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Double connectivity of $SO(3)$ group manifold

+ 6 like - 0 dislike
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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.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Roopam
asked Feb 3, 2014 in Theoretical Physics by Roopam (145 points) [ no revision ]
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Didn't you just ask the question last week?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Isidore Seville
@IsidoreSeville- No.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Roopam
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?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Qmechanic
Related question by OP: physics.stackexchange.com/q/96542/2451 Related: physics.stackexchange.com/q/13787/2451 , physics.stackexchange.com/q/65767/2451 and links therein.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Qmechanic
@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.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
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Dear WetSavannaAnimal aka Rod Vance- "...*have established that there are integer and half integer spins*..." I have not quite understood this fact too. So I re-edited my questions. How is this topological fact related to the two types of representations of SO(3), a class labeled by half-integer j values and another class labeled by integral j values? This is the first issue which I'm stuck at.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Roopam
physics.stackexchange.com/questions/147/…

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Kostya

1 Answer

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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 $U$.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
answered Feb 5, 2014 by Valter Moretti (2,025 points) [ no revision ]
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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?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Roopam
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...

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
@V.Moretti- Okay, I understand. Now it is absolutely clear. Thanks.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Roopam
I made precise the relevant topological relation in the text just expanding a comment.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
@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.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
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@WetSavannaAnimal aka Rod Vance Thank you for your comment. Yes, I wrote in a hurry and I wrote something quite confused "discrete normal subgroup with two elements" (<<with two elements>> does not make much sense: Instead you wrote correctly: "a discrete normal subgroup containing only two elements". Thanks.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
@Roopam Look at top of page 70 of this (the theorem is stated at the bottom of p69, and you should be able to see that the method of this little gem of a proof works for any connected Lie group even though Stillwell cites "path connected matrix Lie group") ohkawa.cc.it-hiroshima.ac.jp/AoPS.pdf/MathTextBook/…

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance

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