# What is the importance of $\pi_{i}G$?

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 $SO(5)$?

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Is that symplectic group supposed to have an even dimension? – Will Sawin Jul 22 '12 at 2:17
I see the problem, corrected. – Kerry Jul 22 '12 at 2:21
Incidentally, $\pi_2 G = 0$ for any Lie group; this is a theorem of Bott. See mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups for a very nice discussion or Milnor's book "Morse Theory." – Akhil Mathew Jul 22 '12 at 2:25
You mean simple ones, or any? I did read Bott's collective papers but never seen this theorem. I shall check the link. – Kerry Jul 22 '12 at 2:27
The vanishing of $\pi_2$ was discussed at least once in MO: google might be of help finding the thread, which was very enlightening! – Mariano Suárez-Alvarez Jul 22 '12 at 4:51

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.

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A cool application of the fact $\pi_2G = 0$ is that any principal bundle (and hence any vector bundle) over $S^3$ is trivial, which if I remember correctly was first noticed by Serre. – Paul Reynolds Jul 22 '12 at 11:26
@Paul, how did he prove it? – Mariano Suárez-Alvarez Jul 22 '12 at 19:24
@Mariano, I can't remember exactly where I read the Serre citation, it was quite a while ago. There are various ways to see it though, the easiest probably as follows. Rank $k$ (smooth) vector bundles over $S^3$ are determined by a clutching map on the equator, i.e. by a homotopy class of maps $S^2 \to O(k)$. – Paul Reynolds Jul 22 '12 at 20:11
@Mariano, it was bugging me that I couldn't remember where I read the Serre comment so I typed "noticed by J.P.Serre" into Google and miraculously it came up with exactly one book, a book I am always reading. I had not remembered the situation correctly (it seemed too easy a fact to require a Serre). Remark 9.57 on p249 of Einstein Manifolds by Besse states: "There exist locally trivial fibre bundles whose structural group cannot be reduced to a Lie group, for any family of local trivialisations (this was noticed by J.P.Serre). For example, any principal $G$-bundle over $S^3$ is trivial... – Paul Reynolds Jul 23 '12 at 15:30
... if $G$ is a Lie group (since $[S^3,BG] = \pi_2(G) = 0$, a result due to E.Cartan [Car 11]), whereas $S^3$ does admit some non-trivial sphere bundles (see for example [An-Bu-Ka]). For these bundles the total space admits no Riemannian metric such that the projection becomes a Riemannian submersion with totally geodesic fibres." Not totally relevant to the question but an interesting fact none-the-less. – Paul Reynolds Jul 23 '12 at 15:35