# Finding holomorphic map on Riemann surface from a map between two Riemann surfaces

I have a non-constant degree two map between Riemann surfaces $R$ and $S$, $f: R \to S$. I'm trying to find a holomorphic homeomorphism $\tau: R \to R$ such that $f(\tau) = f$ and $\tau^2$ is the identity.

I think I can define the correct map but I'm struggling to show it is a homeomorphism and holomorphic. Here is what I have so far:

• $f$ is degree 2 so from the degree formula this means that any point $s \in S$ can have at most 2 pre-images.
• I think we should say given $r \in R$, $\tau$ should send $r$ to it's pre-image friend $r'$ (where $f(r) = f(r')$) and if $r$ is a ramification point then $\tau$ should do nothing. This would certainly satisfy the properties required of $\tau$ and it's defined on the whole of $R$.
• To show this $\tau$ is holomorphic I understand we need to show it is holomorphic as a map in local coordinates. So we pick a neighbourhood $U$ small enough of $r$ so that $f$ can be written in local form in $U$. I was thinking of somehow using $f$ to help us define a local coordinate in which $\tau$ is holomorphic and homeomorphic but I'm struggling to get anywhere with this.

Thanks for any help

By construction, $\tau$ is its own inverse, so when we've shown that $\tau$ is holomorphic, it follows that $\tau$ is a homeomorphism.

Consider a point $r\in R$ such that $f^{-1}(f(r)) = \{ r, r'\}$ has two elements. Using charts around $r,r'$ and $f(r)$, we find that there are open neighbourhoods $U,V$ of $r$ and $f(r)$ respectively, such that $f\lvert_U \colon U \to V$ is biholomorphic, and there are neighbourhoods $U', V'$ of $r'$ resp. $f(r') = f(r)$ such that $f\lvert_{U'}\colon U' \to V'$ is biholomorphic. By restricting the neighbourhoods, we can assume that $V' = V$. Then on $U$ we have

$$\tau\lvert_U = (f\lvert_{U'})^{-1}\circ (f\lvert_U),$$

which shows that $\tau$ is holomorphic on $U$ (and a local biholomorphism).

It remains to consider ramification points. If $r$ is a ramification point, there are local charts around $r$ and $f(r)$ in which $f$ corresponds to $z \mapsto z^2$. It follows that on such a chart around $r$, $\tau$ corresponds to $z \mapsto -z$, which again is holomorphic.

• Thank you - this is really helpful. When we choose the charts around $r$ and $r'$ how can we ensure $f$ is biholomorphic there? it seems that points near $x$ and $x'$ all map into the same point and $f$ is 2-1 there? I'm also struggling to see why $-z$ isn't a different point to x in the local coordinates you described? – Wooster May 14 '15 at 14:41
• Is it correct that the local form on $U$ and $U'$ is $z$ which is a biholomorphism? Even though the global form of $f$ is 2-1 around there? – Wooster May 14 '15 at 14:43
• For the case where $f(r)$ has two distinct preimages, we start by picking two disjoint neighbourhoods $U_0$ of $r$ and $U'_0$ of $r'$. Applying the open mapping theorem, we see there is a neighbourhood $V$ of $f(r)$ such that every $s\in V$ is attained in $U_0$ and in $U'_0$. Let $U = f^{-1}(V) \cap U_0$ and $U' = f^{-1}(V) \cap U'_0$ to see that $f$ is injective on the neighbourhoods $U$ of $r$ and $U'$ of $r'$ [since $f$ is globally two-to-one, counting multiplicities], and $f(U) = f(U') = V$, and bijective holomorphic maps are biholomorphic. – Daniel Fischer May 14 '15 at 14:58
• For the ramification points, choose the charts so that $r$ and $f(r)$ correspond to $0$. $f$ cannot be injective on any neighbourhood of $r$, or $r$ wouldn't be a ramification point. Since $f$ is two-to-one, the local form of $f$ is $\psi(f(\varphi^{-1}(z))) = z^2$, where we can pick our charts so that the neighbourhoods correspond to the unit disk. Then $z \mapsto -z$ corresponds to a map - $p \mapsto \varphi^{-1}(-\varphi(p))$ - fixing $r$ and no other point in a (small enough) neighbourhood of $r$. – Daniel Fischer May 14 '15 at 15:02
• Okay I understand the case where we have two distinct pre-images, thanks! So for a ramification point our map is $\tau(r) = r$ - and in a neigbhourhood of $r$ this should not be the identity because ramification points are isolated. Now I'm struggling to see why (in local coordinates) our map is forced to be $z \to -z$? – Wooster May 14 '15 at 15:25