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).


1 Answer 1


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.

  • $\begingroup$ Alright, but why do both spaces have the same dimension? $\endgroup$
    – none
    Commented Sep 14, 2012 at 1:19
  • $\begingroup$ You are using the $n$-form right? Then this should follow by definition. $\endgroup$ Commented Sep 14, 2012 at 1:36
  • $\begingroup$ I'm sorry, but I don't follow. Are you using Poincare duality here? $\endgroup$
    – none
    Commented Sep 14, 2012 at 1:37
  • $\begingroup$ The way I see it, injectivity of this map is the hard part of the question. $\endgroup$
    – none
    Commented Sep 14, 2012 at 2:13
  • 3
    $\begingroup$ 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, $\endgroup$
    – Matt E
    Commented Sep 14, 2012 at 3:15

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