# Nash-Tognoli Theorem

The Nash-Tognoli theorem says that every closed, smooth manifold is diffeomorphic to a real algebraic variety.

Suppose I wanted to study the Ricci curvature of some class of manifolds.

Is there a "good reason" to look at manifolds as algebraic varieties?

I.e. is there an example where geometric or differential questions can be answered using algebra (by viewing the manifold as a variety) that cannot be (perhaps as easily) answered through "regular" geometry?

• I can't think of any advantage of studying a manifold as the zero set of a possibly large set of polynomials. There might be some specific questions that are more easily addressed using this approach, but I think most are not. It's somewhat analogous to efforts to study a Riemannian manifold by first embedding it isometrically in Euclidean space, using Nash's theorem. This rarely led to anything useful. It, however, was used by Schoen and others to be able to work with harmonic maps of low regularity. – Deane Yang Feb 27 at 21:00
• I thought the Nash-Tognoli theorem is only about manifolds without a Riemannian metric. Or can you realize any riemannian Manifold as the zero set of polynomials? – Michael Bächtold Feb 27 at 22:41
• Hm, it seems like, in Kahler geometry in particular, you can show that Kahler manifolds are complex algebraic varieties, and in fact, (maybe with some conditions) they are projective. I believe this is including their Kahler metrics and so on. This in turn gives people a way to construct, count, and deform them. So because we know Kahler manifolds with c_1 = 0 are Ricci flat, it's sort of like studying the curvature. – pictorexcrucia Mar 1 at 16:35