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Let $f : T^2 \to S^3$ be the smooth map of a 2-torus into $S^3$, therefore $$\int_{T^2}f^*\omega = 0.$$ There is a closed $2$-form $\beta$ on $T^3 = S^1 \times S^1 \times S^1$ and a map $g : T^2 \to T^3$ such that $$\int_{T^2}g^*\beta \neq 0.$$ Use the given information to show $S^3$ and $T^3$ are not diffeomorphic.

So I got that $H^2(S^3)$ is a zero class hence all closed forms are exact. But I couldn't see $H^2(T^3)$ is nontrivial.

Also, it sounds familiar but why trivial and nontrivial cohomology are not diffeomorphic..?

WishingFish
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  • This basically means $H^2_{dR}(T^3)$ is nontrivial while $H^2_{dR}(S^3)$ is trivial hence $S^3$ and $T^2$ cannot be diffeomorphic. – Roy Han Aug 09 '13 at 04:47
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    @WishingFish: Please show us your work, because I don't know where/why you are stuck, and this forum is not a homework solutions manual. Also, I noticed that you have been posting a bunch of these questions, and I believe you are rushing through the material (you're doing integration now but were stumped on definitions involving differential forms?) -- I strongly advise slowing way down and going back to the beginning to make sure you are actually understanding and mastering the material. Sorry for this coming off as a lecture, I just don't want you to end up hurting your math endeavors. – Chris Gerig Aug 09 '13 at 04:55
  • Hi @RoyHan, thank you. I got closer, but still haven't straighten out. So I got that $H^2(S^3)$ is a zero class hence all closed forms are exact. But I couldn't see $H^2(T^3)$ is non trivial. By the way, may I ask what does your $dR$ mean? – WishingFish Aug 09 '13 at 05:07
  • Also, it sounds familiar but why trivial and nontrivial cohomology are not diffeomorphic..? Thanks @RoyHan. – WishingFish Aug 09 '13 at 05:09
  • Hey @ChrisGerig, shown above is my work! I did the first 2 problem, and stuck at the last one. So I gave the conclusion of first 2 problem as given, and ask about the last one. But I will add something on the third one. – WishingFish Aug 09 '13 at 05:10
  • And for your following remark @ChrisGerig, it is too late to drop the course, and I do have an exam in one week! But I appreciate your concern. – WishingFish Aug 09 '13 at 05:13
  • @WishingFish This is basically definition: Consider $g^:H^2(T^3)\to H^2(T^2)$ and $[\beta]\in H^2(T^3)$, then the given condition means $[g^ \beta] \in H^2(T^2)$ cannot be trivial, which implies $[\beta]\neq 0$. Now we find a non-trivial class of $H^2(T^3)$. If $S^3$ is diffeomorphic to $T^3$, then its de Rham cohomology should be ring-isomorphism, especially group isomorphism in every dimension which is clearlynot in the 2-dimension case. – Roy Han Aug 09 '13 at 05:17
  • Got it, best answer, thank you @RoyHan. – WishingFish Aug 09 '13 at 05:38

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I will write the very detailed explanation for your second question (it is not proof however, for most mathematics students this is the 'obvious' level problem). For the first part, please refer to the comment.

If smooth manifolds $M,N$ is diffeomorphic, then the graded algebra $H^*_{dR}(M)=H^*_{dR}(N)$

[proof] Let $f:M\to N$ be a diffeomorphism and $g:N\to M$ its inverse, then since $f\circ g=id_N$ and $g\circ f=id_M$, we find $id=g^*\circ f^*:H^*(N)\to H^*(N)$ and $id=f^*\circ g^*:H^*(M)\to H^*(M)$.

Roy Han
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