# Proof of equivalence of conformal and complex structures on a Riemann surface.

I am trying to understand the fundamentals of Riemann surface theory and so far I have the following:

--Definition 1. A conformal structure on a Riemann surface $\Sigma$ is an equivalence class of metrics $$[g]=\{e^{2u}g \colon u\in \mathcal{C}^\infty(\Sigma)\}$$

--Definition 2. A complex structure on a Riemann surface $\Sigma$ is an equivalence class of complex atlases, where two atlases are considered equivalent iff their union forms a new complex atlas.

Note 1. Each complex structure has a canonical representative given by the maximal atlas

---Riemann's Uniformization Theorem. In any given conformal structure, there exists a unique metric with constant curvature of either $1,0$ or $-1$.

Note 2 This gives a means of choosing a canonical representative for each conformal structure.

My questions are these:

(I) Which theorem tells us that there is a bijection between these two definitions of conformal and complex structures?

(II) What is meant by the "Riemann moduli space"?

Many Thanks, A.

EDIT: In establishing the claim (I), it appears to be neccessary to take as assumption that a surface is orientable.

Equivalence of conformal structure and complex structure:

1).From complex structure to conformal structure is easy. Given a complex structure, we have $T_p M \simeq \mathbb{C}$. With complex number $i$ on tangent space, there is a natural orthogonality $<1,i>=0$.The orthogonality property is independent of the chart chosen because the transition map is holomorphic (Rienmann Cauchy equation). So the inner product structure is well-defined at the tangent spaces of every point of M which yield a Riemannian metric on $M$ hence a conformal structure.

2). On the other hand , if one is given a conformal structure on $M$, suppose one metric in this conformal equivalence is $ds^2=Edx^2+2Fdxdy+Gdy^2$. One can check this can be written in complex notation as $ds=\lambda(z)|dz+\mu(z)d \overline{z}|$ for some positvie function $\lambda(z)$ and complex valued function $\mu(z)$ with $|\mu(z)|<1$. To say the coordinate map $\phi_k: U_k \rightarrow \mathbb{C}$ is conformal, it's the same as saying this diffeomorphism preserves angles: $$\frac{\phi_k^{\ast} d\eta^2 (u,v)}{(\phi_k^{\ast}d\eta^2(u,u))^{\frac{1}{2}}{(\phi_k^{\ast} d\eta^2(v,v))^{\frac{1}{2}}}}=\frac{ds^2 (u,v)}{(ds^2(u,u))^{\frac{1}{2}}{ (ds^2(v,v))^{\frac{1}{2}}}}$$ where $|d\eta|$ is the Euclidean metric on $\mathbb{C}$.

Equivalently, this is to say $|dz+\mu(z)d\overline{z}|$ is proportional to $|d\eta|=|d\phi_k(z)|=|{\phi_{k}}_{z}dz+{\phi_k}_{\overline{z}}d\overline{z}|.$ Hence a solution of the Beltrami equation ${\phi_{k}}_{\overline{z}}=\mu(z){\phi_{k}}_{z}$ yield the existence of such a conformal map. Click here

Notice up to this point, we are only talking about the angle-preserving property of a conformal map from the perspective of a metric other than the holomorphic(analytic) property of a conformal map! But we are heading there. Because of this angle-preserving property, one knows the coordinate map preserves complex structure. Remember it's nothing more a rotation by $\frac{\pi}{2}$ on the plane.
Click here $$J{\phi_{k}}_{\ast}v={\phi_{k}}_{\ast}J v$$ However, $${\phi_{k}}_{\ast}(J \frac{\partial}{\partial{x}})={\phi_{k}}_{\ast}(\frac{\partial}{\partial y})={u}_y +i {v}_y$$ Where $\phi_k(x,y)=u(x,y)+iv(x,y)$

Similarly, $$J ({\phi_{k}}_{\ast} \frac{\partial}{\partial x})= J(u_x+iv_x)=i u_x- v_x$$

So $${u}_y +i {v}_y=i u_x- v_x$$

Thus we at last reach Cauchy Riemann which means the atlas is compex analytic.

• Nice explantation! It hadn't occurred to me that constructing an atlas with angle and orientation preserving transition functions (i.e. a complex structure), given a conformal class, is essentially the same as constructing isothermal coordinates (hence the Beltrami equation). Dec 18, 2017 at 14:06
• Regarding part (1), however, I think one or two more words should be given to show why this gives a well defined conformal structure. Indeed your definition of orthogonality makes reference to a specific choice of coordinates about the point $p$ (you need to specify a chart to choose which vectors in $T_p\Sigma$ will represent the elements 1 & $i$), and it is only due to the fact that transition functions between any coordinates will be holomorphic, that any other choice of coordinates would give rise to an equivalent choice of inner-product at that point (modulo rescaling). Dec 18, 2017 at 14:11
• Also, for the final part of (2), don't we need to use that the surface is orientable? Dec 18, 2017 at 14:15
• Yes, for part1), this is a point I should have added. For part 2), complex manifold is orientable, the angle preserving equality implicitly contains the information that it's counterclockwise rotation and counterclockwise makes sense because the surface is oriented.
– Dai
Dec 19, 2017 at 18:23
• "So the inner product structure is well-defined" -- only up to a positive factor Sep 12, 2019 at 14:25

As Dmitry Zaitsev correctly noted, the proof in Part (1) of the accepted answer is incomplete: The (complex) isomorphisms $$T_pM\to {\mathbb C}$$ are well-defined only pointwise: Unless the tangent bundle $$TM$$ is trivial, you cannot even make choices of these isomorphisms continuous with respect to $$p$$. What an (almost) complex structure on the tangent bundle $$TM$$ gives you is the orientation on $$M$$ and the notion of an angle between tangent vectors. It does not give a well-defined notion of length for tangent vectors. The standard way to deal with this problem is as follows:

Let $$\{(U_\alpha, \phi_\alpha)\}_{\alpha\in A}$$ be a locally finite system of holomorphic coordinates on the given Riemann surface $$X$$, i.e. the cover $${\mathcal U}= \{U_\alpha\}_{\alpha\in A}$$ of $$X$$ is locally finite (every point is covered by finitely many charts). Why such a cover exists is a tricky issue: I will simply assume that $$X$$ is paracompact. (It is a theorem due to Rado that every Riemann surface is paracompact. See the discussion here.) Paracompactness means that every open cover admits a locally finite subcover.

Now, given this, we proceed as follows. Let $$\{\eta_\alpha: \alpha\in A\}$$ be a partition of unity subordinate to the open cover $${\mathcal U}$$.

For each $$\alpha\in A$$ equip $$U_\alpha$$ with the pull-back metric $$g_\alpha= \phi_\alpha^*(|dz|^2)$$, obtained by the pull-back of the Euclidean metric from $${\mathbb C}$$ via $$\phi_\alpha$$. Set $$h_\alpha= \eta_\alpha g_\alpha$$ and extend $$h_\alpha$$ by zero to the rest of $$X$$. The result is a semi-Riemannian metric $$h_\alpha$$ on $$X$$ (it is only positive semidefinite rather than definite on tangent spaces).

Lastly, set $$h=\sum_{\alpha\in A} h_\alpha.$$ Since $${\mathcal U}$$ is locally finite, this sum is a smooth and semi-Riemannian; it is a Riemannian metric since $$\{\eta_\alpha: \alpha\in A\}$$ is a partition of unity. Conformality of this metric on $$X$$ follows from the following observation:

Observation. Let $$g=\rho(z)|dz|^2$$ be a conformal Riemannian metric on an open subset $$U\subset {\mathbb C}$$. Let $$f: V\to U$$, a biholomorphic map from an open subset $$V\subset {\mathbb C}$$. Then the pull-back metric $$f^*(g)$$ equals $$\rho(f(w))|f'(w)|^2 |dw|^2$$ and, hence, is again conformal. Thus, conformality of a Riemannian metric on a Riemann surface is independent of the local holomorphic chart. In particular, a finite sum of conformal metrics is again conformal.

Lastly, here is an amusing and little-known fact (due to Robert Gunning and Raghavan Narasimhan). Suppose that $$X$$ is a connected noncompact Riemann surface. Then $$X$$ admits a conformal Riemannian metric of zero curvature. (Such a metric is typically incomplete.)

• Thanks Moishe for the detailed response. This point about the complex structure only giving a notion of orthogonality was raised at the time (see my comment), before the answer was accepted. It is nice to see this careful argument using a partition of unity though, so thank-you for the input! I am curious as to whether a locally finite cover is necessary for the proof. I had some ideas of how the conformal structure could be defined by an extension procedure (see my response to Dietrich's answer), but I never wrote this up, so it may be flawed. I will keep thinking :) Mar 6, 2020 at 9:20
• Also interesting theorem about non-compact surfaces!! Does this mean that my statement of the uniformization theorem from the original question (specifically the uniqueness part) is wrong, or needs to be supplimented with compactness? Mar 6, 2020 at 9:22
• @AerinmundFagelson: Instead of saying "compact" in the statement of the Uniformization Theorem you should say that the metric is complete. Then everything is OK. See my answer here. As for a locally finite cover: Pretty much yes. The reason is that one can ask if a smooth surface (without a given complex structure and assuming only Hausdorffness) admits a Riemannian metric. This turns out to be equivalent to paracompactness of the surface. Mar 6, 2020 at 10:51
• @AerinmundFagelson The bottom line is that your extension procedure will not work without paracompactness. Mar 6, 2020 at 13:34
• I guess, I will wear the down-vote on this answer as a badge of honor. Mar 8, 2020 at 3:08

(1) Specifying a complex structure completely specifies the conformal structure, and vice-versa. One might see this from the following Theorem:

Theorem: Let $R$ and $S$ be Riemann surfaces induced by oriented $2$-dimensional Riemannian manifolds $(M,ds^2)$ and $(N,ds_1^2)$ respectively. Then the map $f\colon (M,ds^2)\rightarrow (N,ds_1^2)$ is conformal if and only if $f\colon R\rightarrow S$ is biholomorphic.

(2) Riemann's moduli space $R_g$ is the space of analytic equivalence classes of Riemann surfaces of fixed genus $g$ (see also here).

• In (1), how does one define a map between Riemannian manifolds to be biholomorphic without first choosing complex structures? It seems like we want to show two Riemann surfaces have metrics in the same conformal class if and only if their corresponding atlases are in the same complex class. But the question then becomes: how do we associate any complex atlas to a corresponding conformal class and vise-versa? Apr 2, 2016 at 9:12
• I've been thinking about how to construct a correspondence straight from the fact that in 2 dimensions holomorphic maps are precisely those conformal maps which preserve orientation. How does this sound? Given a surface $\Sigma$ with complex structure $c$, let $z_1:U_1\to\mathbb{C}$ be some chart and define a metric on this coordinate patch in real coordinates $x_1=Re(z_1)$, $y_1=Im(z_1)$ by $ds^2=dx_1^2+dy_1^2$. Now if $z_j:U_j\to\mathbb{C}$ is any other chart, then the transition function $z_j\circ z_1^{-1}:z_1(U_1\cap U_j)\to z_j(U_1\cap U_j)$ is holomorphic and thus conformal Apr 3, 2016 at 10:59
• Therefore, we may define a function $\rho_j:U_j \to \mathbb{R}$ so that $ds^2=dx_1^2+dy_1^2=\rho_j(x_j,y_j)(dx_j^2+dy_j^2)$ on the overlap $U_1\cap U_j$, where $x_j=Re(z_j)$,$y_j=Im(z_j)$. Thus we can extend our metric to $U_1\cup U_j$ by defining it on the patch $U_2$ by $ds^2=\rho_j(x_j,y_j)(dx_j^2+dy_j^2)$. Repeating this extension process, we can construct a metric $g$ on the entire surface $\sigma$. We see that the conformal class of $g$ will be independent of the initial choice of chart $z_1$ and of the choice of patching functions $\rho_j$. Apr 3, 2016 at 11:07
• Conversely, we have that if $V_1,V_2\subset \mathbb{R}^2$ and $x,y$ are cartesian coordinates on $\mathbb{R}^2$, then $\psi:V_1\to V_2$ is holomorphic with respect to the complex coordinate $z=x+iy$ if and only if it is conformal AND det($D\psi)>0$. Thus given a conformal structure on an ORIENTED surface, we may construct an atlas whose transition functions are both conformal and orientation preserving. Choosing some local isothermal coordinates $x,y$ and defining a complex coordinate $z=x+iy$ we can then patch together to obtain a corresponding complex atlas. Apr 3, 2016 at 11:18