I am trying to see why the Chevalley groups (not limited to the adjoint group) over $\mathbb R$ are without compact factors in order to use the Borel density theorem.
I've been told in another thread that $G(\mathbb R)$ splits over $\mathbb Q$ and so over $\mathbb R$, but I could not understand that from the construction of the group, why is that.
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3$\begingroup$ When hyou write "Chevalley group" it is helpful to indicate whether it goes back to the 1955 Chevaley construction over an arbitrary field (of the adjoint group) or to the Yale lectures by Steinberg in 1967-68. Simplicity over the real field suggests the first source. In either case the group is split over the rational or real numbers, as advertised. $\endgroup$– Jim HumphreysCommented Jul 6, 2019 at 1:09
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$\begingroup$ I edited my question: I did mean the Chevalley groups as Steinberg lectured about and as you said the simplicity only apply to the adjoint ones. I just have no clue how to show the split. $\endgroup$– AmiCommented Jul 6, 2019 at 1:46
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1$\begingroup$ Note that the main thing one gets from Steinberg's generality is that, for example, SL$_n$ becomes a "Chevalley group" (not just the adjoint group PGL$_n$). Both are split over the field $\mathbb{Q}$, but the simplicity gets weakened to having finitely many central subgroups (themselves finite groups). $\endgroup$– Jim HumphreysCommented Jul 6, 2019 at 11:25
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There are probably multiple ways to see that $G(\mathbb{R})$ is non-compact when $G$ is a Chevalley group (in either the narrow sense of Chevalley or the broader sense of Steinberg's lectures). One way is sketched in the next paragraph. I did try to find a clear statement of this in the sources, but couldn't find an expicit formulation.
Anyway, one argument is that the Chevalley group is generated by copies of the additive group of the field (in the form of root groups relative to a maximal algebraic torus). These root groups are closed, hence would be compact in the usual topology (which they obviously aren't) if the real points of the group were compact. Since the group of points of a Chevalley group over a field like $\mathbb{R}$ is almost-simple, this shows that the Lie group does not have compact factors. Of course, one has to start with the Zariski topology and compare with the finer usual topology.
There are not many sources for the comparison of real semisimple Lie and algebraic groups, though at the end of his life Borel was quite interested in explaining this connection. See his lectures for Hong Kong, which he couldn't deliver in person: here.
See also the standard Russian 1988 text (published by Springer in an English translation in 1990), Lie Groups and Algebraic Groups by Onishchik and Vinberg, in particular, Section 5.2. See also the short Chapter 5 in Steinberg's 1967-68 Yale Lectures on Chevalley Groups, republished in 2017 in typeset and edited form by AMS (part of a series on sale this month). The Onishchik-Vinberg book is unusual in that much of the theory is developed in a series of "problems" with solutions sketched.
answered Jul 6, 2019 at 19:19
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$\begingroup$ Maybe it'll be simpler to prove that the sublattices of $G(\mathbb Z)$ are Zariski dense in $G(\mathbb R)$ without using the Borel density theorem? $\endgroup$– AmiCommented Jul 7, 2019 at 20:02
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$\begingroup$ @Ami: Correct, but it does depend on what tools you already have. $\endgroup$ Commented Jul 8, 2019 at 13:28
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$\begingroup$ @YCor: Thanks for the help with proofreading! I don't always put enough effort into this (for example, my "expicit" near the end of the first paragraph). Macular degeneration in my eyes has been kept under control for years by regular Avastin injections, but in the past year I've had to start on Eliquis (a new blood thinner) due to afib, which counteracts that improvement in vision. $\endgroup$ Commented Jul 8, 2019 at 13:32