The universal cover $\widetilde{\mathbf{SL}}(2,\mathbf{R})$ of $\mathbf{SL}(2,\mathbf{R})$ has no faithful finite-dimensional representations.

Question

In this answer https://math.stackexchange.com/q/129656, he try to prove that $\widetilde{\mathbf{SL}}(2,\mathbf{R})$ has no faithful finite-dimensional representations.

I have three question among that answer.

1. $\widetilde{\mathbf{SL}}(2,\mathbf{R})$ is simply connected
2. $\widetilde{\mathbf{SL}}(2,\mathbf{R})$ is simply connected implies that $\Phi$ and $\Psi\circ\pi$ are the same representation
3. $\Phi$ factors over $\pi$ implying non-trivial kernel

Thank you!

Answer 1

1. By definition $\widetilde{\mathbf{SL}}(2,\mathbf{R})$ is the universal covering space of $\mathbf{SL}(2,\mathbf{R})$ and thus, by definition of universal covering space, it is simply connected.
2. This is a consequence of the Lie group-Lie algebra correspondence. Specifically the Homomorphism theorem which stablishes that if $G$ and $H$ are Lie groups with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$, respectively, if $G$ is simply connected and if $\phi:\mathfrak{g}\to\mathfrak{h}$ is a Lie algebra homomorphism, then there exists an unique Lie group homomorphism $\Phi:G\to H$ such that $\phi=d\Phi(1)$. Thus since in your case $d\Phi=d(\Psi\circ\pi)$, the uniqueness part of this theorem yields $\Phi=\Psi\circ\pi$.
3. A straighforward computation shows that since $\Phi=\Psi\circ\pi$, we have $\ker(\pi)\subseteq \ker(\Phi)$ and as $\ker(\pi)\cong \mathbf{Z}$, you obtain thta $\ker(\Phi)$ is non trivial.

ADDED: I am going to justify why $\ker(\pi)\cong \mathbf{Z}$. This has to do with the facts that $\pi_1(\mathbf{SL}(2,\mathbf{R}),1)\cong \mathbf{Z}$ and that $\pi$ is a covering projection.

It is easy to see, for example by using Iwasawa decomposition, that $\mathbf{SL}(2,\mathbf{R})$ has the same homotopy type of $\mathbf{SO}(2)$ (actually $\mathbf{SO}(2)$ is a deformation retract of $\mathbf{SL}(2,\mathbf{R})$) and $\mathbf{SO}(2)$ is isomorphic (as a Lie group) to the circle group $S^1=\{z\in\mathbf{C} \; | \; |z|=1\}$. Thus $\pi_1(\mathbf{SL}(2,\mathbf{R}),1)\cong \mathbf{Z}$. Now, $\ker(\pi)$ is the fiber over $1$ of the covering projection $\pi$ and since $\mathbf{SL}(2,\mathbf{R})$ acts by monodromy simply transitively on any fiber (this is from the theory of covering spaces), so there is a natural bijection $\mathbf{Z}\cong \pi_1(\mathbf{SL}(2,\mathbf{R}))\to \ker(\pi)$, and it is easy to see that this bijection is a group homomorphism.