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I read the result that every compact $n$-manifold is a compactification of $\mathbb{R}^n$.

Now, for surfaces, this seems clear: we take an n-gon, whose interior (i.e., everything in the n-gon except for the edges) is homeo. to $\mathbb{R}^n$, and then we identify the edges to end up with a surface that is closed and bounded.

We can do something similar with the $S^n$'s ; by using a "1-gon" (an n-disk), and identifying the boundary to a point. Or we can just use the stereo projection to show that $S^n-\{{\rm pt}\}\sim\mathbb{R}^n$; $S^n$ being compact (as the Alexandroff 1-pt. -compactification of $\mathbb{R}^n$, i.e., usual open sets + complements of compact subsets of $\mathbb{R}^n$). And then some messy work helps us show that $\mathbb{R}^n$ is densely embedded in $S^n$.

But I don't see how we can generalize this statement for any compact n-manifold. Can someone suggest any ideas?

Thanks in Advance.

  • 1
    Here's a guess that I don't know how to make work: put any Riemannian metric $g$ on $M$ and let $p\in M$ be arbitrary. Define $B = \{v \in T_pM | $ $exp_p(tv)$ is minimizing for $0\leq t < 1\}$. Then, one can show $B$ is diffeomorphic to a ball and $exp|_B$ is a chart. I'd guess that $exp(B)$ is dense in $M$, but I'm not sure how to prove it.2011-01-19
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    @Jason: Neat! That can be made to work without too much trouble.2011-01-21
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    @George: Would you mind showing how it can be made to work without too much trouble? Perhaps as an answer?2011-01-21
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    @Jason: Just did. It requires some basic facts about the cut locus.2011-01-22
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    You mean compact and *connected*, of course.2011-01-22

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