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Let $g$ be the genus of a closed Riemann surface, what can be said about $g$ if the tangent bundle $T$ of that surface is trivial?

From the formula for the degree of a tangent bundle, $\deg(T)=2-2g$, I guess I can see that it's true for $g=1$. Is that it or am I understanding something wrong here?

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What can you see? – Mariano Suárez-Alvarez Feb 18 '13 at 6:48
I think the degree of a trivial tangent bundle was $0$, is that right? – user62057 Feb 18 '13 at 6:56
Indeed. ${}{}{}$ – Mariano Suárez-Alvarez Feb 18 '13 at 7:01
So that's the answer? Maybe I'm just over-thinking this question a bit... – user62057 Feb 18 '13 at 7:09
I changed $deg(T)$ to $\deg(T)$, coded as \deg(T), with a backslash. The not only makes it non-italicized, but also results in proper spacing in things like $3\deg T$. It is standard usage. – Michael Hardy Feb 18 '13 at 13:52

Just to make the above observations into an answer:

The genus $g(M)$ of a closed (oriented) Riemann surface $M$ is related to the Euler characteristic $\chi(M)$ by the formula $2-2g(M) = \chi(M)$. Moreover for oriented manifolds, the Euler characteristic is a characteristic class, which means that in particular $\chi(M) = \chi(TM) = 0$ if $TM$ is a trivial bundle. Hence in your situation you see, that the only closed oriented Riemann surface with trivial tangent bundle is the torus $T^2$.

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A Riemann surface is a complex manifold so its tangent bundle has a complex structure. If the tangent bundle is also trivial then its first Chern class must be zero. By Chern-Weil theory the first Chern Class is represented by 1/2pi times the curvature 2 form of any Levi-Civita connection.

For the sphere with the standard metric its integral is 2 and for surfaces of genus 2 and higher it is negative since the surface can be given a metric of constant negative curvature. For the flat torus the curvature 2 form is identically zero. So the first Chern class is zero only on the torus and all other closed compact Riemann surfaces must have non-trivial tangent bundle.

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