Let $M$ be compact, connected and orientable manifold without boundary. Let $\sigma_1,\ldots, \sigma_n$ be generators of $H_1(M,\mathbb Z)$. Why is the map $I:H_{dR}^1(M) \to \mathbb R^n$ given by $$I([a]) = \left(\int_{\sigma_1} a , \ldots, \int_{\sigma_n} a\right)$$ an isomorphism?

I believe the proof involves Hodge theorem but I do not know where.

Observation: $H_{dR}^1(M)$ is the first De Rham cohomology group and $H_1(M,\mathbb Z)$ is the first singular homology group with coefficients on $\mathbb Z$.

  • 1
    $\begingroup$ This is not true in the presence of torsion: Consider $\mathbb{RP^2}$. $\endgroup$ – Thomas Rot Feb 22 '18 at 14:55
  • $\begingroup$ Ah ok, I forgot to add that $M$ is orientable. $\endgroup$ – Hugocito Feb 22 '18 at 15:09
  • $\begingroup$ This is also false on $RP^3$ I think. $\endgroup$ – Thomas Rot Feb 22 '18 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.