Let be an algebraic variety over ; that is to say, the zero locus of a bunch of polynomials with complex coefficients. We will consider this zero locus as a topological space using the ordinary topology on . One of the main themes of algebraic geometry in the last century has been learning how to study the topology of in terms of the algebraic properties of the defining equations.
In this post, I will explain that there are intrinsic limits to this approach; things that cannot be computed algebraically. In particular, I want to explain how from a categorical point of view, we can’t even compute the homology . And, even if you don’t believe in categories, you’ll still have to concede that we can’t compute . This is a very pretty example and it should be more widely known.
Absolutely none of the ideas in this post are original; I think most of them are due to Serre. (Thanks to Attila Smith in comments for the reference.)
What does it mean that something cannot be computed algebraically? I could give a general definition, but let me just explain the kind of examples we will be looking at. Let be a finite Galois extension of . Suppose that all of the equations defining have coefficients in . Let be the new variety obtained by replacing all the coefficients of by their Galois conjugates, for some . Since the Galois action preserves the truth of every algebraic statement; clearly we will not be able to use algebraic methods to distinguish and . Thus, if we can come up with cases where and have different topology, then the manner in which they differ will be something that cannot be seen by algebraic methods.
If we were looking at topology of real points, it would be easy to build examples. For example, let and let be the elliptic curve , so is the elliptic curve . Looking at real solutions, has two connected components and has only one. In the figure below, is the solid black curve and is the dashed blue curve.
Looking at complex solutions, however, both and are two dimensional tori. (More precisely, tori with a single point removed; the “point at infinity”.) So one might think that determining the real geometry of a variety requires some subtle ideas, but that complex geometry is purely algebraic. In fact, this is not so.
Consider two lattices in the complex plane. The first, , will be spanned by and . The second, , Will be spanned by and .
Let be the elliptic curve and let . Notice that both and are taken into themselves under multiplication by . This multiplication descends to holomorphic endomorphisms of and , which we will denote and .
Since is an elliptic curve, it can be written as for some and . (Plus a point at infinity.) It turns out that the endomorphism imposes enough rigidity on the situation to see that and can be taken to be algebraic numbers; I will sketch an explanation for this below. In fact, we can take and to be in . Moreover, the map can be written as , where and are rational functions whose coefficients are algebraic — explicitly, in .
I wanted to compute , , and for you*, but I couldn’t find them in any books or websites and it seemed a bit difficult. I was able to work out that the invariant of is
There is a theory of elliptic curves with unusual endomorphisms — the technical term is curves with complex multiplication. Here is one consequence of this theory: Let and let be the Galois automorphism where and . Then the curve is given by . Moreover, the automorphism is given by applying to the coefficients of .
So, why do I say that “from a categorical point of view, we can’t even compute the homology .” Of course, is naturally isomorphic to , and unnaturally isomorphic to ; similarly, is also unnaturally isomorphic to . So you could say that and were the same, if you are willing to engage in unnatural acts.
But homology is a functor! Computing it should means computing what it does to morphisms, not just what it does to objects! Looking at how multiplication by acts on and . we see that the module is generated by a single element, where as the module is not. So there is an element of so that and span , but there is no so that and span . In short, the action of an endomorphism on cannot be computed algebraically.
A clever trick, for those who mistrust categories: If you believe in the categorical philosophy, you are done now. But, if you are a classical sort, you might want to see two varieties and which are not homeomorphic.
If we were living in the topological category, we could take the mapping cylinder of and get a map whose topology reflects the map . But in the algebraic category, mapping cylinders don’t exist. In the remainder of this post, I’ll sketch a trick to get around that, due to again to Serre.
Just like abelian curves come from lattices in , abelian varieties come from lattices in higher dimensional complex vector spaces. Just as has nonisomorphic modules, so does for certain values of . A more complicated version of the above argument constructs an abelian variety , with endomorphism , such that , and such that there is a Galois automorphism for which is topologically different from . I’ll write for the dimension of ; for the curious, .
Now, let be simply connected, with a free action of . (And let this variety and action be given by rational coefficients, so it is unaffected by .) Consider . This is , where acts diagonally. It is extremely plausible, and not hard to prove, that . Now, some basic topology tells us that and are both semidirect products of acting on . However, in the case of , this action is by the action of on , and in the case of it is by the other action. These are nonisomorphic groups, so we see that even those who mistrust categories must admit that cannot be computed algebraically.
Sketch of why these numbers are algebraic. We can write down the condition that and give an automorphism of , whose square is . This is a whole lot of equations in , and the coefficients of and ; all the coefficients of these equations are rational. Now, an elliptic curve will only have such an automorphism if is closed under multiplication by . It is a good exercise to show that any such lattice is, up to dilation and rotation, either or . So these equations will only have two solutions. (Actually, they will have two solutions up to automorphism, see the footnote. But this is a sketch!) Any set of equations with finitely many solutions, and coefficients in a field , has all of its solutions in ; this is a corollary of the Nullstellensatz.
*There is a technical point I am glossing over: and are not quite well defined. Specifically, the elliptic curve is, as a complex variety, isomorphic to . Moreover, even if , and are all in , the isomorphism will involve coefficients in . This is called twisting, and is important if you want to get all the details right, but I will not deal with it.