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Why does the universal cover of $SL_{2}(\mathbb{R})$ have no finite dimensional representations?

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It hasn't? I could perhaps believe it has no faithful finite-dimensional representation, though. – Henning Makholm Apr 9 '12 at 13:52
up vote 13 down vote accepted

Here's a slightly different way of seeing that $\DeclareMathOperator{\SL}{SL(2,\mathbb{R})}\DeclareMathOperator{\USL}{\widetilde{SL}(2,\mathbb{R})}\DeclareMathOperator{\SLC}{SL(2,\mathbb{C})}\DeclareMathOperator{\sl}{\mathfrak{sl}(2,\mathbb{R})}\DeclareMathOperator{\slc}{\mathfrak{sl}(2,\mathbb{C})}\USL$ has no faithful finite-dimensional representations:

Start with a representation $\Phi$ of $\USL$ on a finite-dimensional real vector space. The associated representation $\phi=d\Phi(1)$ of $\sl$ can be complexified to yield a representation of $\slc$. Since $\SLC$ is simply connected, this corresponds to a representation $\Psi_{\mathbb{C}}$ of $\SLC$. Since $\SL$ is a subgroup of $\SLC$, we get a representation $\Psi$ of $\SL$. The covering projection $\pi:\USL \to \SL$ yields another representation $\Psi\circ\pi$ of $\USL$. It is straightforward to check that $\phi = \psi$ ($= d(\Psi\circ\pi)(1)$). Since $\USL$ is simply connected, this means that $\Phi$ and $\Psi\circ\pi$ are actually the same representation. This shows that $\Phi$ factors over $\pi$, and hence has non-trivial kernel.

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To show that $d\Phi(1) = d(\Psi \circ \pi)(1)$, is it just $d(\Psi \circ \pi)(1) = d\Psi(1)\circ d\pi(1) = d\Psi(1) = d\Phi(1)$ where the last equality is because of how $\Psi$ is defined? – h.j.s Apr 9 '12 at 23:39
Yes, exactly.${}$ – t.b. Apr 9 '12 at 23:41
By the way the same argument applies to every non trivial cover of $SL(2\mathbb{R})$, since morphisms of connected Lie groups are determined by their differential at the identity (even without simply connectedness). – Lor Apr 19 '13 at 16:06

I don't have a deep answer -- just a proof.

First, I assume you want to know why $\widetilde{SL}_2(\mathbb{R})$ has no finite-dimensional faithful representations, i.e. why it cannot be expressed as a matrix group.

The proof is that the Lie algebra of $\widetilde{SL}_2(\mathbb{R})$ is $\mathfrak{sl}_2(\mathbb{R})$, so any representation of $\widetilde{SL}_2(\mathbb{R})$ must induce a representation of $\mathfrak{sl}_2(\mathbb{R})$ at the tangent space to the identity. Moreover, since $\widetilde{SL}_2(\mathbb{R})$ is connected, the representation of $\widetilde{SL}_2(\mathbb{R})$ is entirely determined by the representation of $\mathfrak{sl}_2(\mathbb{R})$.

Now, the representations of $\mathfrak{sl}_2(\mathbb{R})$ have been completely classified, and it is easy to check that none of them induce a faithful representation of $\widetilde{SL}_2(\mathbb{R})$. So that's that.

On a more philosophical level, there's no reason to think that every Lie group would be a matrix group, so it's not surprising that $\widetilde{SL}_2(\mathbb{R})$ turns out not to be. This happens a lot for covers of matrix groups. For example, the universal covers of the symplectic groups are the metaplectic groups, which also have no faithful finite-dimensional representations.

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Instead of using the classification of representations you can complexify to a representation of $\mathfrak{sl}_2(\mathbb{C})$ and use that $SL_2(\mathbb{C})$ is simply connected to get a representation of $SL_2(\mathbb{C})$. This restricts to a representation of $SL_2(\mathbb{R})$ and you can precompose this new one with the covering projection of $\widetilde{SL}_2(\mathbb{R}) \to SL_2(\mathbb{R})$. Then you can check on the level of Lie algebras that nothing happened, that is every representation of $\widetilde{SL}_2(\mathbb{R})$ factors over a representation of $SL_2(\mathbb{R})$. – t.b. Apr 9 '12 at 14:02
@t.b. Yes, that is a nice way to see it. Why not post it as your own answer? – Jim Belk Apr 9 '12 at 14:06
Okay, done. I haven't checked right now, but I think similar arguments work in the other non-matrix groups you mentioned. – t.b. Apr 9 '12 at 14:25
It may be worth pointing out that every compact Lie group is a matrix group, but that the proof is nontrivial. – Jason DeVito Apr 9 '12 at 14:41

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