Orientation of the Riemann surface corresponding to $w=\sqrt{\lambda-z^2}$

91 Views Asked by At

I am trying to describe the Riemann surface $F_{\lambda}$ corresponding to the function $w=\sqrt{\lambda-z^2}$ where $\lambda\in \mathbb{C}, \lambda\ne 0$ over $|z|\le 2$. What I have so far are the following:

Its branch points are $\pm \sqrt{\lambda}$. By removing a branch cut that connects $-\sqrt{\lambda}$ and $\sqrt{\lambda}$, we obtain a domain $D$ of $z$ so that two copies of $D$ correspond to two single-valued functions on $D$ coming from the function $w=\sqrt{\lambda-z^2}$, where one is negative of the other at the same input $z$. These two copies of $D$ are open subsets of $F_{\lambda}$.

enter image description here

To glue these two copies (drawn as two squares on the left of figure), observe that if we analytically continued $w$ on one copy of $D$ along a closed path $\Delta$ that encloses the branch cut, we still stay in the same $D$. However, if analytically continued $w$ along $\nabla$ that passes through the branch cut (e.g. we let $z$ runs from $-2i$ to $2i$ in the figure), we will move from one $D$ to another as soon as we pass the branch cut. Therefore, if we open up the branch cut into $\Delta$, the two copies can be glued along $\Delta$ to create a cylinder as in figure. Note that in order to match $\nabla$ correctly and to match the branch points $\pm \sqrt{\lambda}$ when glueing two copies, we first need to flip one copy of $D$ across the real axis, as indicated in the figure.

My question is: Say I want to assign an orientation to $F_{\lambda}$ as in the figure, did I get the correct induced orientation on $D$ as in the figure? Or in other words, am I being consistent with my choice of orientation in the figure?

Any help would be much appreciated!

1

There are 1 best solutions below

0
On BEST ANSWER

Briefly, the orientation in your diagrams looks correct. In case it's of interest, it's possible to depict the Riemann surface in real three-space by graphing the real parts of the branches of square root. The blue mesh is the sheet where the real part is positive in the upper half-plane and negative in the lower half-plane; the slit from $-\sqrt{\lambda}$ to $\sqrt{\lambda}$ is opened into a literal circle. The gray-shaded surface is the other sheet. The $z$-plane is shown by its real and imaginary axes; vertical projection is the two-sheeted branched cover.

The Riemann surface of a complex circle