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Sep 28, 2010 at 5:25 vote accept D. Savitt
Sep 28, 2010 at 5:25 comment added D. Savitt Accepting this answer because what Ryan suggested (and Jim encouraged) is what I will actually end up doing. I'm taking the discussion here as suggesting that the answer to the precise question that I asked is "no", even if no one has fully committed to that.
Jul 16, 2010 at 2:36 comment added D. Savitt Victor -- just noticed your comment. Here's connectedness for GL(n,C), to give you the flavor. By Jordan decomp, it's enough to see separately that semisimple elts and unipotent elts are in the conn comp of the identity. A unipotent element is of the form exp(n) where n is nilpotent and in the Lie algebra, so lies on the path exp(tn) through the identity. A semisimple element is conjugate to an element in (C^\times)^n, which is connected and contains the identity.
Jul 2, 2010 at 4:37 comment added Victor Protsak For connectedness in classical Lie groups, you can get away with polar decomposition (e.g. orthogonal by symmetric positively definite for $GL(n,\mathbb{R})$). That reduces everything to the case of compact classical groups, which are then done by hand. I don't see how Jordan decomposition would help with connectedness, actually (or connectedness with motivating nilpotent Lie algebras).
Jul 1, 2010 at 16:43 comment added Jim Humphreys It's hard to deal insightfully with such issues just for classical groups, except maybe connectedness. Anyway, Jordan decomposition can be used effectively in the structure theory of (complex) semisimple Lie algebras and is essential for semisimple algebraic groups in arbitrary characteristic. But far less so in traditional approaches to semisimple Lie groups. The intrinsic nature of the decomposition in $G$ relies mainly on the algebraic group structure, where Borel's general argument is hard to improve on in most special cases.
Jul 1, 2010 at 16:08 comment added D. Savitt and considering the semisimple and unipotent parts separately. (Then when you eventually come to the definition of a nilpotent Lie algebra, you've already seen that it's a helpful idea.) Before posting my question I'd convinced myself that the argument in Borel/Jim's book was just a bit more technical than I wanted at this stage, and that if there was a smart way of doing it by hand then that's what I'd prefer. (Leaving the general argument for later, not leaving it out!!) But perhaps I should give it another look -- or perhaps there's no choice, if the answer to my question is "no"!
Jul 1, 2010 at 15:56 comment added D. Savitt Hi guys, thanks for this. First, let me say what I'm up to. I've decided for pedagogical reasons that after discussing the foundations of Lie groups (a la Warner) and before launching into basic structure theory of Lie algebras, I'd like to spend a couple of days working concretely with classical groups. The point is to motivate some of the general theory by running up against concrete instances of issues that will be addressed in general later. In particular, if it turns out to be reasonable I'd like to address connectedness of these groups by applying Jordan decomposition (cont'd)
Jun 30, 2010 at 15:55 comment added Jim Humphreys The context seems to be real or complex "classical" linear Lie groups, where algebraic group techniques are often quite natural (as Borel's papers demonstrate) even if the heavier algebraic geometry needed for quotient spaces and such is avoided. The hybrid approach is important, whereas a thorough treatment of either Lie groups or algebraic groups is almost impossible in a graduate course; ditto for modern algebraic geometry. Specific problems and examples are essential for motivation, but using too many ad hoc arguments won't help students to go farther in any direction.
Jun 30, 2010 at 15:04 comment added Ryan Reich Is D. Savitt asking about general Lie groups? He writes "Lie group" but the title says "classical group" and he seems to be using groups defined by equations in $GL_n$. Either way, your/Borel's proof does give a more precise version of his "lowbrow" argument even if he does go on to talk about stabilizers of lines in spaces of bilinear forms.
Jun 30, 2010 at 14:37 comment added Jim Humphreys The arguments in my book are based on Borel's, which give a clear unified version for affine algebraic groups and their Lie algebras of older algebraic group proofs: Kolchin, Borel 1956, Chevalley seminar 1956-58 (Expose 4, written up by Grothendieck). Algebraic groups (not Lie groups) are treated, but not much affine algebraic geometry is used. On the plus side, this works over any algebraically closed field; but for real linear groups you have to adapt the results from the complex case. Like others I'm doubtful about what can be understood from special cases without general ideas
Jun 30, 2010 at 14:08 history answered Ryan Reich CC BY-SA 2.5