Re: Conformal structure on a Riemannian manifold

Nikos Apostolakis <absent@xxxxxxxxxxxxxxxxxxx> writes:

>"Ignat Soroko" <ignat.soroko@xxxxxxxxx> writes:
>
>> It is known that on any 2-dimensional Riemannian manifold there exists
>> a conformal structure induced by the Riemannian metric. This is usually
>> done by choosing the so-called isothermal coordinates. The question is:
>>
>> Can any compact Riemann surface be obtained as a 2-dimensional
>> submanifold in R^3, complex structure being induced by the the
>> Riemannian metric of R^3?
>>
>
>No. For example the standard flat torus does not embed in R^3.

That's irrelevant; the non-embeddability of a (not, as you implicitly
observe below, "the") torus-with-a-flat-Riemannian-metric into R^3
does not imply the non-embeddability of a torus-with-a-conformal-
structure-or-equivalently-a-complex-structure-(any-one-of-which-
comes-from-many-different-Riemannian-metrics,-some-flat,-some-not)
into R^3. Think about the many inequivalent Riemannian metrics
on the 2-sphere, all of which give it (once it's oriented) its
one and only complex structure.

Another interesting question is as follows. Note (I learned
this from Dennis Sullivan, who got it from someone else, I
don't know who) that, just as every *Riemannian* structure
on an oriented 2-manifold gives the 2-manifold a conformal
(equivalently, a complex) structure, so does every metric
*polyhedral* structure on an oriented 2-manifold: in the
middle of a face, and even at any point in the middle of an
edge, there is a natural flat metric and therefore a natural
conformal structure; at a vertex, the geometry is that of a
cone with some angle (not necessarily 2\pi), and such a cone
has a natural conformal structure too, which is compatible
off the cone point with the other structure. In particular,
every polyhedral surface in R^3 has a natural complex
structure. Question: does every complex structure on a
compact oriented 2-manifold arise in this way from some
polyhedral surface in R^3?

Lee Rudolph

.