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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-Reimann 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?

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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. –  Ryan Budney May 26 '11 at 17:40
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. –  cduston May 26 '11 at 19:09
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. –  Andy Putman May 26 '11 at 19:19
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. –  cduston May 26 '11 at 20:07
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1 Answer

up vote 4 down vote accepted

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.

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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. –  Tom Goodwillie May 26 '11 at 18:42
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? –  Andy Putman May 26 '11 at 19:03
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