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Nov 25, 2011 at 17:16 history edited Vitali Kapovitch CC BY-SA 3.0
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Nov 25, 2011 at 17:13 comment added Vitali Kapovitch @ Martin: yes, it's completely clear. $i\alpha(z)=log(\frac{g(z)}{z})$ and its branch is well defined since $g$ is is fixed point free.
Nov 25, 2011 at 17:03 comment added Martin Brandenburg Is it completely clear that we can choose $\alpha$ to be continuous?
Nov 25, 2011 at 16:14 comment added Joseph O'Rourke @Vitali: Thanks!!! And apologies re $\mathbb{Z}^2$ vs. $\mathbb{Z}_2$.
Nov 25, 2011 at 15:59 history undeleted Vitali Kapovitch
Nov 25, 2011 at 15:59 history edited Vitali Kapovitch CC BY-SA 3.0
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Nov 25, 2011 at 15:41 history deleted Vitali Kapovitch
Nov 25, 2011 at 15:39 history answered Vitali Kapovitch CC BY-SA 3.0