I have a piecewise linear (PL) surface transversely immersed in $\mathbb{R}^3$; is this a Riemann surface in the sense that I can describe it with a local coordinate $z\in \mathbb{C}$? My basic argument is that in 2 dimensions I think the PL and smooth categories coincide, so the question reduces to "can any smooth immersed surface in $\mathbb{R}^3$ be a Riemann surface?" My first instinct is that this is false; the added complex structure would chance the nature of the surface (ie the Cauchy-Riemann equations must now be satisfied). However, you can put an almost complex structure on any even-dimensional real manifold so I'm thinking the statement might actually be true.


Any ideas?



Are transversely immersed PL surfaces Riemann surfaces?
This question seems very confused. It is true that every PL surface can be given a canonical smooth structure. It is also true that a surface $X$ that is smoothly immersed in $\mathbb{R}^n$ can be given a canonical Riemann surface structure -- pulling back the Euclidean metric to $X$ gives a Riemannian metric on $X$ and thus a conformal structure, and it is known that (in real dimension $2$) there is a natural bijection between conformal structures and Riemann surface structures.


However, if $X$ is a PL surface and $Y$ is the smooth surface obtained (in a canonical way) by smoothing $X$ and $\phi : X \rightarrow \mathbb{R}^n$ is a PL immersion, then I don't think there is any canonical choice of a smooth immersion $\phi' : Y \rightarrow \mathbb{R}^n$. You can certainly choose $\phi'$ to be close to $\phi$ in any reasonable sense of the word close, but that's not good enough.


