# Commutativity of the fundamental group of any Lie Group [closed]

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

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@Marina: by and large, MO is for questions of research interest. For questions at this level we have math.stackexchange.com. –  Qiaochu Yuan May 25 '12 at 13:58
I think the more important issue is that this is an exact dubplicate: mathoverflow.net/questions/35868/… –  Sean Tilson May 25 '12 at 14:24
@Sean: the other question was phrased less offensively (to me) -- the OP wanted to know "why" this is true. The current OP simply asks us to do her homework for her (I know this is homework because this was the first slightly nontrivial exercise on fundamental groups I remember doing when I was a student (I believe it is in Massey's book). It is extremely important to keep this sort of junk out of MO. –  Igor Rivin May 25 '12 at 14:41
Igor, I'm not a mathematician and this is not my homework. It came when I was reading Rubikov on field theory. I'm very sorry if my lack of background offended you. Please close the question if you feel you need to. –  user14210 May 25 '12 at 14:48

## closed as too localized by Igor Rivin, Qiaochu Yuan, Dmitri Pavlov, HJRW, André HenriquesMay 25 '12 at 15:26

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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.

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Were were you when I needed to do my homework? –  Igor Rivin May 25 '12 at 14:46
@Igor Rivin, I thought it might be homework, but then I spent long enough thinking about the problem that I decided to write it up anyway. –  MTS May 25 '12 at 14:59
Thanks for fixing the latex, Henrik! For some reason the cases environment wasn't rendering properly when I was typing up the answer. –  MTS May 25 '12 at 15:00
@MTS you did a wonderful job, my comment is meant to be negative on the question, but positive on the answer –  Igor Rivin May 25 '12 at 16:52
@Igor, thank you! I only wish I knew a good way to draw the picture that I had in my head in latex somehow. –  MTS May 25 '12 at 17:35

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.

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It's certainly the nicest proof of this result I've ever seen! –  Stefano V. May 25 '12 at 10:30
This is terrific. –  Steven Landsburg May 25 '12 at 12:11
"...by the Eckmann-Hilton argument." –  Qiaochu Yuan May 25 '12 at 12:20
the point is that the group law in the fundamental group of a topological group is a group homomorphism; a group whose law is a homomorphism is easily checked to abelian (I knew this argument but not with the interpretation by group objects). –  Yves Cornulier May 25 '12 at 13:18
Nice! I knew there was some clever way of doing it. –  MTS May 25 '12 at 14:58
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.