# 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?

## 1 Answer

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 $z\in [-1,1]$.

For a beautiful proof, see Goluzin, Geometric theory of functions of a complex variable, (English translation: AMS 1969), Chapter 11.

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.

The answer to the other two questions is no. The procedure that you describe is called the Schiffer variation of conformal structure, and it indeed changes the conformal structure (of the resulting compact surface). See for example, S. Nag, Schiffer variation of complex structure and coordinates on Teichmuller spaces, Proc. Indian Acad. Sci., 94 (1985) 2-3, p. 111-122. The original source is M. Shiffer and D. Spencer, Functionals of finite Riemann surfaces. Princeton University Press, Princeton, N. J., 1954.