Homotopy group of the conformal group

Question

I would like to know which are the first three homotopy groups of the conformal group SO(4,2):

$$\pi_n(SO(4,2))=? \quad n=1,2,3$$

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

Answer 1

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

Comments

What is an $n-1$-connected map?

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

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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.

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

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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)?

This post imported from StackExchange Mathematics at 2016-05-25 15:22 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance

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@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).

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

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@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.

This post imported from StackExchange Mathematics at 2016-05-25 15:22 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance