1
$\begingroup$

My reference is Abikoff's book "Real analytic theory of Teichmuller spaces.

Let $X$ be a Riemann surface with two boundaries, we can construct a mirror surface $\bar{X}$ defined to be the same underlying topological surface, but using the complex coordinate $\bar{z}$ for an open prievously charted by $z$. We can glue $X$ and its mirror $\bar{X}$ to get a borderless Riemann surface X^\mathbb{C}, which inherits the charts $z$, $\bar{z}$ above, but also has coordinates on the former borders, this surface is called the complex double of $\mathbb{X}$. (The construction can be found in full details at pag. 45 of the book.)

Consider a disk $D$ in the complex plane, and remove from it a smaller disk $D'$, so that we get an annulus. Using an appropriate map, I can also see the annulus as the upper halfplane from which a disk $D''$ has been removed. I would like to see explicitly the mirror and double construction for $X = D \setminus D'$, which should yield a torus.

Following Abikoff, a painless method to do this is to realise $X$ as a suitable quotient of the upper half plane $U$, this is quickly done realising the annulus as an half-annulus in $U$, on which we identify inner and outer boundary. (More precisely the annulus is obtained by quotienting U by an Fuchsian group which in the upper half plane model is generated by a multiplication by a real number $\lambda$.) If we take the conjugate of this annulus, we get the mirror of $X$ and $X^\mathbb{C} = X \cup \bar{X}$.

However, I tried on my own to construct the double and I got this: Consider $X$ as the upper half-plane minus a circle $C$, and consider the conjugate circle $\bar{C}$. We take $L \setminus \bar{C}$ to be another copy of the topological surface underlying $X$ and construct a map sending $V \in X$ to $\bar{V} \in L \setminus \bar{C}$. We chart $\bar{V}$ with $\bar{z}$, thus defining the Riemann surface $\bar{X}$.

The double $X^\mathbb{C}$ is then $\mathbb{C} \setminus (C \cup \bar{C})$ modulo an identification of conjugate points in $\partial{C}$ and $\partial{\bar{C}}$?

Edit: an extra question

If $z$ is the obvious coordinate on $\mathbb{C} \setminus (C \cup \bar{C}),$ the condition for a meromorphic differential

$\omega = \phi(z) dz,$

to give a meromorphic differential on $X^\mathbb{C}$ is that if $z \in \partial C $ then $\overline{\phi(z)} = \phi(\bar{z})$ or $\phi(z) = \phi(\bar{z})$?

$\endgroup$
  • 1
    $\begingroup$ The answer to the last sentence is "yes". $\endgroup$ – Alexandre Eremenko Mar 3 '18 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.