Harold Williams, Pablo Solis, and I were chatting and the following question came up.
In Lie group land (where you're doing differential geometry), given a finite-dimensional Lie algebra g, you can find a faithful representation g → End(V) by Ado's theorem. Then you can take the group generated by the exponentiation of the image to get a Lie group G⊆GL(V) whose Lie algebra is g. I think this is correct, but please do tell me if there's a mistake.
This argument relies on the exponential map, which we don't have an the algebraic setting. Is there some other argument to show that any finite-dimensional Lie algebra g is the Lie algebra of some algebraic group (a closed subgroup of GL(V) cut out by polynomials)?
Best Answer
A Lie subalgebra of $\mathfrak{gl}(n,k)$ which is the Lie algebra of an algebraic subgroup of $GL(n,k)$ is called an algebraic subalgebra. Apparently there are Lie subalgebras which are not algebraic, even in characteristic zero. If $\mathfrak{g}$ is the Lie algebra of an affine algebraic group then it must be ad-algebraic, ie. its image in $\operatorname{End}(\mathfrak{g})$ under the adjoint representation must be an algebraic subalgebra. An example of a non-ad-algebraic Lie algebra is given on pg. 385 of Lie Algebras and Algebraic Groups, by Tauvel and Yu.