# Spin and fundamental group

1. Aug 17, 2011

### vanirof

How does fundamental group determines number of possible quantum statistics?
Why is number of possible statistics equal to number of different possible paths?

2. Aug 17, 2011

### homeomorphic

Hmmm...

I'm not an expert on the spin-statistics theorem, but I think I will take a stab and say what I know. I'm a topologist, so I perk up at the mention of the fundamental group.

If you have n particles, their configuration space is like M cross M cross ... cross M, where M is the configuration space of one of them. And then, if you want to impose the condition that the particles can't occupy the same spot, you take the out set that has repeated coordinates.

At this point, I'm confused about whether I want to mod out by the action of the symmetric group on the factors. In 2 dimensions, the fundamental group will be the pure braid group if you don't mod out by permutations, otherwise, just the braid group. The picture here is that the world-lines of particles can braid around each other. A path in configuration space is really a braid because there is one path for each particle.

If you are in 3 dimensions or higher, if you mod out by permutations, the fundamental group will be trivial because there is no braiding. If you cross with the time axis, you have 4-dimensions, so an under-crossing is the same as an over-crossing. If you don't mod out by permutations, you get the symmetric group because you can switch particles however you want. So, I think maybe you don't want to mod out by permutations for our purposes, but I'm not sure.

Anyway, I think the idea is this. Naively, if you want to form the Hilbert space for n particles, you might guess that it should be the tensor product, by super-position. But that's not quite right because it doesn't have the right symmetries. In particular, you ought to be able to switch the particles around. In other words, the Hilbert space should be a representation of the symmetric group. Furthermore, because it's a symmetry of the system, it has to send a state vector to a scalar multiple of itself. The only representations that qualify are the trivial representation (bosons) and the one that multiplies by the sign of the permutation (fermions).

But, maybe we are being too naive again. So, the idea is that, maybe we shouldn't be able to just switch the particles however we want, using the symmetric group. Maybe it matters how we do the switching. And, rather than just the symmetric group, what that suggests is that we should use the fundamental group of the configuration space--which basically is the world-lines of the particles. So, the Hilbert space ought to be a representation of that group, in order to have the appropriate symmetries. And nature seems to bear this out. Mostly, what we see, in our 3 dimensions is fermions and bosons, whose Hilbert spaces are representations of the symmetric group. And in systems that are essentially 2-d, we get anyons, whose state spaces are representations of the (pure?) braid group.

3. Aug 18, 2011

### vanirof

Thank you for quick and detailed reply.
About moding out,did you ment taking symmetrized or antisymmetrized tensor product of M^n configuration space?
I read that fundamental group in space with dimension different from 2 is cyclic of order 2 because there is two different classes of paths (loops) ,and for n=2 is infinite cyclic group.This is because group of physical objects is Poincare group wich contains SO(n).
Still is hard for me to see dirrect connection between number of paths and number of possible statistics.
When constructing the physical spaces of n particles why scalar that multiplies the vector,and sends it to himself must be real? As i remember in physic all vector are same wich differ up to phase factor.
Why for example there is no classification of statistics according to cyclic group of order 3?Is it becaouse of number of different paths or is it because they can be decomposed to a permutation of order 2?
There is one more thing i wander about.If we make condition that all swithing of particles are in same time, than for example in permutation group cyclic permutation of order 3 cannot be decomposed to cyclic permutation of order 2.

4. Aug 18, 2011

### homeomorphic

No. You can't really take symmetric/anti-symmetric products of the configuration space because it's just a manifold. What I mean is taking the quotient space. So (for n=2) you identity a point like (a, b) with (b,a) (remember a and b are distinct point in the configuration space, M of one particle). That's what I mean by modding out by the symmetric group. The symmetric group acts by permuting the coordinates, so you are "modding out" by that.

Yes. That agrees with what I said. I said the fundamental group is the symmetric group, if you don't mod out by its action. For n = 2, the symmetric group is just the cyclic group of order two.

That doesn't sound right. I'm not thinking relativistically here, and I am not sure if it's necessary--on some level, it probably is, but not for my heuristic arguments here. But the relevant group here isn't the Poincare group--it's the fundamental group of the configuration space. More or less unrelated to the Poincare group as far as I can tell.

It's not exactly paths--it's homotopy classes of loops. What is a path in configuration space? That is the question you have to answer. A path in configuration space is a map from the interval [0,1] into the configuration space. And what is the configuration space of n particles? Well, it's M cross M cross ... cross M. So, a path in the configuration space is really n paths in M. And those paths are not allowed to intersect because we removed all the points (a, a) with repeated coordinated from the space. So, it's a braid if you see it in terms of the world-lines of the particles. And, as I was trying to explain, you can consider the paths as symmetries of the system that switch the different particles, keeping track of how we did the switching (if you want them to switch the particles, though, you do have to mod out by the symmetric group action--otherwise, since the fundamental group consists of LOOPS that come back to their starting position, you can't get any non-trivial permutations because each particle returns to its starting position). That's the connection.

Last edited: Aug 18, 2011
5. Aug 18, 2011

### vanirof

So nice to finaly have someone to talk about this,who could clearify this subject to me.
http://en.wikipedia.org/wiki/Anyon
section Topological basis,
Group of physical system is a direct product of Poincare and permutation group in relativistic ,or Galilean and permutation group in nonrelativistic formulation. Both Poincare and Galilean group contains subgroup of 3 dimensional rotations,which could be represented by phi ball( Hypersphere of rotations),on wich rotations for 180 and -180 are identified,and because of that there are 2 types of loops,so fundamental group is Z2 for three dimensional rotations and also for Poincare/Galilean group.
I am not sure what is then fundamental group of product of permutation group and Rotation group,but 2 Z2 arise from two different subgroup of physical system.Think there are maybe 4 one dimensional irreducible representations of direct product of these, but i have two calculate this.
Why just + or -1 when you sayd that that scalar could be any phase factor?Is it because other one dimensional representation of permutation group are equivalent to these ?
For other Z2 that arises from rotations there are also two one dimensional ireps.
This equal time cyclic permutation of order 3 cannot be decomposed to a paths (which are permutations of order 2),so how imposing equal time condition affects the fundamental group of such system? Z2 is not enough to describe such system as fundamental group.Is it Z3 then fundamental group?And if number of particles is big then (fallowing that logic for any cyclic permutation) asimptoticaly fundamental group would be also braid group for 3 dimensional space as it is for 2 dimensional.

6. Aug 18, 2011

### vanirof

Dirrect product of two Z2 is K2 which has 4 nonequivalent irreducible representations.
I am not sure whether the fundamental group of direct product of groups is dirrect product of fundamental groups.

7. Aug 18, 2011

### homeomorphic

It may take me a while to respond to all the points.

We are talking about different spaces. My fundamental groups are fundamental groups of the classical configuration space. Your fundamental groups are those of Lie groups SO(3) and friends that arise as symmetries of the system. I realize now, I was answering a related, but slightly different question, than the one you were asking. In my setting, the fundamental group is the symmetric group, which depends on what n you choose. In your setting, it's always cyclic of order 2. The two viewpoints are related, but I'm not completely sure how.

Here's something that I think is closer to what you were asking. It has to do with covering-spaces. A cover is a certain kind of map. The general idea is that for an n-fold cover, each point should have n points in its pre-image. So, you can thinking of it as a space that wraps around n times over the space that it is covering. It turns out that covering spaces are classified by the fundamental group of the space that they are covering. They correspond to the subgroups of the fundamental group. If the fundamental group was cyclic of order 6, you get a cover corresponding to the subgroup of order 2, for example. The number of times it covers is the number of cosets. So, in this case, you would get a 3-fold cover. So, the biggest cover you can have would correspond to the trivial subgroup, and it would be a 6-fold cover. We call that biggest cover the universal cover.

So, when your fundamental group has order 2, the only cover you have is a 2-fold cover (and the 1-fold cover, but that is just the space itself). These other covers could potentially be symmetry groups of your system. So, if you have SO(3) lurking around, another symmetry group that you can easily cook up is its double cover SU(2). And sure enough, bosons have SO(3) symmetry because if you rotate by 360, you get back to where you started, and fermions have SU(2) symmetry because a 360 rotation doesn't take you back to where you started. Instead, it multiplies by -1. It's what we call a spinorial rotation.

From this perspective, paths in these groups are like rotations. In SO(3), a loop is a time-varying rotation that comes back to where it started. In SU(2), there's only one loop (up to homotopy), which would be like a rotation by 4 pi, rather than two pi, and it's a time-varying spinorial rotation.

At any rate, the number of statistics ought not to be equal to the order of the fundamental group from this vantage point, except in this case because it's such a silly little group. If it were a more complicated fundamental group, maybe the number of statistics would be the number of covering spaces, which is different. At least that's what it looks like to me.

By the way, it is true for path-connected spaces that the fundamental group of the product spaces is the direct product. However, permutation groups are discrete groups, so they are not path-connected. Actually, fundamental groups are only defined for path-connected spaces.

8. Aug 18, 2011

### homeomorphic

Yes.

It is when they are defined, but they are only defined if the space is path-connected, which isn't the case here. Permutation groups are discrete groups, so they don't have a fundamental group (or if you consider path-components, it would be trivial for all path-components because each path-component consists of only one point).

9. Aug 19, 2011

### vanirof

Ok.Let me summerize what confuses me.
Separations of elementary particles is on bosons ,which have integer spin and fermions with halfinteger spin.
This classification is derived on two separate ways:
1.from behavior of system of n identical particles upon permutations which gives phase factor + or - 1 based on in which onedimensional irep they are in
2.From properties of Lie group of rotations, which give their spin as a maximal weight and divide particles based on the identity rotation which is for fermions 4 pi and for bosons 2 pi

Important fact is that the symmetry of physical system of n identical particles is direct product of permutation group of order n and Poincare(Galilean) group.
So why is somethimes said that division on bosons and fermions is just because of 1. and othertimes just because of 2.
Shouldn't be there 4 kindes of particles because of the product of these symmetries?Two for each spin.
I think that two kindes fall out becaouse of microcausality condition that is demanded in quantum field theory,which leaves only bosons and fermions.As example Dirac equation discribes relativistic particle and give correct value in nonrelativistic limit for its spin.This part is based on theory of 2.Second part is based on 1.it is imposing comutation relations for field of that particle.Commutation relations are derived from second quantization,and one of them breaks microcausality,and is rejected.

10. Aug 19, 2011

### homeomorphic

I am not sure. I thought the two approaches were redundant, hence they both give the same two types of particles. That's my hunch.

At any rate, there are more than two approaches, and these are more like just for intuition, at least in the form we are talking about here. I don't know very much QFT--I'm just trying to learn it when I can, so my expertise lies more on the topology side of the question (I work on topological quantum field theory stuff). I don't really have much of an idea how the spin statistics theorem is proved. I just happen to have some insight into the topology aspects of it, just enough physics to be able to relate it to the topology on some level, and that's about it.