5
$\begingroup$

Let $K$ be an arbitrary field and $\mathfrak{g}$ a finite-dimensional Lie $K$-algebra.

Let $\mathfrak{nil}_n\leq\mathfrak{sol}_n\leq\mathfrak{gl}_n$ be the Lie algebras of all ((strictly) upper-triangular) $n\!\times\!n$ matrices over $K$. By the Ado-Iwasawa theorem, $\mathfrak{g}$ admits an embedding into some $\mathfrak{gl}_n$, and by Engel's theorem $\mathfrak{g}$ is nilpotent iff it admits an embedding into some $\mathfrak{nil}_n$.

Does every finite-dimensional solvable Lie $K$-algebra admit an embedding into some $\mathfrak{sol}_n$?

If not, how about if we assume $K$ is algebraically-closed with positive characteristic? The Lie's theorem states that this holds true over algebraically-closed fields of zero characteristic.

$\endgroup$
3
  • 4
    $\begingroup$ See also here. Lie’s theorem indeed still holds in positive characteristic provided that $\dim (V)<char(K)$. $\endgroup$ Apr 7, 2015 at 8:14
  • 2
    $\begingroup$ In char. 0 every finite-dimensional solvable Lie $K$-algebra embeds into upper triangular matrices over some finite extension of $K$ (but not always over $K$, for instance the 3-dimensional Lie algebra of the group of isometries of the Euclidean plane is a counterexample); when it can be embedded into upper triangular matrices over $K$, it is often called $K$-triangulable. $\endgroup$
    – YCor
    Apr 7, 2015 at 10:10
  • 1
    $\begingroup$ PS: for any non-algebraically closed field $K$ there is a non-$K$-triangulable f.d. solvable Lie $K$-algebra: just take a non-$K$-trigonalizable $n\times n$ square matrix and consider the semidirect product $K^n\rtimes K$ defined by this matrix. In characteristic zero a solvable Lie algebra is triangulable iff all eigenvalues of its elements in the adjoint representation are in $K$. $\endgroup$
    – YCor
    Apr 7, 2015 at 10:13

2 Answers 2

7
$\begingroup$

The shaded question obviously has a negative answer for an arbitrary field $K$ (e.g., if the characteristic is 0 but the field fails to be algebraically closed). Maybe it's better to rewrite the question? In any case, it's essential to start with $K$ algebraically closed. Moreover, you might as well assume $K$ has prime characteristic $p$: Lie's Theorem already takes care of characteristic 0 in the algebraically closed case. But in prime characteristic the study of solvable Lie algebras (and their finite dimensional representations) gets extremely hard to organize, as one sees for example in papers of Strade and in the textbook by Strade-Farnsteiner. The old example of a solvable Lie algebra consisting of $p \times p$ matrices which can't be put in triangular form (due I think to Jacobson) illustrates the negative side quite well, since for example its derived algebra fails to be nilpotent.

On the positive side, if you start with a solvable Lie algebra $\mathfrak{g}$ embedded in some $\mathfrak{gl}(V)$ with $p > \dim V$, the usual proof of Lie Theorem's goes through and allows you to realize $\mathfrak{g}$ by upper triangular matrices relative to some basis of $V$. (This is an exercise in my old book on Lie algebras and has been known for a long time.) There are also some positive results in the literature for very special cases such as Lie algebras of Borel subgroups of algebraic groups.

In terms of building structure, classification, and representation theories in characteristic $p$, there has been little success in the study of solvable Lie algebras taken in isolation, especially those which are not the Lie algebras of algebraic groups. The study of simple Lie algebras (especially the restricted ones) and auxiliary solvable subalgebras has gone much further (in work of Block-Wilson, Premet-Strade in particular) for $p>3$, but nothing here is easy. Although Jacobson did establish the viability of the purely algebraic theory of Lie algebras in all characteristics, his work also showed some of the obstacles ahead.

$\endgroup$
6
$\begingroup$

This statement for a given field $K$ is equivalent to Lie-Kolchin for that field. Thus it's false over any field that has a counterexample, which includes all fields of positive characteristic (see Wikipedia).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.