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$$ \newcommand{\R}{\mathbb{R}} \newcommand{\geu}{g_{\text{Eu}}} \newcommand{\X}{\mathcal{X}} \newcommand{\iX}{\mathring{\X}}$$

Let $\X$ be a closed bounded convex set in some Euclidean space. Its relative interior $\iX$ is a non-compact bounded contractible smooth manifold; endow it with some Riemannian metric $g$.

Let $N: \iX \hookrightarrow \R^n$ be a Nash embedding, i.e. an isometric embedding of $\iX$ in a sufficiently large Euclidean Riemannian manifold $(\R^n, \geu)$ (which always exists). So $N(\iX)$ is a non-compact contractible Riemannian submanifold of $(\R^n, \geu)$, with $g = N^*\geu$.

Is it in general possible to find a compact submanifold $(\R^n, \geu)$ containing $N(\iX)$?

As an example, consider the following case:

$\X = \{x \in \R^2: x_1 + x_2 = 1, x_1 \geq 0, x_2 \geq 0\}$ is the standard simplex in $\R^2$; its relative interior $\iX$ is a segment without endpoints in $\R^2$, so a 1-dimensional non-compact bounded contractible smooth manifold.

Endow the positive orthant of $\R^2$ with the following metric

$$g(x) = \frac{1}{x_1} dx^1 \otimes dx^1 + \frac{1}{x_2} dx^2 \otimes dx^2$$ and let $\iX$ inherit this metric from the ambient space.

Consider the following diffeomorphism of the positive orthant onto itself:

$$N(x) = 2\sqrt{x}$$ It's easy to check that $N: (\iX,g) \hookrightarrow (\R^2, \geu)$ is a Nash embedding, and that the image of $\iX$ under $N$ is the portion of the 1-sphere $S^1$ lying in the positive orthant. Thus, the 1-sphere $S^1$ is a compact manifold that contains $N(\iX)$.

The metric used in this example is known as Shahshahani metric [1]; the Nash embedding is due to Akin [2].

[1] 1. Shahshahani, S. A New Mathematical Framework for the Study of Linkage and Selection. (American Mathematical Soc., 1979).

[2] 1. Akin, E. The differential geometry of population genetics and evolutionary games. in Mathematical and Statistical Developments of Evolutionary Theory 1–93 (Springer Netherlands, 1990).

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    $\begingroup$ Hint: Look for a 1-dimensional counter example. $\endgroup$
    – Moishe Kohan
    Commented Mar 15 at 15:59
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    $\begingroup$ It seems to me that your question is really a purely topological one. I don't see where the Riemannian metric plays a role. If the open submanifold can be extended to a compact submanifold, then the Riemannian metric induced by the Euclidean metric extends the Riemannian metric of the original open submanifold to the compact submanifold. $\endgroup$
    – Deane Yang
    Commented Mar 15 at 17:23
  • $\begingroup$ @DeaneYang I'd argue that the Riemannian metric plays a role in this way: the open submanifold $N(\mathring{\mathcal{X}})$ that needs be extended to a compact submanifold is the image of the isometric embedding $N$, that does depend on the original metric $g$ on $\mathring{\mathcal{X}}$. $\endgroup$
    – DavideL
    Commented Mar 18 at 11:31
  • $\begingroup$ If the metric didn't matter, one could always extend the initial $n$-dimensional open convex manifold $\mathring{\mathcal{X}}$ to the $(n+1)$-sphere $S^{n+1}$ (identify $\mathring{\mathcal{X}}$ with an open $n$-ball, and identify the ball with an open hemisphere of $S^{n+1}$. ) But nothing would guarantee that the metric $g$ on $\mathring{\mathcal{X}}$ extends to $S^{n+1}$, since $g$ is not induced by an ambient Euclidean metric. $\endgroup$
    – DavideL
    Commented Mar 18 at 11:31
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    $\begingroup$ First, the embedding can be extended to an embedding of a compact manifold only if the embedding is bounded. If the embedding is bounded, the question is whether there exists an extension of the open submanifold to a compact submanifold. That's a purely topological question. If such an extension exists, then the Riemannian metric induced on the compact submanifold is an extension of the Riemannian metric on the open subset. That step is automatic. $\endgroup$
    – Deane Yang
    Commented Mar 18 at 18:10

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