In arXiv:1005.0583 the authors wrote that in two dimensional space the configuration space of n particles is multiply-connected and therefore the fundamental group of the configuration space is the braid group.

Further, in the case of three dimensional space and when the particles are indistinguishable, then the fundamental group is equal to the permutation group. This means that the clockwise and anticlockwise exchange of two particles is equal and therefore we have only fermions and bosons in this case.

Is there a mathematical proof that in two(three) dimensional space the fundamental group of the configuration space is braid(permutation) group? And how I can show, that in three dimension the braid group breaks down to the permutation group and why then clockwise and anticlockwise exchange of two particles is equal?


Intuitively this actually is very plausible, though you have to work your intuition a bit; fix a point in configuration space, this can be seen as a single point in something that locally looks like $\Bbb R^{2n}$ or $\Bbb R^{3n}$, but it is more convenient to think of $n$ distinct points of (something that locally looks like) $\Bbb R^2$ or $\Bbb R^3$.

A path through configuration space is a collection of paths based at these points. In the case of indistinguishable particles, we may also have paths starting at one point and ending at another, as long as at each point a path starts and a path ends. The only restriction is that the paths don't intersect for any given $t$, if $t$ is the (common) parameter of the paths. If the parameter runs from 0 to 1, the paths form a subset of $\Bbb R^2\times [0,1]$ and $\Bbb R^3\times [0,1]$. Two paths (in configuration space) are homotopic if there is a homotopy between all individual paths in $\Bbb R^2$ or $\Bbb R^3$ that leaves all end points fixed. In $\Bbb R^2$ it is quite easy to see that this gives the braid group.

Now for $\Bbb R^3$: you may be aware that in $\Bbb R^4$ there are no knots: all knots are homotopic to the circle. This may seem counterintuitive because we cannot visualize it, but in reality it's utterly trivial: it is the equivalent of having a circle in the plane with a point inside, it is not possible to continuously move the point outside the circle without intersecting it. However, when you add a dimension it is clear that this is not the case. Likewise for a circle in the $x-y$-plane around the $z$-axis is $\Bbb R^4$. This can be continuously untangled. Namely, move you circle in the $x-y$-plane into the parallel plane where $z = 0$, $w = 1$, if $w$ is the fourth coordinate. Every point of the circle has $w$-coordinate equal to 1. Move it within the $w = 1$, $z = 0$ plane to some large $x$. At every moment the $w$-coordinate of every point of the circle was 1, so it never intersected the $z$-axis.

Anyway, in 3D the braid of paths lives in $\Bbb R^4$ and can be untangled. Two paths in configuration space (i.e. braids of paths in $\Bbb R^3 \times [0,1]$) are homotopic to each other exactly when they connect the same initial and final points, i.e. when they define the same permutation. Paths twisting around eachother can all be transformed continously into each other.

  • $\begingroup$ Thanks. From a geometrical point of view it was fully clear to me. So it seems that there is just this geometrical description but no algebraic proof, like why for example in 3D space dimensions $B_{k}=B_{k}^{-1}$ is true, where $B_{k}$ is a generator of the braid group. $\endgroup$ – Lars Milz Jun 11 '15 at 6:04
  • $\begingroup$ @LarsMilz If by an algebraic proof you mean one in terms of generators and relations, and you already assume that you know that the geometric braid group is generated by crossing the $k$-th string over the $k+1$-st string, which I guess is what you refer to as $B_k$ (after having ordered the points in some way), then you only need to show that the relation $B_k = B_k^{-1}$ holds. (continued) $\endgroup$ – doetoe Jun 11 '15 at 12:18
  • $\begingroup$ This part cannot be done by algebra of course, you will need some geometry: you could take explicit representatives of the generators and their inverses and show they are homotopic, use the fact that a knot complement in 4-space is simply connected (or adapt a proof thereof) etc. Once you have this relation, you can construct an isomorphism by sending $B_k\mapsto (k\ \ (k+1))$. $\endgroup$ – doetoe Jun 11 '15 at 12:18

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