0
$\begingroup$

This is a curiosity question.

$M$ and $N$ are smooth manifolds.

A closed curve $x:[0,1] \to M$ is a smooth curve such that $x(0) = x(1)$.

A smooth map $f : M \to N$ is an embedding if for all $p \in M$ differential $D_p f$ is injective and $M \cong f(M)$ in topological sense, where $f(M)$ has a subspace topology.

It seems that in $\mathbb{R}^2$ every compact hypersurface can be represented by an image of some closed plane curve. Moreover, I can say that image of every closed curve which is also an embedding (assuming flat torus topology on $[0,1]$ i. e. 0 = 1) is a compact hypersurface of $\mathbb{R}^2$.

Are there any similar characteristic of compact hypersurface in $\mathbb{R}^d$ for $d > 2$

I am not interested in purely topological characteristics like finite subcover for each cover, or existence of converging subsequence for every sequence.

I was thinking about something like hypersurface $H$ is compact if and only if every maximal geodesic in $H$ is a closed curve and an embedding.

1 Answers 1

1

So basically you are asking for a classification of compact hypersurfaces in $R^d$ which is a very broad and hard question. Now you have completely ignored the geometry and you are asking for non-topological characterization. This is indeed strange.

Consider your guess. This is even true for torus with flat metric, i.e. when you consider torus as a quotient of $R^2$. There are geodesics which are not closed.

Even with very specific geometry (i.e. some curvature condition on the metric) this is hard question. To give a perspective, the Poincare conjecture is classification of simply connected compact hypersurface without boundary in $R^4$.

  • 0
    I think I got your example. Flat torus has a non closed geodesic generated by curve in $\mathbb{R}^2$ and Flat torus is an hypersurface of $\mathbb{R}^3$ by Nash embedding theorem. Probably, the condition maximal geodesics can be reduced to having compact images in the hypersurface. Class of compact manifolds with boundary seems to diverse as it includes a closure for any bounded hyper surfaces which is very waste. I'm not sure about curvature and connectedness. Thanks for the answer.2017-01-29
  • 0
    Can this answer be understood as no?2017-01-29