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

Question

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.

Answer 1

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.

Answer 2

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.)

Answer 3

(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).