# If $A$ is the ring of continuous functions on a genus $g$ surface, can the genus of $X$ be seen by simple algebra in $A$?

I was describing to a friend the result that a compact Hausdorff space is determined up to homeomorphism up to by its ring of continuous functions, and he asked how one could see the genus of a surface algebraically. I think it's a good question, and I don't know the answer.

Two parts to this question:

1. Suppose it is known that $A$ is the ring of continuous functions of a closed surface $X$ (topological manifold, real 2-dimensional) of genus $g$. Where in the algebraic structure of $A$ is the genus of $X$? (There are comments from other users about $K$-theory below... I want to emphasize that I am dreaming of a simple algebraic characterization, analogous to the relationship between idempotents and connected components.)

2. Does the answer to 1 say something interesting for more general classes of rings (ex, those that are rings of continuous functions on a compact Hausdorff space, or more general rings...)

Any ideas? (Analogous to the idempotents / connected components dictionary.)

Here are some thoughts of my own:

1. Picking a smooth structure on $X$ amounts to passing to some subring of differentiable functions. In this setting, results about the existence of vector fields with zeros of prescribed index could be interpreted as the existence of derivations with some prescribed zeros (by choosing some arbitrary Riemannian metric and translating the results about vector fields) - though its not clear how to interpret the index of the zero algebraically, or how to describe these subrings.

2. Consider all ring homomorphisms to $C(X) \to C(S^1)$ up the homotopy relation induced by factoring through $C(S^1 \times I)$ in the right way. This crudely recaptures the fundamental group, but I don't see how to turn it into an algebraic condition.

3. Vector bundles are the same as finite projective modules via Swan's theorem - so certain sections of the vector bundle are elements of the corresponding module? Maybe one could get at the Poincare-Hopf computation of the genus that way, but again I don't see how to compute the index. (Also as far as I know, one still needs a smooth structure to use Poincare-Hopf.)

I would be happy to just be able to distinguish $S^2$ and $T^2$ via their ring of continuous functions.

• You may be interested in Morse Theory. – Aloizio Macedo Jan 9 '16 at 17:09
• K-theory is no good here. You can't tell oriented surfaces apart with their (topological) complex K-theory rings. – Mike Miller Jan 9 '16 at 21:14
• @MikeMiller I'm completely ignorant of $K$-theory, so I don't see the connection. However, I think that accessing the full structure $C^*$ algebra structure (in particular, the norm) requires passing back to topological space first. I really am dreaming of something inherently algebraic, like the description of $H_0$ in terms of idempotents. – Lorenzo Najt Jan 10 '16 at 2:43
• Given the ring $A$, one recovers the space $X$ as the MaxSpec of $A$, and then you can compute the genus (or any other topological invariant) of $X$ any way you like --- for example, via Cech cohomology. The two steps --- going from $A$ to $X$ and going from $X$ to its genus --- are easily described, so it's easy to describe the composition. "Easily described" is not the same thing as "transparent", but I don't see a reason to expect transparency. – Steven Landsburg Jan 28 '16 at 22:03
• Real line bundles on $X$ are classified by $H^1(X,\mathbb{Z}/2\mathbb{Z})$, so assuming you mean continuous real-valued functions, the number of isomorphism classes of rank $1$ projective modules over $A$ is $4^g$. – Julian Rosen Jan 28 '16 at 22:22

I'll assume we're talking about complex functions; if real, tensor with $\mathbb{C}$. Now pass to the group of units. With the topology given by spectral radius (this is an algebraic description of the C-* topology), the group of connected components of the group of units is $H^1(X, \mathbb{Z})$ which of course knows the genus.

If you really like idempotents then you should try learning about K-theory. There's a very concrete and elementary way to define $K^0$ in terms of idempotents in the matrix algebras $M_n(A)$. You don't need the C*-structure of $A$. But on the other hand $K^0$ doesn't know the genus of a surface. $K^1$ does but it's a bit trickier to define, I think.

Edit: $H^1(X, \mathbb{Z})$ can be isolated a bit more algebraically too. To extract $H^1(X, \mathbb{Z})$ from the group of units it suffices to isolate the connected component of the identity. The elements in this connected component can be distinguished by the fact that they are divisible.

Edit #2: So, purely for the sake of distinguishing $S^2$ and $T^2$, you can say the following: on $S^2$, every unit has a square root, but not on $T^2$.

• @Arul: take the equivalence relation generated by the relation "two units are connected by a straight line consisting of units." I think this is the same as the same-component relation but I haven't thought about it in detail. – Qiaochu Yuan Jan 28 '16 at 22:59
• @QiaochuYuan Yes, you are right. Obviously, if two units are connected by a line of units, they induce the same map $H_1(X) \to H_1(\mathbb{C}^{\ast})$. Conversely, if $u$ and $v$ are maps $X \to \mathbb{C}^{\ast}$ inducing the same map on homology, then $u v^{-1}$ has a continuous logarithm $w$, and then $v \exp(a w)$ is a continuous path from one to the other, for $a \in [0,1]$. – David E Speyer Jan 29 '16 at 16:38
• @David: but not a straight line, right? Although maybe you're saying that path can easily be approximated by straight lines? (Which is the sort of argument I had in mind.) I wanted to avoid mentioning exponentials to keep things as algebraic as possible: talking about straight lines only requires a real vector space structure. – Qiaochu Yuan Jan 29 '16 at 16:40
• Oh, sorry. I am confused by what you mean by "straight line". You mean literally straight in the vector space. That sounds plausible but I need to think a bit about whether it is actually true. – David E Speyer Jan 29 '16 at 16:42
• Choose $N$ large enough that $|w|/N < \pi$. Then $\exp(w/N)$ is disjoint from the negative real line, so $a \exp(w/N) + (1-a)$ is a unit for all $a \in [0,1]$. Thus $a v \exp((k+1) w/N) + (1-a) v \exp(k w/N)$ is a unit for all such $a$, and we can take a piecewise linear path through the points $v \exp(k w/N)$ for $0 \leq k \leq N$. – David E Speyer Jan 29 '16 at 16:56

The torus has two functions $f$ and $g$ which are (1) relatively prime, (2) each have two square roots, and (3) whose product has $4$ square roots. For instance take two functions which vanish on disjoint loops which are not null-homologous. There sphere does not have two such functions because (1) implies that $V(f)$ and $V(g)$ are disjoint and (2) implies that the complement of $V(f)$ and $V(g)$ are connected, and (3) implies that the complement of $V(fg)$ is not connected. These conditions are impossible to all satisfy at once on the sphere.

• I think a similar technique should work for surfaces in general, the idea being to use the Mayer-Vietoris sequence to convert a question about the dimension of $H^1$ into a question about the dimension of some $H^0$'s which counts connected components, which can in turn be detected by numbers of square roots. – Philip Engel Jan 29 '16 at 14:19
• Thank you! This is a very elegant and simple idea. (I guess you are thinking of the real valued functions here, at least that is what I am visualizing.) – Lorenzo Najt Jan 31 '16 at 8:23