Question

I often see conditons like $\pi_{2}G\not=0$ in reading old papers on Lie groups(no, my memory is wrong, they asked if $\pi_{1}G$ is free). I want to ask why we need this condition and how the higher dimensional homotopy group matters. I have `read' most of available introductory graduate level textbooks on representation theory of Lie group and Lie algebra like Fulton and Harris, and I seldom(if ever) see higher diemsional homotopy groups enters into the discussion. I remember when I ask this to my undergraduate advisor, he responded that this is a "common condition" so we should not worry about it. Given the fact that most Lie groups are of huge dimension, it seems reasonable not to worry about if it is 2-connected. But now thinking in retrospective, I am wondering if the matter is this simple. Also, for practical purposes is there a practical way to compute it by its representations? For example $Sp(14,\mathbb{C})$ or $SO(5)$?

Answer 1

I don't know whether I really get you point. The importance of $\pi_i G$ for higher $i$'s is that they determine whether a $G$-principal bundle can be trivialized.

Remember a $G$-principal bundle $E\rightarrow B$ is trivial if and only if we can find section $s: B\rightarrow E$. It is natural to ask $B$ has a cell decomposition and the section $s$ always exists on the $0$-skeleton of $B$(which are points in $B$). It is easy to see that the section can be extended to $1$-skeleton if and only if the image of the points are in the same connect component of $G$.

Now we can move on by induction: if we have built a section on $i$-skeleton, can we extend it to $i+1$-skeleton? Similar to the $0$-skeleton case, we can see that the section can be extended if and only if the corresponding map $S^i\rightarrow G$ is homotopic to the constant map.

So if $\pi_i G$ are all zero, we can always extend the map and get a section, hence the principal $G$-bundles are always trivial. If $\pi_i G$ are not all zero, it requires more work.

Hopefully this makes things clear.