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I read the answer to this question and am very intrigued by its simple and elegant explanation of the emergence of anyon, boson & fermion statistics.

@Trimok basically says:

In a space-time with one time dimension and $D$ spatial dimensions, finding possible different statistics is equilalent to look at the fundamental group (first homotopy group) of $SO(D)$

For $D=1$, the fundamental group is trivial.

For $D=2$, the fundamental group is $\mathbb{Z}$.

For $D>=3$, the fundamental group is $\mathbb{Z}_{2}$

So, it explains, why, in 3 spatial dimensions, there are only 2 kinds of statistics (fermions and bosons), while the situation is different with 2 spatial dimensions.

Unfortunately I do not know how to find the fundamental group for $SO(D)$, and its physical significance.

Could someone elaborate on how to show/prove what he says please?

Qmechanic
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SuperCiocia
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2 Answers2

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I studied algebraic topology from a mathematical point of view, so I can only try to explain the physical interpretation.

Beginning with the mathematics. The fundamental group associated with a pointed topological space is a set of equivalence classes of closed loops (under homotopy). But what does this mean? Let's start with basics.

  • A topological space $X$ is a general mathematical structure that is equipped with the notion of continuity and convergence (and etc.).

  • A pointed topological space is just the old topological space $X$, now equipped with a chosen point $a\in X$. One write this as $\left(X,a\right)$.

  • Why do we need a pointed topological space? Because then we can look on loops starting and ending at $a$. Those are continuous functions $\gamma :[0,1]\rightarrow X$ with $\gamma(0)=\gamma(1)=a$.

Those loops are interesting, because you can connect them one after another

$$(\gamma_{1}\ast\gamma_{2})(t)=\left\{\begin{matrix}\gamma_{1}(2t) & 0\leq t\leq\frac{1}{2}\\ \gamma_{2}(2t-1) & \frac{1}{2}< t\leq 1\end{matrix}\right.$$

and you can find their inverse (caution! this is incorrect, see below)

$$\gamma^{-1}(t)=\gamma(1-t)$$

You also have an identity element

$${\rm id}(t)=a$$

Therefore, you have here (almost) a group structure. For this to be a group, you can't really look at all the loops as different - you must consider two loops $\gamma_{1}$ and $\gamma_{2}$ to be the same, if you can deform $\gamma_{1}$ into $\gamma_{2}$ continuously. This is known as an homotopy. A great illustration of this is available in this Wikipedia page. I attach it also here

enter image description here

Mathematicians call this kind of "considering two objects to be the same" by the name equivalence relation, and this results in equivalence classes. Let's now denote by $[\gamma_{1}]$ all the loops that are equivalent to $\gamma_{1}$ under the homotopy relation. The set of all those loops

$$\pi_{1}(X,a)=\left\{[\gamma]|\gamma : [0,1]\rightarrow X,\: \gamma(0)=\gamma(1)=a\right\}$$

is called the fundamental group. Now returning to the inverse of a loop, try to think why

$$\gamma^{-1}\ast\gamma\neq{\rm id}$$

but in fact

$$[\gamma^{-1}]\ast[\gamma]=[{\rm id}]$$

Some examples

  • Let $X=\mathbb{R}^3$, i.e. the 3D space, and choose $a=0$ to be the origin. You can imagine that you can shrink every loop in $\mathbb{R}^{3}$ to a point, so all the loops are equivalent. This means that $\pi_{1}(\mathbb{R}^3,0)=\{{\rm id}\}$ is trivial.

  • Let $X=S^{1}$, the unit circle, and choose $a=(1,0)$. Note that every loop is characterized by how many times it circles the origin. That's an integer known as the winding number. So for instance, you have a loop that circles the origin once counter-clockwise ($=1$) and you have a loop that circles the origin twice, but clockwise ($=-2$) and so on. Therefore $\pi_{1}(S^{1},(1,0))=\mathbb{Z}$.

More relevant examples

  • Note that ${\rm SO}(1)=\{1\}$, so of course $\pi_{1}({\rm SO}(1))=\{{\rm id}\}$ is trivial.

  • For ${\rm SO}(2)$, you can argue that it is the same as $S^{1}$, in the sense that rotations in 2D are equivalent to $e^{i\theta}$, which form the unit circle. Thus $\pi_{1}({\rm SO}(2))=\mathbb{Z}$.

  • For ${\rm SO}(3)$, every rotation can be achieved by giving the axis of rotation, with is a point on the sphere $S^{2}$, and the rotation angle, which is in $S^{1}$. Note, however, that $(\boldsymbol{n},\theta)\sim(-\boldsymbol{n},-\theta)$ are equivalent rotations. This turns out to be equivalent to something known as the real projective space ${\rm RP}^{3}$, which has $\pi_{1}({\rm SO}(3))=\pi_{1}({\rm RP}^{3})=\mathbb{Z}_2$.

Now to the physics! What does the physics care about loops in ${\rm SO}(D)$? Let $t$ denote the time. How can we describe the exchange of two particles? The answer is by a curve $\gamma : [0,1]\rightarrow {\rm SO}(D)$ that describes the rotation of the two particles in the center-of-mass frame. This is a loop, because a full rotation of $2\pi$ is equivalent to doing nothing. Therefore, the number of topologically distinct loops is in fact the number of the different statistics possible. See also the following beautiful illustration taken from this Wikipedia page.

Anticlockwise rotationClockwise rotation

Anticlockwise rotation $\qquad\qquad\qquad$ Clockwise rotation

To quote the caption of this figure from Wikipedia

Exchange of two particles in $2 + 1$ spacetime by rotation. The rotations are inequivalent, since one cannot be deformed into the other (without the worldlines leaving the plane, an impossibility in $2D$ space).

eranreches
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  • This is brilliant, thank you. Physically/intuitively, how can I think that there are fewer loops in 3D than in 2D, when the "space" that I can explore (i.e. the number of rotations) is much larger? – SuperCiocia Dec 14 '18 at 21:42
  • @SuperCiocia That's a great question. The larger your space is, the easier it is to deform loops one into another. Think of a plane $\mathbb{R}^2$ with a hole in it. This is the same case as with $S^1$. A loop that encloses the hole cannot be shrank into a point. However, when the space is larger, for instance $\mathbb{R}^3$ with a hole, then every loop can be shrank into a point. Now you can bypass the hold from above/below; you couldn't do that before. – eranreches Dec 14 '18 at 21:48
  • Thanks. Is there a relation between the fact that the fundamental group for $SO(3)$ is $\mathbb{Z}_2$ and the fact that the rotation to $(n, \theta)$ is equal to the one to $(-n, -\theta)$? If yes, how can I think of the $2D$ case? – SuperCiocia Dec 14 '18 at 21:55
  • In the $2D$ case you have only one axis, so we just need an angle. This also simplify things as we don't need to worry about 'minus the axis', as in the $3D$ case. – eranreches Dec 14 '18 at 21:58
  • What's the differnece of taking the fundamental group $\pi_1$ of a space like $S^1$ or $\mathbb{R}^3$, and taking the fundamental group of a group like $SO(3)$? Do I have to specify a vector space over which $SO(3)$ is acting? – SuperCiocia Dec 14 '18 at 23:34
  • Those spaces have different structures. The first two are easy to realize, while the third is much more challenging. As I said, you can think of ${\rm SO}(3)$ as a ball of radius $\pi$ with antipodal points on $S^2$ identified. Try to think about the loops possible in this space. As for your second question, you don't have to specify such a vector space. The fundamental group depends only on the topology of ${\rm SO}(3)$. The fact that the latter also admits a group structure and can act on a vector space is irrelevant. – eranreches Dec 15 '18 at 15:09
  • Is that because Lie groups are both groups and differentiable manifolds? – SuperCiocia Dec 15 '18 at 15:24
  • Yes. Note, though, that this is more than what we need. A differential manifold is a specific type of topological manifold which itself is a specific type of a topological space. – eranreches Dec 15 '18 at 15:31
  • So what's the exact (rigourous) relatio between $SO(3)$ and $S^2$? – SuperCiocia Dec 15 '18 at 15:37
  • Let us continue this discussion in chat. – eranreches Dec 15 '18 at 15:41
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The answer by @eranreches was posted before I posted this one. I have only one thing to add here, namely a different way of deducing that the fundamental group of $SO(D)$ has exactly two elements when $D\geq 3$. The approach used here is also connected to physics in an interesting way, because it relies on properties of the covering group of $SO(D)$, which describes how spinors transform. (This answer doesn't use spinors, but it does use the covering group.)

Throughout this answer, $D$ is assumed to be $D\geq 3$.

Loosely speaking, the fundamental group of a manifold $M$ is the set of continuous maps from the circle $S^1$ into $M$, modulo continuous deformations, equipped with a group structure as explained in detail by eranreches. Saying that the fundamental group of $SO(D)$ has exactly two elements is the same as saying that every loop in $SO(D)$ can be continuously deformed into either of two representative loops, and that these two representative loops cannot be continuously deformed into each other. In the usual representation of $SO(D)$ by matrices of size $D\times D$, we can take these two representative loops to be $$ L_1: e^{i\theta}\rightarrow \text{identity matrix} $$ and $$ L_2: e^{i\theta}\rightarrow \text{Rotation through angle $\theta$ in the $x$-$y$ plane}. $$ Here, $e^{i\theta}$ denotes an element of the circle $S^1$, parameterized by $0\leq\theta< 2\pi$. The map $L_1$ sends every point of $S^1$ to the same point of $SO(D)$, namely to the identity matrix. Both maps describe closed loops, because they both start ($\theta=0$) and end ($\theta=2\pi$) at the identity matrix in $SO(D)$.

  • First Claim: Every loop in $SO(D)$ can be continuously deformed to either $L_1$ or $L_2$.

  • Second Claim: $L_2$ cannot be continuously deformed to $L_1$.

Together, these two claims imply that the fundamental group of $SO(D)$ has exactly two elements. (Recall that $D\geq 3$ throughout this answer.) To deduce both of these claims, consider the covering group Spin($D$) of $SO(D)$. A way to construct this covering group is outlined in another post:

The important facts here are:

  • The covering is two-to-one. Given a faithful matrix representation of Spin($D$), the elements $\pm I$ in Spin($D$) both correspond to $I$ in $SO(D)$, where $I$ denotes the respective identity matrices.

  • Every loop in $\text{Spin}(D)$ can be continuously deformed to a point. In other words, $\text{Spin}(D)$ is simply connected.

Because of the two-to-one relationship of $\text{Spin}(D)$ to $SO(D)$, any closed loop in $SO(D)$ corresponds to either

  • Case A: A closed loop in $\text{Spin}(D)$; in other words, a path from $g$ to $g$ in $\text{Spin}(D)$. The map $L_1$ obviously has this form.

  • Case B: A path from $g$ to $-g$ in $\text{Spin}(D)$. The map $L_2$ has this form, as shown in the post cited above.

Here, $g$ denotes some matrix in the given faithful matrix representation of $\text{Spin}(D)$. The fact that Spin($D$) is simply connected immediately establishes the First Claim, so the fundamental group of $SO(D)$ cannot have any more than two elements. (By the way, if $D=2$, then the argument fails because the simply-connected covering group of $SO(2)$ is not a two-to-one covering.)

To establish the Second Claim, we need to show that Case B cannot be continuously deformed to Case A. This follows from the fact that the matrix $g$ in Spin($D$) cannot be the zero matrix, so the endpoints of the path in Case B must remain distinct from each other in Spin($D$).

Altogether, this shows that the fundamental group of $SO(D)$ has exactly two elements for every $D\geq 3$. The key to the argument is that $SO(D)$ has a two-to-one covering that is simply connected. A way of constructing this covering explicitly is described in the post cited above.

Chiral Anomaly
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  • Thanks for the great answer. The physicist in me is trying to get an intuitive picture. Why are there only two "fundamental" loops in $SO(D)$, which all loops can be deformed into? Because, thanks to the double covering, a certain rotation can be mapped to $(n, \theta)$ or $(-n, -\theta)$? But then in $2D$, shouldn't I just have one fundamental loop...? – SuperCiocia Dec 14 '18 at 22:03
  • @SuperCiocia For $D=2$, the simply con'd cov'g group (the additive group $\mathbb{R}$ of real $\theta$) doesn't have any nontrivial loops, nor does Spin($D$) for $D\geq 3$. The key is to count the number of paths (not necessarily loops) in the covering group that are mapped to loops in $SO(D)$ and that can't be continuously deformed to each other without breaking the corresponding loop in $SO(D)$. For $D=2$, any interval $0\leq \theta < 2\pi n$ for integer $n$ maps to a loop in $SO(2)$, and these intervals in $\mathbb{R}$ can't be deformed to each other without breaking the loop in $SO(2)$. – Chiral Anomaly Dec 14 '18 at 22:13
  • I think my confusion is addressed here (http://www.damtp.cam.ac.uk/user/examples/D18S.pdf), I am trying to get my head around figure 2. – SuperCiocia Dec 14 '18 at 22:16
  • @SuperCiocia Figure 2 shows a disk with antipodal boundary points identified, as a lower-dimensional analog of a space like $SO(3)$. The figure is trying to show that if you have a loop that intersects the boundary in two points A and B (which are each equivalent to their antipodal points A' and B'), then you can continuously bring those two points together until they meet and then pull the loop away from the boundary (so it disappears on the antipodal side.) The result is a loop that doesn't intersect the boundary, so it can shrink to a point. – Chiral Anomaly Dec 14 '18 at 22:24
  • Yeah I get it now. I think I am starting to understand this. Quick question then. I am asking this in the context of bosons and fermions in $3D$. The typical short derivation of "there are only two distributoins" is the following. You start with a wavefunction which depends on two sets of variables $\psi(1,2$. You define the permutation operator $P$ which swaps variables. So if I do $P^2$ I should get back to the original point, hence I require $P^2 =1$, and hece $P = \pm 1$, one being fermions and the other one being bosons, when $\psi$ is (anti)symmetric upon label change. – SuperCiocia Dec 14 '18 at 22:28
  • WHERE in this derivation have I assumed that I am in 3D? How can I show that the 2D case would give more resutsl?? – SuperCiocia Dec 14 '18 at 22:29
  • @SuperCiocia I won't pretend to understand that typical short derivation, but I suspect that the idea of a continuous transition from $\psi(1,2)$ to $\psi(2,1)$ is important in the full derivation. I suspect that the key is to consider continuously moving the two particles, maintaining spacelike separation between them, until they have swapped positions, and then requiring that the resulting wavefunction is proportional to (and therefore physically equivalent to) the original. The proportionality factor should be the same no matter how they were continuously swapped. (continued...) – Chiral Anomaly Dec 14 '18 at 22:49
  • @SuperCiocia So for $D\geq 3$ the proportionality factor must be $\pm 1$, but for $D=2$ it can be a more general complex number. I'm not sure about the details, but I think it goes something like that. – Chiral Anomaly Dec 14 '18 at 22:50
  • Here, https://en.wikipedia.org/wiki/Exchange_operator#Construction, is what I call the "typical derivation". I do not understand where or what is the assumption that makes it solely applicable to 3D. – SuperCiocia Dec 14 '18 at 22:59
  • @SuperCiocia As far as I can tell, that derivation isn't really legit even for $D\geq 3$, because it doesn't explain why we should have $P^2|1,2\rangle=|1,2\rangle$ instead of only $P^2|1,2\rangle\propto|1,2\rangle$, which is physically equivalent. To deduce that the factor must be $1$, it's essential to consider continuously swapping the particles and require that the factor vary continuously with the path used to swap them. So the reason the short "derivation" doesn't require specifying $D\geq 3$ is because it's not really a legitimate derivation in the first place; it omits a key piece. – Chiral Anomaly Dec 14 '18 at 23:05
  • Nice, yes. Great – SuperCiocia Dec 14 '18 at 23:06
  • Hey sorry again. Looking at Fig. 1a in the link above. Why is that only applicable to SO(3)? What if that were a just a 2D circle, like in SO(2) - wouldn't the loops still be contractible? – SuperCiocia Dec 14 '18 at 23:58
  • Actually sorry I thikn I was interpreting the figure wrong. I thought the black point was the centre of the the sphere. It's actually a point on the surface. – SuperCiocia Dec 15 '18 at 00:04