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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.

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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 $PGL_2(\mathbf{R})$]. –  YCor Sep 19 '12 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 '12 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 '12 at 15:58
    
@Wildberd: So you say that the additional conclusion does not hold, yes? –  Petra Schwer Sep 19 '12 at 16:02
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@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. –  YCor Sep 19 '12 at 17:31
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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.

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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 '12 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 '12 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)? –  user28172 Feb 8 '13 at 14:52
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