Let $G$ be a connected semisimple Lie group with trivial centre and $\mathfrak{g}$ its Lie algebra. The adjoint representation of $G$ defines an isomorphism of $G$ onto the connected component of the identity of $Aut(\mathfrak{g})$. The latter is a real algebraic group, but its identity component (in the Lie group topology) needs not be. However, in many cases $G$ is indeed isomorphic to a real algebraic group.

Is there an example of a connected semisimple Lie group $G$ with trivial centre which is not isomorphic (as a Lie group) to a real algebraic group?

Here it's really important that $G$ be centreless, otherwise it's easy to find counterexamples (such as the universal cover of $SL_2\mathbb{R}$).

The situation is complicated by the following fact. The real algebraic group $SO(n,1)$ is connected in the Zariski topology, while it has two components as a Lie group; thus, its identity component $G_n$ might be a good candidate for a counterexample, since it is centreless and doesn't inherit a real algebraic structure from $SO(n,1)$. However, at least for $n\leq 3$, $G_n$ happens to be isomorphic to a real algebraic group in an unrelated way (namely $\mathbb{R}$, $\mathbb{P}SL_2\mathbb{R}$ and $\mathbb{P}SL_2\mathbb{C}$ for $n=1,2,3,$ respectively).

There are several other questions around on the relationship between algebraic and Lie groups, but none of them seems to be relevant in this case...


1 Answer 1


You basically answered your own question: The connected component $G$ of $SO(2,1)$ is not real algebraic. As such, $G$ should be isomorphic to (the real points of) $PSL(2,\mathbb R)$. But it isn't. As algebraic groups $PSL(2,\mathbb R)$ and $PGL(2,\mathbb R)$ are isomorphic (in fact, the natural homorphism between them is an isomorphism over $\mathbb C$ and therefore over $\mathbb R$). But $PGL(2,\mathbb R)$ clearly has two components distinguished by the sign of the determinant.

  • $\begingroup$ I don't think I get what you mean... The identity component of SO(2,1) is isomorphic to SL_2(R)/{I,-I} as a Lie group. And since SL_2(R) is real algebraic, so is SL_2(R)/{I,-I}. It's a special case of 5.5.10 in Springer's book: one constructs an algebraic representation of SL_2(R) with Zariski-closed image H and kernel {I,-I}; the same representation gives a Lie group isomorphism between SL_2(R)/{I,-I} and the real algebraic group H. What am I missing? $\endgroup$ Jun 15, 2016 at 19:42
  • $\begingroup$ See the question mathoverflow.net/questions/16145/…, it might help to clear up some confusion. $\endgroup$
    – Guntram
    Jun 15, 2016 at 20:31
  • $\begingroup$ I think the problem is that 5.5.10 in Springer's book fails over R since it relies on the image of a morphism of algebraic groups being closed and one needs an algebraically closed field for this... $\endgroup$ Jun 16, 2016 at 11:17
  • 2
    $\begingroup$ You should not confuse a real algebraic group with its group of real points. For me it is helpful to think of a real algebraic group as a complex group plus extra structure namely complex conjugation. The real points are then just the fixed points. This way, it is perfectly possible for a homomorphism $G_1\to G_2$ to be surjective without the corresponding map of real points to be surjective. This is exactly what happens here: $SL(2,\mathbb C)\to PGL(2,\mathbb C)$ is surjective but $SL(2,\mathbb R)\to PGL(2,\mathbb R)$ is not. This phenomenon has been studied under the name Galois cohomology. $\endgroup$ Jun 16, 2016 at 11:30
  • $\begingroup$ I think I was using the terminology "real algebraic group" improperly then... I simply meant "a Zariski-closed subgroup of $GL_n\mathbb{R}$". So I guess the correct phrasing of my question is: "is there a connected semisimple Lie group with trivial centre which is not isomorphic to the group of real points of a real algebraic group?". Does this change things? $\endgroup$ Jun 16, 2016 at 14:18

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