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Let $F$ be an algebraically closed field and let $L$ be a simple Lie algebra of dimension $n$ over $F$. Let $ad: L\longrightarrow End_F(L)$ denote the adjoint representation of $L$. If $F$ has characteristic zero, then it is well-known that, for every linear transformation $f\in ad(L)$, the semisimple and nilpotent parts of $f$ are also contained in $ad(L)$ (see e.g. Section 5.4 in the Humphreys's book "Introduction to Lie algebras and representation theory").

In general, the same conclusion is not true if $F$ has characteristic $p>0$. Is it possible to have the extreme case in which $ad(L)$ does not contain any nonzero semisimple linear transformation of the vector space $L$?

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up vote 14 down vote accepted

Any finite dimensional simple Lie algebra over an algebraically closed field of characteristic $p>3$ contains a nonzero element $x$ such that $ad(x)$ is semisimple. This is a nontrivial fact, and the only proof I can think of relies on the classification. For $p>3$, the simple Lie algebras split into three families: Lie algebras of simple algebraic groups and their quotients, filtered Lie algebras of Cartan type, and the Melikian algebras (which only occur when $p=5$). The first class is easy to sort out, of course. The third class has a very explicit presentation which is easy to find in the literature. Then one can see straight away that each Melikian algebra $\mathcal{M}(m,n)$ admits two commuting linearly independent elements $t_1,t_2$ sich that $ad(t_i)^p=ad(t_i)$ for $i=1,2$.

The real issue with this problem (as well as with many similar problems) is the existence of filtered Lie algebras of Cartan type $H$ corresponding to Hamiltonian forms of the first kind. Such Hamiltonian forms involve parameters which are sometimes organised as sets of matrices in canonical Jordan form. Sadly, the commutator relations in the corresponding Lie algebras depend on these parameters as well, which makes computations an unpleasant experience.

However, the good news is that any finite dimensional simple Cartan type Lie algebra $L$ contains a unique maximal subalgebra of smallest codimension. It is called the $standard\ maximal\ subalgebra$ of $L$ and often denoted $L_{(0)}$. Skryabin proved in [Comm. Algebra, 23 (1995), 1403-1453] that under our assumptions on $p$ the subalgebra $ad(L_{(0)})$ of $\mathfrak{gl}(L)$ is closed under taking $p$-th powers; see Theorem 2.1 in $loc.\ cit$. By maximality, $L_{(0)}$ does not consist of $ad$-nilpotent elements. Since $ad(L_{(0)})$ is closed under taking $p$-th powers, it is straightforward to see that there is a nonzero element $x\in L_{(0)}$ such that $ad(x)$ is semisimple (in fact, one can say a lot more than that).

I have no idea as to what happens when $p$ is $2$ or $3$ as we have no classification in these characteristics. This adds to a long list of open problems of which my favourite is: does there exist a finite dimensional simple Lie algebra $L$ with a finite automorphism group. This never happens when $p>3$ and there is a rather short proof of this fact in the literature.

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@Sasha: It's good to get this kind of clear commentary from an expert participant in the recent work. (I've corrected a minor misprint in the last paragraph.) –  Jim Humphreys Jun 12 '12 at 21:11
    
@ Alexander Premet: thank you very much for your clear and satisfactory answer. –  Rocky Smith Jun 12 '12 at 23:08
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The question is ambiguous in prime characteristic, since the meaning of "semisimple element" isn't straightforward for a simple Lie algebra consisting of matrices. Possibly what's meant here is "semisimple as a matrix in the given realization of L". but there's no reason for this property to be intrinsic to the Lie algebra itself as it is in characteristic 0 (thanks indirectly to Weyl's theorem on complete reducibility). Maybe "semisimple" refers instead to the adjoint action? Anyway, it's ambiguous. Jordan decomposition works best when it propagates automatically to other matrix representations of the given Lie algebra.

On the other hand, Engel's theorem holds in arbitrary characteristic: if a Lie algebra consists of nilpotent (or ad-nilpotent) matrices, it must be nilpotent and therefore not "simple" when that's defined to exclude abelian Lie algebras.

But in general there's no reason to expect a simple Lie algebra of matrices to contain the semisimple and nilpotent parts of its elements. Moreover, even for restricted Lie algebras there is not yet a complete classification of the simple ones in characteristic 2 or 3. So the question asked needs to be more carefully focused to have a yes/no answer.

ADDED: Given the clarification and comment, maybe it would be helpful to comment briefly on relevant literature (which is technical but not too plentiful in this area). For the structure theory of modular Lie algebras, with emphasis on the restricted case, there is a monograph by G.B. Seligman (Springer, 1967), especially V.7, and related papers by D.J. Winter such as the one in Acta Math. (1969). Then there is a harder-to-find Dekker book by H. Strade and R. Farnsteiner (1988), especially Chapter 2; this book was at first typeset readably, according to one of the authors, but Dekker insisted on replacing it with a typed photocopy to fit their style. Later work on classification of simple Lie algebras (especially restricted ones) appears in papers by Block-Wilson and later Premet-Strade, who also wrote monographs. As elsewhere, this type of classification tries to exploit classical ideas involving "toral" subalgebras.

At any rate, in restricted Lie algebras which are simple and live over suitable fields, one eventually knows that Jordan decomposition makes sense and propagates under restricted homomorphisms, while nonzero toral subalgebras as well as nonzero nilpotent elements exist. The focus on these Lie algebras is due of course to the fact that the Lie algebra of linear algebraic group has a natural restricted structure when viewed as an algebra of derivations.

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Jim Humphreys: Thanks for your remark. I edited the questions and hope that it is clearer now. –  Rocky Smith Jun 1 '12 at 17:50
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Professor Humphreys: if $L$ is restricted and simple then the adjoint representation $L⟶ad(L)$ is a restricted isomorphism (the p-map on $ad(L)$ is the ordinary p-power of linear transformations). If $F$ is perfect then for any element $x\in L$ we have the Jordan-Chevalley decomposition $x=x_s+x_n$, where $x_s$ is a semisimple element and $x_n$ is a p-nilpotent element of $L$, and $[x_s,x_n]=0$. It is easy to see that $ad(x)=ad(x_s)+ad(x_n)$ is just the Jordan decomposition of the linear transformation $ad(x)$,so $ad(L)$ contains the semisimple and nilpotent parts of its elements in this case. –  Salvatore Siciliano Jun 1 '12 at 18:19
    
Of course, the nilpotent and semisimple parts of the elements of $ad(L)$ are contained in the restricted subalgebra $H$ generated by $ad(L)$ in $End_F(L)$. It is also clear that $H$ does contains semisimple elements, otherwise it would be nilpotent by Engel Theorem, which is a contradiction. But what about $ad(L)$? –  Rocky Smith Jun 1 '12 at 20:42
    
@unknown: I also expect the answer to your reformulated question to be No (at least for the known simple Lie algebras). Probably this is implicit in the methods used by Premet and Strade, which I'm not familiar enough with; it may be worthwhile to ask one of them directly. Working out the structure and classification is very difficult but also not yet well integrated into a wider context involving applications. As a result there are not yet many experts around. –  Jim Humphreys Jun 2 '12 at 14:10
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