The fundamental lemma of calibrated geometry states that calibrated submanifolds are absolutely volume minimising in their homology class. In the proof, homology equivalent is used synonymously with cobordant. With respect to which homology theory are they minimising? Are they volume minimising with respect to singular homology?
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I don't know what proof you're referring to that uses cobordant. The proof I know uses singular homology (say with piecewise-smooth chains) to deduce that we have $\int_M \phi = \int_{M'}\phi$ when $\partial M=\partial M'$, $M$ is homologous to $M'$, and $\phi$ is closed.
answered Oct 21, 2019 at 16:01
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1$\begingroup$ Sorry, I think I should have been more specific. To simplify, let us assume that $M$ and $M'$ have no boundary. The proof I know assumes that there is a smooth submanifold $N$ such that $\partial(N)=M-M'$. Then we can use Stokes theorem: $$\int_M \phi-\int_{M'} \phi=\int_{\partial N} \phi=0.$$ I don't know how to infer this without assuming such an N. If $M$ and $M'$ are homologous in singular homology, there is a singular simplex (or a sum of them) whose boundary is $M-M'$. But this needn't be a smooth submanifold. This is the point I don't understand. $\endgroup$– deepfloeOct 23, 2019 at 8:37
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2$\begingroup$ As I said, you want to use homology with piecewise-smooth chains, not continuous chains. Effectively, the point is that deRham's Theorem tells you that singular cohomology is isomorphic to deRham cohomology. See, for example, this mathoverflow post or discussion, as I recall, in Warner's manifolds book, deRham's book, or problem 14 in Chapter 11 of Spivak's A Comprehensive Introduction to Differential Geometry, volume 1. $\endgroup$ Oct 23, 2019 at 16:46