Universal cover of $SL_2(\mathbb{R})$ and Metaplectic group I am having trouble to understand the group theoretic structure of the coverings of $SL_2(\mathbb{R})$ and I would like some help.
More precisely why is the center of its universal cover is $\mathbb{Z}$. I know that the fundamental group of $SL_2(\mathbb{R})$ is isomorphic to $\mathbb{Z}$ so the universal cover has $\mathbb{Z}$ as a central subgroup. I think equality comes from the fact that $PSL_2(\mathbb{R})$ has the same universal cover but is centerless (cannot see why this implies equality though...).
I would also like to know and I think the answer of the first question should help, why is the center of the double cover of $SL_2(\mathbb{R})$ isomorphic to $\mathbb{Z}/4\mathbb{Z}$. I am not even sure I understand the definition of double cover.
Finally I would like to understand why this group is not linear (therefore gives an example of semisimple group with finite center that is not linear).
Thank you!
 A: Recall a fact from general group theory: $\phi(Z(G))\leq Z(\phi(G))$ for any group homomorphism $\phi\colon G\to H$.  That gives $Z(G)\leq \phi^{-1}(Z(\phi(G))$.  In the particular case of $\widetilde{SL_2(\mathbb{R})}\to PSL_2(\mathbb{R})$ this gives $Z\left(\widetilde{SL_2(\mathbb{R})}\right)\leq\ker\left(\widetilde{SL_2(\mathbb{R})}\to PSL_2(\mathbb{R})\right)$.
Moreover, it is a general fact that a discrete normal subgroup $N$ of a connected group $G$ is central (gist of proof: show that the centraliser of an element $h\in N$ contains a neighbourhood of the identity, hence equals to $G$.).  So $Z:=Z\left(\widetilde{SL_2(\mathbb{R})}\right)=\ker\left(\widetilde{SL_2(\mathbb{R})}\to PSL_2(\mathbb{R})\right)$.
Now it is easy to compute $\pi_1(PSL_2(\mathbb{R}))=\mathbb{Z}$.
The (connected) double cover of $SL_2(\mathbb{R})$ is what happens if you quotient by smaller subgroups of $Z$
$$
\widetilde{SL_2(\mathbb{R}))}\to\color{red}{\frac{\widetilde{SL_2(\mathbb{R}))}}{Z^4}
\to\frac{\widetilde{SL_2(\mathbb{R}))}}{Z^2}}=SL_2(\mathbb{R}).
$$
(There is of course a not-connected double cover $SL_2(\mathbb{R})\times C_2$ but that is boring.)
Now any finite-dimensional representation of $\widetilde{SL_2(\mathbb{R})}$ comes from lifting a representation of $\mathfrak{sl}_2(\mathbb{R})$ (general fact: representation of simply-connected Lie group and its Lie algera are the same thing.  For nonsimply-connected ones you have representation of the group inducing a representation of the Lie algebra but not necessarily conversely).  But the irreducible $\mathfrak{sl}_2(\mathbb{R})$-representations all come from $SL_2(\mathbb{R})$ by explicit verification (the $(n+1)$-dimensional irrep $x^iy^j\mapsto (ax+by)^i(cx+dy)^j$, $i+j=n$ obviously comes from $SL_2(\mathbb{R})$).  Therefore, all finite-dimensional representations of $\widetilde{SL_2(\mathbb{R})}$ factors through $SL_2(\mathbb{R})$ and thus is not faithful.  The (connected) double cover of $SL_2(\mathbb{R})$ is similarly not linear.
