I believe that I have run across the statement that if $X$ is a compact smooth manifold and $CX$ is the cone on $X$, i.e. $[0,1] \times X$ modulo $(0,x)\sim(0,y)$ for all $x,y \in X$, then $CX$ admits the structure of a smooth manifold with boundary, with $\left\{1\right\} \times X$ smoothly immersed in it as the boundary, just when $X$ is diffeomorphic to a sphere in Euclidean space with its standard smooth structure. I would like a reference, or a reference to a similar statement.
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ if $CX$ is smooth, the intersection homology will egale to singular homology. Hence, all the Betti number of $X$ is 0 except the 0 and top degree. Thus...by Poincare conjecture, $X$ is a sphere. But I think there should be a simple argument. $\endgroup$– shuCommented Nov 1, 2013 at 14:54
-
$\begingroup$ @shu : The Poincare conjecture also requires $X$ to be simply-connected. $\endgroup$– SamCommented Nov 1, 2013 at 15:31
-
1$\begingroup$ @sam, you are right I forgot it. But is there a proof without using Poincare conjecture? $\endgroup$– shuCommented Nov 1, 2013 at 15:42
-
$\begingroup$ @shu : I doubt it. $\endgroup$– SamCommented Nov 1, 2013 at 15:43
-
1$\begingroup$ Without the Poincare conjecture you can say $X$ is a homotopy-sphere, that's what your argument gives you. $\endgroup$– Ryan BudneyCommented Nov 3, 2013 at 12:33
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
There is no need for smoothness here : if $X$ is a compact topological $n$-manifold and $CX$ is the cone on $X$, then $CX$ is a topological manifold if and only if $X$ is homeomorphic to a sphere. This is trivial for $n=1$, so assume that $n \geq 2$. The backward implication is trivial, so assume that $CX$ is a topological manifold. Let $p_0 \in CX$ be the cone point. The local homology groups $H_{k}(CX,CX-p_0)$ are then $\mathbb{Z}$ for $k=0,n+1$ and $0$ otherwise. Looking at the long exact sequence for the pair $(CX,CX-p_0)$, we then get that $H_{k}(X)$ is $\mathbb{Z}$ for $k=0,n$ and $0$ otherwise. In other words, $X$ is a homology $n$-sphere. Next, since $n+1 \geq 3$ and $CX$ is an $(n+1)$-manifold, the space $CX$ must satisfy the following condition : for all point $q \in CX$ and all neighborhoods $U$ of $q$, there must be a neighborhood $V$ of $q$ such that $V \subset U$ and $V-q$ is simply connected. Around the cone point $p_0$, it is easy to see that this condition implies that $X$ must be simply-connected. The Poincare conjecture thus implies that $X$ is homeomorphic to an $n$-sphere.
-
5$\begingroup$ Is it true that, under smoothness assupmtion, X is diffeomorphic to a sphere? (I'm not a specialist, but for instance, if I take the cone of an exotic 7-sphere, does it admit a smooth structure?) $\endgroup$ Commented Nov 1, 2013 at 16:29
-
6$\begingroup$ @user40911 Not in high dimensions: if $C\Sigma$ had a smooth structure for which $\Sigma \hookrightarrow C\Sigma$ were a smooth embedding, then removing a small ball from $C\Sigma$ we obtain an $h$-cobordism from $\Sigma$ to the sphere, so $\Sigma$ is not exotic. $\endgroup$ Commented Nov 1, 2013 at 17:29
-
10$\begingroup$ One the other hand, the double suspension of a homology sphere is homeomorphic to a sphere. So the cone of a non-manifold can be a manifold. See J. W. Cannon: Shrinking cell-like decompositions of manifolds. Codimension three, Ann. Math. 110 (1979), 83-112. R. D. Edwards: The double suspension of a certain homology 3-sphere is S^5, Notices AMS 22 (1975), A-334. for details. $\endgroup$ Commented Nov 2, 2013 at 1:15