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Suppose $G$ is a connected real Lie group. The quotient $G/Z(G)$ is the image of the adjoint representation, so a linear group. Is it known for which groups this quotient is Lie isomorphic to an algebraic group? Is there a classification?

Compact groups are all algebraic (Chevalley's theorem), so we can assume $G$ is not compact. If $G$ is nilpotent, this is always the case, because linear connected nilpotent Lie groups are a direct product of a torus and a simply connected group that is the image of a polynomial map (the exponential). Moreover, as remarked in the comments, $G/Z(G)$ is actually simply connected, when $G$ is nilpotent. If $G$ is semisimple, $G/Z(G)$ is a centerless direct product of simple groups. Some relevant discussion: Centreless semisimple Lie group that is not real algebraic
As noted in the comments, some non-compact simple Lie groups are algebraic (e.g. $\text{SL}_3(\mathbb{R})$) and some other ones are not (e.g. $\text{PSL}_2(\mathbb{R})$).

What about solvable groups? Do you know an example of a simply connected solvable group that is not a central extension of a linear algebraic group?

Luis
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    What about the group of $3 \times 3$ upper-triangular matrices with diagonal entries positive? This is a connected component of an algebraic group but I don't think is an algebraic group itself, and the same thing is true mod center. – Will Sawin Mar 09 '22 at 03:34
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    The image of $G/Z(G)$ in the automorphisms of the Lie algebra of $G$ is not always algebraic, so that proof strategy won't work, but the counterexample I have for that has $G / Z(G)$ itself algebraic: It's $ \mathbb R^3 \rtimes \mathbb R^2$, with the action of $\mathbb R^2$ on $\mathbb R^3$ by a rank two subtorus of the diagonal matrices defined by a non-algebraic equation. – Will Sawin Mar 09 '22 at 03:35
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    How about Allen Knutson's example https://mathoverflow.net/a/23594/297 : $\mathbb{R}^4 \rtimes \mathbb{R}$ with $\theta$ acting by $\left[ \begin{smallmatrix} \cos \theta & -\sin \theta && \ \sin \theta & \cos \theta && \ && \cos (a \theta) & - \sin(a \theta) \ && \sin(a \theta) & \cos (a \theta) \ \end{smallmatrix} \right]$ for irrational $a$? This is simply connected and, if I am not mistaken, centerless. – David E Speyer Mar 09 '22 at 03:48
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    Actually, even for semi-simple Lie groups this is false: the connected component of identity of SO(1,n) is not an algebraic group. – Venkataramana Mar 09 '22 at 05:21
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    The answer is negative for $\mathrm{SL}_2(\mathbf{R})$ itself: $\mathrm{PSL}_2(\mathbf{R})$ is not isomorphic to the group of $\mathbf{R}$-points of any algebraic $\mathbf{R}$-group. – YCor Mar 09 '22 at 11:19
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    @YCor: you are right. But is it not the case that $PSL_2(\mathbb R)$ the connected component of identity of $SO(1,2)$? – Venkataramana Mar 09 '22 at 11:23
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    @Venkataramana yes of course it's the connected component of $\mathrm{PGL}_2(\mathbf{R})$ (which is the group of real points of $\mathrm{PGL}_2$). In any case as you pointed out, the assertion "If $G$ is semisimple, this is always the case" of the OP is wrong. – YCor Mar 09 '22 at 11:24
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    @YCor: of course you are well aware of all this. Mine was a "tongue in cheek" comment. – Venkataramana Mar 09 '22 at 11:26
  • Thanks for all the examples – Luis Mar 09 '22 at 11:57
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    You should fix the question to take the counterexamples into account: to cover the semisimple groups, you need to relax the algebraicity definition (e.g., being the unit component in the Lie group of real points). Also for the nilpotent case, the point is that the quotient of every connected nilpotent Lie group (which may be non-linear) by its center is always simply connected. – YCor Mar 09 '22 at 13:50
  • @YCor About the simple case: is it a dichotomy compact/non-compact? – Luis Mar 09 '22 at 19:04
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    @Luis no: consider $\mathrm{SL}_3(\mathbf{R})$. – YCor Mar 09 '22 at 22:21
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    @YCor. Thanks. Since you suggested before to fix the question, I added these examples too. – Luis Mar 09 '22 at 22:58

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