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Assume $M^m,N^n$ are two oriented compact connected manifold, if $f:N\to M$ is a smooth embedding, then the fundamental class $[N]\in H_n(N,\mathbb{R})$ is send to a homology class $f_\ast[N]$ in singular homology group $H_n(M,\mathbb{R})$. Under Poincaré duality, this corresponds a cohomology class $S$. We also know that from Poincaré duality in de Rham cohomology, there exists a cohomology $\eta\in H^{m-n}_{dR}(M)$ such that for any $\zeta\in H^{n}_{dR}(M)$, $$ \int_{N}f^*\zeta = \int_{M}\zeta\wedge\eta. $$ Does $S$ corresponds to $\eta$ in de Rham cohomology?

Sun peilain
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1 Answers1

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Yes; this is more or less the Thom isomorphism. The Poincaré dual of $f_*[N]$ is (represented by) the normal bundle of $N$ in $M$ (or more precisely the Thom class). The equality you mentioned is the Thom isomorphism when stated appropriately.

My favorite reference for this is Glen Bredon's "Topology and Geometry," particularly section VI.11.

pancini
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  • Thanks for your answer, and I have one more question. Over a closed connected manifold $N$, if $\omega\in H^n_{dR}(N,\mathbb{R})$ represent a cohomology class $[\omega]$ in $H^n(N,\mathbb{R})$, why the pairing $\langle[N],[\omega]\rangle$ equal to $\int_{N}\omega$? – Sun peilain Oct 01 '25 at 14:20
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    That's by definition; the angle brackets are short for the pairing which evaluates a cohomology class on a homology class. In the de Rham context, this means evaluating a differential form on a submanifold. – pancini Oct 01 '25 at 16:26
  • But if I consider $\langle[N],[\omega]\rangle$ as the result of the singular cohomology class $[\omega]$ acting on the singular homology class, that is, $\omega$, then does it equal to $\int_{N}\omega$? – Sun peilain Oct 02 '25 at 06:30
  • I mean yes but as an abuse of notation. I.e. a singular cohomology class is not a differential form. Even a de Rham cohomology class is not technically a differential form (it is an equivalence class of them). So it's good to be careful at first, but in spirit many people will interchangeably use $\langle[N],[\omega]\rangle$, $\omega$, and $\int_n\omega$. I'd say the second one is least common. – pancini Oct 02 '25 at 07:54
  • Actually even now I am overcomplicating. They are equal in the literal sense that they all are the same number. But it's still worth understanding that "evaluating a singular cohomology class on a singular homology class" and "integrating a differential form on a submanifold" are of course totally different contexts. The fact that they agree is the magic of the De Rham theorem. – pancini Oct 02 '25 at 08:02
  • Not strictly speaking，the isomorphism from De Rham cohomology to singular cohomology is given by the integral over the singular simplexes. So does there exist a family of singular simplexes whose Linear combination represents the fundamental class $[N]$ in the sense of homology, nearly "cover" the whole manifold $N$? – Sun peilain Oct 02 '25 at 09:20
  • Yes that's what a fundamental class is. – pancini Oct 02 '25 at 11:09
  • Thank you very much for your patient answers. – Sun peilain Oct 02 '25 at 11:19