How to prove the real points of affine variety defined over the reals is a differentiable manifold?

146 Views Asked by At

Let $V$ be an affine algebraic set defined as a zero set of real polynomials. Then $V \cap \mathbb{R}^n \backslash V^*$ defines a differentiable manifold according to Wikipedia https://en.wikipedia.org/wiki/Differentiable_manifold. Here $V^*$ is the set of singular points. I have been taking this fact for granted, but realized not sure how to actually prove it. I would greatly appreciate comments and references.

PS I would greatly appreciate a reference for this also!

1

There are 1 best solutions below

0
On BEST ANSWER

The non-singular real points of $V$ form a Nash manifold (of dimension $\dim(V)$) which is a stronger notion. A reference is Proposition 3.3.11 of Bochnak, Coste, Roy, Real algebraic geometry. As mentioned in the comments, the argument is plugging the definition of non-singular points into the implicit function theorem.