Let $M$ be a smooth oriented $n$-dimensional manifold and denote by $A \in H_n(M;\mathbb{Z})$ the fundamental class of $M$ (a generator of singular homology consistent with the orientation of $M$). Consider the following two maps from the top de-Rham cohomology group $H^n_{\mathrm{dR}}(M)$ to $\mathbb{R}$:

  1. Regular integration of $n$-forms: $\omega \mapsto \int_M \omega$. This is defined using partition of unity and descends to cohomology by Stokes's theorem.
  2. Represent $A$ as a smooth chain $A = [\sum a_i \sigma_i]$ where $\sigma_i : \Delta^n \rightarrow M$ are smooth $n$-simplices and integrate $\omega$ by $$ \omega \mapsto \int_{\sum a_i \sigma_i} \omega := \sum a_i \int_{\Delta^n} \sigma_i^*(\omega). $$ This is well defined and independent of the representation of $A$ again by Stokes's theorem for chains.

Both maps are $\mathbb{R}$-linear maps from a one dimensional real vector space to $\mathbb{R}$ and so are a real multiple of one another. Why are they equal?

This should probably involve some careful tracing of definitions, identifications and dualities, but I can't put my finger on what is the crux of the matter.

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    obviously, there is no problems for triangulable manifolds – Andrey Ryabichev May 20 '17 at 09:22
  • In short, because the volume density on $M$ defines a measure which, in particular, is countably additive. Integrating $n$-forms then is equivalent to integrating functions against the measure. – Moishe Kohan May 24 '17 at 03:00

1 Answers1


We may assume that $A=[\sum\sigma_i]$ is actually just represented as the sum of all the $n$-simplices in some oriented smooth triangulation of $M$. Since the two integration maps differ by a constant multiple, it suffices to compare them on just a single $n$-form $\omega$ with nonzero integral. Pick $\omega$ to be supported on the interior of $\sigma_1$ such that $\int_{\Delta^n} \sigma_1^*(\omega)=1$. We then have $\int_{\Delta^n} \sigma_i^*(\omega)=0$ for $i\neq 1$, so your second integral sends $\omega$ to $1$.

We now compute $\int_M \omega$ as follows. Note that $\sigma_1$ itself is an oriented smooth chart of $M$, when restricted to the interior of $\Delta^n$. We can now construct a partition of unity on $M$ subordinate to a covering by oriented charts which has as one of its functions a bump function $f$ supported on the interior of $\sigma_1$ which is $1$ on the entire support of $\omega$. When we compute $\int_M\omega$ using this partition of unity, all the terms vanish except the one corresponding to $f$, since $f$ is the only one that does not vanish on the support of $\omega$. By definition, we then have $$\int_M\omega=\int_{\Delta^n}\sigma_i^*(f\omega)=\int_{\Delta^n}\sigma_i^*(\omega)=1.$$ Thus both integrals send $\omega$ to $1$, so they are equal.

Eric Wofsey
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