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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!

WrabbitW
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1 Answers1

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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.

user10354138
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  • Thank you for your answer. I am not sure I get the last argument for double cover representations. I feel that since it is not simply connected we cannot use the same argument with the Lie algebra? – WrabbitW Nov 03 '18 at 21:40
  • Recall every representation of a quotient $G/H$ lifts to a representation of $G$ by $\tilde{\rho}(g)=\rho(gH)$, so a representation of the double cover lifts to $\widetilde{SL_2(\mathbb{R})}$, but that representation factors through $SL_2(\mathbb{R})$ so you cannot get any (finite-dim) faithful representation. – user10354138 Nov 03 '18 at 21:47
  • Yes but lifting do not preserve faithfulness so I would not be able to conclude? – WrabbitW Nov 03 '18 at 23:55
  • No, the lifting shows that $\rho$ needs to factor through $SL_2(\mathbb{R})$ (because $\tilde{\rho}$ does), which you can conclude from universal property of quotient. – user10354138 Nov 04 '18 at 01:04
  • I think I know what bothered me. The theorem saying that $\mathfrak{sl}_2(\mathbb{R})$ is completely reducible do not only concern faithful representations but all of the finite dimensional ones. So if I have a representation of the double cover I lift it to the universal cover to get a representation of it. Now I know that it decomposes into irreducible factors and all of them come from a representation of $SL_2(\mathbb{R})$ (because it is the same as a representation of $\mathfrak{sl}_2(\mathbb{R})$) and therefore my representation could not be faithful to start with. – WrabbitW Nov 06 '18 at 16:22
  • Also when you wrote $\widetilde{SL_2(\mathbb{R})} \rightarrow \widetilde{SL_2(\mathbb{R})}/2\mathbb{Z}=SL_2(\mathbb{R})$ did you mean $\widetilde{PSL_2(\mathbb{R})} \rightarrow \widetilde{PSL_2(\mathbb{R})}/2\mathbb{Z}=SL_2(\mathbb{R})$ ? – WrabbitW Nov 07 '18 at 17:34
  • $\widetilde{PSL_2(\mathbb{R})}$ and $\widetilde{SL_2(\mathbb{R})}$ are the same thing, but no, we don't really want to work with $PSL_2(\mathbb{R})$ here (since we are only interested in $SL_2$ and its covers) once we know $Z$ is cyclic. – user10354138 Nov 07 '18 at 17:56
  • but then we would have $\widetilde{SL_2(\mathbb{R})}/2\mathbb{Z} \simeq SL_2(\mathbb{R}) \simeq \widetilde{SL_2(\mathbb{R})}/\mathbb{Z}$? Isn't $\widetilde{SL_2(\mathbb{R})}$ a double cover of $\widetilde{PSL_2(\mathbb{R})}$? – WrabbitW Nov 07 '18 at 18:10
  • No, $\widetilde{SL_2(\mathbb{R})}$ is both the universal cover of $SL_2(\mathbb{R})$ and of $PSL_2(\mathbb{R})$, since the universal cover is the unique (under mild assumptions which are satisfied here) path-connected simply-connected (base-point preserving) covering space (up to (covering) isomorphism) and the composition $\widetilde{SL_2(\mathbb{R})}\to SL_2(\mathbb{R})\to PSL_2(\mathbb{R})$ is a covering. Also, be careful when you write $\mathbb{Z}$ without specifying which copy of $\mathbb{Z}\subseteq Z$. – user10354138 Nov 07 '18 at 18:29
  • Okay I think I get it, $\widetilde{SL_2(\mathbb{R})}/Z \simeq PSL_2(\mathbb{R})$ and $\widetilde{SL_2(\mathbb{R})}/2Z \simeq SL_2(\mathbb{R})$. What confused me is that you usually see the following exact sequence $$1\rightarrow \mathbb{Z} \rightarrow \widetilde{SL_2(\mathbb{R})} \rightarrow SL_2(\mathbb{R}) \rightarrow 1$$ but in this case what is unsaid is $\mathbb{Z}≃2Z$ – WrabbitW Nov 08 '18 at 15:24