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Let $X$ be a $C^\infty$ manifold, compact oriented and connected of dimension $n$. How do you prove that the integration map $$\int_X: \omega \mapsto \int_X \omega $$ from $H^n_{DR}(X)$ to $\mathbb{R}$ is an isomorphism? (Without using Poincare duality).

none
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1 Answers1

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Here are the steps to do this:

  1. Show that the formula you gave is well-defined (that is, show that it doesn't depend on the class of $w$). This is just Stokes' Theorem, assuming your manifold has no boundary. Linearity is obvious.

  2. For surjectivity, since $M$ is oriented there exists a $n$-form $\omega_0$ such that $\int_M \omega_0 = c > 0$. Now just multiply $\omega_0$ by the appropriate constant.

  3. Now note that both spaces have the same dimension, so integration must be an isomorphism.

student
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  • Alright, but why do both spaces have the same dimension? – none Sep 14 '12 at 01:19
  • You are using the $n$-form right? Then this should follow by definition. – Bombyx mori Sep 14 '12 at 01:36
  • I'm sorry, but I don't follow. Are you using Poincare duality here? – none Sep 14 '12 at 01:37
  • The way I see it, injectivity of this map is the hard part of the question. – none Sep 14 '12 at 02:13
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    Dear student, Is step 3 really as simple as noting something? You have to somehow show that if $\omega$ is an $n$-form for which $\int_X \omega = 0,$ then $\omega$ is a coboundary. Regards, – Matt E Sep 14 '12 at 03:15
  • @MattE,None Sorry, I was thinking about a way to fix this, but wasn't able to. For a moment there, I thought it was trivial, but I guess it isn't. – student Sep 15 '12 at 13:48
  • I looked it up in Lee's book on Smooth Manifolds. There, it amounts to a standard-seeming argument, using partitions of unity to show that if a top form integrates to zero then it is exact. Is there any easier way? – none Sep 17 '12 at 02:32