Failure of Jacobson-Morozov in positive characteristics The Jacobson-Morozov theorem that any nilpotent element $e$ in the Lie algebra of a simple algebraic group $G$ can be embedded in an $\mathfrak{sl}_2$-triple, has a restriction (in terms of the Coxeter number) on the characteristic of the underlying field (assumed to be algebraically closed). This restriction is also required for the "uniqueness" of the triple, up to $C_G(e)$-action. (This result is due to Kostant.) In his 1980 paper, Pommerening had removed the restriction on the characteristic in Jacobson-Morozov's theorem, up to very small exceptions (i.e., characteristic is "bad"). Does the uniqueness as in Kostant's result also hold with this weaker restriction? If it does, then where does Jacobson-Morozov along with uniqueness result of Kostant fail in positive characteristics?
 A: Sorry to necro-post, but I have been meaning to point out for a while that I have a reasonably complete answer to the OP in some work with Adam Thomas, 'The Jacobson--Morozov Theorem and complete reducibility of Lie subalgebras' just made available in Proc. LMS, here. We were interested in $G$-complete reducibility of subalgebras of the Lie algebra $\mathfrak g$ of the reductive algebraic group $G$. A subalgebra is $G$-completely reducible if whenever it is in a parabolic subalgebra $Lie(P)$ of $\mathfrak g$ it is in a Levi subalgebra $Lie(L)$ of $Lie(P)$, a definition essentially due to Serre. It turned out that this definition matches up nicely with the Jacobson--Morozov theorem in that $G$-complete reducibility of all subalgebras guarantees you the uniqueness. We had to do a lot of case-by-case analysis, unfortunately, but we did get out the following:
Let $h(G)$ be the Coxeter number of $G$, $p$ the characteristic of the algebraically closed field $k$.


*

*Any nilpotent element $e$ can be extended to an $\mathfrak{sl}_2$ triple $(e,h,f)\in \mathfrak{g}\times \mathfrak{g}\times \mathfrak{g}$ uniquely up to simultaneous conjugation by $C_G(e)$, if and only if $p>h(G)$.


Let $b(G)$ be the largest prime amongst those for which the Dynkin diagram contains an $A_{p-1}$ subdiagram and bad primes for $G$.


*The number of nilpotent orbits and conjugacy classes of $\mathfrak{sl}_2$-subalgebras is the same if and only if $p>b(G)$. Moreover a bijection can be realised in a natural way by sending an $\mathfrak{sl}_2$-subalgebra to the nilpotent orbit of largest dimension meeting it.


The quantitative difference between the two results above essentially arises from the qualitative fact that when $p\leq h(G)$, then an $\mathfrak{sl}_2$-subalgebra can contain two non-conjugate nilpotent elements.
We also established, using the classification of nilpotent orbits, for the bad primes $p=3$ and $p=5$ in the exceptional groups, which nilpotent elements can be extended to $\mathfrak{sl}_2$-triples. All can except the exceptional orbit with label $A_2^{(3)}$ in $G_2$ when $p=3$.
A: Sasha has answered concisely the basic question here with a counterexample involving Lie type $A$, where all primes are good but need not be very good (meaning that $p$ should not divide $n$ for $\mathrm{SL}_n$).   
At the risk of being less concise, let me add some wider perspective to the question by asking why one wants a result like Jacobson-Morozov in the first place?   For a semisimple Lie or algebraic group over an algebraically closed field of characteristic 0, the Lie algebra reflects quite well a lot of the group structure; it is also semisimple and doesn't depend on the isogeny class of the group.    So the Lie algebra and the adjoint action of the group of it (or its adjoint action on itself) becomes a basic tool in further study of structure such as nilpotent orbits or in the study of representations, often simplifying matters.
In prime characteristic the classification of the groups (in the Borel-Chevalley theory) leads to much the same list as in characteristic 0, but the Lie algebras can behave badly: for some simple groups, the Lie algebra fails to be simple, and the structure of the Lie algebra can vary with isogeny type.   As noted in the question and in Sasha's example, attempts to imitate Jacobson-Morosov embedding (with suitable uniqueness up to conjugacy) run into problems in certain characteristics.   Often the embeddings into simple 3-dimensional subalgebras do exist, but under some restrictions on the prime or on the degree of nilpotency of the given nilpotent element: see for instance Carter's 1985 book, section 5.3.   
Chevalley himself largely abandoned Lie algebra methods in his classification seminar, but later work (for instance in Hogeweij's old Utrecht thesis) at least clarifies the precise structure of all the Lie algebras coming from simple algebraic groups.   And in representation theory, Jantzen presents in his book Representations of Algebraic Groups a scheme-theoretic substitute for the Lie algebra in the form of Frobenius kernels.    The classification of unipotent classes and nilpotent orbits (along with centralizers) has been developed carefully over the years in all characteristics: see the recent AMS monograph by Liebeck and Seitz.  Substitutes in the algebraic group setting for Jacobson-Morozov are worked out in the many papers by George McNinch and Donna Testerman, former students of Seitz.  Much can be done, though it gets more sophisticated and uses more algebraic geometry.
In good characteristic (excluding the primes 2, 3, 5 for some Lie types) the basic idea of Jacobson-Morosov survives in the treatment of Pommerening, which was done largely to make the Bala-Carter classification more uniform.   It's always worth asking what use is to be made of the Lie algebra in prime characteristic.  Certainly in characteristic 2 the idea of using $\mathfrak{sl}_2$ triples gets less interesting.    
A: The uniqueness can break down very badly in positive characteristic. Supose $G=SL_p$ where $p$ is the characteristic of the base field. Take a regular nilpotent element $e$ in $\mathfrak{g}=\mathfrak{sl}_p$. Then there is a nilpotent element $f\in\mathfrak{g}$ such that $e$, $f$ and $h=[e,f]$ form an $\mathfrak{sl}_2$-triple with the property that  $h^p=h$. Note that the identity matrix $I$ is in 
$\mathfrak{g}$. It is easy to see that there is $f_0\in\mathfrak{g}$ such that $[e,f_0]=I$ (many lecturers find this fact useful when explaining that Lie's theorem can fail in characteristic $p$). 
Let $\lambda$ be a scalar such that $\lambda^p\ne \lambda$.
Since $h$ commutes with $I$ and $ad\ h$ is semisimple, we may assume further that $[h,f_0]=-2f_0$.
Then  $(e,h+\lambda I, f+\lambda f_0)$ is another $\mathfrak{sl}_2$-triple containing $e$. If the spans  $\mathfrak{s}_1$ and $\mathfrak{s}_2$ of the   triples are conjugate under $G$, then restricting the $p$-dimensional vector representaion of 
$\mathfrak{sl}_p$ to $\mathfrak{s}_1$ and $\mathfrak{s}_2$  we would get equivalent representations of $\mathfrak{sl}_2$.
However, the representation we get from $\mathfrak{s}_1$ is restricted whereas the one we get from
$\mathfrak{s}_2$ is not. So the triples are not conjugate under $C_G(e)$.
One can replicate this example inside any Lie algebra of a reductive group $\widetilde{G}$ whch contains $G$ as a closed subgroup. 
