0
$\begingroup$

I was stuck on showing the following problem:

For $G$ a real connected solvable Lie group, the commutator group $[G,G]$ is nilpotent.

There are a few ways I thought about this problem:

Approach 1: By Ado's theorem, we can assume that the Lie algebra $\mathfrak{g}$ of $S$ is a matrix Lie algebra. Lie's theorem tells us there exists a complex vector space $V$ and a representation $\mathfrak{g} \rightarrow \mathfrak{gl}(V)$ with respect to which the elements of $\mathfrak{g}$ are upper triangular. It then follows that the derived algebra $[\mathfrak{g}, \mathfrak{g}]$ is nilpotent.

The part that makes me worry is that I am assuming that $[G,G]$ has $[\mathfrak{g}, \mathfrak{g}]$ its Lie algebra.

Approach 2: Someone suggested that I can think of $G$ as a subgroup of the real linear group in which case we do get that $\mathfrak{g}$ consists of upper triangular matrices. I am unsure why we can assume that $G$ is linear

$\endgroup$
1
$\begingroup$

Ado's theorem is hard, but it's much easier to use the adjoint representation instead. The argument shows that, denoting by $\mathfrak{z}$ the center of $\mathfrak{g}$ (= kernel of adjoint representation), the linear Lie algebra $[\mathfrak{g}/\mathfrak{z},\mathfrak{g}/\mathfrak{z}]$ is nilpotent. Since $[\mathfrak{g},\mathfrak{g}]$ is a central extension of $[\mathfrak{g}/\mathfrak{z},\mathfrak{g}/\mathfrak{z}]$ (by the central kernel $\mathfrak{z}\cap[\mathfrak{g},\mathfrak{g}]$), it follows that $[\mathfrak{g},\mathfrak{g}]$ is nilpotent.

This argument also applies directly to the connected Lie group, once you know that the kernel of its adjoint representation is central.

$\endgroup$
  • $\begingroup$ Do you think the Adjoint representation could be used to show that $G$ is a real linear matrix group ? $\endgroup$ – user135520 Mar 25 at 13:47
  • 1
    $\begingroup$ @user135520 first, not every connected Lie group has a faithful continuous finite-dimensional representation (what you call "being a real linear matrix group"). If you're asking for Lie algebras, Ado's theorem is notoriously hard to prove, so in general the answer is no. Still, it applies in many cases, notably when the center is trivial, which already encompasses the semisimple case, and in general it reduces to finding a linear representation that is faithful on the center, which in some cases is easy. $\endgroup$ – YCor Mar 25 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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