# Gluing Riemann surfaces

Let $X$ be a compact Riemann surface with boundary $\partial X$. Assume (for simplicity only) that $\partial X$ has a single connected component. Let us fix an orientation preserving diffeomorphism $\phi\colon S^1\tilde\to \partial X$ with the standard unit circle $S^1$ which is the boundary of the standard unit disk $D$. Consider the closed topological surface $Y:=X\cup_\phi D$ obtained from $X$ by gluing the disk $D$ along $\phi$.

Q1. Is it true that $Y$ has a unique complex structure such that the interiors of $X$ and of $D$ are complex subspaces of $Y$ with their original complex structures?

Q2. If the answer to Q1 is 'yes', is it true that if one chooses another orientation preserving diffeomorphism $\phi'\colon S^1\tilde\to \partial X$ then the new Riemann surface $Y'$ (as in Q1) is isomorphic (as a complex manifold) to $Y$?

Q3. Let $X_0\subset X$ be another Riemann surface which is obtained from $X$ by a small deformation of the boundary of $X$ (thus topologically $X_0$ is a deformation retract of $X$). Let $Y_0$ be obtained from $X_0$ as in Q1. Are $Y$ and $Y_0$ isomorphic?

The answer to the first question is "yes". This is called conformal gluing, and the proof is based on the following lemma due to Lavrentiev: Let $\phi$ be an increasing diffeomorphism of $[-1,1]$ onto itself. Then there is a simple curve in the unit disk connecting $-1$ and $1$ breaking the disk into two domains $D^+$ (above) and $D^-$ (below), so that there exist conformal maps $f_1$ from the upper half-disk to $D_+$ and $f_2$ from the lower half-disk to $D^-$ which satisfy $f_1\circ\phi(z)=f_2(z)$ for $z\in [-1,1]$.
The condition that $\phi$ is a diffeomorphism can be substantially relaxed (See Ahlfors, Lectures on quasiconformal mappings, Chap IV, where this is called $M$-condition, modern term is "quasisymmetric"), but for arbitrary homeomorphisms this is not true.