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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 $1$.

Colo
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    If you assume the group structure is compatible with the topology of Riemann surfaces, then its fundamental group has to be abelian. Hence only torus can has such group structure. check this for more details http://en.wikipedia.org/wiki/Fundamental_group#Lie_Groups –  May 26 '14 at 10:11
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    For example, every topological group is homogeneous. – Tomasz Kania May 26 '14 at 10:11
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    I don't think your statement is correct. For example $\mathbb{C}$ is non-compact, a Riemann-surface, and carries a group structure.

    Probably the statement you mean is: If $X$ (your Riemann surface) is compact, then it admits a group structure if and only if the genus is $1$.

    – jmc May 26 '14 at 10:18
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    If you want a Lie group structure, then it must parallelizable, i.e. the tangent bundle must be trivial. – Thomas Rot May 26 '14 at 10:18
  • @jmc I modified the question as you suggested. – Colo May 26 '14 at 12:42

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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.

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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.)

user39082
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  • That seems like too strong a result to quote to rule out the $2$-sphere. Under the additional assumption that what is required is a Lie group structure you only need the hairy ball theorem. – Qiaochu Yuan May 26 '14 at 19:29