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)?



Is every finite-dimensional Lie algebra the Lie algebra of an algebraic group?
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](http://books.google.com/books?id=ntKhAutD8I0C&pg=PA385&lpg=PA385&dq=Lie+Algebras+and+Algebraic+Groups,+by+Tauvel+and+Yu.&source=bl&ots=4V46-zVx1n&sig=Wiq7bS5rJocA-k09uJVYWXDdqTw&hl=en&ei=ZHbXSvmiG4HysgOx2cCLBg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA4Q6AEwAA#v=onepage&q=&f=false), by Tauvel and Yu. 


