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 2connected. 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 $SO(5)$?

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. 

