Question

How do we formally prove that the fundamental group of any Lie group is always commutative?

Answer 1

As Vahid says, it is true for any topological group. Here is a proof. I'm sure there are nicer, more conceptual ones out there, but here goes.

Let $G$ be your topological group. Take two loops $\sigma$ and $\gamma$ in $G$, based at the identity of $G$, which we will denote by $e$. Let $\sigma \cdot \gamma$ be the concatenation of the two loops. This is given by $$ (\sigma \cdot \gamma) (t) =
\begin{cases} \sigma(2t) & \quad \text{ if } 0 \le t \le 1/2 \\ \gamma(2t-1) &\quad \text{ if } 1/2 \le t \le 1 \end{cases} $$ (Sorry, couldn't manage to format that any better. Feel free to edit if you know how to put a nice brace bracket to the left of that definition.)

The idea is this. We will show that $\sigma \cdot \gamma$ is homotopic to to the loop given by the pointwise product of $\sigma$ and $\gamma$. Let's call that loop $\rho$, so $$ \rho(t) = \sigma(t)\gamma(t).$$

Now define an auxiliary function $P : [0,1] \times [0,1] \to G$ by $$ P(s,t) = \begin{cases} \sigma\left( \frac{2t}{1+s} \right) & \quad \text{ if } 0 \le t \le \frac{1+s}{2} \\ e &\quad \text{ if } \frac{1+s}{2} \le t \le 1 \end{cases}$$

At $s=0$, this function does the whole loop $\sigma$ as $t$ goes from $0$ to $1/2$, then sits at $e$. In other words, at $s=0$ this is the first half of the loop $\sigma \cdot \gamma$. As $s$ gets larger, $P$ does the whole loop $\sigma$ as $t$ goes from $0$ to $\frac{1+s}{2}$. At $s=1$, $P$ does the loop $\sigma$ at normal speed.

Then similarly define a function $Q : [0,1] \times [0,1] \to G$ by $$ Q(s,t) = \begin{cases} e & \quad \text{ if } 0 \le t \le \frac{1-s}{2} \\ \gamma \left( \frac{2t-1+s}{1+s} \right) &\quad \text{ if } \frac{1-s}{2} \le t \le 1 \end{cases}$$

At $s=0$ this is just the second half of the loop $\sigma\cdot\gamma$, while at $s=1$ it is exactly the loop $\gamma$.

So finally, define $$ H(s,t) = P(s,t) \cdot Q(s,t). $$ At $s=0$ this is $\sigma \cdot \gamma$, while at $s=1$ it is the pointwise product loop $\rho$. $H$ is clearly continuous, and $H(s,0) = e = H(s,1)$ for all $s$, so this is a homotopy of loops between $\sigma \cdot \gamma$ and $\rho$.

Now we can redo that process and show that $\rho$ is homotopic to the other concatenation $\gamma \cdot \sigma$. So this shows that $\pi_1(G)$ is abelian.

Answer 2

One-sentence explanation: because the fact that a topological group $G$ is a group object in topological spaces makes its fundamental group $\pi_1(G)$ a group object in groups, and this is an abelian group.

Answer 3

Geometric proof: A connected Lie group $G$ is homotopy equivalent to a maximal compact subgroup, so we may assume $G$ is compact. Being compact, $G$ admits a bi-invariant Riemannian metric with respect to which it is a symmetric space, the symmetry $s$ at the identity being just the inversion map. Now a homotopy class in $\pi_1(G,1)$ can be represented by a closed geodesic $\gamma$ (of minimal length in its homotopy class, by a shortening process). Since the differential of $s$ at $1$ is minus identity, $s$ sends $\gamma$ to itself parametrized backwards. It follows that the homomorphism induced by $s$ on the $\pi_1$-level is inversion. However, the inversion map in a group is a homomorphism if and only if the group is Abelian.

Answer 4

It is actually true for all topological groups. Topological groups possess a structure which makes them H-spaces and fundamental group of every H-space is abelian. The formulation and the proof is given in Algebraic Topology, Homotopy and Homology, by Switzer Pages 14-16.