# Can one show the equivalence of the abstract and classical Jordan decompositions for simple Lie algebras without complete reducibility?

The following fact is basic in the theory of complex Lie algebras:

Theorem. Let ${\mathfrak g} \subset {\mathfrak gl}_n({\bf C})$ be a simple Lie algebra, and let $x \in {\mathfrak g}$. Let $x = x_s + x_n$ be the Jordan decomposition of $x$, thus $x_s, x_n \in {\mathfrak gl}_n({\bf C})$ are commuting elements that are semisimple and nilpotent respectively. Then $x_s,x_n$ also lie in ${\mathfrak g}$.

The standard proof of this theorem (e.g. Proposition C.17 of Fulton-Harris) proceeds, roughly speaking, by using complete reducibility of ${\mathfrak g}$-modules to reduce to the case when ${\bf C}^n$ is irreducible, at which point Schur's lemma may be applied. This fact allows one to show that the abstract Jordan decomposition for simple (or semisimple) Lie algebras is preserved under linear representations.

My (somewhat informal) question is whether the above theorem can be established without appeal to the complete reducibility of ${\mathfrak g}$-modules. I would also like to exclude any use of Casimir elements or the Weyl unitarian trick (since these give completely reducibility fairly quickly), and also wish to avoid using deeper structural facts about Lie algebras, such as the theory of Cartan subalgebras or root systems. I am happy to use the Killing form and to assume that it is non-degenerate on simple Lie algebras (and more generally, to assume Cartan's criteria for solvability or semisimplicity); I am also happy to use the closely related fact that all derivations on a simple Lie algebra are inner.

The best I can do without complete reducibility is to show that $x_s = x'_s + h$, $x_n = x'_n - h$, where $x'_s \in {\mathfrak g}$ is ad-semisimple, $x'_n \in {\mathfrak g}$ is nilpotent, and $h \in {\mathfrak gl}_n({\bf C})$ is nilpotent and commutes with every element of ${\mathfrak g}$; one can also show that $h$ vanishes when restricted to ${\mathfrak gl}(W)$ for every ${\mathfrak g}$-irreducible module $W$ of ${\bf C}^n$ (this is basically a Schur's lemma argument after observing that $h$ has trace zero on $W$), which shows that $h$ vanishes if one assumes complete reducibility. But it seems remarkably difficult to conclude the argument without complete reducibility.

Alternatively, I would be happy to see a proof of complete reducibility that did not use any of the other ingredients listed above (Casimirs, the unitary trick, or root systems). (My motivation for this is that I have been trying to arrange the foundational theory of finite-dimensional complex Lie algebras in a way that moves all the "elementary" theory to the front and the "advanced" structural theory to the back, at least according to my own subjective impressions of the elementary/advanced distinction. I had thought I had achieved this to my satisfaction in these notes, until someone recently pointed out a gap in my proof of the above theorem, which I have thus far been unable to bridge without using complete reducibility.)

• Do you consider Cartan's criterion elementary enough? I always found its proof an unmotivated (if short and very readable, by now) tour-de-force. – darij grinberg Sep 15 '13 at 1:31
• One can give a proof using just basic notions for linear algebraic groups over any field $k$ of char. 0 (no unitarian trick or root systems or serious structure theory lurking in the shadows). It isn't suitable for your purposes, but doesn't use anything about the theory of simple Lie algebras. Since char($k$)=0 and $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ by simplicity, Corollary 7.9 in Ch. II of Borel's book on algebraic groups implies $\mathfrak{g}={\rm{Lie}}(G)$ for a (smooth) Zariski-closed $k$-subgroup $G \subset {\rm{GL}}_n$. Use 4.4 in Ch. I (valid in any characteristic) to conclude. – Marguax Sep 15 '13 at 4:22

Here are a few extended comments. First, it's always desirable to re-examine basic material as more of it accumulates and makes the research frontier look impossibly remote to students. The handy book What Every Young Mathematician Should Know (But Didn't Learn in Kindergarten) gets harder to write every year.

It's never easy to say what ingenious new approaches are possible, but the question in the header has already been tackled (unsuccessfully) in a moderately paced textbook by Karin Erdmann and her former student Mark Wildon Introduction to Lie Algebras (Springer 2006, reprinted with some corrections in softcover format). The authors maintain a list of corrections to both versions, including a further correction to Theorem 9.16 on preservation of Jordan decomposition. They tried to avoid Weyl's theorem on complete reducibility but tripped over a hidden obstacle.

As this cautionary example indicates, it is tempting to simplify proofs but not always easy. The rigorous approach taken by Bourbaki (and Serre) to such matters is reliable though not always user-friendly. In any case, semisimple Lie algebras can be studied over an algebraically closed field of characeristic 0 using just some linear algebra and basic abstract algebra. It's not at all necessary, except for motivation, to deal with complex coefficients, Lie groups, or linear algebraic groups. Even so, there are sophisticated arguments including Cartan's criterion for solvability which seem hard to avoid.

Historically the classification of simple Lie algebras over a field such as $\mathbb{C}$ doesn't involve notions like Chevalley-Jordan decomposition or Weyl's complete reduciblity theorem. It's at least partly a matter of taste which approach to take, but Casimir operators and such do come up naturally if one gets into representation theory. My main concern about restructuring the foundations is that it should be done rigorously and in a way that doesn't force students to go back and start over again if they decide to pursue the subject further.

P.S. Some other relevant questions on MO can be found by searching "Jordan decomposition".

This is how I do this in my third year course on Lie algebras: Since we may assume that the Killing form $\kappa$ of $\mathfrak g$ is is non-degenerate, we can make use of the direct sum decomposition ${\rm End}({\mathfrak g})= {\rm ad}({\mathfrak g})\oplus M$, where $M=({\rm ad}({\mathfrak g}))^\perp$. The subspace $M$ has the property that $[{\rm ad}({\mathfrak g}),M]\subseteq M$, where the commutator brackets are taken in ${\rm End}({\mathfrak g})=\mathfrak{gl}({\mathfrak g})$. Hence we can write any $D\in{\rm Der}({\mathfrak g})\subset {\rm End}({\mathfrak g})$ as $D={\rm ad}\,x+m$ for some $x\in {\mathfrak g}$ and $m\in M$. For any $y\in{\mathfrak g}$ we then have $[D,{\rm ad}\,y]=[{\rm ad}\,x,{\rm ad}\,y]+[m,{\rm ad}\,y]$. Since $[D,{\rm ad}\,y]={\rm ad}(Dy)\in {\rm ad}({\mathfrak g})$ and $[m,{\rm ad}\,y]\in M$, it follows that $D={\rm ad}\,x$ (one should keep in mind here that the centre of $\mathfrak g$ is trivial). We thus conclude that all derivations of $\mathfrak g$ are inner, and this does not rely on Weyl's theorem on complete reducibility.

At this point one can use the fact that the semisimple and nilpotent parts, $D_s$ and $D_n$, of the Jordan decomposition of the endomorphism $D\in\mathfrak{gl}(\mathfrak{g})$ are again derivations of $\mathfrak g$. This is a nice (and elementary) exercise which can be found in Jacobson's book on Lie algebras, and one can replace $\mathfrak g$ by any finite dimensional algebra $A$, not necessarily associative or Lie. By the above, $D_s={\rm ad}\,x_s$ and $D_n={\rm ad}\,x_n$ for some $x_s,x_n\in {\mathfrak g}$ which are uniquely determined since $\mathfrak{z}(\mathfrak{g})=0$. The elements $x_s$ and $x_n$ commute as so do $D_s$ and $D_n$ in $\mathfrak{gl}(\mathfrak{g})$. The fact that the Jordan decomposition exists in $\mathfrak{gl}(V)$ has to be proven, of course.

• Let $x = x_s' + x_n'$ be the Jordan decomposition of $x$ in $\mathfrak{gl}(V)$. So by your argument, the restriction of $\operatorname{ad}_{\mathfrak{gl}(V)} x_s'$ to $\mathfrak{g}$ equals $\operatorname{ad}_{\mathfrak{g}} x_s$, where $D_s = \operatorname{ad} x_s$ as in your answer. So $x_s - x_s'$ lies in the centralizer of $\mathfrak{g}$ in $\mathfrak{gl}(V)$, but I do not see how we get $x_s' \in \mathfrak{g}$. – spin Aug 22 '18 at 7:27
• I am not sure if I am missing something, but in the first edition of the book by Karin Erdmann and Mark Wildon they tried to do something like in your answer to avoid Weyl's complete reducibility theorem, but their approach had some problems (see the answer by Jim Humphreys). – spin Aug 22 '18 at 7:30
• For us V=g and the Jordan decomposition of any element in gl(g) is unique. We know that for D=ad x the derivations D_s and D_n of g are inner (since Der(g)=ad(g)). This gives us commuting inner derivations ad(x_s) and ad(x_n) and hence unique commuting x_s,x_n in g such that x=x_s+x_n. This is all we need really. – Alexander Premet Aug 22 '18 at 10:51

My motivation for this is that I have been trying to arrange the foundational theory of finite-dimensional complex Lie algebras in a way that moves all the "elementary" theory to the front and the "advanced" structural theory to the back[…]

While the statement of the theorem is simple, the Chevalley-Jordan decomposition is very subtle, because it is not an infinitesimal structure—i.e. a structure that you can define just by using the Lie algebra—instead, it is a relique of “the” group whose Lie algebra is under consideration. (In your example, it is really the group, which is a subgroup of $\mathbf{GL}$.)

Let us look at the “true” statement of the Theorem:

Theorem'. Let $𝔤⊂\mathfrak{gl}_n(C)$ be a Lie algebra, and let $x∈𝔤$. Let $x=x_s+x_n$ be the Jordan decomposition of $x$, thus $x_s,x_n∈\mathfrak{gl}_n(C)$ are commuting elements that are semisimple and nilpotent respectively. If $\mathfrak{g}$ is the Lie algebra of a subgroup $G$ of $\mathbf{GL}$ then $x_s$ and $x_n$ are in $\mathfrak{g}$.

(You can replace $\mathbf{GL}$ by any group, of course.)

To see the importance of groups, just look at the following example: consider all algebraic groups whose Lie algebra is $C^n$. In the Lie algebra of the torus, each element is semi-simple, while in the Lie algebra of $C^n$ each element is nilpotent. If you look at intermediate fundamental groups, the Chevalley-Jordan decomposition is “mixed”.

The Lie algebra of the smallest subgroup $\Gamma$ of $\mathbf{GL}$ whose Lie algebra contains $x$ is $\mathfrak{c}^2(x_s) \oplus Cx_n$ (double centraliser). (This Lie algebra is the “algebraic hull” of $x$, according to Chevalley.) Showing this is enough to show the Theorem' since you have $\Gamma\subset G'$.

There is an “elementary” proof for Theorem' but it is quite intricated and uses deformations in the Grassmann variety, local diffeomorphisms (i.e. the exponential of the group) and the algorithm to compute Hermite normal form of a matrix (i.e. to compute limits in Grassmann varieties).

I think that Theorem' (or Theorem) is much more subtle than it looks, so the option that choosed Fulton and Harris, assume $G$ simple and take advantage of this, might be the most interesting for you. The proof with the “unitarian trick” is probably the fairest one, because it really sets emphasis on the group whose Lie algebra you study.

• I'm confused. If $G = \{ \begin{pmatrix} 1 & t & 0 \\ 0 & 1 & 0 \\ 0 & 0 & e^t \end{pmatrix} \} \subset GL_3$, then $G$ is a subgroup with Lie algebra $\mathbb{R} x$ for $x = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$. In this example, $x_s$ and $x_n$ are not in $\mathfrak{g}$ – DES-SupportsMonicaAndTransfolk Jun 3 '14 at 17:58
• @DavidSpeyer This is not an algebraic subgroup. (In my answer, a subgroup is meant to be an algebraic subgroup.) – Michael Le Barbier Grünewald Jun 3 '14 at 21:13
• Okay. So this is in fact not a relic of the group either, it is a relic of algebraic structure. – DES-SupportsMonicaAndTransfolk Jun 4 '14 at 0:25
• @DavidSpeyer In the context of the book by Fulton & Harris a group is an algebraic group, so I did not emphasize this, but yes it a relic of the structure of algebraic group. – Michael Le Barbier Grünewald Jun 4 '14 at 5:32

I doubt this will be suitable for you, but I think the "right proof" is:

*) first prove abstract Jordan decomposition for a pro-finite dimensional algebra (this is easy, since the commutant of the left action is the right action and vice-versa).

*) thus, $x_s$ and $x_n$ are well-defined as elements of the universal enveloping algebra completed at the kernel of all finite dimensional representations.

*) check that they are primitive elements; this is just the fact that for any $V\otimes W$ and operators $a\colon V\to V$ and $b\colon W\to W$, the semi-simple and nilpotent parts of $a\otimes 1+1\otimes b$ are $a_s\otimes 1+1\otimes b_s$ and $a_n\otimes 1+1\otimes b_n$.

It may be there's a way to "down-blend" this proof that I'm not seeing. The idea that the compatibility with tensor product forces the Jordan decomposition to lie in the Lie algebra is a simple one, but I'm not seeing it at the moment.

• This looks like a very appealing argument, but I wonder how the hypothesis that ${\mathfrak g}$ is simple (or semisimple) comes into this, since of course the theorem is false in general (e.g. for one-dimensional Lie algebras). – Terry Tao Sep 15 '13 at 15:10
• Well for a one-dimensional Lie algebra, the universal enveloping algebra is $\mathbb C[x]$, the completion is $\prod_{\lambda \in \mathbb C} \mathbb C[[x-\lambda]]$. $x_s$ and $x_n$ are well-defined in this algebra, but primitivity fails. I think the primitivity criterion works for elements of the universal enveloping algebra but can fail for its completion? Then the semisimplicity has to be used in showing that the Jordan decomposition lies in the universal enveloping algebra, not just its completion. – Will Sawin Sep 15 '13 at 18:18