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2 Fixed some math to make it clearer

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 $\pi_k(G) = \pi_{k-1}\Omega G$

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) 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 geodesicsgeodesics**. 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 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

1

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 $\pi_k(G) = \pi_{k-1}\Omega G$

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 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 $t$.