# 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 $n>4$, $\pi_3(SO(n))=\mathbb{Z}$, 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 $n>1$ $$\pi_n(SO(4,2))=\pi_n(SU(2))\times \pi_n(SU(2))=\pi_n(S^3)\times \pi_n(S^3),$$ 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 $S^3$.

• It would be probably useful to emphasize the fact that ellipses in your last formula denote groups which we probably do not know. Jun 1, 2016 at 4:15
• @marRrR I have updated the answer, which previously contained a mistake. I am not a topology expert, so it would be nice if someone could check the current version. Jun 11, 2016 at 23:33

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.

• What is an $n-1$-connected map?
– Danu
May 24, 2016 at 22:13
• Why would "Euclideanizing" preserve the homotopy groups? It doesn't preserve other topological properties like compactness. Also, you didn't actually give the homotopy groups, stating they are Abelian does not uniquely identify them. May 24, 2016 at 22:24
• Do you mean that the fundamental group of Lie groups (not algebras: the algebra is of course the simply connected $\mathbb{R}^N$) is Abelian (as an aside: a property of general topological groups)? May 25, 2016 at 0:23
• @WetSavannaAnimalakaRodVance -- yes, the fundamental group of any topological group is abelian (the fundamental group functor takes group objects in the category of spaces to group objects in the category of groups because it preserves products). May 25, 2016 at 4:43
• @WillO Indeed. See Thm 14.6 and 14.7 on my page herre for an elegant and very simple proof of this equivalence of products. Unfortunately the originator of this proof hasn't published it, and the blog where first I read about it is gone. May 25, 2016 at 6:06