From: Joerg Winkelmann Newsgroups: sci.math.research Subject: Re: Classification of Lie groups/algebraic groups Date: Sat, 27 Jun 1998 19:47:22 +0100 Greg Kuperberg wrote: > The only compact Lie groups which are not semisimple are tori and direct > products of tori and semisimple compact Lie groups. Almost. You have to allow for quotients of direct products T \times S by diagonally embedded finite central subgroups. In this case the group itself as not a direct product, although a finite covering is. > The semisimple real algebraic groups are all semisimple real Lie groups > but with some restrictions on the fundamental group. I'm not sure, > but I think the restriction may simply be that the remaining center > has to be finite. It is good that you are not sure :-) The precise criterion is the following: The inclusion of the real Lie algebra into its complexification yields a Lie group homomorphism between the simply connected semisimple real Lie group to the corresponding simply-connected complex semisimple Lie group. This map has a kernel, which is a discrete central subgroup of the simply-connected real Lie group. A quotient of the simply-connected real Lie group is algebraic if the discrete central group divided out contains at least this kernel. For instance, SL(2,R) is algebraic. Since its fundamental group is Z, there are finite coverings of SL(2,R) corresponding to the subgroups nZ of Z. These coverings are real semisimple Lie groups which are not real algebraic, but nevertheless have finite centers. Joerg Winkelmann ============================================================================== From: greg@matching.math.ucdavis.edu (Greg Kuperberg) Newsgroups: sci.math.research Subject: Re: Classification of Lie groups/algebraic groups Date: 27 Jun 1998 17:06:03 -0700 In article <3595302a.0@maser.urz.unibas.ch>, Joerg Winkelmann wrote: >Greg Kuperberg wrote: >> The only compact Lie groups which are not semisimple are tori and direct >> products of tori and semisimple compact Lie groups. >You have to allow for quotients of direct products T \times S >by diagonally embedded finite central subgroups. >In this case the group itself as not a direct product, although a >finite covering is. Oh right, I forgot. Spin^c(n), the non-trivial extension of a circle by Spin(n), is a lately popular example. This reminds me of one of my favorite math puzzles: How many Lie groups (up to isomorphism) are topologically a pair of circles? >> The semisimple real algebraic groups are all semisimple real Lie groups >> but with some restrictions on the fundamental group. I'm not sure, >> but I think the restriction may simply be that the remaining center >> has to be finite. >It is good that you are not sure :-) The precise criterion is >the following: The inclusion of the real Lie algebra into its >complexification yields a Lie group homomorphism between the simply >connected semisimple real Lie group to the corresponding simply-connected >complex semisimple Lie group. This map has a kernel, which is a >discrete central subgroup of the simply-connected real Lie group. >A quotient of the simply-connected real Lie group is algebraic if >the discrete central group divided out contains at least this kernel. In other words, saying that a real Lie group is algebraic means that you can complexify it, and you can only do that when its center is one occurs for a complex Lie group with the complexified Lie algebra. That makes sense. -- /\ Greg Kuperberg (UC Davis) / \ \ / Visit the xxx Math Archive Front at http://front.math.ucdavis.edu/ \/ * The tapestry of mathematics is woven with the free exchange of ideas *