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I am asked to construct two 3-manifolds $M_1$,$M_2$ both covered by two open sets $U$,$V$ (different for each manifold) s.t. the intersection is diffeomorphic to $\mathbb S^1 \times\mathbb R^2$ but the second de Rham group of $M_1$ and $M_2$ is trivial and not trivial respectively. My thought so far is to take $M_1=\mathbb S^2 \times\mathbb R$ and $M_2=\mathbb S^1 \times\mathbb R^2$ but I can't work out the part about the intersection.

amWhy
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gene wolfe3
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    Can you solve a similar problem in dimension 2? Do you know the Mayer-Vietoris sequence? – Moishe Kohan Mar 23 '24 at 19:58
  • I do but i dont see why that helps to prove that U intersect V is diffeomorphic to \mathbb S^1 \times\mathbb R^2 @MoisheKohan – gene wolfe3 Mar 23 '24 at 20:07
  • What do you mean “to prove”? You have to make constructions with that property. – Ted Shifrin Mar 23 '24 at 20:08
  • Judging by your comment, I think you misunderstood the problem. It is to find an example with particular homological properties. Before you are doing something hard in dimension 3, try to do its analogue in dimension 2, where the intersection will be an annulus. Once you solved the problem in dimension 2, I can give you another hint, for dimension 3.... – Moishe Kohan Mar 23 '24 at 20:11
  • @TedShifrin I mean that the examples i cound find for the two manifolds satisfy the de Rham property but i cant find suitable covers U,V. Also my reasoning was that since R is contractible then S2xR is homotopic to S2, is that correct? – gene wolfe3 Mar 23 '24 at 20:12
  • I think you should follow @MoisheKohan’s advice. Your idea for $M_2$ makes no sense to me. – Ted Shifrin Mar 23 '24 at 20:16

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For $M_2$, finding $U$ and $V$ is pretty obvious so I guess you need help for $M_1$. For this, consider a "latitude map" $z : \mathbb{S}^2 \rightarrow [-1,1]$ sending the south pole on $-1$, the north pole on $1$ and any point of the equator on $0$. Then $U = \{z < 1/2\} \times \mathbb{R} \subset M_1$ and $V = \{z > -1/2\} \times \mathbb{R}$ fit because $U \cap V = \{-1/2 < z < 1/2\} \times \mathbb{R}$ and $\{-1/2 < z < 1/2\} \subset \mathbb{S}^2$ is a strip around the equator, thus homeomorphic to $\mathbb{S}^1 \times \mathbb{R}$.

I suggest you make a draw of what happens on $\mathbb{S}^2$, it is pretty obvious when you visualize it.

Cactus
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  • I got stuck trying to find correct U,V and to write them down but together with your comment and Moishe Kohan's suggestion to try for 2-dimensional manifolds i finally got it. Thank you very much. Is my reasoning for calculating the de Rham groups correct though? because i didnt use mayer-vietoris – gene wolfe3 Mar 23 '24 at 20:34
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    I don't know how you calculated it but it is pretty easy to do once you know that, by de Rham theorem, for all manifold $M$ and integer $k$, $H_{\mathrm{dR}}^k(M) \cong H^k(M,\mathbb{R}) \cong H_k(M,\mathbb{R})$. You conclude by saying that $\mathbb{S}^2 \times \mathbb{R}$ is homotopic to $\mathbb{S}^2$ and $H_2(\mathbb{S}^2,\mathbb{R}) = \mathbb{R}$. – Cactus Mar 23 '24 at 20:49