# The Unapologetic Mathematician

## Integrals over Manifolds (part 1)

We’ve defined how to integrate forms over chains made up of singular cubes, but we still haven’t really defined integration on manifolds. We’ve sort of waved our hands at the idea that integrating over a cube is the same as integrating over its image, but this needs firming up. In particular, we will restrict to oriented manifolds.

To this end, we start by supposing that an $n$-form $\omega$ is supported in the image of an orientation-preserving singular $n$-cube $c:[0,1]^n\to M$. Then we will define

$\displaystyle\int\limits_M\omega=\int\limits_c\omega$

Indeed, here the image of $c$ is some embedded submanifold of $M$ that even agrees with its orientation. And since $\omega$ is zero outside of this submanifold it makes sense to say that the integral over the submanifold — over the singular cube $c$ — is the same as the integral over the whole manifold.

What if we have two orientation-preserving singular cubes $c_1$ and $c_2$ that both contain the support of $\omega$? It only makes sense that they should give the same integral. And, indeed, we find that

$\displaystyle\int\limits_{c_2}\omega=\int\limits_{c_2\circ c_2^{-1}\circ c_1}\omega=\int\limits_{c_1}\omega$

where we use $c_2^{-1}\circ c_1$ to reparameterize our integral. Of course, this function may not be defined on all of $[0,1]^n$, but it’s defined on $c_1\left([0,1]^n\right)\cap c_2\left([0,1]^n\right)$, where $\omega$ is supported, and that’s enough.

September 5, 2011 - Posted by | Differential Topology, Topology

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3. […] Remembering that diffeomorphism is meant to be our idea of what it means for two smooth manifolds to be “equivalent”, this means that is either equivalent to or to . And I say that this equivalence comes out in integrals. […]

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