The group $SO(n) \subset SO(n+1)$ by an $n-1$-connected map. Consequently for $k < n-1$ $\pi_k(O(n+1)) = \pi_k(O(n))$. I am euclideanizing $SO(4,2) \rightarrow SO(6)$, and not considering for the time the hyperbolic aspects. So all we have to consider is the fundamental group $\pi_1(SO(4,2))$ The Serre fibration
$$
SO(n) \rightarrow SO(n+1) \rightarrow SO(n+1)/SO(n) \sim S^n
$$
gives the sequence of homotopies
$$
\pi_k(SO(n)) \rightarrow \pi_k(SO(n+1)) \rightarrow \pi_k(SO(n+1)/SO(n))
$$
has $\pi_k(S^n) = 0$ this demonstrates the equality. I will now state that it is known that the fundamental group of Lie algebras are abelian.

To continue this, sorry I had to post due to interruption, I now appeal to Bott periodicity. I now use the fact from Bott periodicity theorem that $\pi_k(Sp) = \pi_{k+4}(O)$. Now we can focus in on $\pi_1(sp(2))$ and the knowledge that $sp(2) \sim U(1)$. The homotopy is abelian, which means it is equal to its homology group, which for the circle is $\mathbb Z$

As for not going hyperbolic, it is the case with physics problems that one looks at the Euclidean case first.

This post imported from StackExchange Mathematics at 2016-05-25 15:22 (UTC), posted by SE-user Lawrence B. Crowell