem 7: Let be a field with , and a curve. Then,

Before we give the proof, let us quickly remind ourselves what is. Recall that for a curve of genus , there is an abelian variety, the Jacobian of , denoted , representing the relative degree Picard functor (for more information, see Kleiman’s famous notes). Then, for any abelian variety we have the associated Tate module  defined as , with the obvious Galois action (the Galois group acts additively!). This is a free -module of rank . Thus, upon tensoring with we obtain a -dimensional -adic Galois representation of . This is an object which, ostensibly complicated, is fairly well-understood.

Proof: Let us begin with the most important case of . Now, above we computed that as a -module was which, by the precious paragraph, is just . Thus, upon passing to the limit, we obtain that, as a -module we have that is just , and so tensoring with gives

Then, finally, we appeal to one last abstract nonsense fact from -adic cohomology. Namely, for any smooth projective variety , the -module

is known as . It then follows (essentially from a version of the projection formula) that . So, finally we see that

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

and that (i.e. Jacobians are self-dual).

Remark: Martin Olsson pointed out to me that there is an easier way to prove that . There is always a map of -modules

Taking , , and gives us an injective map of -modules

But, these two -spaces have the same dimension since, if one forgets about the action of , (i.e. as sheaves on one has a NON -equivariant isomorphism ). Thus, it is actually an isomorphism of -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 -action. In fact, the above Theorem 7 is usually stated, but one forgets the -action, essentially just stating that . This is proven by actually computing as above, and doing the identification (i.e. ). This, as remarked above, is totally fine if all you care about is the structure as a -space. In fact, if you only care about these Betti numbers, then you can just reduce the computation to the case over (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 -module since you, having done it before, just willy-nilly make this identification. I wish that this comment, and more generally a focus on the -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: . 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. ). For lack of motivation, I will not record here the more ‘standard way’ of computing the -module . The key is to use the fact that the -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 , where is a number field, using a somewhat different technique. Namely, we’ll figure out what the action of Frobenii are on 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 -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 .

Recall first that if is any scheme over , then there is a morphism called the absolute Frobenius of . This is the map which is the identity on the underlying topological space of , and which on an affine open corresponds to the ring map . Note that if is a scheme over , with , then is not a morphism of -schemes. Indeed, the composition corresponds precisely to the usual Frobenius map on which is not the identity. That said, one can easily see that is a morphism of -schemes.

This absolute Frobenius then allows us to define four different types of Frobenius on where is some variety. Indeed, we can take , we can take , we can take , or we can take . 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 , so . Then, takes to . The relative Frobenius takes to the , but leaves the coefficients alone. The arithmetic Frobenius acts by taking the coefficients to the -power, and the geometric Frobenius acts by taking -roots of the coefficients.

Now, the important thing to note is that all the Frobenii are (iso)morphisms of schemes , and so induce (iso)morphism of . The only two important facts we’ll need are:

1. The (iso)morphisms and induce the same (iso)morphisms of .
2. The (iso)morphism induced by is the one corresponding to the action of the geomtric Frobenius in on (definitionally).

We then have the following famed theorem of Grothendieck:

Theorem(Grothendieck-Lefschetz formula): Let be a smooth projective scheme. Denote by the morphism . Then, for all , we have the following equality:

Remark: Note that is the set of fixed points of the map on , 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 , a variety , and any map , then the number of non-degenerate fixed points is counted by a similar formula.

So, let’s begin our computation of by computing what the Betti numbers of are. This is easy. One can either use standard Mayer-Vietoris arguments from -adic cohomology. Or, after choosing an embedding , use the fact that

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

Now, what this automatically tells us is that is semi-simple! Indeed, any one (or zero) dimensional space is. Thus, we can entirely get our hands on if we can understand how acts for all , and for all .

So, now, let us begin by reducing the task of understanding on , to a scenario for which Grothendieck’s trace formula comes into play. Namely, begin by considering . This is clearly smooth and projective. And so, assuming , we see by smooth proper base change, we have an isomorphism in such a way which intertwines the map . So, we see that acts on in the same way that (here , and this is the standard map ) acts on –in particular, they have the same trace. But, with the obvious -action coincides, as a -representation, with with the action coming from an embedding .

Thus, we see that we’ve reduced our computation to understanding how works on , where we assume that (where ) which is what we now do.

So, begin by noting that the Betti numbers  are the same as the ones computed above (remember our smooth proper base change computation!). Thus, if we denote by , for , then necessarily we have that for . Moreover, if we compute the traces of , these are clearly just .

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

for all . 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 . Now, we can uniquely identify a line by choosing a non-zero point on it, of which there are 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 . Thus, the left hand side is nothing but

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

Lemma 8: Suppose that is a field, and we have elements such that for all the following holds

and is not a root of unity. Then, for .

Proof: Using Newton’s identities, since you know that . Thus, each is a root of

whose roots are precisely . Thus, we have that . Now, since is not a root of unity, the left hand side has exactly elements. So, now, if , this would be a contradiction since the right hand side would the have at most elements. Thus, for some .

Thus, the Grothendieck trace formula and this lemma allow us to deduce that for each . That said, from the Weil cohomology axioms, we actually can deduce much more. For example, using the orientation isomorphism, one can see that . But, the real important thing we can deduce is that . Indeed, identify each with . Then, the cup-product gives us a -morphism . One can check that this map is non-zero (just check it on one cover!). Thus, the -equivariance tells us precisely that .

So, now we claim that this implies that for all . Indeed, note that by the above, and so it suffices to prove that . But, by the lemma we know that for some . Consider . This is an everywhere strictly increasing function on . So, the fact that (by checking of the Grothendieck trace formula!) we conclude that necessarily .

Thus, with the above, we can finally compute :

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

Proof: Our calculation of Betti numbers, as above, shows that for the claim is correct. Thus, we’re left with the case .

Now, since admits a smooth projective model over all of (just !), our discussion in the first section (the proof of Theorem 1), shows that is unramified outside of primes of lying over (remember, we’re ignoring the inifnite places here). Thus, by Chebotarev density, it suffices to understand , where , , and is some embedding.

But, as we mentioned before, the trace of acting on is the same as the trace of (once again, acting on . But, since is either or -dimensional, one can quickly check that the trace of the action of is the reciprocal of the trace of the action of . But, we calculated this to be . Thus, we see that .

But, note that also has as the trace of for all . Thus, by Chebotarev density, and Brauer-Nesbitt, it follows that .

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

Thus, we see that by applying the power of Brauer-Nesbitt/Chebotarev density, and just the basic properties of -adic cohomology (the ones one would ‘want’ it to satisfy: Weil cohomology axioms, and trace formula), we can tease out the -representation .

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

Now, while the computation of 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 by computing traces of Frobenii.

Even if we don’t expect to be able to do this for all curves , one case seems like a particularly good candidate for such a technique. Namely, if happens to be an elliptic curve , then (the non-trivial cohomology group) has something internal to do with . Namely, since in this case, we’d hope that we could do some internal-to- computation which would give the result. So, let’s run with this idea and see what happens.

Now, as discussed above, our elliptic curve has a smooth model , for some dense open subscheme , which is necessarily of the form , for some finite set of primes, which, without loss of generality, we assume contain the primes dividing . So, for each , and for each embedding , let’s compute , where is our Galois representation.

To do this, we begin by consider , which is smooth and projective. Now, by smooth proper base change, we know that we have an isomorphism of -spaces

which intertwines the map . Moreover, the -representation is the same as the -representation given by the composition induced by the inclusion . Thus, we see that is the same as (as usual, ) where we denote by the action of on .

That said, a quick calculation, by the definition of , and the geometric Frobenius map one checks that ‘s action on is the same as ‘s action on . But, by the Grothendieck trace formula, we have that

Then, just by the Weil cohomology theory axioms, we know that (as -modules–i.e. the trivial action), and as -module. So, since ‘s action on is the same as ‘s action on , which is trivial, we see that

Similarly, ‘s action on is the same as action on which, by the very definition is , so

Thus, solving, we find

On the other hand, it’s a super common fact (see, for example, Silverman’s text) that the action of on is intertwined by the action of on , which has trace .

Thus, we have shown that for all :

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 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 is semi-simple! A posteriori we do, since it’s equal to (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 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 , usually denoted , is isomorphic to . 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.