I was thinking how to embed a Riemannian manifold in the Euclidean space. I had an idea, then I found the Nash embedding theorem but I was expecting something different, this is what I thought:
Because the invariant (in $2$ dimension) is $$ds^2=(a)dx^2+(b)dy^2$$ where $(a)$ and $(b)$ are space functions, we can describe this flat 2 dimensional space in 3d space:
$$f(x,y)=z, \ \ \text{ where }dz/dx=a^{1/2} \text{ and } dz/dy=b^{1/2},$$
so translating in words and if I am correct:
A particle who moves experiences an elongation due to the curvature which is $a^{1/2}$ if is moving on $x$ axis.
So I believed that for every $2$ dimensions we needed an extra dimensions which codified the curvature but the Nash embedding theorem says we need:
- $n$ (with $n \le m(3m+11)/2$ if $M$ is a compact manifold, or
- $n\le m(m+1)(3m+11)/2$ if $M$ is a non-compact manifold.
instead of something like $n!$. can somebody tell me why my idea was wrong?
It's not a question about Nash's theorem, is about my idea.