0

Suppose $\mathfrak{g}$ is a Lie algebra, let $\mathfrak{g}^* $ be its dual. I need $\mathfrak{g}^* $ to be an affine algebraic variety. How is it one?

I suppose the only important thing is that $\mathfrak{g}^* $ is a vector space, and the additional structure doesn't matter. But even for vector spaces how I don't know how that works: I have read here that given a vector space $V$, one can construct an affine scheme by taking the spectrum of the symmetric algebra of its dual $V^* $. But in my case I have that $\mathfrak{g}^* $ itself is a scheme. Do I just fix a basis of $\mathfrak{g}^* $ so that $g^* $ is isomorphic to some $\mathbb{A}^n$ and then go from there?

toyr99
  • 41
  • Well, $k^n$ is both a vector space and an affine algebraic variety, so you could just "transport structure" by isomorphisms. But there is indeed also a basis-free way of doing it. – Zhen Lin Jun 11 '23 at 13:33
  • Are you referring to the construction of a scheme starting from a classic variety, where we associate to every open set $U$ the closed irreducible subsets which have non empty intersection with $U$? – toyr99 Jun 11 '23 at 13:40
  • For the purposes of this discussion, it is better not to think about schemes. Especially if your field $k$ is not algebraically closed. – Zhen Lin Jun 11 '23 at 13:54
  • I must think about schemes because I actually need to construct the jet scheme of $\mathfrak{g}^*$ – toyr99 Jun 11 '23 at 13:58
  • First you need to understand the classical version. Then you can generalise to the scheme (schematic...?) version. – Zhen Lin Jun 11 '23 at 13:59

0 Answers0