5
$\begingroup$

For concreteness let's focus on $SO(3)$. A projective representation is such that, $$U(g_1) U(g_2) = e^{i\theta(g_1, g_2)} U(g_1g_2), \ \ g_1, g_2 \in SO(3).$$ What I'm struggling to fully appreciate is the relation between the phase factor $\theta(g_1, g_2)$ and the fundamental group $\pi_1(SO(3)) = \mathbb Z_2$. Reading this answer here, there seems to be a connection between $\pi_1(SO(3)) = \mathbb Z_2$ and the kind of projective representations $SO(3)$ can have. I'm not asking how to construct $\pi_1(SO(3))$, but rather how knowing $\pi_1(SO(3))$ we can know something about $\theta(g_1, g_2)$. The relationship seems to suggest the following, which are also my questions about this:

  1. Can we write $\theta(g_1, g_2)$ as the integration of some quantity over a line connecting $g_1$, and $g_2$? Or more formally $\theta(g_1, g_2) = {\int_{\gamma} A}$, such that $\gamma$ is a path connecting $g_1$, and $g_2$.
  2. If the above is true, then we would need to show that the integral is invariant under a smooth change of $\gamma$, or that $\text{curl} \ A = 0$. How to show this?

I have seen arguments about how projective representations of $SO(3)$ has to correspond to linear representations of the universal cover $SU(2)$, a statement that I'm also not sure how to prove.

$\endgroup$
5
  • $\begingroup$ That's a question in algebraic/differential topology better suited for the Math.SE. $\endgroup$
    – DanielC
    Commented May 5, 2021 at 14:40
  • $\begingroup$ I was fearing that if I posted in Math.SE I'll get a super abstract answer that I would not be able to follow. I was hoping for an answer that someone with a physics background (and not very hardcore on the math) could understand (if any exists). $\endgroup$
    – A. Jahin
    Commented May 5, 2021 at 14:50
  • 1
    $\begingroup$ It's more a cohomology issue than a homotopy issue. $\endgroup$
    – mike stone
    Commented May 5, 2021 at 15:03
  • 1
    $\begingroup$ Since it's a self-answered Q&A of mine, I don't want to mod-close this question as duplicate, but physics.stackexchange.com/q/203944/50583 has probably everything you need to know. $\endgroup$ Commented May 5, 2021 at 15:37
  • 2
    $\begingroup$ I think the link with the fundamental group (as in the post linked by OP) is still to be made, so it's not a duplicate $\endgroup$ Commented May 5, 2021 at 15:48

0

You must log in to answer this question.

Browse other questions tagged .