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.

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    $\begingroup$ 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. $\endgroup$ Jan 19 '11 at 6:10
  • $\begingroup$ @Jason: Neat! That can be made to work without too much trouble. $\endgroup$ Jan 21 '11 at 18:05
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    $\begingroup$ @George: Would you mind showing how it can be made to work without too much trouble? Perhaps as an answer? $\endgroup$ Jan 21 '11 at 19:45
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    $\begingroup$ @Jason: Just did. It requires some basic facts about the cut locus. $\endgroup$ Jan 22 '11 at 13:19
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    $\begingroup$ You mean compact and connected, of course. $\endgroup$ Jan 22 '11 at 17:55

Assuming $M$ is a differentiable manifold, this can be shown by choosing normal coordinates about any point $P\in M$. The following is expanding on the method suggested by Jason DeVito in the comments.

Putting a metric on the manifold, we can define the exponential map $\exp_P\colon T_PM\to M$ and then, choosing an orthonormal basis $e_1,\ldots,e_n$ for $T_PM$, we get a map $$ \begin{align} f\colon\mathbb{R}^n&\to M,\\ (x^1,\ldots,x^n)&\mapsto\exp_P(x^ie_i) \end{align} $$ (using the summation convention). As the manifold is compact, this is an onto mapping (only completeness is required). The idea, then, is that there is a circled open subset $U\subseteq\mathbb{R}^n$ on which this becomes a coordinate map with dense image. That $U$ is circled (aka balanced) means that the line segment joining any point $x\in U$ to the origin is in $U$. It can be seen that any circled open set is diffeomorphic to $\mathbb{R}^n$ (also see star domain).

Let $D$ be the set of $x\in\mathbb{R}^n$ such that $[0,1]\to M$, $t\mapsto f(tx)$ is a minimizing geodesic. Equivalently, $D$ is the set of $x\in\mathbb{R}^n$ with $d(P,f(x))=\Vert x\Vert$, from which we see that it is closed. Also, let $U$ be the interior of $D$. The set $\partial D\equiv D\setminus U$ is called the cut locus (more precisely, the cut locus is the subset of $T_PM$ corresponding to $\partial D$).

Fixing some $x\in\mathbb{R}^n\setminus\{0\}$, let us consider the line $t\mapsto f(tx)$ for $t\ge0$. Define $r > 0$ to be the maximum real number with $rx\in D$. For any $0 < r_0 < r$, we can show the following.

  1. On $[0,r_0]$, $t\mapsto f(tx)$ is the unique minimizing geodesic joining $P$ to $f(r_0x)$.
  2. The derivative of $f$ at $r_0x$ is invertible, so $f$ is nonsingular at $r_0x$.
  3. $r_0x$ is in $U$.
  4. $f(D)=M$ and $f(\partial D)\subseteq M$ has zero measure.

Property (3) implies that $U$ is circled, so it is diffeomorphic to $\mathbb{R}^n$. Property (1) says that $f$ is one-to-one on $U$ and, by (2), it is a diffeomorphism. Then, by (4), $f(U)\supseteq M\setminus f(\partial D)$ is dense, so $f$ gives a diffeomorphism from $U$ to a dense subset of $M$.

I'll give proofs of these statements now, although they do seem quite standard. See also these notes and, in particular, Lemma 5.3 for the proofs of (1) and (2) and Lemma 5.4 for the proof of (4).

Proof of (1): Consider a minimizing geodesic $\gamma\colon[0,r_0]\to M$ joining P to $f(r_0x)$, which will have length no more than $r_0\Vert x\Vert$. Then, we can extend $\gamma(t)$ to $r_0\le t\le r$ by setting $\gamma(t)=f(tx)$. As this curve joins $P$ to $f(rx)$ and is of length no more than $r\Vert x\Vert$, it must be a geodesic. So, $\gamma(t)=f(tx)$ for all $t\le r$, and $t\mapsto f(tx)$ is a unique minimizing geodesic on $[0,r_0]$.

Proof of (2): Choose a nonzero $y\in\mathbb{R}^n$ and consider the vector field $t\mapsto Y_t$ given by $Y_t=t\nabla_y f(tx)=\frac{\partial}{\partial s}f(t(x+sy))\vert_{s=0}$. By geodesic deviation, this is a Jacobi field. Also, $Y_0=0$ so, if $\nabla_yf(tx)$ was zero, $P$ and $f(r_0x)$ would be conjugate points along $t\mapsto f(tx)$. Then, it is a standard result that $t\mapsto f(tx)$ is not a minimizing geodesic on $[0,r]$ for any $r > r_0$ (see characterization of the cut locus), contradicting the choice of $r$. So, $\nabla_yf(r_0x)\not=0$, and $f$ is nonsingular at $r_0x$.

Proof of (3): If not, then there would be a sequence $x_i\not\in D$ tending to $r_0x$. By the definition of $D$, there exist $y_i\in\mathbb{R}^n$ with $f(y_i)=f(x_i)$ and $\Vert y_i\Vert < \Vert x_i\Vert$. Passing to a subsequence, we can suppose that $y_i$ tends to a limit $y$. So, by continuity, $\Vert y\Vert\le r_0\Vert x\Vert$ and $f(y)=f(r_0x)$. By (1), this means that $y=r_0x$. But, then, setting $a_i=\Vert y_i-x_i\Vert$, (2) contradicts the limit $\nabla_{(y_i-x_i)/a_i}f(r_0x)\sim (f(y_i)-f(x_i))/a_i=0$.

Proof of (4): By completeness, for any $Q\in M$, there is at least one minimizing geodesic joining $P$ to $Q$. So, $f(x)=Q$ for some $x\in D$, and $f(D)=M$. Next, (3) implies that, for any $x\in\mathbb{R}^n\setminus\{0\}$, the radial line $t\mapsto tx$ ($t\ge0$) intersects $\partial D$ at a single point. This means that $\partial D$ has zero measure. As $f$ is locally Lipschitz, it maps zero measure sets to zero measure sets, so $f(\partial D)$ has zero measure.

Finally, I'll admit that the details are a bit trickier than I initially thought when I commented that the method can be made to work "without too much trouble".

  • $\begingroup$ Thanks for writing this - it's the third bullet point I couldn't see how to prove! $\endgroup$ Jan 22 '11 at 16:29

Since the question is tagged differential-topology, I assume we are talking about a differentiable manifolds.

One way to see that $M$ is a compactification of $\mathbb{R}^n$ is the following:

Take $f$ to be a Morse function on $M$ with one minimum.The flow of $\nabla f$ around the minimum will parametrize almost all of $M$ (the missing part will be the unstable manifold of critical points of smaller index).

  • $\begingroup$ @Noz: Do you have a reference or sketch proof why the flow parameterizes almost all of M? $\endgroup$ Jan 21 '11 at 18:10
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    $\begingroup$ Let $\phi_t(x)$ be the flow of $\nabla f$. My claim is that the set $X=\{x\in M\vert lim_{t\rightarrow -\infty}\phi_t(x)=m\}$ is open dense in $M$. In order to this remark that since $M$ is compact any point $x$ has a limit along the flow. However since there is only one local minimum, any point not in $x$ must converge to a critical point of index $>0$. It follows then by Morse lemma that those belong to cells of positive codimensions. This is actually the same idea of Thomas Rot, but the missing point before is that you can find a triangulation with only $1$ $n$-simplex. $\endgroup$
    – Noz
    Jan 21 '11 at 18:30
  • $\begingroup$ @Noz: Ah, yes. You also need a unique local minimum, rather than just a unique minimum. It seems obvious that there is a Morse function with a unique local minimum. Do you have a reference for this though? $\endgroup$ Jan 22 '11 at 3:25
  • $\begingroup$ @Georges Lowther: There are several ways to achieve this. But maybe a way related to your solution is the following: consider the square of the distance function from a point $p\in M$ associated to a Riemannian metric $g$. Then it has only one minimum (all other critical q point cannot be local min because nearby there are points on the geodesic connecting p to q). As critical points seems related to the cut loci of p, I wouldn't be surprise that my $X$ is your $f(U)$. One might be concerned that the function is not Morse, but small perturbation of $g$ will make it Morse. $\endgroup$
    – Noz
    Jan 22 '11 at 8:15
  • $\begingroup$ @Noz: Thanks. I am a bit concerned that making a small perturbation of $g$ will introduce local minima. Intuitively, it does seem that you should be able to smooth it in such a way as to not do this, but I'm not sure of the proof. (btw, it's George, not Georges. Adding the s means that I wasn't notified of your comment). $\endgroup$ Jan 22 '11 at 16:53

Here is the answer in the context of topological manifolds:

  1. For smooth manifolds you already have completely satisfactory answers.

  2. If a compact topological manifold $M$ has dimension $n\ne 4$ then it admits a handlebody decomposition (see the discussion and references here). Once you have a handlebody decomposition $M=H_1\cup_S H_2$, you use the fact that the complements to some $n-1$-dimensional disjoint disks $\Delta_{1,j}, \Delta_{2j}$ in the interiors of $H_1, H_2$ are homeomorphic to $R^n$. Now, find an open $n-1$-disk $D\subset S$ disjoint from the boundaries of the disks disks $\Delta_{1,j}, \Delta_{2j}$. Then the union $$ int(H_1)\cup D \cup int(H_2) $$ is homeomorphic to $R^n$. This union is clearly dense in $M$.

  3. There are 4-dimensional manifolds which do not admit a handle decomposition, but Frank Quinn in

Ends of Maps III: Dimensions 3 and 4, Journal of Differential Geometry vol. 17 (1982)

proved that all noncompact 4-dimensional manifolds are smoothable. Hence, removing a point from such a manifold results in a smoothable manifold which admits an open and dense subset homeomorphic to $R^4$.

The conclusion is then that every compact topological $n$-manifold is homeomorphic to a compactification of $R^n$.


My guess it that you could generalize your example of the n-gon to any triangulation (with higher simplices, instead of only triangles) of your manifold. Any manifold admits a triangulation.

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    $\begingroup$ Just a remark: There are topological manifolds that are not PL and so do not admit a triangulation. en.wikipedia.org/wiki/4-manifold $\endgroup$
    – Sam Nead
    Jan 19 '11 at 13:15
  • $\begingroup$ But if we restrict to the PL category then I believe that you are correct in saying the two-dimensional proof generalizes. $\endgroup$
    – Sam Nead
    Jan 19 '11 at 13:17
  • $\begingroup$ I usually think of a manifold as a smooth manifold, I am not comfortable with (counter) examples outside of this class. Thanks for the precision though! $\endgroup$
    – Thomas Rot
    Jan 19 '11 at 16:58
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    $\begingroup$ Indeed, if the manifold is triangulable, then the union of the interiors of the top dimensional simplices, together with the interiors of the codimension-$1$ simplices in a spanning tree of the adjacency graph with vertices the top dimensional simplices, should do it. $\endgroup$ Jan 22 '11 at 20:41

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