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Nov 10, 2015 at 21:15 comment added HJRW @YCor, oh, sure.
Nov 10, 2015 at 21:14 comment added YCor @HJRW That's right, but it's much more interesting to indeed know that there is no non-identity conjugate to this subgroup, and that 's indeed what Lee Mosher says.
Nov 10, 2015 at 19:31 comment added HJRW @YCor, to show it has no cusp would be to show that no such matrix is conjugate into the subgroup.
Nov 10, 2015 at 19:26 comment added Igor Rivin @YCor Yes, but one has to start somewhere :)
Nov 10, 2015 at 18:54 comment added YCor well, the issue is to show that no such matrix is in the subgroup. That is, the hyperbolic surface defined by this group has no cusp.
Nov 10, 2015 at 15:46 history answered Igor Rivin CC BY-SA 3.0