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Friday, May 25, 2018

About the Fundamental Group of a Topological Group

Proposition. Let $(G,\cdot)$ be a topological group. Let $C_e$ be the path-connected component of the identity $e$. Then

  1. $C_e$ is a topological group.
  2. $\pi_1(C_e,e)$ is abelian.
  3. For any $g\in G$, $\pi_1(G,g)$ is abelian.
Lemma. Let $(G,\cdot,*)$ be a set with two binary operations on it. Also suppose that both operations have a common identity $e$ and for any $g,h,j,k\in G$,
$$(g\cdot h)*(j\cdot k)=(g*j)\cdot(h*k).$$
Then $*=\cdot$ and $*$ is commutative.

Proof of Lemma. If $g,h\in G$ then $$g * h=(e\cdot g)* (h\cdot e)=(e*h)\cdot(g*e)=h\cdot g=(h*e)\cdot (e*g)=(h\cdot e)*(e\cdot g)=h*g.$$
The result follows.

Proof of Proposition. To prove 1. remember that for $u\in G$, $\phi_u:G\to G$ given by $x\mapsto xu$ is a homeomorphism. Also denote $*$ as the product of paths or homotopy classes of paths. Now take paths $\alpha:e\leadsto g$ and $\beta:e\leadsto h$. Then since $\phi_h$ is continuous, $\phi_h\circ \alpha:h\leadsto gh$, so $gh\in C_e$. Also $\phi_{g^{-1}}\circ \alpha: g^{-1}\leadsto e$, so $g^{-1}\in G$. So $C_e\leq G$ and since the restrictions of $*$ and $\square ^{-1}$ are well defined in $C_e$, then $C_e$ is a topological group.

To prove 2. define $[\alpha]\cdot[\beta]=[\alpha\cdot \beta]$ where $(\alpha\cdot\beta)(s)=\alpha(s)\beta(s)$. This operation is well defined on homotopy classes, since $\alpha\simeq_{\text{Rel}0,1}^H\alpha'$ and $\beta\simeq_{\text{Rel}0,1}^J\beta'$ then the function $H\cdot J$ is a homotopy, relative to $0,1$ between $\alpha\cdot\beta$ and $\alpha'\cdot\beta'$.

So now there are two binary operations on $\pi_1(C_e,e)$, given by $*$ and $\cdot$. The homotopy class $[c_e]$ of the constant loop $c_e$ is the identity of $(\pi_1(C_e,e),*)$. But it's also an identity in $(pi_1(C_e,e),\cdot)$. Indeed, if $\alpha$ is a loop starting at $e$, then $$(c_e\cdot \alpha)(s)=c_e(s)\alpha(s)=e\alpha(s)=\alpha(s)=\alpha(s)e=\alpha(s)c_e(s)=(\alpha\cdot c_e)(s).$$

Also, if $\alpha,\beta,\gamma,\delta$ are loops in $e$, then for $0\leq s\leq 1$
$$\begin{align*}(\alpha\cdot \beta)*(\gamma\cdot \delta)(s) &=\left\{\begin{array}{cc}\alpha(2s)\beta(2s) & 0\leq s\leq \frac{1}{2}\\\gamma(2s-1)\delta(2s-1) & \frac{1}{2}\leq s\leq 1\end{array}\right.\\ &= (\alpha *\gamma)\cdot(\beta *\delta)(s).\end{align*}(s)$$
so taking homotopy classes
$$([\alpha]\cdot[\beta])*([\gamma]\cdot [\delta])=([\alpha]*[\gamma])\cdot([\beta] *[\delta]).$$

As a result of the lemma $*=\cdot$ and $*$ is commutative, so $\pi_1(C_e,e)$ is abelian.

To prove 3. note that $\pi_1(G,e)=\pi_1(C_e,e)$ and for $g\in G$, $\phi_{g*}:\pi_1(G,e)\to \pi_1(G,g)$ is an isomorphism. The result follows. 
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