# 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). – Aerinmund Fagelson Dec 18 '17 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). – Aerinmund Fagelson Dec 18 '17 at 14:11
• Also, for the final part of (2), don't we need to use that the surface is orientable? – Aerinmund Fagelson Dec 18 '17 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 '17 at 18:23

(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? – Aerinmund Fagelson Apr 2 '16 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 – Aerinmund Fagelson Apr 3 '16 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 $\rho_j$. – Aerinmund Fagelson Apr 3 '16 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. – Aerinmund Fagelson Apr 3 '16 at 11:18