# Show that the equivalence relation $z\sim w$ iff $z=2^nw$ induces a compact Riemann surface structure on the quotient

Define the equivalence relation $z\sim w$ iff $z = 2^n w$ for some integer $n$. Let $R = \mathbb{C}/\sim$, show that $R$ naturally has the structure of a compact Riemann surface, in fact find $\tau_1, \tau_2$ such that $R$ is conformally equivalent to $\mathbb{C}/\langle\tau_1,\tau_2\rangle$.

Now I can somewhat see how this would be a torus, both the real and imaginary directions are identified in some regular way as one would need for a torus. But I really have no idea how to go about showing this. I have tried to showing this in a similar way as is done on page 36 of Donaldson's notes but believe this method will not work. As surely around any open neighbourhood of a point in $\mathbb{C}$ there will be an infinite number of equivalent points also contained in this neighbourhood so I can't see how we would be able to construct any sort of chart or free action.

I'm pretty lost for ideas as to how else to even approach this problem so any help is appreciated.

• You need to remove the origin of $\mathbb{C}$ before you take the quotient. May 7, 2017 at 14:45
• @MichaelAlbanese ah yes. Is the issue I detailed not still a factor though? Even with the origin removed how do we find an open neighbourhood? May 7, 2017 at 17:33
• Note that your equivalence relation restricted to $\Bbb R^+$ can be recast as $\ln z \equiv \ln w \mod \ln 2$. May 7, 2017 at 18:40

As I pointed out above, we need to remove the origin, so the space we're interested in is $R = \mathbb{C}^*/\sim$. Note that we can view $R$ as the quotient of a group action, namely $\mathbb{Z}\times\mathbb{C}^* \to \mathbb{C}^*$ where $(n, z) \mapsto 2^nz$.

It is not true that an open neighbourhood of $w \in \mathbb{C}^*$ necessarily has infinitely many equivalent points under $\sim$. For every $a \in (0, \infty)$, the half-open annulus $\{z \in \mathbb{C} \mid a \leq |z| < 2a\}$ is a fundamental domain of the group action, i.e. every point in $\mathbb{C}^*$ is equivalent to a unique point in the annulus. So for $w \in \mathbb{C}^*$, choose $a$ such that $\frac{1}{2}|w| < a < |w|$, then $a < |w| < 2a$ so $\{z \in \mathbb{C}^* \mid a < |z| < 2a\}$ is an open neighbourhood of $w$ in which no two points are equivalent.

Let $\pi : \mathbb{C}^* \to \mathbb{C}^*/\sim$ be the quotient map. We put a complex structure as follows: let $(\pi(U), \varphi)$ be charts where $V = \pi(U)$ for some $U$ with no two points equivalent and $\varphi : V \to U$ such that $\varphi_w(\pi(z)) = z$ - note, such a $\varphi$ exists because no two points of $U$ are equivalent so each point in $\pi(U)$ has a unique preimage. One can check that these charts are compatible and define a maximal atlas. Note that $R = \mathbb{C}^*/\sim\, = \pi(\mathbb{C}^*) = \pi(\{z \in \mathbb{C} \mid 1 \leq |z| \leq 2\})$. As $R$ is the continuous image of a compact set, it is itself compact.

It is not hard to see that, topologically, $R$ is a torus as it can be obtained by identifying the inner and outer boundaries of the closed annulus $\{z \in \mathbb{C} \mid 1 \leq |z| \leq 2\}$ - note, this is exactly what the quotient map $\pi$ does. A torus is usually constructed as a parallelogram with opposite sides identified. These two constructions really are the same because by identifying one pair of opposite sides of the parallelogram, we obtain a closed annulus where the unidentified sides become the inner and outer boundaries.

More generally, if $\alpha \in \mathbb{C}^*$ with $|\alpha| > 1$, and we define $\sim$ by $z \sim w$ if and only if $z = \alpha^nw$, then the above argument shows that $R_{\alpha} := \mathbb{C}^*/\sim$ is a compact Riemann surface which is topologically a torus. Therefore $R_{\alpha}$ is of the form $\mathbb{C}/\Lambda$ for some lattice $\Lambda$. We can choose the lattice to be of the form $\langle 1, \tau\rangle$ where $\tau \in \mathbb{H} = \{z \in \mathbb{C} \mid \operatorname{Im}(z) > 0\}$. There should be some correspondence between $\tau$ and $\alpha$, once we've determined what it is, we just need to find the $\tau$ corresponding to $\alpha = 2$.

We want to find a biholomorphism $\mathbb{C}/\Lambda \to R_{\alpha}$ for appropriate choices of $\tau$ and $\alpha$. One way to construct such a map is to begin with a holomorphic map $\mathbb{C} \to \mathbb{C}^*$ which descends to the quotients. A natural candidate is the exponential map.

Consider $f : \mathbb{C} \to \mathbb{C}^*$ given by $f(z) = \exp(-2\pi iz)$. Note that if $m + n\tau \in \Lambda$,

$$f(z + m + n\tau) = f(z)f(m)f(n\tau) = f(z)f(\tau)^n.$$

So if we take $\alpha = f(\tau)$, then $z$ and $z + m + n\tau$ have the same value under the induced map $\mathbb{C} \to R_{\alpha}$ given by $z \mapsto [f(z)]$. Therefore, it descends to a well-defined map $F : \mathbb{C}/\Lambda \to R_{\alpha}$ given by $z + \Lambda \mapsto [\exp(-2\pi iz)]$. In local coordinates, the map $F$ is given by $z \mapsto f(z - \lambda)\alpha^k$ for some $\lambda \in \Lambda$ and some $k \in \mathbb{Z}$. As $f$ is holomorphic, so too is $F$. Note that if $[f(z)] = [f(w)]$ then $f(z) = f(w)f(\tau)^n = f(w + n\tau)$ and therefore $a = w + n\tau + m$, so $z + \Lambda = w + \Lambda$. That is, $F$ is injective and therefore a biholomorphism between $\mathbb{C}/\Lambda$ where $\Lambda = \langle 1, \tau\rangle$ and $R_{\alpha}$ where $\alpha = f(\tau) = \exp(-2\pi i\tau)$.

As $f$ is surjective, for any $\alpha \in \mathbb{C}^*$, we can find an appropriate $\tau$. If $|\alpha| > 1$, then

$$1 < |\alpha| = |\exp(-2\pi i\tau)| = \exp(\operatorname{Re}(-2\pi i\tau)) = \exp(2\pi\operatorname{Im}(\tau)),$$

so $\tau \in \mathbb{H}$. In this particular case, we just need to find a $\tau$ satisfying $\exp(-2\pi i\tau) = 2$. One option is $\tau = \frac{i}{2\pi}\ln 2$.

I have attempted to indicate what $f$ does in the image below. Note that on the left, both pairs of sides (the green and the blue pairs) need to be identified to obtain a torus, while on the right, the green pair has already been identified, so only the blue pair remains.

Note, the green curve on the right from $1$ to $\tau$ can wind in the clockwise or anticlockwise direction.

Consider $\gamma : [0, 1] \to \mathbb{C}$ given by $\gamma(t) = t\tau$. This is a parameterisation of the green line between $0$ and $\tau$ in the left picture. Composing with the map $f$, we get a parameterisation of the green curve on the right. If $\tau = a+bi$, then

$$(f\circ\gamma)(t) = \exp(-2\pi it\tau) = e^{2\pi bt}(\cos(-2\pi at) + i\sin(-2\pi at)).$$

If $a > 0$, then $\sin(-2\pi at) < 0$ for small $t$, so the curve winds clockwise. If $a < 0$, then $\sin(-2\pi at) > 0$ for small $t$, so the curve winds anticlockwise. If $a = 0$, then $\sin(-2\pi at) = 0$ so $\tau$ lies on the real axis and $f\circ \tau$ is a parameterisation of the straight line segment from $1$ to $\tau$.

Moreover, if $a \neq 0$, the curve can also wind multiple times.

Each time the curve winds around the centre circle, it crosses the real axis twice. Therefore, if the solutions to $\operatorname{Re}((f\circ\gamma)(t)) = 0$ for $t \in [0, 1]$ are $0, t_1, \dots, t_k$, then the number of winds is $\left\lfloor\frac{k}{2}\right\rfloor$. Note that $\operatorname{Re}((f\circ\gamma)(t)) = e^{2\pi bt}\sin(-2\pi at) = 0$ is equivalent to the equation $\sin(-2\pi at) = 0$. The zeroes of sine are integer multiples of $\pi$, so we need to determine for how many integers $m$ is there $t \in [0, 1]$ such that $-2\pi at = m\pi$; equivalently, how many integers $m$ such that $t = -\frac{m}{2a} \in [0, 1]$. There are $\lfloor 2a\rfloor + 1$ such integers, so $k = \lfloor 2a\rfloor$ and hence the green curve winds $\left\lfloor\frac{\lfloor 2a\rfloor}{2}\right\rfloor = \lfloor a\rfloor$ times. Note, the sign of the winding indicates the orientation: positive winding is clockwise, while negative winding is counterclockwise.