4

I'm looking at the torus given by $X = \mathbb{C}/\Lambda$ where $\Lambda$ is the lattice spanned by $1$ and $\omega$ where $\omega$ is a primitive cube root of unity.

I've shown that $\sigma(z) = \omega z$ is a well-defined map on the torus and now I've been asked to explain how to put a Riemann surface structure on the set of equivalence classes $Y = \{z, \sigma(z), \sigma^2(z)\}$ such that the natural map $X \to Y$ is holomorphic.

I'm having quite a lot of difficulty with this. I'm not sure how to visualise $Y$ as a space, I can see that if $z$ is a fixed point then $Y$ is just a copy of $X$ and then the map $X \to Y$ is just the identity but I'm struggling to write down explicit local coordinates for $Y$ in general.

Thanks for any help

Wooster
  • 3,775
  • Some hints to get started: A sketch of the fundamental domain for $\Lambda$, and a fundamental domain for the quotient, should help sort out the quotient topologically. Next, where are the ramification points of the quotient map, and what's the order of branching at each? – Andrew D. Hwang Jan 09 '15 at 17:22
  • Thank you for your comment. I'm comfortable with the quotient $\mathbb{C}/\Lambda$ and I can use Riemann Hurwitz to show it is topologically a torus, but my difficulty lies in understanding what the space $Y$ is? – Wooster Jan 09 '15 at 20:44
  • The points of my earlier hints were: Keep quotienting (i.e., subdivide a fundamental domain for $\Lambda$ to get a fundamental domain for $Y$), and find the branch points of the map $X \to Y$ (a.k.a., fixed points of $\sigma$) and their orders. :) Incidentally, there's not much choice for the topology of $Y$: genus of curves is non-increasing under holomorphic surjections by Riemann-Hurwitz. – Andrew D. Hwang Jan 09 '15 at 21:48
  • Ah okay, I think part of my problem is understanding what $Y$ is. because surely $Y$ is just the whole of $X$ because it contains all $z$? – Wooster Jan 09 '15 at 21:52
  • Well...elements of $Y$ aren't elements of $X$, they're (unordered) triples of elements of $X$ (with finitely many exceptions, the fixed points of $\sigma$). Here's an easier question to get some practice with the ideas: Let $X = \mathbf{C}/\Lambda$ be a torus, define $\sigma(z) = -z$, and let $Y = X/\sigma$ be the quotient. Can you sketch a fundamental domain for $Y$ (as a polygon with edge gluings) and identify $Y$ topologically? (Hint: The fundamental domain for $X$ comprises two copies of the fundamental domain for $Y$.) – Andrew D. Hwang Jan 09 '15 at 22:03
  • After thinking about your example for a while I think $Y$ is a cylinder? Is the Riemann surface structure then just the collapsing map? In my example in the question, do we then consider $Y = X/\sigma$ where $\sigma(z) = wz&? and the question is then finding the fundamental domain for that? – Wooster Jan 09 '15 at 23:16
  • Close. :) The quotient has no boundary (the torus has none, and holomorphic maps are open). The weird-looking feature is, the quotient has corners in the Euclidean metric, but is a topological manifold, and can be given a holomorphic structure even at the "corners". (The same is true for the $Y$ you're actually interested in.) – Andrew D. Hwang Jan 10 '15 at 00:14

1 Answers1

2

$\newcommand{\Cpx}{\mathbf{C}}$Here's one possible diagram of a fundamental domain $R$ for $X = \Cpx/\Lambda$, the images of $R$ under the cyclic group generated by $\sigma$, and a polygon-with-edge-gluing for the quotient $Y = X/\langle\sigma\rangle$:

Fundamental domain for a cyclic quotient of the hexagonal torus

To fill in the remaining details, it may help to notice that the action of $\sigma$ on $R$ amounts to cutting $R$ into equilateral triangles along the segment from $0$ to $1 + \omega$ and rotating each triangle one-third of a turn counterclockwise about its center. (Edges joined in $X$ remain joined after this operation!)

There are three fixed points of $\sigma$ in $X$, namely $0$ and the two centers of the triangles. Consequently, the quotient map $X \to Y$ has three branch points, each of order three. The function $(z - z_{0})^{3}$ may be used as a local coordinate near the image of $z_{0}$ in $Y$, i.e., near each vertex of the fundamental domain of $Y$. (N.B., this fundamental domain has three vertices, since $0$ and $1 + \omega$ are identified.)

Andrew D. Hwang
  • 78,195
  • Thank you for this comprehensive answer, I'm just trying to understand it now. When calculating the fixed points I get $z = \frac{m+n\omega}{1-\omega}$, how did you then deduce the only possibilities were the midpoints of the triangles? Thanks again – Wooster Jan 14 '15 at 17:57
  • Apologies, I have quite a few questions. I can see "intuitively" that the fixed points are branched points, but my definition of a branch point is when the derivative is non-zero in local coordinates. How could we possibly write the quotient map in local coordinates to see this? Or is there another way to see that the fixed points are branch points? – Wooster Jan 14 '15 at 18:19
  • @Wooster: I got the fixed points from geometry, but algebraically you could write $z = a + b\omega$ with $a$, $b$ in $[0, 1)$, set $\sigma(z) \equiv z \pmod{\Lambda}$, and find all (three) solutions. For the branching behavior: If you pick charts for $Y$ as indicated, the map $f$ may be viewed as the cubing map over each branch point, i.e., $f(z - z_{0}) = (z - z_{0})^{3}$. Branching at $z_{0}$ occurs because $f'(z_{0}) = 0$ (n.b., not $\neq 0$). :) – Andrew D. Hwang Jan 14 '15 at 20:42
  • Okay that's clear, thank you! What chart are you using there to come up with the local map? I'm struggling to see how to arrive at that. Apologies, this material is very new to me so I'm having to spell out all the details! – Wooster Jan 14 '15 at 21:06
  • No trouble at all. :) The idea is to use an affine coordinate on $X$, and then near a branch point in $Y$, a.k.a., "the image of a fixed point $z_{0}$ of $\sigma$" or "a vertex of the small rhombus", to use $(z - z_{0})^{3}$ as a local coordinate. One key point about this example is that cubing maps a neighborhood of a branch point bijectively to a disk about $0$ in $\Cpx$, and so acts as a local holomorphic coordinate of $Y$. The local expression for $f$ near $z_{0}$ is now, more or less by definition, $f(z - z_{0}) = (z - z_{0})^{3}$. – Andrew D. Hwang Jan 14 '15 at 21:25
  • So near any point z_0 that isn't a branch point we use the coordinate $\psi: [z] \to z-z_0$ and then around branch points $z_1$ we use the coordinate $\phi: [z] \to (z-z_1)^3$? I'm struggling to see that is a well-defined coordinate and how you made the choice of a cube? – Wooster Jan 14 '15 at 21:41
  • Quick tangential question: Do you have a good intuitive understanding of the complex $n$th power map ($n > 1$ an integer), particularly, its branching behavior at $0$ and the way angles at the origin behave? In case not: Write $z$ in polar form and investigate. :) – Andrew D. Hwang Jan 15 '15 at 00:24