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  Idea of Covering Group

+ 8 like - 0 dislike
  1. $SU(2)$ is the covering group of $SO(3)$. What does it mean and does it have a physical consequence?

  2. I heard that this fact is related to the description of bosons and fermions. But how does it follow from the fact that $SU(2)$ is the double cover of $SO(3)$?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Roopam
asked Jan 31, 2014 in Theoretical Physics by Roopam (145 points) [ no revision ]

3 Answers

+ 7 like - 0 dislike

Great, important question. Here's the basic logic:

  1. We start with Wigner's Theorem which tells us that a symmetry transformation on a quantum system can be written, up to phase, as either a unitary or anti-unitary operator on the Hilbert space $\mathcal H$ of the system.

  2. It follows that if we want to represent a Lie group $G$ of symmetries of a system via transformations on the Hilbert space, then we must do so with a projective unitary representation of the Lie group $G$. The projective part comes from the fact that the transformations are unitary or anti-unitary "up to phase," namely we represent such symmetries with a mapping $U:G\to \mathscr U(\mathcal H)$ such that for each $g_1,g_2\in G $, there exists a phase $c(g_1, g_2)$ such that \begin{align} U(g_1g_2) = c(g_1, g_2) U(g_1) U(g_2) \end{align} where $\mathscr U(\mathcal H)$ is the group of unitary operators on $\mathcal H$. In other words, a projective unitary representation is just an ordinary unitary representation with an extra phase factor that prevents it from being an honest homomorphism.

  3. Working with projective representations isn't as easy as working with ordinary representations since they have the pesky phase factor $c$, so we try to look for ways of avoiding them. In some cases, this can be achieved by noting that the projective representations of a group $G$ are equivalent to the ordinary representations of $G'$ its universal covering group, and in this case, we therefore elect to examine the representations of the universal cover instead.

  4. In the case of $\mathrm{SO}(3)$, the group of rotations, we notice that its universal cover, which is often called $\mathrm{Spin}(3)$, is isomorphic to $\mathrm{SU}(2)$, and that the projective representations of $\mathrm{SO}(3)$ match the ordinary representations of $\mathrm{SU}(2)$, so we elect to examine the ordinary representations of $\mathrm{SU}(2)$ since it's more convenient.

This is all very physically important. If we had only considered the ordinary representations of $\mathrm{SO}(3)$, then we would have missed the "half-integer spin" representations, namely those that arise when considering rotations on fermionic systems. So, we must be careful to consider projective representations, and this naturally leads to looking for the universal cover.

Note: The same sort of thing happens with the Lorentz group in relativistic quantum theories. We consider projective representations of $\mathrm{SO}(3,1)$ because Wigner says we ought to, and this naturally leads us to consider its universal cover $\mathrm{SL}(2,\mathbb C)$.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user joshphysics
answered Jan 31, 2014 by joshphysics (835 points) [ no revision ]
Most voted comments show all comments
@Hunter Yeah, I agree that one could equivalently start with the Lie algebra as starting point (motivated by orbital angular momentum), especially since in quantum mechanics one is mostly concerned with alebras of observables. Ultimately, I feel that having both points of view in mind leads to the richest understanding.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user joshphysics
@Hunter Also, I almost forgot. When you described what you mean when you say that $\mathrm{SU}(2)$ has periodicity $4\pi$, I think it would be more mathematically accurate, and less misleading to say instead that the particular projective representation of $\mathrm{SU}(2)$ which you consider has "periodicity" $4\pi$.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user joshphysics
@joshphysics: I am not being able to reconcile these facts. SU(2) is homeomorphic to the 3 sphere which is simple connected, and has the 0 to $2 \pi$ path can be shrunk to a point, so how does action by $SU(2)$ have periodicity $4 \pi$ in the context mentioned by Hunter above. Secondly, SO(3) is homeomorphic to $S^3 \backslash Z_2$ which is not simply connected and has the $0$ to $4 \pi$ path as the trivial loop, or identity map. How is this consistent with the fact that action by $SO(3)$(rotations in 3D space) have period $2 \pi$?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user ramanujan_dirac
@joshphysics: Sorry, I should probably use the word diffeomorphic rather than homeomorphic in my above comment.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user ramanujan_dirac
@ramanujan_dirac I'm not quite understanding your objections. Hunter is referring to a particular projective representation of $\mathrm{SU}(2)$. Can you explain why you think the diffeomorphism you refer to is inconsistent with the existence of such a representation?

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user joshphysics
Most recent comments show all comments
That's a good question that I haven't thought about since the first time I learnt about spin. I learnt most of this stuff following Griffiths' book before we had been taught the mathematical language of group theory. In our class we first found the commutation relations of the orbital angular momentum $[L_i,L_j] = i \epsilon_{ijk} L_k$.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Hunter
Then we used the method of ladder operators to analyse it, and subsequently using spherical harmonics we found that $m$ cannot take on half-integer values. After that we studied intrinsic angular momentum (i.e. spin) and we took the commutation relations of angular momentum as a postulate and found it agrees with experiments; for spin we are allowed to take into account half-integer values of $m$. Now that I have learnt group theory, and with the message you wrote above,

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user Hunter
+ 6 like - 0 dislike

After the answers by joshphysics and user37496, it seems to me that a last remark remains.

The quantum relevance of the universal covering Lie group in my opinion is (also) due to a fundamental theorem by Nelson. That theorem relates Lie algebras of symmetric operators with unitary representations of a certain Lie group generated by those operators. The involved Lie group, in this discussion, is always a universal covering.

In quantum theories one often encounters a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ such that:

(1) They are symmetric (i.e. defined on a dense domain $D(A_i)\subset {\cal H}$ where $\langle A\psi|\phi\rangle = \langle \psi|A\phi\rangle$)


(2) they enjoy the commutation relations of some Lie algebra $\ell$: $$[A_i,A_j]= \sum_{k=1}^N iC^k_{ij}A_k$$ on a common invariant domain ${\cal D}\subset {\cal H}$.

As is known, given an abstract Lie algebra $\ell$ there is (up to Lie group isomorphisms) a unique simply connected Lie group ${\cal G}_\ell$ such that its Lie algebra coincide with $\ell$. ${\cal G}_\ell$ turns out to be the universal covering of all the other Lie groups whose Lie algebra is $\ell$ itself.

All those groups, in a neighbourhood of the identity are isomorphic to a corresponding neighbourhood of the identity of ${\cal G}_\ell$. (As an example just consider the simply connected $SU(2)$ that is the universal covering of $SO(3)$) so that they share the same Lie algebra and are locally identical and differences arise far from the neutral element.

If (1) and (2) hold, the natural question is:

Is there a strongly continuous unitary representation ${\cal G} \ni g \mapsto U_g$ of some Lie group $\cal G$ just admitting $\ell$ as its Lie algebra, such that $$U_{g_i(t)} = e^{-it \overline{A_i}}\:\: ?\qquad (3)$$

Where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of $\cal G$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is some self-adjoint extension of $A_i$.

If it is the case, $\cal G$ is a continuous symmetry group for the considered physical system, the self adjoint opertors $\overline{A_i}$ represent physically relevant observables. If time evolution is included in the center of the group (i.e. the Hamiltonian is a linear combination of the $A_i$s and commutes with each of them) all these observables are conserved quantities. Otherwise the situation is a bit more complicated, nevertheless one can define conserved quantities parametrically depending on time and belonging to the Lie algebra of the representation (think of the boost generator when $\cal G$ is $SL(2,\mathbb C)$).

Well, the fundamental theorem by Nelson has the following statement.

THEOREM (Nelson)

Consider a set of operators $\{A_i\}_{i=1,\ldots, N}$ on a common Hilbert space ${\cal H}$ satisfying (1) and (2) above. If ${\cal D}$ in (2) is a dense subspace such that the symmetric operator $$\Delta := \sum_{i=1}^N A_i^2$$ is essentially self-adjoint on $\cal D$ (i.e. its adjoint is self-adjoint or, equivalently, $\Delta$ admits a unique self-adjoint extension, or equivalently its closure $\overline{\Delta}$ is self-adjoint), then:

(a) Every $A_i$ is essentially self-adjoint on $\cal D$,


(b) there exists a strongly continuous unitary representation on $\cal H$ of the unique simply connected Lie group ${\cal G}_\ell$ admitting $\ell$ as Lie algebra, completely defined by the requirements: $$U_{g_i(t)} = e^{-it \overline{A_i}}\:\:,$$ where $t\mapsto g_i(t)$ is the one-parameter Lie subgroup of ${\cal G}_\ell$ generated by (the element $a_i$ of $\ell$ corresponding to) $A_i$ and $\overline{A_i}$ is the unique self-adjoint extension of $A_i$ coinciding to $A_i^*$ and with the closure of $A_i$.

Notice that the representation is automatically unitary and not projective unitary: No annoying phases appear.

The simplest example is that of operators $J_x,J_y,J_z$. It is easy to prove that $J^2$ is essentially self adjoint on the set spanned by vectors $|j,m, n\rangle$. The point is that one gets this way unitary representations of $SU(2)$ and not $SO(3)$, since the former is the unique simply connected Lie group admitting the algebra of $J_k$ as its own Lie algebra.

As another application, consider $X$ and $P$ defined on ${\cal S}(\mathbb R)$ as usual. The three symmetric operators $I,X,P$ enjoy the Lie algebra of Weyl-Heisenberg Lie group. Moreover $\Delta = X^2+P^2 +I^2$ is essentially self adjoint on ${\cal S}(\mathbb R)$, because it admits a dense set of analytic vectors (the finite linear combinations of eigenstates of the standard harmonic oscillator). Thus these operators admit unique self-adjoint extensions and are generators of a unitary representation of the (simply connected) Weyl-Heisenberg Lie group. This example holds also replacing $L^2$ with another generic Hilbert space $\cal H$ and $X,P$ with operators verifying CCR on an dense invariant domain where $X^2+P^2$ (and thus also $X^2+P^2 +I^2$) is essentially self adjoint. It is possible to prove that the existence of the unitary rep of the Weyl-Heisenberg Lie group, if the space is irreducible, establishes the existence of a unitary operator from ${\cal H}$ to $L^2$ transforming $X$ and $P$ into the standard operators. Following this way one builds up an alternate proof of Stone-von Neumann's theorem.

As a last comment, I stress that usually ${\cal G}_\ell$ is not the group acting in the physical space and this fact may create some problem: Think of $SO(3)$ that is the group of rotations one would like to represent at quantum level, while he/she ends up with a unitary representation of $SU(2) \neq SO(3)$. Usually nothing too terrible arises this way, since the only consequence is the appearance of annoying phases as explained by Josh, and overall phases do not affect states. Nevertheless sometimes some disaster takes place: For instance, a physical system cannot assume quantum states that are coherent superpositions of both integer and semi-integer spin. Otherwise an internal phase would take place after a $2\pi$ rotation. What is done in these cases is just to forbid these unfortunate superpositions. This is one of the possible ways to realize superselection rules.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
answered Jan 31, 2014 by Valter Moretti (2,085 points) [ no revision ]
+1: Very illuminating! I had never heard of Nelson's theorem; this is why we need more mathematical physicists on physics.SE.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user joshphysics
There is a theorem due to Wigner that establishes that every continuous projective unitary rep. of a Lie group can always be seen as a proper continuous unitary representation of a central extension of the group (maybe of its universal covering I do not remember now), equipped with a certain natural topology and differentiable structure making the central extension a Lie group as well. Galileo's group can only be treated following this way and its central extensions embody the mass of the physical system. All that gives eventually rise to the so called Bargamann superselection rule.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user V. Moretti
Wow even more great information! I just keep learning more and more from every successive thing you write V. Moretti. Thanks.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user joshphysics
+ 4 like - 0 dislike

I'd like to add to Josh's answer, because he didn't really explain what a universal covering group is. Essentially, a space $T$ is a covering space of another space $U$ if, for an open subset of $U$, there's a function $f$ that maps a union disjoint open subsets of $T$ to the subset of $U$. Or, more simply worded, pick a piece of your space $U$, and I'll find you some number of different pieces of $T$ that have a function mapping them onto the piece of $U$. In the case of $T = Spin(3)$ and $U = SO(3)$, there are two disjoint subsets of $Spin(3)$ for every subset of $SO(3)$, so we say that $Spin(3) \cong SU(2)$ is the double cover of $SO(3)$.

Now, the way that this relates to bosons and fermions is where Josh's answer comes in. We want physical states to live in vector spaces that carry (projective) representations of our symmetry groups. The "projective" part means that our states may pick up a phase when transformed to other states -- so, for example, if you rotate a spin-1/2 state 360$^{\circ}$, the state picks up a minus sign. It turns out that, at least in the case of $SO(3)$, we can eliminate the need for the "projective" part of this -- and thus those pesky minus signs -- by considering instead representations of the covering space.

This post imported from StackExchange Physics at 2014-04-12 19:04 (UCT), posted by SE-user user37496
answered Jan 31, 2014 by user37496 (40 points) [ no revision ]

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