Abstract:For any countable group $G$ whatsoever, there is a complete hyperbolic surface whose isometry group is $G$.
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- Daniel Allcock
- Affiliation: Department of Mathematics, University of Texas at Austin, Austin, Texas 78712
- MR Author ID: 620316
- Email: email@example.com
- Received by editor(s): May 4, 2005
- Published electronically: May 5, 2006
- Additional Notes: This work was partially supported by NSF grant DMS-024512.
- Communicated by: Alexander N. Dranishnikov
- © Copyright 2006
American Mathematical Society
The copyright for this article reverts to public domain 28 years after publication.
- Journal: Proc. Amer. Math. Soc. 134 (2006), 3057-3059
- MSC (2000): Primary 51M09; Secondary 20F99
- DOI: https://doi.org/10.1090/S0002-9939-06-08460-7
- MathSciNet review: 2231632