# What manifolds are boundaries of euclidian spaces ?

I would like to know if there are compact (n-1)-manifolds $N$ that are not spheres but such that there is a manifold with boundary $M$ which satisfies the following two properties:

• $\partial M\cong N$

• $M-\partial M\cong \mathbb{R}^n$

I am primarily interested in this question in the category of smooth manifolds but I would be interested to know the answer in the topological case as well.

-
I think this question was answered here: mathoverflow.net/questions/81714 –  Misha Feb 21 '13 at 19:42

$N$ has to be a homotopy-sphere. So as long as it's dimension isn't $4$, there's a proof that it has to be the standard $S^{n-1}$.

These arguments appear in the Kosinski book on smooth manifolds. The basic idea goes like this.

1) $N$ is simply connected. There's the inclusion map $N \to M$, but if $p \in int(M)$ then there's also a retraction map $M \setminus \{p\} \to N$. Provided $n \geq 3$ removal of a point does not affect the fundamental group.

2) $N$ is a homology sphere by Alexander/Poincare duality of the pair $(M,N)$. Part (1) technically gives us this as well but this argument works even if $M$ is not contractible.

So by the Whitehead theorem, $N$ is a homotopy sphere. Moreover, $M$ is a contractible manifold whose boundary is a homotopy sphere. So $M$ is a disc by the h-cobordism theorem, and the disc has a unique smooth structure (not known in dimension $5$ still).

So that resolves all cases except $dim(N)=4$.

-
Great answer, thanks ! –  Geoffroy Horel Feb 21 '13 at 22:50


Edit: I have slightly refined the argument below so that it goes through when the interior of the smooth manifold $M$ is only homeomorphic to $\RR^{n+1}$, not necessarily diffeomorphic. By the way, this weaker assumption makes no difference when $n\neq 3$, as the uniqueness (up to diffeomorphism) of differentiable structures on $\RR^{n+1}$ implies that the interior of $M$ is then actually diffeomorphic to $\RR^{n+1}$. So the easier, more elementary argument below assuming the interior of $M$ is diffeomorphic to $\RR^{n+1}$ is enough when $n\neq 3$.

The final answer is: if a compact smooth manifold $M$ has interior homeomorphic to $\RR^{n+1}$, then its boundary $\partial M$ is diffeomorphic to $S^n$, as long as $n>4$. In dimension $n=4$, we get only a homeomorphism between the boundary and $S^n$ by using Freedman's h-cobordism theorem instead of the smooth h-cobordism theorem in the argument below. We still get a diffeomorphism in dimension $n=3$, by the recently proved original Poincaré conjecture.

The argument goes as follows. It is simpler when the interior is actually diffeomorphic to $\RR^{n+1}$, so let us assume that at first. Let $M$ be a compact smooth manifold with interior diffeomorphic to $\RR^{n+1}$. Then removing the interior of the closed unit disc $D$ from $\RR^{n+1}$ (which we identify with the interior of $M$), we get a smooth cobordism $N=M\setminus(\int D)$ between $\partial M$ and $\partial D=S^n$ (the boundary of the removed disc).

On the one hand, the inclusion of $\partial D=S^n$ into $N$ is a homotopy equivalence, as $N$ is actually homotopy equivalent to $N\setminus \partial M=\RR^{n+1}\setminus(\int D)$ (since $\partial M$ has a collaring in $M$ and thus in $N$).

On the other hand, the inclusion of $\partial M$ into $N$ is also a homotopy equivalence. Again, note that $\partial M$ has a collar in $N$, and the complement of this collar is compact in the interior of $M$, i.e. in $\RR^{n+1}$. So by pushing off to infinity radially within $\RR^{n+1}\setminus(\int D)$ (via a suitable homotopy with bounded support starting at the identity of $\RR^{n+1}\setminus(\int D)$), we obtain a deformation of $N$ into the collar of $\partial M$ which fixes $\partial M$ pointwise. Then the collar of $\partial M$ deformation retracts onto $\partial M$, and these two deformations together give a deformation retraction of $N$ onto $\partial M$.

To modify the above argument when the interior of $M$ is only homeomorphic to $\RR^{n+1}$ via a homeomorphism $\varphi:\int M \to \RR^{n+1}$, make the following changes:

• $N=M\setminus(\int D)$ is now defined to be $M$ minus the interior of a closed disc $D$ smoothly embedded in the interior of $M$. So $N$ is still a smooth cobordism between $\partial M$ and $\partial D$.

• Importantly, note that the image $\varphi(\partial D)$ need not be a standard sphere in $\RR^{n+1}$. However, $\varphi(\partial D)$ is locally flat in $\RR^{n+1}$ (since $\partial D$ is locally flat in $\int M$), so the Shoenflies theorem implies that it is taken to a standard sphere in $\RR^{n+1}$ via some homeomorphism of $\RR^{n+1}$. By modifying $\varphi$, we can then assume that $\varphi(\partial D)$ is the standard sphere $S^n\subset \RR^{n+1}$.

• Since $\varphi(\partial D)=S^n$ is now the standard sphere in $\RR^{n+1}$, we can apply, essentially unchanged, the above two arguments showing that the inclusions of $\partial M$ and $\partial D$ into $N$ are deformation retracts.

In conclusion, $N$ is a smooth h-cobordism between $\partial M$ and $\partial D$, where $\partial D$ is diffeomorphic to the standard sphere $S^n$. Since $S^n$ is simply connected, the h-cobordism theorem implies that $\partial M$ is diffeomorphic to $S^n$ when $n>4$. As stated earlier, in dimension $n=4$, we only get a homeomorphism between $\partial M$ and $S^n$ by Freedman's h-cobordism theorem. In dimension $n=3$ we can use the Poincaré conjecture in dimension 3 (recently proved) to conclude that $\partial M$ is diffeomorphic to $S^3$, since those manifolds are homotopy equivalent to $N$ and thus to each other. In dimensions $n<3$, the classification of closed $2$-manifolds and $1$-manifolds shows we get a diffeomorphism again.

Finally, note that the elementary argument above (before the modification requiring Schoenflies theorem) showing that $N$ is a h-cobordism between $\partial M$ and $\partial D$ holds equally well for a compact topological manifold $M$ whose interior is homeomorphic to $\RR^{n+1}$. Then $\partial M\simeq N\simeq \partial D\cong S^n$. Since the Poincaré conjecture holds topologically in all dimensions, we conclude that the boundary of $M$ is homeomorphic to $S^n$.

-
Thanks for a great answer Ricardo. I have accepted Ryan's answer but it's nice to have details. –  Geoffroy Horel Feb 21 '13 at 22:52
@Geoffroy: No problem! –  Ricardo Andrade Feb 22 '13 at 1:13