This Week’s Finds in Arithmetic Geometry

It has been going around the math blogosphere that in honor of John Baez’s 20 year anniversary of doing This Week’s Finds we all do one in our area. So here’s a brief This Week’s Finds in Arithmetic Geometry. Hopefully this will raise awareness of the blog that essentially pioneered math/physics blogging (and if you’re into arithmetic geometry some papers you might not have caught yet).

Since the content and style of Baez’s This Week’s Finds vary so much, I’ll just copy what Jordan Ellenberg did here and give some papers posted the last week that caught my attention.

The fields of definition of branched Galois covers of the projective line by Hilaf Hasson caught my attention because I just met him and saw him speak on this exact topic a few weeks ago at the Joint Meetings. Although the results are certainly interesting in themselves, the part that a blog audience might appreciate is the set of corollaries to the results.

Recall that a major open problem in number theory is the “Inverse Galois Problem” which asks which groups arise as Galois groups. I even posted an elementary proof that if you don’t care what your fields are, then any finite group arises as {Gal(L/K)}. In general (for example if you force {K=\mathbb{Q}}), then the problem is extremely hard and wide open.

If you haven’t seen this type of thing before, then it might be surprising, but you can actually use geometry to study this question. This is exactly the type of result that Hilaf gets.

Next is New derived autoequivalences of Hilbert schemes and generalised Kummer varieties by Andreas Krug. This topic is near and dear to me because I study derived categories in the arithmetic setting. I haven’t taken a look at this paper in any depth, but I’ll just point out why these types of things are important.

In the classification of varieties one often tries to study the problem up to some type of birational equivalence otherwise it would be too difficult. Often times birational varieties are derived equivalent, but not the other way around. So one could think of studying varieties up to derived equivalence as a slightly looser classification.

When trying to figure out what two varieties that are derived equivalent have in common, a typical sticking point is that you need to know certain automorphisms of the derived category (i.e. autoequivalence) exist to get nice cohomological properties or something. When papers constructing new autoequivalences come out it always catches my attention because I want to know if the method used transfers to situations I work in.

Lastly, we’ll do La conjecture de Tate entière pour les cubiques de dimension quatre sur un corps fini by François Charles and Alena Pirutka. If you don’t know what the Tate conjecture is, then lots of people refer to it as the Hodge conjecture in positive characteristic.

If you’ve ever seen any cohomology theory, then you should be at least passingly familiar with the idea that certain sub-objects (subvarieties or submanifolds etc) can be realized as classes in the cohomology. Sometimes this is due to construction and sometimes it is a major theorem.

The particular case of the Tate conjecture says the following. Consider the relatively easy to prove fact. If you take a cycle on your variety {X/k}, then the cohomology class it maps to (in {\ell}-adic cohomology) will be invariant under the natural Galois action {Gal(\overline{k}/k)} (because it is defined over {k}!). The Tate conjecture is that any Galois invariant cohomology class actually comes from one of these cycles.

The fact that mathematicians can have honest arguments over whether or not the Tate conjecture or the Hodge conjecture (a million dollar problem!) is harder just gives credence to the fact that it is darned hard. If you weren’t convinced, then just consider that this paper is proving the Tate conjecture in the particular case of smooth hypersurfaces of degree {3} in {\mathbb{P}^5} just for the cohomology classes of degree {4}. People consider this progress, and they should.


An Application to Elliptic Curves

Let’s do an application of our theorems about finitely generated projective modules over Dedekind domains. This is another one of those things that seems to be quite well known to experts, but it is not written anywhere that I know of. Suppose {E} and {F} are elliptic curves defined over a number field {K} (this works in more generality, but this assumption will allow us to not break into lots of weird cases), and assume that the {\ell}-adic Tate modules are isomorphic for all {\ell}.

Recall briefly that the {\ell}-adic Tate module is just the limit over all the {\ell^n} torsion points, i.e. {\displaystyle T_\ell(E)=\lim_{\longleftarrow} E(\overline{K})[\ell^n]} as a {G=Gal(\overline{K}/K)}-module. We discussed this before in this post. An isogeny {\phi: E\rightarrow F} defined over {K} induces an action via pushforward {T_\ell (E)\rightarrow T_\ell (F)} which is Galois equivariant. In fact, if {\ell \nmid \text{deg}(\phi)}, then it induces an isomorphism of Tate modules.

First define {Hom(E,F)} to be the set of isogenies over {K} (this could get me in trouble and has been the main delay in this post). If {E} and {F} are isogenous, then the natural action of {End(E)} by composing turns {Hom(E,F)} into a rank {1} projective module over {End(E)}.

The question we want to ask ourselves is how much information do we get from the Tate module. It seems that surely this would not be enough information to recover the curve up to isomorphism, but recall that most elliptic curves do not have complex multiplication. Let’s start with that case. Suppose {E} is non-CM so that {End(E)\simeq \mathbb{Z}}. The only endomorphisms are the isogenies given by multiplication by an integer.

The Tate conjecture formally says that there is an isomorphism {Hom(E,F)\otimes_\mathbb{Z} \mathbb{Z}_\ell \stackrel{\sim}{\rightarrow}Hom_G(T_\ell(E), T_\ell(F))} (proved by Faltings in this case). This tells us that if you have some isomorphism {T_\ell(E)\simeq T_\ell(F)}, then there is an isogeny that induces it (maybe there is a less powerful tool to see this in this case). But we’ve now assumed that {End(E)} is {\mathbb{Z}}, so {Hom(E,F)} is not just a locally free module of rank {1}, but just plain free of rank {1}. All other isogenies are just composing this one with multiplication by an integer.

Let {\phi} be the generator of {Hom(E,F)}. If {deg(\phi)} is {n}, then all isogenies are divisible by {n}. Since we assume the Tate modules are isomorphic for all {\ell}, just pick some {\ell} that divides {n}. Since Tate says there is an isogeny inducing the isomorphism we get a contradiction unless {n=1}. Thus {deg(\phi)=1} and hence the generator is actually an isomorphism. This proves a fact I’ve seen stated, but haven’t seen written anywhere. If {E} and {F} are non-CM elliptic curves with isomorphic Tate modules for all {\ell}, then they must be isomorphic.

This should seem a little strange, because it basically says we can recover the curve up to isomorphism merely from knowing {H_1}. It turns out that weirder things can happen for CM curves, but we can use our structure theory from the last post to figure out what is going on. Suppose now that {End(E)} is the full ring of integers in a quadratic imaginary field (the only other possibility is that it is merely an order in such a field).

It turns out that if {E} and {F} have isomorphic Tate modules for all {\ell}, then we can’t just conclude they are isomorphic. Here is a good way to think about this. We have that {End(E)} is a Dedekind domain, and {Hom(E,F)} is a rank {1} projective module over it, so it is either generated by {1} element and hence free in which case the same type of argument will show {E} and {F} must be isomorphic. The reason we get no information in the case where it is generated by two things is that these degrees can be coprime. In fact, they must be or else the same argument gives an isomorphism again.

This recently came up in something I was working on, and I couldn’t believe that I couldn’t find this fact stated anywhere (but several number theorists confirmed that this was something they knew). It might be because introductory books don’t want to assume the Tate conjecture, and anything that does assume the Tate conjecture assumes you can figure this out for yourself.