For the restricted Lorentz group $SO^+(1,3)$, all of its projective representations are of the form \begin{equation} D(\Lambda_1)D(\Lambda_2) = \pm D(\Lambda_1\Lambda_2) \end{equation}
As far as I know, this is because the fundamental group of $SO^+(1,3)$ is $\mathbb{Z_2}$.
How would we generalize this to the case of the restricted indefinite orthogonal group $SO^+(p,q)$? That is, for the projective representations \begin{equation} D(\Lambda_1)D(\Lambda_2) = c_{\Lambda_1, \Lambda_2}D(\Lambda_1\Lambda_2) \end{equation}
what values can $c_{\Lambda_1, \Lambda_2}$ take?
What I'm guessing is that it is somehow dependent on the fundamental group of $SO^+(p,q)$. So, as the fundamental group of $SO^+(p,q)$ is \begin{array}{|c|c|} \hline \pi_1(SO^+(p,q)) & p = 0 & p = 1 & p = 2 & p\ge 3 \\ \hline q = 0 & 1 & 1 & \mathbb{Z} & \mathbb{Z_2} \\ \hline q = 1 & 1 & 1 & \mathbb{Z} & \mathbb{Z_2} \\ \hline q = 2 & \mathbb{Z}& \mathbb{Z} & \mathbb{Z}\times\mathbb{Z} & \mathbb{Z_2}\times\mathbb{Z} \\ \hline q \ge 3 & \mathbb{Z_2} & \mathbb{Z_2} & \mathbb{Z_2}\times\mathbb{Z} & \mathbb{Z_2}\times\mathbb{Z_2} \\ \hline \end{array}
As $SO^+(p,1)$ and $SO^+(p,0)$ also have $\mathbb{Z_2}$ as its fundamental group for $p\ge 3$, I expect $c_{\Lambda_1, \Lambda_2} = \pm 1$ for these. But, I have no idea how to prove this. Also, I have no idea what $c_{\Lambda_1, \Lambda_2}$ can be for the other values of $SO^+(p,q)$, nor how to go about proving this. So, how does one determine values of $c_{\Lambda_1, \Lambda_2}$ for these cases?