All Questions
1 question
2
votes
0
answers
105
views
Vanishing of Goldman bracket requires simple-closed representative?
Let $\Sigma$ be a connected oriented surface, and $[-,-]\colon \Bbb Z\big[\widehat\pi(\Sigma)\big]\times Z\big[\widehat\pi(\Sigma)\big]\to Z\big[\widehat\pi(\Sigma)\big]$ be the Goldman Bracket. Note ...
- 1,097