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Is there any physical significance of the fact that the group manifold (parameter space) of $SO(3)$ is doubly connected?

EDIT 1: Let me clarify my question. It was too vague. There exists two equivalence classes of paths in the group manifold of SO(3) or in other words, $\Pi_1(SO(3))=Z_2$. This space is therefore doubly connected. There are paths which come back to initial configurations after a rotation of $2\pi$ and others after a rotation of $4\pi$, with proper parametrization of angles.

Using this fact, is it possible to show that such a topology admits the existence of half-integer spins and integer spins? I understand spinors as objects whose wavefunctions pick up a -ve sign after a rotation of $2\pi$, and comes back to itself after a rotation of $4\pi$. Right? But from the topological argument given above, it is not clear to me, that how does it lead to two kinds of wavefunctions, spinor-type $(j=\frac{1}{2},\frac{3}{2},\frac{5}{2}...)$ and tensor-type $j=0,1,2,...$? It is not explicitly clear how these two types of paths in SO(3) group manifold will lead to such transformation properties on "the wavefunctions"?

EDIT 2: http://en.wikipedia.org/wiki/SO%283%29#Topology.

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Didn't you just ask the question last week? –  Isidore Seville Feb 3 at 14:18
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@IsidoreSeville- No. –  Roopam Feb 3 at 14:28
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Comment to the question (v1): You mean apart from the fact that the Lie group $SO(3)$ doesn't have even-dimensional (=half-integer spin) irreps? –  Qmechanic Feb 3 at 14:58
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@IsidoreSeville Roopam Sinha's excellent questions and their excellent answers have established that there are integer and half integer spins and nothing else, essentially and ultimately because a simply connected topological space has no nontrivial coverings (i.e. not homeomorphic to the original space) (see the proof in Massey, "Algebraic Topology", for example). So now he's seeking help to understand the spin statistics theorem, which is quite a distinct next step from that fact and this is not a theorem I feel (as a non QFT specialist) I understand well enough to answer. –  WetSavannaAnimal aka Rod Vance Feb 5 at 0:43
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1 Answer 1

up vote 8 down vote accepted

Just in view of the double universal covering provided by $SU(2)$, $SO(3)$ must a quotient of $SU(2)$ with respect to a central discrete normal subgroup with two elements. This is consequence of a general property of universal covering Lie groups:

If $\pi: \tilde{G} \to G$ is the universal covering Lie-group homomorphism, the kernel $H$ of $\pi$ is a discrete normal central subgroup of the universal covering $\tilde{G}$ of $G= \tilde{G}/H$, and $H$ is isomorphic to the fundamental group of $G$, i.e. $\pi_1(G)$ (wich, for Lie groups, is Abelian) .

One element of that subgroup must be $I$ (since a group includes the neutral element). The other, $J$, must verify $JJ=I$ and thus $J=J^{-1}= J^\dagger$. By direct inspection one sees that in $SU(2)$ it is only possible for $J= -I$. So $SO(3) = SU(2)/\{I,-I\}$.

Notice that $\{I,-I\} = \{e^{i4\pi \vec{n}\cdot \vec{\sigma}/2 }, e^{i2\pi \vec{n}\cdot \vec{\sigma}/2 }\}$ stays in the center of $SU(2)$, namely the elements of this subgroup commute with all of the elements of $SU(2)$. Moreover $\{I,-I\}=: \mathbb Z_2$ is just the first homotopy group of $SO(3)$ as it must be in view of the general statement I quoted above.

A unitary representations of $SO(3)$ is also a representation of $SU(2)$ through the projection Lie group homomorphism $\pi: SU(2) \to SU(2)/\{I,-I\} = SO(3)$. So, studying unitary reps of $SU(2)$ covers the whole class of unitary reps of $SO(3)$. Let us study those reps.

Consider a unitary representation $U$ of $SU(2)$ in the Hilbert space $H$. The central subgroup $\{I,-I\}$ must be represented by $U(I)= I_H$ and $U(-I)= J_H$, but $J_HJ_H= I_H$ so, as before, $J_H= J_H^{-1}= J_H^\dagger$.

As $J_H$ is unitary and self-adjoint simultaneously, its spectrum has to be included in $\mathbb R \cap \{\lambda \in \mathbb C \:|\: |\lambda|=1\}$. So (a) it is made of $\pm 1$ at most and (b) the spectrum is a pure point spectrum and so only proper eigenspeces arise in its spectral decomposition.

If $-1$ is not present in the spectrum, the only eigenvalue is $1$ and thus $U(-I)= I_H$. If only the eigenvalue $-1$ is present, instead, $U(-I)= -I_H$.

If the representation is irreducible $\pm 1$ cannot be simultaneously eigenvalues. Otherwise $H$ would be split into the orthogonal direct sum of eigenspaces $H_{+1}\oplus H_{-1}$. As $U(-1)=J_H$ commutes with all $U(g)$ (because $-I$ is in the center of $SU(2)$ and $U$ is a representation), $H_{+1}$ and $H_{-1}$ would be invariant subspaces for all the representation and it is forbidden as $U$ is irreducible.

We conclude that,

if $U$ is an irreducible unitary representation of $SU(2)$, the discrete normal subgroup $\{I,-I\}$ can only be represented by either $\{I_H\}$ or $\{I_H, -I_H\}$.

Moreover:

Since $SO(3) = SU(2)/\{I,-I\}$, in the former case $U$ is also a representation of $SO(3)$. It means that $I = e^{i 4\pi \vec{n}\cdot \vec{\sigma} }$ and $e^{i 2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ are both transformed into $I_H$ by $U$.

In the latter case, instead, $U$ is not a true representation of $SO(3)$, just in view of a sign appearing after $2\pi$, because $e^{i 2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ is transformed into $-I_H$ and only $I = e^{i 4\pi \vec{n}\cdot \vec{\sigma}/2 }$ is transformed into $I$ by $U$.

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@V.Moretti- Amazing. This answer is very insightful. But I still wonder where does the topology of the SO(3) group manifold (which I mentioned in the question) enter this business? It appears to me that the above answer uses the fact that there is a 2-to-1 homomorphism between SU(2) and SO(3). Or am I missing the connection? –  Roopam Feb 5 at 11:03
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SO(3) is a Lie group with fundamental homotopy group $\mathbb Z_2$. Therefore its universal covering group is a simply connected Lie group and $SO(2)$ is obtained by taking the quotient of the latter and a discrete normal subgroup with two elements. That two is the "topological" information: It remembers the structure of the first homotopy group of the manifold $SO(3)$. Knowing that the universal covering is $SU(2)$ one has to look for a discrete subgroup containing only two elements... –  V. Moretti Feb 5 at 11:05
    
@V.Moretti- Okay, I understand. Now it is absolutely clear. Thanks. –  Roopam Feb 5 at 11:11
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I made precise the relevant topological relation in the text just expanding a comment. –  V. Moretti Feb 5 at 11:35
    
@V.Moretti I'm sure you meant to say this in your comment above but in general you have to look for a discrete normal subgroup containing only two elements. These by Schreier's theorem must then be subgroups of the centre. Only being pedantic because the OP's clearly very interested in the details, I wouldn't have mentioned it otherwise. –  WetSavannaAnimal aka Rod Vance Feb 13 at 10:31
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