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It is known that the n-string braid group $\mathbb{B}_n$ is isomorphic to $GL_n(\mathbb{F}_1[t])$ (This is a result of Kapranov and Smirnov. I would like to ask: What are the advantages(if any) in thinking $\mathbb{B}_n$ as $GL_n(\mathbb{F}_1[t])$?

cat
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    Can you give a reference on this to see what is the meaning of the statement? – Mariano Suárez-Álvarez Oct 08 '11 at 03:42
  • I think the words "known" and "result" are too strong here. It is not clear what the right hand side is exactly. In particular, if it is true, then $GL_n({\mathbb F}_1[t]$, $n\ge3$, does not have Kazhdan property (T). That seems weird. –  Oct 08 '11 at 03:46
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    I think this assertion is less a "result" and more an interpretation of a theorem of Drinfeld. Here is the argument to which I believe you are referring: http://matrix.cmi.ua.ac.be/fun/index.php/kapranovsmirnov.html

    I asked a related question here http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group

     – John Wiltshire-Gordon Oct 08 '11 at 03:57
  • @Mariano This is a link to the paper http://matrix.cmi.ua.ac.be/fun/library/KapranovSmirnov.pdf. @Mark thanks for your comment. I wanted to emphasize that an isomorphism between $GL_n(\mathbb{F}_1)$ and $\mathbb{B}$_n$ was known by Kapranov and Smirnov. However, probably I misunderstood Kapranov-Smirnov's argument and such isomorphism does not exist. On the other hand, certainly both groups are related. – cat Oct 08 '11 at 04:14
  • @Jhon thanks a lot. When I wrote my previous comment I had not seen your comment. You are right. I am referring that argument. – cat Oct 08 '11 at 04:21
  • Does the group $GL_n({\mathbb F}_1[t])$ have torsion? It looks like it is so for $n=1$. Note that the group $B_n$ is torsion-free. –  Oct 08 '11 at 14:05

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