Failure of Jacobson Morozov in positive characteristics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:55:28Z http://mathoverflow.net/feeds/question/105781 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105781/failure-of-jacobson-morozov-in-positive-characteristics Failure of Jacobson Morozov in positive characteristics P-Samuel 2012-08-29T01:16:58Z 2012-08-30T19:30:41Z <p>The Jacobson-Morozov theorem that any nilpotent $e$ in the lie algebra of a simple algebraic group $G$ can be embedded in an $sl_2$-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 (<em>i.e.</em>,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?</p> http://mathoverflow.net/questions/105781/failure-of-jacobson-morozov-in-positive-characteristics/105935#105935 Answer by Alexander Premet for Failure of Jacobson Morozov in positive characteristics Alexander Premet 2012-08-30T12:24:15Z 2012-08-30T13:24:29Z <p>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. </p> http://mathoverflow.net/questions/105781/failure-of-jacobson-morozov-in-positive-characteristics/105985#105985 Answer by Jim Humphreys for Failure of Jacobson Morozov in positive characteristics Jim Humphreys 2012-08-30T19:20:28Z 2012-08-30T19:30:41Z <p>Sasha has answered concisely the basic question here with a counterexample involving Lie type <code>$A$</code>, where all primes are <em>good</em> but need not be <em>very good</em> (meaning that <code>$p$</code> should not divide <code>$n$</code> for <code>$\mathrm{SL}_n$</code>). </p> <p>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.</p> <p>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. </p> <p>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 <em>Representations of Algebraic Groups</em> 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.</p> <p>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 <code>$\mathfrak{sl}_2$</code> triples gets less interesting. </p>