Is there a systematic way to solve equations in the braid groups? In particular, if $B_3$ is the braid group on three strands with the presentation $\{ a,b\ | \ aba = bab \}$, how do I find $x$ so that $xaxbx^{-1}b^{-1}x^{-1} = bax^{-1}b^{-1}x$ (if such an $x$ exists)?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Some terminology: if there is an algorithm to solve systems of equations and inequations in a group G, then the universal theory of G is said to be decidable. $\endgroup$– HJRWCommented Mar 13, 2010 at 21:23
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
2
This group is a central extension of $PSL_2\mathbb{Z}$, which is a virtually free group. There is an algorithm to solve equations in such groups, and parameterize the solutions. Since your equation is degree zero in $a,b,x$, if the lift of the solution in $PSL_2\mathbb{Z}$ to $B_3$ solves the equation for one lift, it should work for any other lift. I'm not quite sure though how to determine this uniformly over all lifts of the solution. The solutions are given by Makanin-Razborov diagrams, and they are parameterized by various automorphisms. So I think you just need to check one solution in each equivalence class coming from each orbit.
answered Mar 12, 2010 at 23:34
-
$\begingroup$ Technical quibble: I believe Dahmani and Guirardel don't do enough to compute the full Makanin--Razborov diagram. Their algorithm tells you whether or not there is a solution, but doesn't compute all solutions. There are very few concrete examples in which one actually knows the full Makanin--Razborov diagram. $\endgroup$– HJRWCommented Mar 13, 2010 at 18:22
-
$\begingroup$ I see, thanks for the correction. I saw Gurardel talk about this a couple of years ago, but I haven't actually read the paper. $\endgroup$– Ian AgolCommented Mar 14, 2010 at 15:57
$\begingroup$
$\endgroup$
2
The braid group is automatic, so has a solvable word problem. You might find this helpful. However, it's not clear to me that this means that equations can actually be solved.
answered Mar 12, 2010 at 22:26
-
1$\begingroup$ You're quite right. Having solvable word problem is much weaker than having decidable universal theory. For instance, solving the word problem in free groups is very easy, and proving that the universal theory of a free group is decidable is quite hard (see Agol's answer for a link with some references). $\endgroup$– HJRWCommented Mar 13, 2010 at 21:28
-
1$\begingroup$ More specifically: here's an example of a fg nilpotent group with undecidable universal theory. ams.org/mathscinet/search/… $\endgroup$– HJRWCommented Mar 13, 2010 at 21:31
$\begingroup$
$\endgroup$
Your equation is $xaxbXBXXbxAB = 1$; thus the substitution $x = b$ answers your particular question. May I ask where this equation comes from?
