Upper half-plane $\overline{\mathbb{H}}$ with two boundary punctures

Question

Consider $\overline{\mathbb H}$ with two puncture $P_1$ and $P_2$ on the real line, with coordinates $z = x_1$ and $z = x_2$, respectively. Consider another copy of $\overline{\mathbb H}$ with two punctures $P_1$ and $P_2$ on the real line, with coordinates $z = x_1'$ and $z = x_2'$, respectively. Are these two surfaces the same Riemann surface?

Idea. I suspect they are, and it would suffice to exhibit the conformal map that takes the punctures into each other while preserving $\overline{\mathbb H}$. But I am not sure how to do this, could anybody help?

Answer 1

The two surfaces are the same Riemann surfaces. The conformal map that takes the punctures into each other while preserving $\overline{\mathbb H}$ depends on whether $(x_2' - x_1')/(x_2 - x_1)$ is positive or negative.

If $(x_2' - x_1')/(x_2 - x_1) > 0$, then the map is given by$${{z' - x_1'}\over{z' - x_1'}} = {{z - x_1}\over{z - x_2}}.$$Using cross multiplication, we can see this has the form of a linear fractional transformation. Solving for $z'$,$$z' = {{(x_1' - x_2')z + (x_2' x_1 - x_1'x_2)}\over{x_1 - x_2}}.$$$\overline{\mathbb {H}}$ is preserved if $ad - bc > 0$. That is, if $(x_2' - x_1')(x_2 - x_1) > 0$. This is true whenever $(x_2' - x_1')/(x_2 - x_1) > 0$ because $(x_2 - x_1)^2$ is positive. We have that $(x_2' - x_1')/(x_2 - x_1) > 0$ was true by assumption.

If $(x_2' - x_1')/(x_2 - x_1) < 0$, then the map is given by$${{z' - x_1'}\over{z' - x_1'}} = -{{z - x_1}\over{z - x_2}}.$$Solving for $z'$,$$z' = {{(x_1' + x_2')z - (x_2'x_1 - x_1'x_2)}\over{2z - (x_1 + x_2)}}.$$$\overline{\mathbb{H}}$ is preserved if $ad - bc > 0$. That is, if $-(x_2' + x_1')(x_2 + x_1) + (x_2'x_1 - x_1'x_2)(2) >0$, which simplifies to $(x_1 - x_2)(x_2' - x_1') > 0$. This is true whenever $(x_2' - x_1')/(x_2 - x_1) < 0$, and $(x_2' - x_1')/(x_2 - x_1) < 0$ was true by assumption.

Geometrically, $(x_2' - x_1')/(x_2 - x_1) > 0$ means that the order of the punctures is preserved while $(x_2' - x_1')/(x_2 - x_1) >0$ means that the order is reversed.

Answer 2

It's enough to consider the case $x_1^{\prime}=0, x_2^{\prime}=\infty$. If $x_1>x_2\in \mathbb{R}$, the map $$ z\mapsto \frac{z-x_1}{z-x_2} $$ sends $x_1$ to $0$ and $x_2$ to $\infty$, and is conformal because the determinant of the corresponding matrix is $x_1-x_2>0$. And since $$ \Im\frac{z-x_1}{z-x_2}=\frac{(x_1-x_2)\Im z}{|z-x_2|^2}$$ it follows that the map sends $\mathbb{H}$ into itself and $\mathbb{R}\cup\{\infty\}$ into itself.

If for instance $x_2=\infty$, then the map $z\mapsto z-x_1$ sends $x_1$ to $0$ and $x_2$ to $\infty$, and clearly preserves $\mathbb{H}$ and $\mathbb{R}\cup\{\infty\}$.