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!
This is Lemma 1 in
Gunning, R. C., Special coordinate coverings of Riemann surfaces, Math. Ann. 170, 67-86 (1967). ZBL0144.33501.
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.
If you are interested in this staff, I suggest reading the entire Gunning's paper.