Let $S$ be a Riemann surface of genus $g \geq 0$ with $n$ punctures, i.e., with $n$ distict points removed. Let $f: S \rightarrow R$ be a quasi-conformal map. Then $R$ is also Riemann surface of genus $g$ with $n$ punctures.

Assume $S'$ and $R'$ are surfaces arising from $S$ and $R$ by filling in the $n$ punctures. Is it always possible to extend $f$ quasi-conformally to $f':S' \rightarrow R'$ such that the filled in points are fixed but not necessarily pointwise.

The question is true for replacing punctures by boundary components. Since then ideal boundary is not empty and there is always an extension for the ideal boundary.

But in case of punctures the ideal boundary is empty.


1 Answer 1


Isolated points are removable singularities for quasiconformal maps, see for instance Theorem 17.3 in Vaisala's book (where higher-dimensional case is proven too).

J. Vaisala, Lectures on n-dimensional quasiconformal mappings, Lecture Notes in Mathematics 229, Springer, Berlin-Heidelberg-New York, 1971.


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.