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?

  • $\begingroup$ Perhaps you mean to ask, "is there a canonical way to put the structure of a Riemann surface on a PL immersed surface?" Because a PL immersed surface isn't literally a Riemann surface in the sense of having a complex atlas. $\endgroup$ May 26, 2011 at 17:40
  • $\begingroup$ Ok Ryan yes I think that is what I want to ask. Basically there nice ways to describe immersed Riemann surfaces (specifically, the Weierstrass formula), and I would like to describe immersed PL surfaces using the same techniques. $\endgroup$
    – cduston
    May 26, 2011 at 19:09
  • $\begingroup$ The Weierstrass formula is for immersed minimal surfaces. The Riemann surface thing has nothing to do with your problem here -- there is no reason that an immersion of a PL or a smooth surface has to be minimal in any sense of the word. $\endgroup$ May 26, 2011 at 19:19
  • $\begingroup$ In fact it's pretty easy to use the Weierstrass formula to extend this idea to arbitrary surfaces...for instance see Friedrich J. Diff. Geom. 28 (1998). As far I understand it, they still need to be complex surfaces but the minimal condition can be dropped. $\endgroup$
    – cduston
    May 26, 2011 at 20:07

1 Answer 1


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.

  • $\begingroup$ I don't believe that every PL surface can be given a canonical smooth structure, not in the strong sense in which I understand canonical (functorial w.r.t. PL homeomorphisms). A triangulation can be made to determine a smooth structure, but that's not quite the same. $\endgroup$ May 26, 2011 at 18:42
  • $\begingroup$ I suspect that's true; I meant canonical in the sense of "any two smoothings are diffeomorphic". However, I don't know a proof that you can't smooth PL surfaces in a functorial way -- do you know one? $\endgroup$ May 26, 2011 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.