TITLE: What topological restrictions are there for a topological space to be a group?
QUESTION [8 upvotes]: 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 $1$.

REPLY [3 votes]: 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 $$\chi(G)=2(g-1)\stackrel{!}{=}0$$

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.