Wikipedia [1] says,

For $n \geq 2$, the universal cover of the special linear group SL($n, \mathbb{R}$) is not a matrix group (i.e. it has no faithful finite-dimensional representations).

What group is it, then, for example for $n=2$? Does it have a name? What index does SL($2, \mathbb{R}$) have in it?


3 Answers 3


The maximal compact subgroup of $SL(n,\mathbb{R})$ is $SO(n)$. Iwasawa decomposition tells us that as a smooth manifold, $SL(n,\mathbb{R}) \cong SO(n)\times \mathbb{E}^k$ (that second factor is some $k$-dimensional Euclidean space), so $$\pi_1(SL(n,\mathbb{R})) = \pi_1(SO(n)) =\begin{cases} \mathbb{Z}_2, & n\geq 3; \\ \mathbb{Z}, & n=2.\end{cases}$$
This implies that $\widetilde{SL}(n,\mathbb{R})$ is a two-sheeted cover for $n\geq 3$ and an infinite cyclic cover for $n = 2$.

As for what group it is, well, it's the universal cover of $SL(n,\mathbb{R})$. Since it's not a matrix group, you probably haven't encountered it before. You'll just have to take it on its own terms.

For an analogy, this is kind of like asking, "What number is the square root of two?" Well, it's $\sqrt{2}$. We just haven't met it before so the only way we know it is because $\sqrt{2}\sqrt{2} = 2.$ Same here: $\widetilde{SL}(n,\mathbb{R})$ is the group which is simply connected and covers $SL(n,\mathbb{R})$. That is its defining characteristic.

  • 1
    $\begingroup$ In the case $n=2$ I agree that there is no extra nomenclature available; $\tilde{SL}(2,\mathbb{R})$ is referred to everywhere I have seen it as "the universal cover of $SL(2,\mathbb{R})$". $\endgroup$
    – Lee Mosher
    May 4, 2013 at 14:37
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    $\begingroup$ However, when $n \ge 3$, we have an isomorphism $\tilde {SL}(n,\mathbb{R}) \approx \tilde{SO}(n,\mathbb{R}) \times \mathbb{E}^k$, the group $\tilde{SO}(n,\mathbb{R})$ is called the "spin group" en.wikipedia.org/wiki/Spin_group, and it is an integral part of quantum mechanics. $\endgroup$
    – Lee Mosher
    May 4, 2013 at 14:40
  • $\begingroup$ Thanks! I think $\tilde{SL}(2,\mathbb{R})$ is good enough name for me. I thought there might be, for example, some isomorphism like the one @Lee Mosher gave for $n \ge 3$. Or some important appearance of this group e.g. somewhere in quantum mechanics. $\endgroup$
    – The Vee
    May 4, 2013 at 15:23
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    $\begingroup$ @LeeMosher Do you mean that $\widetilde{SL}(n,\mathbb{R})$ is isomorphic as a group to $Spin(n)\times \mathbb{E}^k$? $\endgroup$
    – Neal
    May 5, 2013 at 1:21
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    $\begingroup$ What is the center of $\widetilde{SL}(2,\mathbb{R})$? can we obtain ${SL}(2,\mathbb{R})$ from $\widetilde{SL}(2,\mathbb{R})$ mod center? $\endgroup$ Sep 27, 2019 at 3:00

John Rawnsley's paper "On the universal covering group of the real symplectic group" (Journal of Geometry and Physics 62 (2012), 2044-2058) describes the universal cover $\tilde{Sp}(2n,{\mathbb R})$ of $Sp(2n,{\mathbb R})$ for any $n \geq 1$.


I'm learning this topic right now, so I'm not expert, but actually, we can say a lot about $\widetilde{SL}(2,\mathbb{R})$. I'll write about three points of view, one group theoretic, one differential, and one Riemannan.

  1. Group theory

First of all there is a short exact sequence of groups [see Scott, p. 464, for a reference about this discussion] $$0\to\mathbb{Z}\overset{i}{\to}\widetilde{SL}(2,\mathbb{R})\overset{p}{\to} PSL(2,\mathbb{R})\to0$$ and since $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm1\}$ is just a two-fold cover, i think we also have a s.e.s. $$0\to\mathbb{Z}\overset{i}{\to}\widetilde{SL}(2,\mathbb{R})\overset{p'}{\to} SL(2,\mathbb{R})\to0$$ which correspond to the multiplication $\mathbb{Z}\overset{\cdot 2}{\to}\mathbb{Z}$.

Anyways, let $G$ be either $SL(2,\mathbb{R})$ or $PSL(2,\mathbb{R})$, and by a little abuse of notation $p:\widetilde{SL}(2,\mathbb{R})\to G$ any of the two projections. These are central extensions (even though I don't know how to prove it), so there exist a function (of sets!) $$\phi:G\times G\to \mathbb{Z}$$ such that $\widetilde{SL}(2,\mathbb{R})$ is isomorphic to the set $\mathbb{Z}\times G$ endowed with the product law $$(a,f)\cdot_\phi(b,g):=(a+b+\phi(f,g),fg).$$ Of course there are too many possible functions $g\times G\to \mathbb{Z}$, and only some of them will give rise to $\widetilde{SL}(2,\mathbb{R})$, but all those that do it are characterised by the fact that, if we choose a section (of sets!) $s:G\to\widetilde{SL}(2,\mathbb{R})$ such that $p\circ s=id_G$, then $$i(\phi(f,g))=s(f)s(g)s(fg)^{-1}.$$ These last part is just general theory about abelian extensions [see for example Brown, IV.3, or this thread], but I don't know how write explicitly such a function $\phi$.

  1. Differentiable manifolds

Another interesting approach, which I think is related even though I don't know how [see these slides as a sloppy reference], is the following.

The KAN (or Iwasawa) decomposition of $SL(2,\mathbb{R})$ gives us the following isomorphism (of smooth manifolds, but not of groups since, for example, the only normal subgroup of $SL(2,\mathbb{R})$ is $\{\pm1\}$) $SO(2)\times AN(2,\mathbb{R})\cong SL(2,\mathbb{R})$ where $S^1\cong SO(2)=\Big\{\begin{pmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\\\end{pmatrix}:\theta\in\mathbb{R}\Big\}$ and $\mathbb{E}^2\cong AN(2,\mathbb{R})=\Big\{\begin{pmatrix}r & s \\ 0 & r^{-1}\\\end{pmatrix}:r,s\in\mathbb{R}\Big\}.$

It follow that, as an analitic manifold, it's universal cover is $$\widetilde{SL}(2,\mathbb{R})\cong \mathbb{R}\times AN(2,\mathbb{R})$$

  1. Riemannian geometry

I finish pointing out that one might be tempted to write $\widetilde{SL}(2,\mathbb{R})\cong \mathbb{R}\times \mathbb{E}^2$ which for example tells us that $SL(2,\mathbb{R})$ is aspherical, since its universal cover is contractible, but it is in fact even more interesting to write $\widetilde{SL}(2,\mathbb{R})\cong \mathbb{R}\times \mathbb{H}^2$, where $\widetilde{SL}(2,\mathbb{R})$ get the structure of a line bundle over $\mathbb{H}$ of constant curvature $1$. [for this Riemaniann point of view see Thurston, 3.8].

In fact $\widetilde{SL}(2,\mathbb{R})$ is one of the eight $3$-dimensional geometries.

A final note: I don't know if, and in case how, all this blabbering can be extended to the universal cover of $SL(n,\mathbb{R})$, and I'd be glad if anyone could fill the gaps and questions that I opened up.

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    $\begingroup$ I think I found some sources following the last activity on this question 6 years ago, but I forgot again. I'll see if they can be worth sharing for "filling the gaps" if I find them again, but they did help me back then specifically with the $n=2$ case. $\endgroup$
    – The Vee
    Dec 18, 2020 at 8:45
  • $\begingroup$ Minor correction: In the formula for $AN$, you should impose that $r>0$. (Otherwise, $- \mathrm{Id} \in SO(2) \cap AN$). $\endgroup$ Feb 22, 2022 at 19:00

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