What topological restrictions are there for a topological space to be a group?

Question

I'm trying to provide a group structure for some Riemannian surfaces. I heard that the following result holds:

Let $X$ be a compact Riemannian surface. Then $X$ admits a group structure if, and only if it has genus $1$.

Answer 1

A topological group is (tautologically) an H-space and there are several topological obstructions known for a space to be an H-space. For example the fundamental group must be abelian. Or when an n-sphere is an H-space, then there are n independent vector fields, which by Adams is possible only for n=1,3,7.

(The one condition excludey surfaces of genus at least 2, the other compact surfaces of genus 0, so only genus 1 remains.)

Answer 2

Here's a proof for the result you provided (even if you need stronger assumptions):

Claim: If a closed $2$-dimensional smooth manifold admits a Lie group structure, then it is orientable and it has genus $g=1$ (i.e. it is the torus $T^2$).

Proof: If $G$ is a Lie group with $\dim G=n$, then $TG\cong G\times\mathbb{R}^n$, since you can use the action of the group structure by left multiplication to move around a basis you define at a point. This obviously implies that it must be orientable, by considering a global orientation on $TG$ induced by any choice of basis on $\mathbb{R}^n$. At the same time, it implies that you have $n$ never vanishing vector fields defining a basis at every point (corresponding to the standard basis $\partial_i$ of $G\times\mathbb{R}^n$). Thus the Poincaré-Hopf theorem implies that $\chi(G)=0$. In particular, if $G$ is as assumed in the claim, we must have $g=1$, by the formula $$\chi(G)=2(g-1)\stackrel{!}{=}0$$

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We also have a partial converse to this result:

Claim: If a closed, orientable $2$-dimensional smooth manifold $M$ has genus $g=1$, then it admits the structure of a topological group.

Proof: By the classification of closed surfaces, $M$ is homeomorphic to the torus through some homeomorphism $$h:M\to T^2$$ We define a multiplication $*$ on $M$ by $$x*y=h^{-1}(h(x)\cdot h(y))$$ where $\cdot$ denotes the product on $T^2$. This makes $M$ into a topological group.