If $M,N$ are two Riemann surfaces with boundary, then we can glue them along one of each of their boundary component, which is $S$, to form a new Riemann surface with boundary, but for different gluing we may form different Riemann surfaces with boundary, for example, there may be a $S$ twist, intuitively, it is just we rotate one $S$ an angle then glue it with another surfaces, but my question is how can we show the resulting two surfaces (twisted gluing and untwisted gluing) are different Riemann surface with boundary (their differential structures are the same because we can regard it as a kind of connected sum)? Thanks!
[Math] To differently gluing of two Riemann surfaces with boundary we get different surfaces
gt.geometric-topologyriemann-surfaces
Best Answer
Here is a particularly simple example. We will take a cylinder and glue its ends together in two different ways. The cylinder will be $$C:=\{ z: 0 \leq \mathrm{Im}(z) \leq 1 \} / \mathbb{Z},$$ where $\mathbb{Z}$ acts on the horizontal strip by translation.
Let $\theta$ be a real number between $0$ and $1$. We will glue the ends of $C$ together by identifying $x+\mathbb{Z}$ with $(x+i+\theta) + \mathbb{Z}$, for $x$ real. Call the resulting torus $T_{\theta}$.
We claim that all the $T_{\theta}$ are nonisomorphic as Riemann surfaces. Here is a sketch of a proof: Let $\Lambda_{\theta}$ be the lattice in $\mathbb{C}$ generated by $1$ and $i+\theta$. Clearly, $T_{\theta} \cong \mathbb{C} / \Lambda_{\theta}$. Suppose we had holomorphic isomorphism $f:T_{\theta} \to T_{\theta'}$. Then $f$ would lift to a map on the universal covers, $g: \mathbb{C} \to \mathbb{C}$, and there would be some additive map $h: \Lambda_{\theta} \to \Lambda_{\theta'}$ such that $$g(z+\lambda) = g(z) + h(\lambda)\ \mbox{for} \ \lambda \in \Lambda_{\theta}. \quad (*)$$ Let $D$ be a fundamental domain for $\Lambda_{\theta}$. The function $g$ is $O(1)$ on $D$ (compactness), so, applying $(*)$, we can deduce that $|g(z)| = O(\max(|z|, 1)$. Since $g$ is analytic, this implies that $g$ is of the form $z \mapsto az+b$. In particular, multiplication by $a$ must carry $\Lambda_{\theta}$ to $\Lambda_{\theta'}$.
But there is no complex number $a$ such that $a \Lambda_{\theta} = \Lambda_{\theta'}$, a contradiction.