The level sets of a differentiable function is a manifold

I have seen stronger propositions that imply this, but both their statement and their proofs require more advanced tools than I’d like to use in my text, which is aimed at a general scientific audience.

I want to prove: With $fcolon mathbb R^k to mathbb R$ smooth (infinitely differentiable) and $x$ a non-critical point (i.e. $nabla f(x) neq 0$), then the level set $f^{-1}(x)$ is a manifold. Preferrably providing a procedure to construct charts.

It seems that there should be a proof using undergraduate calculus only. Any ideas?

MathOverflow Asked by user8948 on January 31, 2021

1 Answers

One Answer

Proof of the implicit function theorem in several variables calculus requires the contraction mapping theorem, so is probably not suitable for your audience. You need to use an iterative method and take a limit. You can look for a complete proof in Spivak, Calculus on Manifolds.

If you just replace one of the coordinate functions by $f$, you get a chart, but the proof that it is a chart requires the implicit function theorem.

Answered by Ben McKay on January 31, 2021

Add your own answers!

Related Questions

Euler function summation

0  Asked on December 13, 2020 by andrej-leko


Ask a Question

Get help from others!

© 2022 All rights reserved.