Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've often seen it written that $SU(2)$ is a "two-valued representation" of $SO(3)$ (in theoretical physics books mainly). I have a major conceptual issue with this however.

I know there is a Lie group isomorphism $SU(2)/\mathbb{Z}_2=SO(3)$ so we can assign to every matrix $R$ in $SO(3)$ one of two matrices $U$ or $-U$ in $SU(2)$. But surely the definition of a representation forces us to choose $D(I) = I$ so this "two-valued" business breaks down?

Could someone explain where I'm getting stuck? Also would anyone be able to point me to a resource which treats all this material rigorously? I have no background in Representation Theory and don't particularly want to read a really long text, but at the same time I am unhappy with the heuristic arguments of most physics books. Is there any lucid text taking some middle way?

share|improve this question
3  
It sounds like "two-valued representation" is just a garbled attempt to say "double cover". (Or else physicists use "representation" with a broader meaning than do mathematicians). –  Henning Makholm Dec 21 '12 at 14:04
    
Which books have you read this in? Knowing that might help to explain what they mean. –  mt_ Dec 21 '12 at 14:09
    
Costa and Fogli - Introduction to Spacetime and Internal Symmetries, among others. It's also written here –  Edward Hughes Dec 21 '12 at 14:30
    
I've just done a bit of googling, and perhaps the key is in projective representations? Does this make sense to anyone? –  Edward Hughes Dec 21 '12 at 14:48
2  
In response to your reference request at the end of your question, you might check out math.columbia.edu/~woit/QM/fall-course.pdf. These are detailed lecture notes for an undergraduate math class on quantum mechanics at Columbia which just wrapped up. This might be part of the middle ground you're hoping for. –  Paul Siegel Dec 21 '12 at 15:28

2 Answers 2

up vote 8 down vote accepted

"$SU(2)$ is a two-valued representation of $SO(3)$" is a rather bizarre sentence - it isn't totally clear what is meant by "two-valued" or "representation". One possible meaning, as suggested in the comments, is simply that $SU(2)$ is a double cover of $SO(3)$. But I suspect that something slightly different is meant.

$SU(2)$ has a canonical unitary representation $\pi$ on $\mathbb{C}^2$: namely, view an element of $SU(2)$ as a unitary operator on $\mathbb{C}^2$. As is well-known, this does not descend to an ordinary representation of $SO(3)$. One could try to define $\pi' \colon SO(3) \to U(\mathbb{C}^2)$ by $\pi'(g) = \pi(g')$ where $g'$ is a lift of $g$ to $SU(2)$ along the double cover $SU(2) \to SO(3)$, but the problem is there is no way to choose a lift $g'$ of every $g$ in such a way that $\pi'$ becomes a homomorphism. Nevertheless, there are only two choices for a lift of any $g$ and they differ only by a minus sign; thus while it is impossible to make $\pi'$ an actual homomorphism, it is a homomorphism up to sign:

$$\pi'(g_1 g_2) = \pm \pi'(g_1)\pi'(g_2)$$

Thus physicists sometimes like to think of the canonical representation of $SU(2)$ (which incidentally is the spin representation) as a "two valued" representation of $SO(3)$. This can be convenient if you only care about things like the position and momentum of a particle - which only depend on the magnitude of the wavefunction - and not about things like phase and spin.

Mathematicians have their own way to make sense of this. Recall that the projective general linear group of a vector space $V$ over a field $F$ is defined to be $GL(V)/F^\times$ where $F^\times$ is the multiplicative group of $F - 0$. One defines a projective representation of a group $G$ on $V$ to be a homomorphism $G \to PGL(V)$. If $V$ has a Euclidean / Hermitian structure (in the real / complex case), one can also speak of projective orthogonal / unitary representations; these are homomorphisms into the unitary group of $V$ modulo the unit group of the ground field. With these definitions, the discussion above implies that the canonical representation of $SU(2)$ is in fact a projective orthogonal representation of $SO(3)$ (since the unit group of $\mathbb{R}$ is just $\{\pm 1\}$).

So in conclusion, I interpret the sentence "$SU(2)$ is a two-valued representation of $SO(3)$" to mean "the canonical representation of $SU(2)$ is a projective representation of $SO(3)$".

share|improve this answer
    
That's exactly what I wanted to know - thank you so much! –  Edward Hughes Dec 21 '12 at 15:24
    
"since the unit group of R is just {±1}"; not really. But they are the only scalars in the image of $SO(3)$. –  Alex B. Dec 21 '12 at 16:01
    
Just as the projective unitary group is defined to be $U(n)/U(1)$, the projective orthogonal group is usually defined to be $O(n)/O(1)$, and if I'm not mistaken $O(1) = \{\pm1\}$. –  Paul Siegel Dec 21 '12 at 16:36
1  
@Paul I was only objecting to your saying that the unit group of $\mathbb{R}$ is $\pm 1$. The unit group of $\mathbb{R}$ is $\mathbb{R}^\times=\mathbb{R}\backslash 0$. –  Alex B. Dec 21 '12 at 16:41
1  
Ah, well there is disagreement in terminology here. If you're an algebraist I suppose the "unit group" in a ring is the set of all elements with multiplicative inverses. But to differential geometers the unit group in a normed ring often means the multiplicative group of unit vectors. I think this is the standard terminology in the literature on Clifford algebras, for instance (which is the relevant structure here). In any event, I suppose the ambiguity is worth mentioning. –  Paul Siegel Dec 21 '12 at 16:49

The problem with the physics definition is that it is not the standard definition of a Lie group representation (e.g. as given in Fulton and Harris, Hall's Lie Groups, Procesi's Lie groups, or any other mathematical text on representation theory). What they actually mean is that we have a covering space $\Phi : \textrm{SU}(2) \to \textrm{SO}(3)$ where $\Phi$ is as given as in my answer here. When they say say $\Phi$ is two to one, they mean that the cardinality of the fiber $\Phi^{-1}(x)$ for any $x \in \textrm{SO}(3)$ is equal to 2. This is constant on all of $\textrm{SO}(3)$ because it is connected (one way to see this IIRC is that $\exp : \mathfrak{so}_3\longrightarrow \textrm{SO}(3)$ is surjective).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.