3
$\begingroup$

I am trying to reconcile the notions of algebraic groups, linear algebraic groups, Lie groups, and Lie algebras, along with their notions of root systems, maximal tori, etc. To begin, I am trying to draw a sort of Venn diagram relating algebraic groups, linear algebraic groups, and Lie groups. Certainly all linear algebraic groups are algebraic groups. I know that the universal cover of $SL(2, \mathbb{R})$ is an example of a Lie group which is not a linear algebraic group.

Are all linear algebraic groups Lie groups? I know that they can all be realized as subgroups of some $GL(n, k)$, which is a Lie group, but I believe that a subgroup of a Lie group must be closed to be a Lie group itself.

Thanks for the help!

$\endgroup$
  • $\begingroup$ They're always smooth, so I don't see an obstruction to turning them into manifolds really. But I'm mostly thinking over $\mathbf{C}$ so this might be misleading. $\endgroup$ – Hoot Jan 31 '15 at 18:57
  • 1
    $\begingroup$ Please, write down the definition of an algebraic group that you are using. Do you mean the set of real points of a group scheme defined over the real numbers? (One can talk about algebraic groups in relation to any field, say, p-adic numbers, which will make $Q_p$ and algebraic group which is not a Lie group.) $\endgroup$ – Moishe Kohan Jan 31 '15 at 19:10
  • 1
    $\begingroup$ Yes, they can be realized as Zariski closed subgroups of some $GL(n,k)$. Thus they are closed so also Lie groups. It's the other way around ( as you pointed out) that is not true. $\endgroup$ – Orest Bucicovschi Jan 31 '15 at 19:15
  • 1
    $\begingroup$ @Tarnation: No worries! Books on algebraic groups 1. Humphreys, 2. Springer 3. Algebraic Geometry IV: Linear Algebraic Groups, Invariant Theory 4. Lie Groups and Algebraic Groups by Onishchik & Vinberg $\endgroup$ – Orest Bucicovschi Jan 31 '15 at 19:39
  • 4
    $\begingroup$ It's worth noting that over fields of positive characteristic the notion of a Lie group doesn't make any sense. So any such linear algebraic group is an example of a linear algebraic group which is not a Lie group. $\endgroup$ – Jim Jan 31 '15 at 19:47
5
$\begingroup$

This is mentioned in the comments, but it is my belief that answers to questions should be in the answer section.

If you look at the Wikipedia article on linear algebraic groups you'll see they are defined to be (in different terms than they use, but better suited to the discussion) Zariski closed subgroups of the group of invertible $n\times n$ matrices. Over the real or complex numbers, this implies that they are closed in the usual topology on this Lie group. Being a closed subgroup of a Lie group, such a group is itself a Lie group. So all linear algebraic groups are Lie groups over a field for which the general linear group is a Lie group. In fields where the general linear group is not a Lie group (for example over $\mathbb Q$) a linear algebraic group need not be Lie group, so there is an example. However it still may be, for example if it has the discrete topology.

$\endgroup$
  • $\begingroup$ So if $F$ is an algebraic closure of the field $\Bbb{F}_p$ how do you give the group $SL_n(F)$ the structure of a differentiable manifold? $\endgroup$ – Jyrki Lahtonen Feb 3 '15 at 14:23
  • $\begingroup$ @Jyrki I specified this is over the real or complex numbers. $\endgroup$ – Matt Samuel Feb 3 '15 at 14:25
  • $\begingroup$ @Jyrki I put in a more complete answer. $\endgroup$ – Matt Samuel Feb 3 '15 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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