Let all Lie algebras have finite dimension, and let's first assume we are over $\mathbb C$. Then every semisimple Lie algebra $\mathfrak g$ has Borel subalgebras (defined e.g. as maximal solvable subalgebras), and by well-known conjugacy theorems, the isomorphism class of those is uniquely determined by that of $\mathfrak g$. Actually, for any given root space decomposition

$$\mathfrak g = \mathfrak h \oplus \bigoplus_{\alpha \in R} \mathfrak g_\alpha$$

and choice of positive roots, $$\displaystyle\mathfrak b := \mathfrak h \oplus \bigoplus_{\alpha \in R^+} \mathfrak g_\alpha$$ is such a Borel.

Now I wondered, which solvable Lie algebras occur as Borels of semisimple $\mathfrak g$? Of course you can say, I just answered that myself, because root systems / semisimple Lie algebras are classified, so if I go through the classification, I just wrote them all down.

But I am asking if there are some other criteria, in particular ones which are less obviously relying on the semisimple classification, which can identify the "possible Borels" among the solvable Lie algebras (whose general classification, I understand, is "wild").

As an example, all nilpotent Lie algebras are obviously excluded. Also, the above implies the following necessary criterion: For our solvable Lie algebra $L$, there has to exist a root system $R$ such that $\dim [L,L] = \frac12 \lvert R \rvert$ and $\mathrm{codim} [L,L] = rk(R)$. Which is sufficient for $\dim L \le 3$, but apparently already not in dimension $4$. Also, it still relies on our knowledge of root systems. But I am thinking vaguely in that direction, some combinatorial identities relating some invariants of $L$ (size of its nilradical in particular), which apparently must be fairly restrictive.

Another idea would be to look at the automorphism group of the Lie algebra in question, which I imagine might have a pretty "outstanding" structure for the "possible Borels".

As an upgrade of the question, of course we could look at fields different from $\mathbb C$; and, we can instead look at the analogous question for (Lie / algebraic) groups.

  • 1
    $\begingroup$ Somewhat related is this MO-question. We can first consider the nilradical of a Borel subalgebra and ask which Lie algebras can arise this way. $\endgroup$ Commented Jul 28, 2020 at 18:09
  • 1
    $\begingroup$ Related, but not related enough to be an answer. The question of how to decide which $p$-groups are Sylow $p$-subgroups of the finite simple groups is a tricky one. If one restricts oneself to so-called characteristic $p$-type simple groups(basically groups of Lie type in characteristic $p$) then one still can say very little completely abstractly. It was a lucky thing about CFSG that it didn't require too much $p$-group theory, or it probably still wouldn't be solved today. $\endgroup$ Commented Jul 28, 2020 at 22:28
  • 1
    $\begingroup$ A further remark from geometry. Milnor has asked which solvable Lie groups admit a left-invariant affine structure in "On Fundamental Groups of Complete Affinely Flat Manifolds". It turns out that in particular those solvable Lie groups do, whose Lie algebra arises as Borel of a semisimple Lie algebra. $\endgroup$ Commented Jul 29, 2020 at 10:53

1 Answer 1


A general comment: such Lie algebras have the form $\mathfrak{g}=\mathfrak{a}\ltimes\mathfrak{n}$, with the property that the central series is just $(\mathfrak{g},\mathfrak{n},\mathfrak{n},\mathfrak{n},\dots)$. This is already a specific property; this property of the lower central series, to stabilize in 1 step, is equivalently restated as: every nilpotent quotient is abelian.

Moreover it has another restrictive property, namely that there exists an element of $\mathfrak{a}$ that acts on $\mathfrak{n}$ with only positive (real) eigenvalues (this implies the previous property), and even positive integral eigenvalues, and in a diagonalizable way.

Many 3-dimensional solvable Lie algebras are not of this form.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .