Suppose that $D = \{ x \in \mathbb{R}^2 | ||x||_2 \leq 1 \}$ is the closed disc, and let $g$ be any Riemannian metric on $D$.
I'm wondering about the following version of uniformization: Is $(D,g)$ always conformally equivalent to the usual flat metric? If not, what if $g$ is flat?
I'm trying to understand whether regions bounded by a smooth closed curve in the plane are conformally equivalent.
I'm asking as a follow up to my question here: https://mathoverflow.net/questions/304274/what-is-e-zeta-delta-0-for-a-delta-the-laplacian-of-a-manifold/304286?noredirect=1#comment758887_304286
(The motivation for all of this comes from some questions in graph theory... so even some pointers for basic references in conformal geometry would be super helpful to me.)
Answering my question in the case that $g$ is flat:
If $(U, g_U)$ and $(V,g_V)$ are Riemannian 2 manifolds, and $\phi$ is the conformal mapping between them, then we have that $\phi^*g_V$ is conformal to $g_U$. (This is the completely obvious fact that I wasn't thinking of.)
Now $(U, \phi^*(g_V)$ is isometric to $(V,g_V)$, and $(U, g_U)$ is in the same conformal class.
In particular, for any bounded open regions in the plane with smooth boundary, $U$, we can use the Riemann mapping theorem to find a conformal map $\phi : U \to D$, where $D$ is the closed disc. Thus the metric $\phi^* d_{eucl}$ is conformal to $d_{eucl}$ on $U$. Thus $(U,d_{eucl})$ is in the same conformal class as a metric which makes $(U,g)$ isometric to the usual closed disc.
In the context of the Osgood/Phillips/Sarnak result (discussed over on my linked mathoverflow question), this is saying that the conformal equivalence class of all metrics on $D$ with fixed perimeter and non-positive average Gaussian curvature includes all of the metrics that come from open regions in the plane bounded by a smooth closed curve. (Though it is presumably much larger.)
Answering my question in the comment of the linked math-overflow page: yes, the Sarnak/Osgood/Phillips theorem implies that extremizing the zeta regularized laplacian determinant over all regions bounded by a fixed perimeter closed curve results in a disc.
(Though I'm not 100% confident in my deductions about this right now.)