Is there a q-analog to the braid group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:04:06Z http://mathoverflow.net/feeds/question/45651 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group Is there a q-analog to the braid group? John Wiltshire-Gordon 2010-11-11T06:03:23Z 2010-11-15T17:09:34Z <p>The braid group $B_n$ on $n$ strands fits into a short exact sequence of groups:</p> <p>$$1 \longrightarrow P_n \longrightarrow B_n \longrightarrow S_n \longrightarrow 1,$$</p> <p>where $S_n$ is the symmetric group on the strands, and $P_n$ is the normal subgroup of braids that do not permute the strands.</p> <p>Since symmetric groups are, in some sense, general linear groups over the field with one element,'' perhaps there is some corresponding short exact sequence ending with $GL_n(F_q)$ that specializes to the exact sequence above as $q \rightarrow 1$.</p> <p>In the spirit of the exact sequence above, is there a $q$-analog to the braid group?</p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45654#45654 Answer by Mark Sapir for Is there a q-analog to the braid group? Mark Sapir 2010-11-11T06:21:06Z 2010-11-11T06:21:06Z <p>You may want to look at <a href="http://www.math.ucsb.edu/~jon.mccammond/slides/07-marseille-day3.pdf" rel="nofollow">these</a> slides of Jon McCammond's talk. I am not sure he wrote a paper about it, but he did introduce the idea similar to what you want and in more general situation (arbitrary Artin group instead of the braid group), </p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45659#45659 Answer by Romeo for Is there a q-analog to the braid group? Romeo 2010-11-11T07:09:15Z 2010-11-15T17:09:34Z <p>Possibly. 2 ideas in this circle are: (1) The Artin braid groups can be formulated as Weyl groups" (corresponding to the $A_n$ Dynkin diagrams), and Weyl groups have q-analogues. Many references exist, but I'm not sure what the best is (hopefully others will know). (2) A second but related perspective can be found in <em>Braids, Q-binomials and Quantum Groups</em> by M Aguilar.</p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45685#45685 Answer by James Griffin for Is there a q-analog to the braid group? James Griffin 2010-11-11T11:47:16Z 2010-11-11T11:47:16Z <p>I suggest you look at the Iwahori-Hecke algebras of type A. These deform the symmetric group algebras with relations that look like</p> <p>$T_i^2 = q + (1-q)T_i$</p> <p>for generating elements $T_i$. The braid relations</p> <p>$T_iT_{i+1}T_i = T_{i+1}T_iT_{i+1}$</p> <p>still hold though so you get a (surjective) morphism</p> <p>$kB_n\rightarrow \mathcal{H}_n\rightarrow 0$</p> <p>(<strong>From here on I'm less sure of the details</strong>) giving you a `short exact sequence'</p> <p>$0\rightarrow K_n\rightarrow kB_n \rightarrow \mathcal{H}_n\rightarrow 0$.</p> <p>To define the kernel you should look at the coinvariants of $kB_n$ w.r.t. the coalgebra map of $\mathcal{H}_n$. This makes $K_n$ an algebra but not necessarily a Hopf algebra (although it may be a braided Hopf algebra in a suitable category).</p> <p>The Hecke algebra may be the algebra that you want because the q parameter counts the way the Borel double cosets in some $GL_n(k)$ multiply (recall the Bruhat decomposition). When the Borels become trivial then $q$ becomes one.</p> <p>But notice also that this deformation does not require a deformation of the braid group. I have no idea what the algebra $K_n$ looks like and if indeed it is well defined, it may still turn out to be $kP_n$.</p> <p>To offer an answer to your final question: there may very well be q-analogues of the braid groups, but they may not be what you should be looking for.</p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45696#45696 Answer by grok for Is there a q-analog to the braid group? grok 2010-11-11T13:46:22Z 2010-11-11T13:46:22Z <p>I don't see why there would be a unique scheme that "converges" to the braid group as $q\to1$. Here are some groups that could reasonably considered as the $q$-analogues of $B_n$:</p> <ul> <li><p>Consider the non-commutative power series algebra over $\mathbb F_q$ in $n$ variables, usually written $A=\mathbb F_q\langle\langle x_1,\dots,x_n\rangle\rangle$. Its group of automorphisms $B_{n,q}$ is a $q$-analogue of $B_n$; the natural map $A\mapsto A/\langle x_ix_j\rangle$ induces the map $B_{n,q}\to GL(n,q)$.</p></li> <li><p>At least when $q$ is prime, consider the subgroup of $A$ generated by $1+x_1,\dots,1+x_n$. A classical result of Magnus says it's a free group on $n$ generators. Take its closure in $A$ --- that's a free pro-$q$ group. Consider then the subgroup of $B_{n,q}$ that preserves that group. (If $q$ is a prime power, presumably throw in a few more automorphisms to obtain a group mapping onto $GL(n,q)$).</p></li> </ul> <p>There's a lot of theory on the following filtration of $B_n$: it has a subgroup $P_n$, as the poster mentioned; and $P_n$ has a lower central series converging to $1$, for which the structure of the successive quotients are well understood, as $S_n$-modules. A good $q$-analogue should probably have a filtration by $GL(n,q)$-modules with the same shapes and multiplicities in each degree.</p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45713#45713 Answer by Sheikraisinrollbank for Is there a q-analog to the braid group? Sheikraisinrollbank 2010-11-11T15:32:08Z 2010-11-11T15:54:38Z <p>Here is one strategy that has not been suggested: the braid group is the fundamental group of the space $U/W$, where $U$ is the set of points in the (complexified) reflection representation of $W=S_n$ that have trivial fixer. Now $W=GL_n(\mathbb{F}_q)$ is a reflection group over the finite field $\mathbb{F}_q$; let $U$ be the set of points (over an algebraic closure $F$ of $\mathbb{F}_q$) in the reflection representation of $W$ that have trivial fixer and define the "braid group" to be the etale fundamental group of $U/W$. Essentially by definition it has a surjection onto $W$. It's not clear to me in what sense this q-braid group might converge to the usual one as q goes to 1, but it should definitely play an important role in the study of $GL_n(\mathbb{F}_q)$. For instance, one might be able to define a Hecke algebra deforming $F[GL_n(\mathbb{F}_q)]$ via monodromy representations of this braid group. </p> http://mathoverflow.net/questions/45651/is-there-a-q-analog-to-the-braid-group/45795#45795 Answer by Alexei Oblomkov for Is there a q-analog to the braid group? Alexei Oblomkov 2010-11-12T05:58:57Z 2010-11-12T05:58:57Z <p>There is a nice paper by Etingof and Rains about a non-standard q-deformations of S_n:</p> <p><a href="http://xxx.lanl.gov/abs/math/0409261" rel="nofollow">http://xxx.lanl.gov/abs/math/0409261</a></p> <p>They deform not quadratic relations but the braid relations. It seems to be relevant to the discussion.</p>