# Riemann surface with transition functions of the form $z \mapsto az+b$

Let $$\Sigma$$ be a closed Riemann surface and let $$\{U_\alpha,\phi_\alpha:U_\alpha\rightarrow\mathbb C\}$$ be a family of holomorphic coordinate charts. If the transition maps $$\Phi_{\alpha\beta}:\phi_\alpha(U_\alpha\cap U_\beta)\rightarrow\phi_\beta(U_\alpha\cap U_\beta)$$ are of the form $$z\mapsto az+b$$ where $$a,b$$ are complex numbers (of course determined by the charts $$U_\alpha$$ and $$U_\beta$$). Show that $$\Sigma$$ must be a genus one Riemann surface.

I can prove that $$\Sigma$$ must not be genus zero because a holomorphic developing map $$F:\hat\Sigma\rightarrow \mathbb C$$ can be well-defined using the data of the coordinate charts, where $$\hat\Sigma$$ is the universal covering of $$\Sigma$$. The surface $$\Sigma$$ must not be genus zero since there are no nonconstant holomorphic functions on compact Riemann surfaces.

I know this problem can be solved by introducing flat connections in differential geometry, and is related to a famous open problem (Chern's conjecture), but I believe there is a more elementary solution using theory of Riemann surfaces. Thanks for any help!

• Yes you are right of course, $\Bbb{C}^* / (z\mapsto 2z)$ means we don't have an unramified covering $\Bbb{C}\to \Sigma$ Dec 27, 2021 at 15:02

Gunning proves this by observing that the canonical bundle of the Riemann surface $$\Sigma$$ has to have zero 1st Chern class (since the transition functions of the canonical line bundle are constant), hence, $$deg(K_\Sigma)=0$$. Since for a general compact connected Riemann surface $$X$$, $$deg(K_X)=2g-2$$, in our case the genus of $$\Sigma$$ has to be 1.