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Nov 12, 2011 at 3:53 vote accept Yaping Yang
Nov 10, 2011 at 16:20 comment added Yaping Yang Yes, Todd! You are right. Thank you for correcting!
Nov 9, 2011 at 13:40 comment added Todd Trimble I took the liberty of changing the form of the presentation of the first by replacing relations $(T_iT_j)^{m_{ij}} = 1$ by $T_iT_jT_i\ldots = T_jT_iT_j\ldots$, because if we simply remove the equations $T_i^2 = 1$ as stated, the group presented by $(T_iT_j)^{m_{ij}} = 1$ is different from the group presented by $T_iT_jT_i\ldots = T_jT_iT_j\ldots$. (I think this was my fault, not yours.)
Nov 9, 2011 at 13:35 history edited Todd Trimble CC BY-SA 3.0
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Nov 9, 2011 at 12:25 answer added Adrien timeline score: 2
Nov 9, 2011 at 3:49 history edited Yaping Yang CC BY-SA 3.0
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Nov 9, 2011 at 3:31 comment added Yaping Yang Thank you, Todd! That's exactly my question, and I edited my question according to your rephrasing. Thank you again!
Nov 9, 2011 at 3:26 history edited Yaping Yang CC BY-SA 3.0
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Nov 9, 2011 at 2:27 comment added Todd Trimble Since Jim Humphreys thinks the question is stated loosely, could I try to rephrase? Let's suppose that $W$ is the Coxeter group attached to a Coxeter matrix with entries $m_{ij}$, with generators $T_i$ and relations $T_{i}^2 = 1$ in addition to the relations $(T_i T_j)^{m_{ij} = 1$ for $i \neq j$, and suppose $B_W$ is the corresponding braid group obtained by removing the relations $T_{i}^2 = 1$. Then there is a canonical surjective homomorphism $B_W \to W$, and you want a presentation of the kernel of this homomorphism. Is that the question?
Nov 8, 2011 at 23:31 answer added Jim Humphreys timeline score: 3
Nov 8, 2011 at 18:10 history edited Adrien
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Nov 8, 2011 at 16:57 history asked Yaping Yang CC BY-SA 3.0