$SU(2)$ Representation of $SO(3)$

Question

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?

Answer 1

"$SU(2)$ 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 $SO(3)$".

Answer 2

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).