Several times I've heard the claim that any Lie group $G$ has trivial second fundamental group $\pi_2(G)$, but I have never actually come across a proof of this fact. Is there a nice argument, perhaps like a more clever version of the proof that $\pi_1(G)$ must be abelian?

1$\begingroup$ Isn't for semisimple Lie Groups isn't it true that $\pi_3(G)=\mathbb{Z}$? I would be happy to see explanations of this may be as a byproduct of this discussion. I suppose this is important to see why the "level" of ChernSimons theory is quantized. $\endgroup$ – Anirbit Jun 30 '10 at 14:47

11$\begingroup$ @Anirbit: that cannot be true, for the product of semisimple Lie groups is semisimple,and the $\pi_3$ of a product is the product of the $\pi_3$s. It is true for simple groups, though. $\endgroup$ – Mariano SuárezÁlvarez Feb 18 '13 at 5:19

8$\begingroup$ The proof of $\pi_3=\mathbb Z$ can be found in Bott's paper "An application of the Morse theory to the topology of Liegroups" (1956). $\endgroup$ – André Henriques Aug 23 '14 at 20:15

2$\begingroup$ @AndréHenriques's reference, clickably: Bott  An application of the Morse theory to the topology of Lie groups (MSN), also available in ICM proceedings (MSN). $\endgroup$ – LSpice Dec 27 '19 at 1:47
I don't know of anything as bare hands as the proof that $\pi_1(G)$ must be abelian, but here's a sketch proof I know (which can be found in Milnor's Morse Theory book. Plus, as an added bonus, one learns that $\pi_3(G)$ has no torsion!):
First, (big theorem): Every (connected) Lie group deformation retracts onto it's maximal compact subgroup (which is, I believe, unique up to conjugacy). Hence, we may as well focus on compact Lie groups.
Let $PG = \{ f:[0,1]\rightarrow G  f(0) = e\}$ (I'm assuming everything is continuous.). Note that $PG$ is contracitble (the picture is that of sucking spaghetti into one's mouth). The projection map $\pi:PG\rightarrow G$ given by $\pi(f) = f(1)$ has homotopy inverse $\Omega G = $Loop space of G = $\{f\in PG  f(1) = e \}$.
Thus, one gets a fibration $\Omega G\rightarrow PG\rightarrow G$ with $PG$ contractible. From the long exact sequence of homotopy groups associated to a fibration, it follows that $\pi_k(G) = \pi_{k1}\Omega G$
Hence, we need only show that $\pi_{1}(\Omega G)$ is trivial. This is where the Morse theory comes in. Equip $G$ with a biinvariant metric (which exists since $G$ is compact). Then, following Milnor, we can approximate the space $\Omega G$ by a nice (open) subset $S$ of $G\times ... \times G$ by approximating paths by broken geodesics. Short enough geodesics are uniquely defined by their end points, so the ends points of the broken geodesics correspond to the points in $S$. It is a fact that computing low (all?...I forget)* $\pi_k(\Omega G)$ is the same as computing those of $S$.
Now, consider the energy functional $E$ on $S$ defined by integrating $\gamma^2$ along the entire curve $\gamma$. This is a Morse function and the critical points are precisely the geodesics**. The index of E at a geodesic $\gamma$ is, by the Morse Index Lemma, the same as the index of $\gamma$ as a geodesic in $G$. Now, the kicker is that geodesics on a Lie group are very easy to work with  it's pretty straight forward to show that the conjugate points of any geodesic have even index.
But this implies that the index at all critical points is even. And now THIS implies that $S$ has the homotopy type of a CW complex with only even cells involved. It follows immediately that $\pi_1(S) = 0$ and that $H_2(S)$ is free ($H_2(S) = \mathbb{Z}^t$ for some $t$).
Quoting the Hurewicz theorem, this implies $\pi_2(S)$ is $\mathbb{Z}^t$.
By the above comments, this gives us both $\pi_1(\Omega G) = 0$ and $\pi_2(\Omega G) = \mathbb{Z}^t$, from which it follows that $\pi_2(G) = 0$ and $\pi_3(G) = \mathbb{Z}^t$.
Incidentally, the number $t$ can be computed as follows. The universal cover $\tilde{G}$ of $G$ is a Lie group in a natural way. It is isomorphic (as a manifold) to a product $H\times \mathbb{R}^n$ where $H$ is a compact simply connected group.
H splits isomorphically as a product into pieces (all of which have been classified). The number of such pieces is $t$.
(edits)
* it's only the low ones, not "all", but one can take better and better approximations to get as many "low" k as one wishes.
** I mean CLOSED geodesics here

21$\begingroup$ I love your mouthwatering picture of sucking spaghetti ! Sei di origine italiana ? $\endgroup$ – Georges Elencwajg Dec 15 '09 at 8:38

2

3$\begingroup$ While I have a great grandfather from Italy, I'm from the US. All the credit for the proof SHOULD go to Milnor  I just copied his ideas ;) $\endgroup$ – Jason DeVito Dec 15 '09 at 13:46

1$\begingroup$ "H splits isomorphically as a product into pieces (all of which have been classified). The number of such pieces is t." > Does each piece correspond to a simple group? I hope I can argue that $t=1$ if and only if $G$ is simple. $\endgroup$ – Student Sep 13 '19 at 21:06
The elementary proof that $\pi_1$ is abelian applies more generally to Hspaces (spaces $X$ with a continuous multiplication map $X \times X \to X$ having a 2sided identity element) without any assumption of finite dimensionality, but infinitedimensional Hspaces can have nontrivial $\pi_2$, for example $CP^\infty$ (which can be replaced by a homotopy equivalent topological group if one wants, as Milnor showed). Thus finitedimensionality is essential, so any proof would have to be significantly less elementary than for the $\pi_1$ statement. It is a rather deep theorem of W.Browder (in the 1961 Annals) that $\pi_2$ of a finitedimensional Hspace is trivial.
Hopf's theorem that a finitedimensional Hspace (with finitelygenerated homology groups) has the rational homology of a product of odddimensional spheres implies that $\pi_2$ is finite, but the argument doesn't work for mod p homology so one can't rule out torsion in $\pi_2$ so easily. It's not true that a simplyconnected Lie group is homotopy equivalent to a product of odddimensional spheres. For example the mod 2 cohomology ring of Spin(n) is not an exterior algebra when n is sufficiently large. For SU(n) the cohomology ring isn't enough to distinguish it from a product of spheres, but if SU(n) were homotopy equivalent to a product of odddimensional spheres this would imply that all odddimensional spheres were Hspaces (since a retract of an Hspace is an Hspace) but this is not true by the Hopf invariant one theorem. There are probably more elementary arguments for this.

1$\begingroup$ Thanks for correcting my statement! I have ammended my answer below. This explains why I did not remember the proof! I had misremembered the theorem :( $\endgroup$ – José FigueroaO'Farrill Dec 16 '09 at 1:34

3$\begingroup$ This is very fascinating. Is there a known example of a manifold $M$ such that $\pi_2(Diff(M))$ is nontrivial? $\endgroup$ – Mircea Mar 17 '12 at 7:07

1$\begingroup$ @Mircea $\text{Diff}(S^1\times S^2)\simeq O(2)\times O(3)\times \Omega SO(3)$ has $\pi_2(\text{Diff}(S^1\times S^2))\simeq \pi_3(S^3)\simeq \mathbb{Z}$ $\endgroup$ – Espen Nielsen Feb 13 '15 at 13:29

1$\begingroup$ @Ben Wieland: The vector space needs to be of countable dimension in order to have $CP^\infty$ as its projectivization. However, if the rational functions had a countable basis, this would imply that only countably many complex numbers occurred as poles of rational functions, a contradiction. This is almost counterintuitive: rational functions depend upon only countably many complex parameters (their coefficients) yet as a vector space their dimension is uncountable. $\endgroup$ – Allen Hatcher Mar 28 '16 at 18:30

1$\begingroup$ Let me correct and consolidate my comments. There are two easy constructions of simplicial complex topological groups homotopy equivalent to $CP^\infty$. (1) The free abelian group on $S^2$ is nice because it is $P(C(x))$ (with the right topology), a projective space, though a different one. (2) The bar construction $BS^1$ can be easy to construct and we proved with difficulty that it is homeomorphic to $CP^\infty$. I conjecture that everything is homeomorphic. $\endgroup$ – Ben Wieland Apr 9 '16 at 16:52
Here's another proof based on the structure of the flag variety $G/T$ of $G$. A compact Lie group $G$ has a maximal torus $T$, and $G$ is a principal $T$bundle over the quotient $G/T$. Borel showed that $G/T$ is a complex manifold, and gave a CW decomposition of it with no odddimensional cells. (This is not deep but still astonishing, and the start of a long story; I like the context given by Hirzebruch's eulogy for Borel, available on page 9 here.)
Since $\pi_2(T) = 0$, we have an exact sequence
$$0 \to \pi_2(G) \to \pi_2(G/T) \to \pi_1(T)$$
We can conclude immediately that $\pi_2(G)$ is torsionfree, since $\pi_2(G/T) = H_2(G/T)$ is a free group on the 2cells in $G/T$. After Allen's answer (Hopf's theorem) this shows $\pi_2(G) = 0$.
With a little more Lie theory one can show directly that the connecting map $\pi_2(G/T) \to \pi_1(T)$ is injective. The group $\pi_1(T)$ has a linearly independent subset of simple coroots, and the 2cells in $G/T$ are indexed by simple roots. The connecting homomorphism matches these up in the natural way, which one can see by considering rank 1 subgroups (subgroups of the form $SU(2)$ or $PSU(2)$) of $G$. As a consequence you get a formula for $\pi_1(G)$ in terms of roots and coroots.

1$\begingroup$ It's hard to pick but I think that's the best answer on this thread. Proving that $\pi_2(G)$ is torsion is somehow not very difficult and can be done in several ways. The argument above completes the argument in the most simple way possible I believe. The fact that $G/T$ has evendimensional cell decomposition is the only deep theorem here and even this can be proven by several different ways (morse theory / bruhat decomposition etc...). So I would say this is the most elementary (for me the most "convincing") argument. In particular there are no infinite dimensional spaces here at all. $\endgroup$ – Saal Hardali Dec 26 '17 at 22:50
There's a proof that $\pi_2(G)$ is trivial for compact semisimple Lie groups in section 8.6 of Pressley and Segal's Loop Groups. They say "This proof is in essence the same as Bott's Morse theory proof" but it has some differences in treatment even if the substance is the same. The major difference is that they don't approximate $\Omega G$ by a finite dimensional manifold but by an infinite dimensional one.
In slightly more detail, the idea of the proof is to find a Grassmannian model for $\Omega G$. This is done by considering the action of $G$ on $L^2(S^1;\mathfrak{g}_\mathbb{C})$ and then taking the restricted Grassmannian of this space. Within that, one can identify a subGrassmannian that is diffeomorphic to $\Omega G$. They then find a cell decomposition of this Grassmannian and analyse that. One of the important pieces is to consider the subgroup of polynomial loops in $\Omega G$. This corresponds to a certain subGrassmannian and it's easy to see that for this Grassmannian then all the cells are of even dimension, whence $\pi_1(\Omega_{\operatorname{pol}} G)$ is trivial. The final step is thus to show that the two Grassmannia (corresponding to $\Omega G$ and $\Omega_{\operatorname{pol}} G$) are homotopy equivalent. Then $\Omega G$ and $\Omega_{\operatorname{pol}} G$ are homotopy equivalent and so $\pi_1(\Omega G)$ is trivial.
Thus $\pi_2(G)$ is trivial since $\pi_2(G) = \pi_1(\Omega G)$ (incidentally, one doesn't need the long exact sequence for fibrations to see that $\pi_k(X) = \pi_{k1}(\Omega X)$; that's either by definition or by using the adjunction $[\Sigma X, Y] \cong [X, \Omega Y]$).
As I said, Pressley and Segal say that this is in essence the same as Bott's proof; meaning that it proceeds by a cell decomposition based on "energy". However, it treats the infinite dimensional spaces as infinite dimensional spaces so I like it! Also, Grassmannia are more obviously structured so the cell decomposition may be simpler to see and understand in the Grassmannian model for the loop group.
For the classical Lie groups, I think that an easy way to obtain the result is through the fibrations:
$SO(n1)\to SO(n)\to S^{n1}$,
$SU(n1)\to SU(n)\to S^{2n1}$,
$SP(n1)\to SP(n)\to S^{4n1}$
and the homotopy long exact sequence and $\pi_m(S^n) = 0$ for $m$ less than $n$, and $\pi_2(SO(2)) = \pi_2(SU(2))=0$ and the isomorphism of $SP(2)$ and $SO(5)$.

9$\begingroup$ One can also extend this idea to $G_2$ and $F_4$. $G_2$ sits in the sequences $SU(3)\rightarrow G_2\rightarrow S^6$ and $F_4$ sits in the sequence $Spin(9)\rightarrow F_4\rightarrow CaP^2$, where $CaP^2$ is the Cayley plane. I don't know of similar fibrations for $E_6, E_7,$ or $E_8$ though. $\endgroup$ – Jason DeVito Dec 16 '09 at 2:46

4$\begingroup$ The Cayley plane $\mathbb{C}aP^2$ is given also as a homogeneous space of $E_6$ mod a parabolic subgroup $P$ of $E_6$. This argument can be used to prove that $E_6$ is also connected with the above method. To be more presice, we have $\mathbb{C}aP^2=E_6/U(1)\times Spin(10)$. $\endgroup$ – 314159. Feb 1 '12 at 2:45

$\begingroup$ In fact, one can reduce the problem to these cases with some big theorems (inc. classification of simple Lie groups). see math.binghamton.edu/somnath/Notes/GSS5.pdf $\endgroup$ – Ma Ming Apr 27 '12 at 9:27
This is proved in the book Representations of compact Lie groups by Bröcker and tom Dieck and reviewed here. It is Proposition 7.5 in Chapter V. The proof is for compact, connected Lie groups, but any connected Lie group has the homotopy type of its maximal compact subgroup. (Everything here is for finitedimensional Lie groups, of course.)
Edit: Maybe I should add, given that two of the other answers mention similar proofs, that the one in this book does not use Morse theory. It uses only basic covering space techniques once it is shown that $\pi_2(G)$ is isomorphic to $\pi_2(G_r)$, where $G_r$ are the regular elements, itself not a difficult lemma.
Edit: The following is wrong! It is only true rationally, which is why I didn't remember having seen a proof of the general case :)
Also if you believe that simplyconnected compact Lie groups have the homotopy type of a product of odd spheres, then this follows. It is fairly easy to see that this is the case rationally, but do not remember whether the general statement is hard to prove.
There is a sketch of the proof using the maximal torus as an exercise in GreubHalperinVanstone's book.