Where in fundamental physics are Lie groups actually important (and not just Lie algebras)?

Question

I was wondering where in fundamental physics the global structure of a Lie group actually makes a difference.

Most of the time physicists are sloppy and don't distinguish groups and algebras properly. However, although we talk about groups all the time I wasn't able to come up with an instant where we don't actually care only about the corresponding Lie algebra.

As an example, physicists usually talk about the Poincare group. However, the thing we are really interested in is the corresponding complexified Lie algebra, which actually belongs to the universal covering group

$$ SL(2,\mathbb{C}) \ltimes \mathbb{R}(3,1). $$

Now, the other kind of symmetry that is important in fundamental physics is gauge symmetry. However, again the global structure doesn't seem to be important. To quote from Witten's Physics and Geometry:

“Experiment tells us more directly about the Lie algebra of G than about G itself. When I say that G contains the subgroup SU(3) X SU(2) x U(1), I really mean only that the Lie algebra of G contains that of SU(3) X SU(2) X U(1); there is no claim about the global form of G. For the same reason, in later comments I will not be very precise in distinguishing different groups that have the same Lie algebra.”

Answer 1

An example, where actual elements of the Lie group play a role, are the (non-infinitesimal) gauge transformations in gauge theory. For simplicity, let's assume that $G \subset GL(n, \mathbb C)$ is a matrix group. The gauge field is given by Lie-algebra $\mathfrak g$ valued one form $A_\mu{}^a{}_b$. In addition we can consider a complex mater field with values in the defining representation $\psi^s$.

General changes of $G$-gauge are given by functions $\varphi: \mathcal M \to G$ from the base manifold $\mathcal M$ to $G$. The corresponding gauge transformations are the given by: $$ \psi^s \mapsto \psi^{\tilde s} := \varphi^{\tilde s}{}_{s} \psi^s$$ and $$ A_\mu{}^a{}_b \mapsto A_{\mu}{}^{\tilde a}{}_{\tilde b} := \varphi^{\tilde a}{}_{a} A_\mu{}^a{}_b \varphi^{-1}{}^{b}{}_{\tilde b} + \varphi^{\tilde a}{}_c \partial_\mu \varphi^{-1}{}^{c}{}_{\tilde b}$$.

Note, that $\mathfrak g = \mathfrak g'$ does not imply that $G = G'$. An important example in physics is given by the Lorentz group $SO(1,3)$ and the Spin group $\operatorname{Spin}(1,3) \cong SL(2, \mathbb C)$. While $\mathfrak{so}(1,3) \cong \mathfrak{sl}(2, \mathbb C)$, we have $SO(1,3) \neq SL(2, \mathbb C)$.

This leads to the famous fact, that a spinor has to be rotated by $4 \pi$ to return to its original position.

Mathematically, the reason is that the map $p: SL(2, \mathbb C) \to SO(1,3)$ is $2$ to $1$. Consider a path $\gamma: [0, 2\pi] \to SO(1,3)$ corresponding to full rotation of "$360^\circ$", in particular, $\gamma(0) = \gamma(2\pi) = \mathbb I$. Let $ \gamma':[0,2\pi] \to SL(2, \mathbb C)$ be the lift of $\gamma$, i.e., $\gamma = p(\gamma')$, it turnes out that $\gamma'(2\pi) = - \mathbb I \in SL(2, \mathbb C)$. Therefore, one has to "rotated by $4\pi$" such that $\gamma'(2\pi) \circ \gamma'(2\pi) \psi = \psi$.

Answer 2

A typical example where the Lie group matters is the action of physical rotations.

The group of physical rotations is $SO(3)$ and not $SU(2)$, though they have the same Lie algebra. Here the Lie group matters more than its Lie algebra.

In classical physics, indeed, we use $SO(3)$ to represent rotations. The situation is apparently different in quantum physics, but just apparently different.

We use $SU(2)$ to describe physical rotation on quantum systems because pure quantum states are unit vectors up to phases and this, in addition to the Wigner theorem, permits the use of unitary projective representations of $SO(3)$ to describe the action of rotations. In turn, these projective representations are unitary representations of $SU(2)$. (For mathematical convenience it is better to handle unitary representations than projective unitary reprsentations.) Without this complicate interplay where the definition of quantum states plays a fundamental role (in fact in classical physics the situation is completely different), we would be forced to deal directly with $SO(3)$ in all cases.