I have a question motivated by some examples in section 4.2 of Schlag's A Course in Complex Analysis and Riemann Surfaces.
Example 1. Any smooth, orientable two-dimensional submanifold of $\mathbb{R}^3$ is a Riemann surface
Example 2. Any polyhedral surface $S ⊂ \mathbb{R}^3$ is a Riemann surface.
The first statement says that all smooth orientable surfaces are Riemann surfaces. The second suggests that the converse to the first statement is false.
My question is then:
- Do all orientable topological manifolds embedded in $\mathbb{R}^3$ admit a Riemann surface structure?