Homotopy group of the conformal group I would like to know which are the first three homotopy groups of the conformal group SO(4,2):
$$
\pi_n(SO(4,2))=? \quad n=1,2,3
$$
According to ncatlab, the maximal compact subgroup of (the connected component of) $SO(4,2)$ is $SO(4)\times SO(2)$. Any connected Lie group retracts onto its maximal compact subgroup, so your question is about $SO(4)\times SO(2)$. 

Since $SO(2)$ is just a circle, we have $\pi_n(SO(2))=\mathbb{Z},0,0,\ldots$.

$SO(4)$ is a semi-simple group, $SO(4)\simeq (SU(2)\times SU(2))/\mathbb{Z_2}$, so it is easy to compute the first homotopy groups. 

Indeed because the second homotopy of any simple group is trivial, and the third is $\mathbb Z$, we have $\pi_2(SO(4))=0$ and $\pi_3(SO(4))=\mathbb{Z}^2$ (for $n&gt;4$, $\pi_3(SO(n))=\mathbb{Z}$, since then $SO(n)$ is actually simple). 

The fundamental group of $SO(n)$, $n\geq 3$ is $\mathbb{Z}_2$, as can be seen in this specific case from the isomorphism above.

Thus we find finally (for the connected component)
$$
\pi_n(SO(4,2))=\mathbb{Z}_2\times \mathbb{Z},0,\mathbb{Z}^2,\mathbb{Z}_2^2,\mathbb{Z}_2^2,\mathbb{Z}_{12}^2,\ldots,
$$
or more generally for $n&gt;1$
$$
\pi_n(SO(4,2))=\pi_n(SU(2))\times \pi_n(SU(2))=\pi_n(S^3)\times \pi_n(S^3),
$$
where the homotopy groups of spheres can be found on Wikipedia. Note that we do not know all the homotopy groups of $S^3$ in full generality, as noted by Mariano in the comments.

Edit: fixed the 3rd homotopy group and added the general relation to $S^3$.
The group $SO(n) \subset SO(n+1)$ by an $n-1$-connected map. Consequently for $k &lt; n-1$ $\pi_k(O(n+1)) = \pi_k(O(n))$. 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.