6
$\begingroup$

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...

$\endgroup$

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.