Previously I mentioned in Automorphism group of a Lie group $G$ vs that of a double covering group $\tilde{G}$: same or not? that the automorphism group of a Lie group 𝐺 may be the same as that of its double covering group $\tilde{𝐺̃}$. So $1 \to \mathbf{Z}/2 \to \tilde{G} \to G \to 1,$ either if it is (1) a double cover or (2) a universal double cover ($\pi_1(\tilde{𝐺̃})=0$).
More concretely, let us consider the indefinite orthogonal group $O(p,q)$. There are as many as 8 different double covers of $O(p, q)$.
For $p, q \neq 0$, which correspond to the extensions of the center (which is either $\mathbf{Z}/2 \times \mathbf{Z}/2$ or $\mathbf{Z}/4$) by $\mathbf{Z}/2$. Only two of them are pin groups—those that admit the Clifford algebra as a representation. They are called $Pin(p, q)$ and $Pin(q, p)$ respectively. I suppose are these $Pin(p, q)$ and $Pin(q, p)$ are universal double covers, because they are both simply connected.
Am I correct that all the others are also simply connected? Thus, they are all universal double covers?
Are the automorphism group (including Inn, Out and Aut) of indefinite orthogonal Lie group $G=O(p,q)$ is the same as that of a double covering group $\tilde{G}$?
