Welcome to MathOverflow

Question

2

Assume you have an almost connected simple Lie group G with trivial center. (In particular excluding non-algebraic examples such as the universal cover of SL_2(R).)

Such a group should automatically be an algebraic group over the reals resp. the complex numbers.

Is this true and why?

Can we in addition conclude (EDIT: under a good choice of the field and possibly additional assumptions?) that G is absolutely almost simple as an algebraic group?

EDIT: Asking this I do not want to regard a complex Lie group as a real algebraic group.

edited Sep 19 at 16:08

asked Sep 19 at 15:01

Petra Schwer
128●6

I guess that by "be an alg. group over reals" you mean "is topologically isomorphic to the group of real points of a real algebraic group" (or equivalently, to avoid algebraic groups language, is topologically isomorphic to a Zariski closed subgroup in some $SL_n(\mathbf{R})$). Then $PSL(2,\mathbf{R})$ is a counterexample, although it has index two in $PGL(2,\mathbf{R})$ [the notation is dangerous: if you define $PSL_2(\mathbf{R})$ as the group of real points of the group scheme $PSL_2$, you get $PGL_2(\mathbf{R})$]. – Yves Cornulier Sep 19 at 15:26

Also look at $PGL(2,{\bf C})$ viewed as a real Lie group. The complexification of its Lie algebra is not simple. – Wilberd van der Kallen Sep 19 at 15:40

@Yves Cornulier: can we resolve this problem in general by passing to a finite cover of the group under consideration? Hence maybe editing the question such that we ask for some finite cover of the Lie group to be always algebraic? – Petra Schwer Sep 19 at 15:58

@Wildberd: So you say that the additional conclusion does not hold, yes? – Petra Schwer Sep 19 at 16:02

1

@Petra: for many purposes, to have $G$ as a Zariski dense, open (for the ordinary topology) subgroup of finite index in the group of real points of some algebraic groups, is enough: you can invoke Zariski topology, etc. Otherwise if $G$ is connected for the ordinary topology, it indeed admits a finite covering, that consists of the real points of the simply connected covering (in the algebraic sense) of the corresponding algebraic group. – Yves Cornulier Sep 19 at 17:31

show 4 more comments

Vesselin Dimitrov · Accepted Answer · 2012-09-19 16:18:49Z UTC

4

The answer is yes for complex Lie groups, and follows from the classification. (Root data are in fact defined over $\mathbb{Z}$: a complex semisimple group has not only an underlying algebraic, but even an arithmetic structure). For a more direct explanation, see Theorem 6.3 in the book "Lie Groups and Lie Algebras III" by Onischik-Vinberg: any connected complex Lie group satisfying $G = [G,G]$ and admitting a faithful linear representation (which for semisimple groups is automatic), has a unique underlying complex algebraic structure.

For real Lie groups this is not quite true, as noted above.

edited Sep 19 at 16:18

answered Sep 19 at 15:56

Vesselin Dimitrov
1,046●6●16

	Thanks for the reference! I'll have a look at it. Do you know how to make a precise statement for the real case? – Petra Schwer Sep 19 at 16:09
	@Petra: You can find the same theorem in the book "Lie Groups and Algebraic Groups" by Onishchik and Vinberg, Section 3.3.4, Theorem 5 on page 125. – Mikhail Borovoi Sep 19 at 16:29
	@Petra: If $G$ is an adjoint connected semisimple $\mathbf{R}$-group and $G^{\rm{sc}} \rightarrow G$ is the simply connected central cover in the algebraic sense then $G^{\rm{sc}}(\mathbf{R})$ is connected (and a finite cover of the centerless $G(\mathbf{R})^0$). If we let $H$ vary through the connected semisimple $\mathbf{R}$-groups that are simply connected in the algebraic sense then $H \rightsquigarrow H(\mathbf{R})$ is fully faithful from the category of such $H$ to the category of (connected) Lie groups. Is this what you want to understand (or maybe Borovoi's reference explains it)? – nosr Feb 8 at 14:52

Are certain simple Lie groups linear algebraic groups?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.