3
$\begingroup$

Richard Schwartz, in Mostly Surfaces (Vol. 60. American Mathematical Soc., 2011), defines (on p.14) a translation surface as "a Euclidean cone surface, all of whose 'angle errors' are integer multiples of $\pi$." And a Euclidean cone surface is "a surface that is flat except at finitely many cone points." The 'angle error' is the angle deficit, $2 \pi - \theta(p)$ at the cone point $p$.

My question is:

Q. What role does "integer multiples of $\pi$" play in the theory surrounding translation surfaces? Would half-integer multiples of $\pi$ be less interesting? Rational multiples of $\pi$?

I ask this without understanding much of the theory of translation surfaces. Thanks for any insights the more knowledgeable can provide!

$\endgroup$

1 Answer 1

6
$\begingroup$

Translation surfaces are usually only allowed to have integer multiples of $2\pi$ as cone angles. A translation surface (minus the singularities) admits an atlas of charts whose transition functions are all translations, hence the name. This induces a holomorphic 1-form and an oriented foliation on the surface. Note that the holonomy around the cone points of a translation surface is trivial.

Translation surfaces come up naturally in studying the dynamics of billiards on polygonal tables with corners having rational multiples of $\pi$.

You might enjoy reading Anton Zorich's survey on the subject: http://arxiv.org/pdf/math/0609392.pdf

With integer multiples of $\pi$ at cone singularities you get a half-translation surface, where transition functions are compositions of translations and half-turns. The surface inherits an unoriented foliation and a holomorphic quadratic differential. This sort of structure arises naturally in the solutions to extremal problems on surfaces such as Teichmuller's problem.

$\endgroup$
1
  • $\begingroup$ Thank you for the pointer to Zorich's survey, which amounts to a compact book! $\endgroup$ Feb 9, 2015 at 0:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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