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Let $\mathfrak{g} \subset \mathfrak{gl}_n$ be one of the classical real or complex semisimple Lie algebras. If $g \in \mathfrak{g}$, then $g$ has a Jordan decomposition $g = g_s + g_n$ with $g_s$ semisimple and $g_n$ nilpotent, and $[g_s,g_n]=0$.

The elements $g_s,g_n$, which a priori are just in $\mathfrak{gl}_n$, are both in $\mathfrak{g}$ again. There are various middle-brow general ways to see this (for one, use that $\mathfrak{g}$ is algebraic), but for concrete choices of $\mathfrak{g}$ it's basically elementary, as follows. One knows from the construction of the Jordan decomposition that $g_s,g_n$ are both polynomials in $g$ (different polynomials for different $g$, of course), and (EDIT) you can rig the construction so that these polynomials are odd. The Lie algebra $\mathfrak{g}$ is the subspace of $\mathfrak{gl}_n$ cut out by conditions like $\mathrm{trace}(g)=0$, or $Jg = -g^{t} J$ for some matrix $J$, and so forth. The condition $\mathrm{trace}(g)=0$ is always true for $g_n$, so it's true for $g_s$ if true for $g$. The condition $Jg=-g^t J$ is visibly true for odd $p(g)$ if true for $g$, so if true for $g$ then it's true for both $g_s$ and $g_n$. Thus $g_s$ and $g_n$ visibly satisfy whatever conditions $g$ is required to satisfy, and so are contained in $\mathfrak{g}$.

(This might seem lowbrow but in fact I think this is basically the idea of the proof that Fulton-Harris give for general semisimple Lie algebras.)

Now suppose instead that $G$ is a real or complex linear Lie group with Lie algebra $\mathfrak{g}$. This time the Jordan decomposition is $g = g_s g_u$ with $g_u$ unipotent, and indeed $g_s$ and $g_u$ are still in $G$. But if you try to make the same lowbrow argument as in the Lie algebra case, it appears to die horribly (a condition like $g^t = g^{-1}$ certainly need not be preserved by taking a polynomial in $g$). My question is, is there an elementary way to rescue it? (In particular, something other than just the general argument for algebraic groups.) Obviously you're fine for elements $g$ in the image of the exponential map, so the issue is passing to the whole group. A caveat is that I do $\textit{not}$ want to assume that $G$ is connected.

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Why do you prefer a case-by-case proof, specific moreover to the "classical" cases, instead of the straightforward proof in the general case (no semisimplicity hypotheses on the Lie algebra, etc.) as in Borel's book, which works over any algebraically closed field (and then any perfect field by Galois theory)? Recall that Borel's device is a description of $G$ as the stabilizer of a line in some representation of ${\rm{GL}}(V)$, which replaces the "description" with traces and bilinear forms? Trying to get it from the Lie algebra also makes no sense for disconnected $G$; consider finite $G$. – Boyarsky Jun 30 '10 at 13:01
Uniqueness of Jordan decompositions for endomorphisms means it is functorial. Can you use that to turn your conditions for classical groups into linear ones, e.g. looking at $\mathfrak g$ inside $\mathfrak{gl}(End(V))$ turns the condition $g^t = g^{-1}$ into $g(I) = I$, which then polynomials in $g$ will also satisfy. I guess this would really only be a spelling out of the general argument, but it would be explicit, which might be what you want? – Kevin McGerty Jun 30 '10 at 13:03
By the way, the case of finite $G$ is only interesting in positive characteristic, of course (as otherwise everything is semisimple). – Boyarsky Jun 30 '10 at 13:03
@Boyarsky ("Why do you prefer...?"): for a graduate course in Lie groups/Lie algebras where arguments invoking algebraic geometry would not be appropriate. [Though in fact, I think there's nothing wrong with seeing a "lowbrow" argument for some concrete groups before one does something more general -- you might understand the general case better if you can see how the argument is not so different from something you've already done by hand in specific cases -- as long as what you've does by hand is reasonably thoughtful.] – D. Savitt Jun 30 '10 at 13:29
@D.Savitt: Up to scaling conditions, all of the classical groups are characterized by preserving a line in a space of bilinear forms. So try to use Borel's argument directly in all of those cases, modulo the scaling aspects, and then take care of scalings via triviality of "algebraic characters" on unipotent points (which you can prove directly, since in characteristic zero) to take care of the trivial determinant condition when necessary. As for the trick for $g_u$, it seems that it could be unpleasant to prove by hand that the trick really works (e.g., $g_u$ commutes with $g$). – Boyarsky Jun 30 '10 at 14:12
up vote 2 down vote accepted

Give the proof in Humphreys' "Linear Algebraic Groups". It is essentially a context-free version of the argument you give, and hinges only on the fact that if $\rho_g$ is right-translation by $g$ in $k[\operatorname{GL}_n]$ and $I$ is the ideal defining $G$ in $\operatorname{GL}_n$, then $g \in G$ if and only if $\rho_g(I) \subset I$. It is a simple fact of linear algebra that the semisimple and unipotent parts of $\rho_g$ stabilize any subspace which $\rho_g$ itself stabilizes. The intuition, of course, is that $I$ consists of all the "equations" defining $G$ in $\operatorname{GL}_n$.

This may be the general argument you said you didn't want, in which case I think you should reconsider it as being just the right amount of generality on top of what you have done for Lie algebras.

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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 – Jim Humphreys Jun 30 '10 at 14:37
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. – Ryan Reich Jun 30 '10 at 15:04
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. – Jim Humphreys Jun 30 '10 at 15:55
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) – D. Savitt Jul 1 '10 at 15:56
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. – D. Savitt Jul 16 '10 at 2:36

The proof of the Jordan decomposition for algebraic groups over perfect fields has two parts:

(a) Linear algebra: An automorphism of a vector space has a unique multiplicative Jordan decomposition, which is compatible with maps and tensor products...

(b) Some baby Tannakian stuff.

Most proofs in the literature mix the two parts, making the proof seem more difficult than it is. If you accept the baby Tannakian stuff, which everyone should know anyway, one is left with some easy linear algebra (see, for example, I Section 9 of my online notes).

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