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Dec 14 at 1:06 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Mar 18 at 22:09 history edited Carlo Beenakker CC BY-SA 4.0
typo in title corrected
S Mar 18 at 17:07 history bounty started AMath91
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S Mar 6 at 11:03 history bounty ended CommunityBot
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Feb 27 at 14:35 comment added LSpice TeX note: $\dot{\nabla}'$ \dot{\nabla}' puts the dot too high. You can force TeX to forget the extra height of the dot by using \smash: $\smash{\dot{\nabla}}'$ \smash{\dot{\nabla}}'. I have edited accordingly.
Feb 27 at 14:34 history edited LSpice CC BY-SA 4.0
`\dot{\nabla}'` -> `\smash{\dot{\nabla}}'`
Feb 27 at 14:07 answer added Carlo Beenakker timeline score: 0
S Feb 27 at 9:11 history bounty started AMath91
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Feb 27 at 9:10 history edited AMath91 CC BY-SA 4.0
added 496 characters in body
Feb 25 at 13:27 comment added abx Oops, you are right, I missed that. But what is the problem? Do you have doubts on Goldman's computation?
Feb 25 at 10:28 comment added AMath91 I'm not so sure: there he says the multiplicative factor is -8. I guess it's not so clear even for you...
Feb 25 at 8:04 comment added abx This is spelled out in the paper by Goldman, The symplectic nature of fundamental groups of surfaces., Adv. in Math. 54 (1984), no. 2, 200-225. In each case there is a natural choice, and there is no multiplicative factor.
Feb 25 at 2:31 history edited LSpice CC BY-SA 4.0
Teichmuller -> Teichmüller
S Feb 24 at 21:01 history suggested C7X CC BY-SA 4.0
MathJaxify
Feb 24 at 20:40 review Suggested edits
S Feb 24 at 21:01
Feb 24 at 19:28 history asked AMath91 CC BY-SA 4.0