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Nov 7, 2010 at 14:13 history edited user6976 CC BY-SA 2.5
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Nov 7, 2010 at 4:31 comment added Kate Juschenko @Mark: thank you so much for the paper!
Nov 7, 2010 at 4:10 vote accept Kate Juschenko
Nov 7, 2010 at 4:10 vote accept Kate Juschenko
Nov 7, 2010 at 4:10
Nov 7, 2010 at 3:32 history edited user6976 CC BY-SA 2.5
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Nov 7, 2010 at 3:17 history edited user6976 CC BY-SA 2.5
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Nov 7, 2010 at 1:58 comment added user6976 Note that in your other example with $2$ instead of $4$, the angles are $2\pi/5$, the sum is $6\pi/5\gt \pi$ and the theorem does not apply.
Nov 7, 2010 at 1:56 comment added user6976 You need to compute the Gersten-Stallings angles at each corner of the triangle. For example two sides have $\langle a_0\rangle$ and $\langle a_1 \rangle$ on the edges. Then compute the shortest word in $a_0, a_1$ that is equal to 1. In your case, I am sure, it is 7 (the length of the defining relation involving these two letters). Hence the angle is $2\pi/7$. The sum of all three angles is then $6\pi/7\lt \pi$. Hence by the Gersten-Stallings-Haefliger... the complex of groups is developable, and the B-S group embeds.
Nov 7, 2010 at 1:52 comment added Kate Juschenko Thank you for your suggestions. I have to note that the following: $a_0 a_1 a_0^{-1}=a_1^2$; $a_1 a_2 a_1^{-1}=a_2^2$; $a_2 a_0 a_2^{-1}=a_0^2$ is trivial group, so it is essential that we have power 4. It might be that 3 will also work
Nov 7, 2010 at 1:38 history edited user6976 CC BY-SA 2.5
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Nov 7, 2010 at 1:08 history answered user6976 CC BY-SA 2.5