Nontrivial Phase for projective $SO(3)$ representation? Simple connectedness and its relation with cocycle

Question

A symmetry group $\mathcal{G}$ may be represented on the physical Hilbert space by unitary operators $U(g)$ such that it satisfies the composition rule $$U(g_1)U(g_2)=e^{i\phi(g_1,g_2)}U(g_1g_2).\tag{1}$$ If $\phi(g_1,g_2)$ is of the form $$\phi(g_1,g_2)=\alpha(g_1g_2)-\alpha(g_1)-\alpha(g_2)\tag{2}$$ the projective representations in (1) with such a phase (2) can be replaced by an ordinary representation by replacing $U(g)$ with $$\tilde{U}(g)=e^{i\alpha(g)}U(g).\tag{3}$$

There is a theorem which (in negative sense) tells that if a group is not simply-connected, it will have intrinsically projective representation on the Hilbert space. Therefore, $SO(3)$ will have intrinsically projective representation but not $SU(2)$. In I understand it correct, then this in turn implies, all phases $\phi(g_1,g_2)$ of $SO(3)$ representation doesn't satisfy the relation (2), and therefore, cannot be reduced to an ordinary representation by defining something like (3).

1. Can we write down the phase $\phi(g_1,g_2)$ for a projective $SO(3)$ representation i.e., as a function of $g_1$ and $g_2$ (any two $SO(3)$ group elements)?

2. What is the relation between simple-connectedness of a group and the phase $\phi(g_1,g_2)$. In particular, what is the connection between paths in a group manifold and the phases $\phi(g_1,g_2)$.

3. Why is it that if two paths are continuously deformable into each other or closed loops can be continuously shrunk to a point, then $\phi$ always satisfy (2) and if not, $\phi$ may not satisfy (2)? If possible, please do not use too many technical jargons because I've just started learning these stuffs on my own.

Answer 1

If I understand it correct, then this in turn implies, all phases $\Phi(g_1,g_2)$ of $SO(3)$ representation doesn't satisfy the relation (2), and therefore, cannot be reduced to an ordinary representation by defining something like (3).

It implies that for that projective representation of $SO(3)$, you can not find a function $\alpha$ so that (2) holds for all $g_1, g_2$.

Can we write down the phase $\Phi(g_1,g_2)$ for a projective $SO(3)$ representation i.e., as a function of $g_1$ and $g_2$ (any two $SO(3)$ group elements)?

Of course, if you know the projective representation (explicitly), you can read off the phase. Take for example the spinor representation of $SO(3)$. Let $R_\pi$ be a rotation by 180 degrees, $R_\pi R_\pi = 1$. As we know, $$ U(R_\pi) U(R_\pi) = \mathrm e^{\mathrm i\pi} U(1) \;, $$ i.e. $\Phi(R_\pi, R_\pi) = \pi$.

What is the relation between simple-connectedness of a group and the phase $\Phi(g_1,g_2)$. In particular, what is the connection between paths in a group manifold and the phases $\Phi(g_1,g_2)$.

First, let me state the full version of the theorem you were citing in your question: Non-trivial projective representations can arise in exactly two ways.

1. Algebraically, if the central charge of the Lie algebra can not be removed. (I will not go into this further).
2. Geometrically, if the group is not simply connected.

In other words, if the central charge can be removed and the group is simply connected, all possible phases $\Phi(g_1,g_2)$ are in the trivial two-cocycle.

And to answer your question, we can make the statement a bit more precise: If the central charge can be removed, then the set of possible phases (up to trivial two-cocycles) equals the fundamental group of the Lie group.

Why is it that if two paths are continuously deformable into each other or closed loops can be continuously shrunk to a point, then $\Phi$ always satisfy (2) and if not, $\Phi$ may not satisfy (2)? If possible, please do not use too many technical jargons because I've just started learning these stuffs on my own.

Weinberg (The Quantum Theory of Fields, Vol 1) dedicates a whole chapter (2, Appendix B) to this proof. In the end, it boils down to the Poincaré lemma: In a simply connected space, if a vector field $v$ satisfies $\partial_b v_a = \partial_a v_b$, then it can be written as the gradient of some potential: $v_a = \partial_a \varphi$. With a suitable choice (see Weinberg) of $v$, this allows us to prove the statement.

Answer 2

1. If you present the rotations as Euler angles $(\alpha,\beta,\gamma)$, the phase $\phi$ is the function that yields $1$ as long as the (naive) composition of the two rotations does not exceed $2\pi$ in any angle, and $-1$ for every angle that does exceed $2\pi$. Note that this is a clearly discontinuous function.

2. I discuss this in the last section of this answer of mine. The upshot is that if $G$ is not simply connected, it has a universal covering group $\tilde{G}$ which is a central extension of $G$ by $\pi_1(G)$. Unitary representations $\tilde{\rho} : \tilde{G}\to \mathrm{U}(\mathcal{H})$ of central extensions always descend to projective representations $\rho : G\to\mathrm{PU}(\mathcal{H})$ of the original group since the central part has to be mapped into the center $\mathrm{U}(1)\subset\mathrm{U}(\mathcal{H})$, and therefore becomes trivial when passing to the quotient $\mathrm{PU}(\mathcal{H})$, since all preimages $\pi^{-1}(g)$ of an element $g\in G$ under the projection $\pi:\tilde{G}\to G$ get mapped to the same point in $\mathrm{PU}(\mathcal{H})$, so $\rho = \tilde{\rho}\circ\pi^{-1}$ is well-defined.

3. This has not really to do with the paths as such, but with the general theory of coverings. A sufficiently nice topological space $G$ always has a universal covering $\tilde{G}$ which sits in an exact sequence $\pi_1(G)\to \tilde{G} \to G$, if $G$ is a Lie group, $\pi_1(G)$ is Abelian and its image is central in $\tilde{G}$, so the universal cover is a central extension. Conversely, given a central extension $\mathbb{Z}_n\to\tilde{G}\to G$, we have that $\tilde{G}\to G$ is a covering, and the general classification of coverings says that such coverings are in bijection to subgroups of $\pi_1(G)$. So if $G$ is simply-connected, it has no such discrete central extensions, and if it isn't, there's a universal cover whose linear representations are the same as $G$'s projective representations if there are no smooth central extensions by $\mathrm{U}(1)$ (i.e. no continuous non-trivial cocycles).

   Note that the statement you ask for, i.e. "If $G$ is simply-connected, then $\phi$ is trivial (satisfies your eq. 2)" is false: Even simply connected groups can have non-trivial cocycles (equivalently non-trivial central extensions), they just can't have discrete central extensions or equivalently discontinuous cocycles. However, the physics literature usually ignores this until it becomes relevant (e.g. for the Virasoro algebra/group) since the semi-simple Lie groups we usually deal with have $H^2(G,\mathrm{U}(1)) = 0$ for the group of smooth cocycles modulo coboundary relations by Whitehead's second lemma.