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Let $G$ be a compact Lie group, and $g$ its associated Lie algebra.

In what ways do the higher homotopy groups $\pi_{n}(G)$ with $n>1$ appear in the representation theory of $G$?

As an example, note that there is at least one place where the fundamental group is significant, namely in distinguishing representations of $g$ from those of $G$ itself. Indeed, if $\tilde{G}$ is the simply connected universal cover of $G$ then all representations of $g$ can be integrated to representations of $\tilde{G}.$ However, if $G=\tilde{G}/H$ ($H$ a discrete subgroup of $\tilde{G}$) then a necessary and sufficient condition for a representation of $g$ to integrate to a representation of $G$ is that $\pi_{1}(G)=H$ acts trivially in the representation of $\tilde{G}$.

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"if G˜ is the simply connected universal cover of G then all representations of g can be integrated to representations of G". You should include "finite-dimensional" somewhere in that sentence. The Lie algebra $\mathbb R$ acts on $\mathcal C^\infty(I)$, where $I$ denotes the open unit interval $I = (0,1)$, by sending the basis vector to $\frac{\partial}{\partial x}$, but this representation is not integrable to a representation of $\mathbb R$ on $\mathcal C^\infty(I)$. –  Theo Johnson-Freyd Apr 18 '13 at 3:50

3 Answers 3

up vote 13 down vote accepted

I don't know if this is the sort of thing you are looking for, but the higher rational homotopy groups appear naturally in representation theory. For example, if $G$ is a simply connected compact Lie group, then the algebra

$Ext^\ast_{\mathcal U (\mathfrak g)}(\mathbb C, \mathbb C) = H^\ast(\mathfrak g)$

is a model for the cochain algebra, so it encodes the rational homotopy groups by rational homotopy theory. Using this, one can show that such group is rationally homotopic to a product of odd spheres.

EDIT: I thought maybe I should add some extra detail.

The cohomology algebra $H^\ast(G; \mathbb C) = H^\ast(\mathfrak g; \mathbb C)$ is a graded commutative Hopf algebra, and by a theorem of Hopf must be a symmetric algebra $S(V)$ on a graded vector space $V$ in odd degrees ("symmetric algebra" here is meant in the graded sense; as an ungraded algebra it looks like an exterior algebra). Sullivan's rational homotopy theory says that

$\pi_n(G) \otimes \mathbb C = V^\ast$.

Moreover, the vector space $V$ can be explicitly computed as follows. The cohomology of the classifying space $H^\ast(BG)$ is given by the $W$ invariants in $H^\ast(BT)$, where $T$ is a maximal torus and $W=N_G(T)/T$ is the Weyl group. The classifying space $BT$ is a product of $\mathbb CP^\infty$'s, thus $H^\ast(BT)$ is a polynomial algebra on the vector space $\mathfrak t^\ast$ in degree $2$. By a theorem of Chevalley, $H^\ast(BT)^W$ is also polynomial algebra on some graded vector space in even degrees. This vector space precisely $V^\ast[-1]$.

In particular, this shows that $\pi_2(G)$ is torsion. On the other hand, consider the flag variety $G/T$. This is a compact complex manifold with a complex cell structure. Thus its cohomology is only in even degrees. In particular $\pi_2(G/T)$ is free, and by the LES of a fibration $\pi_2(G)$ injects into $\pi_2(G/T)$ and is torsion free, so zero.

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If $G$ is compact, then we can think about its cohomology with Lie algebra cohomology, so be doing some kind of representation theory. And of course the first nontrivial homology is determined by the the first nontrivial homotopy. (Since $G$ is smooth, we can relate its homology and cohomology, too.)

Specifically, if $G$ is simply connected (and compact), then $H^1 = H^2 = 0$, and $\dim H^3 =$ the number of simple factors. The $H^2$ result, one interprets as the statement that $G$ has no nontrivial central extensions. The $H^3$ (or $\pi_3$) result one usually interprets as saying that $LG$ (free loops) does have nontrivial central extensions.

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When $n = 2,3$:

For any simply-connected finite-dimensional Lie group $G$, $\pi_2(G) = 1$ is trivial. So it can have no applications.

If $G$ is simply-connected, then $\pi_3(G) = \mathrm{H}_3(G,\mathbb Z)$ by Hurewicz's theorem. This is a copy of $\mathbb Z$ whenever $G$ is simple. Classes here can be used to construct central extensions of the loop group of $G$.

I do not know of applications of higher $\pi_n$.

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No applications? Isn't $\pi_2$ being trivial related to the Whitehead lemma? –  Qiaochu Yuan Apr 18 '13 at 4:35
    
Well, I mean, there are many applications of the general statement that $\pi_2$ is trivial. But for no $G$ can elements of $\pi_2(G)$ be applied to some particular construction, the way that $\pi_1$ or $\pi_3$ can. –  Theo Johnson-Freyd Apr 19 '13 at 1:22

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