Suppose that $S$ and $T$ are two smooth manifolds and '$ \Re$' be the reals with the normal manifold structure. And here I use '$=$' to mean diffeomorphism. Is the statement below true?

$ S \times \Re = T \times \Re \Rightarrow S = T$.

What if '$=$' meant homeomorphism?

What if $S$ and $T$ are compact?

PS: Bartnik's splitting theorem states that if a space-time has a COMPACT cauchy hypersurface and is geodesically complete and has non negative ricci curvature every where then it can be split isometrically as a 3-manifold with a riemannian metric and $ \Re$ with the metric $-d^2t$.

It has been shown that such space-times can be splitt smoothly, so if there is a unique splitting then one can reduce the problem to something like:

if such and such conditions hold can the section $s:S \times \Re \longrightarrow E(S) \oplus E(\Re)$ be written as $s1 \oplus s2$ where $s1: S \longrightarrow E(S)$ and $s2: \Re \longrightarrow E(\Re)$.

i know that it really won't get transfered to this exactly. for example $f(x,y)dx + g(x,y)dy$ can be written in the form $F(U)dU + g(V)dV$ iff $ \frac{\partial{f}}{\partial y } = \frac{\partial{g}}{\partial x }$.

(i know that this is a very crude reasoning. for example i don't know for sure whether to reduce the problem to this do i really need it to know whether there is more than 1 splitting or not. or on the other hand if i reduce it to the above, is that any kind of improvement at all or not)

(any opinion on my whole approach is more than welcome.)

that is how i came across this question.


Just to write out Ryan's answer: Let $S$ be the sphere with three closed disks removed. Let $T$ be the torus with one closed disk removed. Note that $T$ is non-planar. Thus $S$ is not homeomorphic to $T$. $\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\cross}{\times}$

On the other hand, let $S' = S \cross \RR$ and let $T' = T \cross \RR$. Then $S'$ and $T'$ are both diffeomorphic to the open, genus two handlebody. Thus $S'$ and $T'$ are diffeomorphic to each other.

I agree that this is homework, but it is good homework! What if $S$ and $T$ are required to be compact?

  • $\begingroup$ sorry if it looks like a homework. i really came across this when thinking about an open problem. I think this shows that I am weak at manifold theory a little bit. I think it is a good idea to add a post script and give some detail about the problem. $\endgroup$ – Ramin Asadi Mar 29 '13 at 21:36
  • $\begingroup$ now if you read the PS you'll notice that compactness is given so ... but if that is a homework too just tell me so, i'll try to figure it out myself. $\endgroup$ – Ramin Asadi Mar 29 '13 at 22:04

No. (Whitehead manifold) $\times \mathbb{R}$ is homeomorphic to $\mathbb{R}^4$. See Rolfsen's "Knots and Links", for example. There are uncountably many other examples in a similar vein.

  • $\begingroup$ Oops - I see Martin beat me to it by seconds. $\endgroup$ – Scott Taylor Mar 29 '13 at 18:00
  • $\begingroup$ No worries, I removed my comment :-) $\endgroup$ – Martin Mar 29 '13 at 18:01
  • $\begingroup$ thank you very much. is what you say true for the diffeomorphism too? $\endgroup$ – Ramin Asadi Mar 29 '13 at 18:07
  • 4
    $\begingroup$ There's much simpler answers. You can take $S$ and $T$ to be 2-dimensional manifolds. But this is a standard homework problem. $\endgroup$ – Ryan Budney Mar 29 '13 at 20:19
  • $\begingroup$ for diffeomorphism case right???? that it what really matters $\endgroup$ – Ramin Asadi Mar 29 '13 at 20:21

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