Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.