# Topology that Algebra can’t see

Let $X$ be an algebraic variety over $\mathbb{C}$; 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 $\mathbb{C}$. One of the main themes of algebraic geometry in the last century has been learning how to study the topology of $X$ 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 $H_1( \ , \mathbb{Z})$. And, even if you don’t believe in categories, you’ll still have to concede that we can’t compute $\pi_1( \ )$. 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 $K$ be a finite Galois extension of $\mathbb{Q}$. Suppose that all of the equations defining $X$ have coefficients in $K$. Let $X^{\sigma}$ be the new variety obtained by replacing all the coefficients of $X$ by their Galois conjugates, for some $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$. Since the Galois action preserves the truth of every algebraic statement; clearly we will not be able to use algebraic methods to distinguish $X$ and $X^{\sigma}$. Thus, if we can come up with cases where $X$ and $X^{\sigma}$ 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 $K = \mathbb{Q}[\sqrt{10}]$ and let $X$ be the elliptic curve $y^2 = x^3 - \sqrt{10} x +1$, so $X^{\sigma}$ is the elliptic curve $y^2 = x^3 + \sqrt{10} x +1$. Looking at real solutions, $X$ has two connected components and $X^{\sigma}$ has only one. In the figure below, $X$ is the solid black curve and $X^{\sigma}$ is the dashed blue curve.

Looking at complex solutions, however, both $X$ and $X^{\sigma}$ 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, $M$, will be spanned by $1$ and $\sqrt{-5}$. The second, $N$, Will be spanned by $2$ and $1+\sqrt{-5}$.

Let $E$ be the elliptic curve $\mathbb{C}/M$ and let $F = \mathbb{C}/N$. Notice that both $M$ and $N$ are taken into themselves under multiplication by $\sqrt{-5}$. This multiplication descends to holomorphic endomorphisms of $E$ and $F$, which we will denote $s$ and $t$.

Since $E$ is an elliptic curve, it can be written as $y^2 = x^3 + ax + b$ for some $a$ and $b$. (Plus a point at infinity.) It turns out that the endomorphism $s$ imposes enough rigidity on the situation to see that $a$ and $b$ can be taken to be algebraic numbers; I will sketch an explanation for this below. In fact, we can take $a$ and $b$ to be in $\mathbb{Q}(\sqrt{5})$. Moreover, the map $s$ can be written as $(x,y ) \mapsto (f(x,y), g(x,y))$, where $f$ and $g$ are rational functions whose coefficients are algebraic — explicitly, in $\mathbb{Q}(\sqrt{5}, \sqrt{-5})$.

I wanted to compute $a$, $b$, $f$ and $g$ 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 $j$ invariant of $E$ is
$\displaystyle{ -632000 + 282880 \sqrt{5} =2^8 * \sqrt{5}^3 * (4+\sqrt{5})^3 * (2-\sqrt{5})^4}.$

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 $K = \mathbb{Q}(\sqrt{5}, \sqrt{-5})$ and let $\sigma$ be the Galois automorphism where $\sqrt{5} \mapsto - \sqrt{5}$ and $\sqrt{-5} \mapsto \sqrt{-5}$. Then the curve $F$ is given by $y^2 = x^3 + a^{\sigma} x + b^{\sigma}$. Moreover, the automorphism $t$ is given by applying $\sigma$ to the coefficients of $s$.

So, why do I say that “from a categorical point of view, we can’t even compute the homology $H_1*( \ , \mathbb{Z})$.” Of course, $H_1(E, \mathbb{Z})$ is naturally isomorphic to $M$, and unnaturally isomorphic to $\mathbb{Z}^2$; similarly, $H_1(F, \mathbb{Z})$ is also unnaturally isomorphic to $\mathbb{Z}^2$. So you could say that $H_1(E, \mathbb{Z})$ and $H_1(F, \mathbb{Z})$ 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 $\sqrt{-5}$ acts on $M$ and $N$. we see that the $\mathbb{Z}[\sqrt{-5}]$ module $M$ is generated by a single element, where as the $\mathbb{Z}[\sqrt{-5}]$ module $N$ is not. So there is an element $\gamma$ of $H_1(E, \mathbb{Z})$ so that $\gamma$ and $s_* \gamma$ span $H_1(E, \mathbb{Z})$, but there is no $\delta$ so that $\delta$ and $t_* \delta$ span $H_1(F, \mathbb{Z})$. In short, the action of an endomorphism on $H_1$ 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 $X$ and $X^{\sigma}$ which are not homeomorphic.
If we were living in the topological category, we could take the mapping cylinder of $s: E \to E$ and get a map whose topology reflects the map $s$. 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 $\mathbb{C}$, abelian varieties come from lattices in higher dimensional complex vector spaces. Just as $\mathbb{Z}[\sqrt{-5}]$ has nonisomorphic modules, so does $\mathbb{Z}[e^{2 \pi i/n}]$ for certain values of $n$. A more complicated version of the above argument constructs an abelian variety $A$, with endomorphism $z$, such that $z^n = \mathrm{Id}$, and such that there is a Galois automorphism $\sigma$ for which $z^{\sigma} : A^{\sigma} \to A^{\sigma}$ is topologically different from $z: A \to A$. I’ll write $g$ for the dimension of $A$; for the curious, $g = \phi(n)/2$.

Now, let $B$ be simply connected, with a free action of $\mathbb{Z}/n$. (And let this variety and action be given by rational coefficients, so it is unaffected by $\sigma$.) Consider $X := A \times_{\mathbb{Z}/n} B$. This is $(A \times B)/(\mathbb{Z}/n)$, where $(\mathbb{Z}/n)$ acts diagonally. It is extremely plausible, and not hard to prove, that $X^{\sigma} = A^{\sigma} \times_{\mathbb{Z}/n} B$. Now, some basic topology tells us that $\pi_1(X)$ and $\pi_1(X^{\sigma})$ are both semidirect products of $\mathbb{Z}/n$ acting on $\mathbb{Z}^{2g}$. However, in the case of $A$, this action is by the action of $z_*$ on $H_1(A, \mathbb{Z})$, and in the case of $A^{\sigma}$ it is by the other action. These are nonisomorphic groups, so we see that even those who mistrust categories must admit that $\pi_1$ cannot be computed algebraically.

Sketch of why these numbers are algebraic. We can write down the condition that $f$ and $g$ give an automorphism of $y^2 = x^3 + ax +b$, whose square is $[-5]$. This is a whole lot of equations in $a$, $b$ and the coefficients of $f$ and $g$; all the coefficients of these equations are rational. Now, an elliptic curve $\mathbb{C}/\Lambda$ will only have such an automorphism if $\Lambda$ is closed under multiplication by $\sqrt{-5}$. It is a good exercise to show that any such lattice is, up to dilation and rotation, either $M$ or $N$. 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 $K$, has all of its solutions in $K^{\mathrm{alg}}$; this is a corollary of the Nullstellensatz.

*There is a technical point I am glossing over: $a$ and $b$ are not quite well defined. Specifically, the elliptic curve $y^2 = x^3 + ax +b$ is, as a complex variety, isomorphic to $y^2 = x^3 + D^2 ax + D^3 b$. Moreover, even if $a$, $b$ and $D$ are all in $\mathbb{Q}$, the isomorphism will involve coefficients in $\mathbb{Q}[\sqrt{D}]$. This is called twisting, and is important if you want to get all the details right, but I will not deal with it.

## 15 thoughts on “Topology that Algebra can’t see”

1. Attila Smith says:

Dear David,
the relevant article by Serre might be:

Exemple de variétés projectives conjuguées non homéomorphes.

Your explanations are friendly and clear: bravo!

2. Thanks for the reference and the complement; I’ll update the article.

3. Jesse Kass says:

Dear David,
Nice post!

You may have already seen the paper, but James Milne and Junecue Suh have some recent work on this topic too.

http://arxiv.org/abs/0804.1953

4. Wesley Calvert says:

Russell Miller and I treated a rather different formalization of the same intuitive problem (can one compute topological invariants from elementary data) in a paper soon to appear in the proceedings of Unconventional Computation 2009.

We showed that if a manifold is “effectively given,” in the sense that we have algebraically computable functions (in the sense of Blum-Shub-Smale) to describe the transition functions, then

a) We cannot, in most cases, decide if a loop is nullhomotopic, but

b) There is a canonical system of loops which suffices to compute a presentation of pi_1.

I don’t have the preprint available online (yet), but can provide it upon request.

Thanks for the interesting post!

5. If you view your curve as a real algebraic set in $\mathbb{R}^4$ then you certainly can compute its Betti numbers algebraically (well, semi-algebraically). See e.g. http://arxiv.org/abs/0806.3911 for abundant references on this.

6. I’m not sure whether you are disagreeing or just discussing a different, but related, idea. So I’ll explain the difference in perspective between Serre’s result and the one you link to.

To my mind, pure algebra does not include testing inequalities, and Galois theory is designed to study things that can be computed algebraically. So semialgebraic methods are outside this paradigm. For example, in Galois theory we consider the field automorphism which takes $+ \sqrt{2}$ to $- \sqrt{2}$. This automorphism preserves field structure but not inequalities: it takes a positive number to a negative one. Even worse, we are allowed to take a real number like $2^{1/4}$ to an imaginary one like $i 2^{1/4}$.

The amazing thing, then is that there is so much that can be computed purely algebraically — including betti numbers! (You only need to read the first section of the linked reference to get my point, although the whole thing is excellent.) It is data like the action of endomorphisms on cohomology, or like the fundamental group, which requires nonalgebraic techniques.

Finally, although this is not my field, I’ll point out that there are lots of subtle issues in semi-algebraic computation. For example, it is not known whether there is an efficient way to test whether an algebraic number is positive!

7. Danny Calegari says:

Dear Wesley – from the sound of it, you might be interested in the papers of Nabutovsky and Weinberger, which have a similar flavor (if you are not already aware of them . . .) For example, see Weinberger’s nice book:

http://www.ams.org/mathscinet-getitem?mr=2109177

Best,

Danny

8. David,
to me it seems that computations involving real algebraic numbers can still be counted as algebraic, although it’s a matter of taste.
By the way, computing the fundamental group becomes doable, as one would be able, starting from a variety, to construct a finite simplical complex with the same fundamental group. Then you can run something along the lines of http://www.maths.qmul.ac.uk/~leonard/fundamental/

To comment on your claim that testing positivity is not known to be efficient, it seems to me that you misread the part of the Lipton’s post you are referring to. Indeed, he talks about SSP, i.e. testing inequalities of the form $\sum_{m=1}^n \sqrt{a_m}\leq k.$ This way you’re effectively leaving the polynomial “universe”. An algebraic way to encode this would need $n$ variables $x_m,$ and a system involving nonlinear polynomial equations and inequalities in all of them. Needless to say, already the task of testing solvability of a system of $n$-variate polynomial equations is also NP-hard, even if you allow complex solutions.

On the other hand, if you work with the encoding of a real algebaric number by a defining univariate polynomial $f\in\mathbb{Z}[t]$ and a rational interval to locate the particular root (or what is called Thom encoding, i.e. the signs of the derivatives of $f$ at the root) then everything becomes efficient, and these procedures are dating back from 19th century.

(The complexity of SSP is related to the complexity semidefinite programming, and as such is indeed quite important. But this is another story.)

9. David,

I’ve been thinking about a problem, only tangentially related to this post, but maybe you’ll enjoy it.

Suppose I hand you an algorithm. It can be run on your household computer. [Formalize this as you will.] This algorithm spit out, one at a time, digits for a positive real number (say less than 1).

Is there another algorithm (call it the Decider) which will decide whether or not the first algorithm is describing 0?

I believe the answer is no. However, I’m a little weak on how one defines an algorithm, in ZFC set theory say.

Application: Given an explicit function (say an L-function attached to an explicit elliptic curve with integer coefficients), is there an algorithm to determine if there is more than a double pole at the origin?

Hi David,
Is it true that when you are looking at “real” or “complex” points in a scheme defined over rationals this incompatibility creeps in.
Consider the following reformulation :
If K1 and K2 are conjugate schemes defined over some algebraic extension of rationals, and L is some Galois extension containing K1 and K2, then for any locally constant sheaf F the cohomologies over K1 and K2 are isomorphic.
But this looks like a obvious base change…

11. David Speyer says:

For a locally constant sheaf with torsion coefficients, sure, that’s base change. If you are going to consider locally constant sheaves like $\mathbb{Z}$, then you have to worry about the fact that the etale and analytic topologies don’t give the same results for such sheaves.

Indeed, the singular cohomology of $X(\mathbb{C})$ with coefficients in $\mathbb{Z}$ is naturally isomorphic to the cohomology of the locally constant sheaf $\mathbb{Z}$ in the analytic topology. The examples above show that $H_{an}(K_1, \mathbb{Z})$ and $H_{an}(K_2, \mathbb{Z})$ need not be naturally isomorphic, but of course you only said isomorphic. I’d need to think a bit to see if I could find some $F$ for which these groups were not isomorphic at all.

13. If my understanding of your question is correct, take the family $y^2z=x(x-z)(x-\lambda z)$ on $\mathbb{P}^2\times \mathbb{A}^1$. For $\lambda\neq 0,1$, you have a smooth elliptic curve, which is topologically $S^1\times S^1$. For $\lambda=0,1$, you have a nodal cubic, which is a sphere with the north and south poles identified, which is not the same topology.