# The Unapologetic Mathematician

## Integrals are Independent of Parameterization

If $c:[0,1]^k\to M$ is a singular $k$-cube and $\omega$ is a $k$-form on the image of $c$, then we know how to define the integral of $\omega$ over $c$:

$\displaystyle\int\limits_{c([0,1]^k)}\omega=\int\limits_{[0,1]^k}c^*\omega$

On its face, this formula depends on the function $c$ used to parameterize the region of integration. But does it really? What if we have a different function $d:[0,1]^k\to M$ with the same image? For convenience we’ll only consider singular $k$-cubes that are diffeomorphisms onto their images — any singular $k$-cube can be broken into pieces for which this is true, and we’ll soon deal with how to put these together.

Anyway, if $c([0,1]^k)=d([0,1]^k)$, then given our assumptions there is some diffeomorphism $f=c^{-1}\circ d:[0,1]^k\to[0,1]^k$ such that $d=c\circ f$. If $J_f$ is everywhere positive, then we say that $d$ is an “orientation-preserving reparameterization” of $c$. And I say that the integrals of $\omega$ over $c$ and $d$ are the same. Indeed, we calculate:

\displaystyle\begin{aligned}\int\limits_{c\circ f}\omega&=\int\limits_{[0,1]^k}(c\circ f)^*\omega\\&=\int\limits_{[0,1]^k}f^*c^*\omega\\&=\int\limits_fc^*\omega\\&=\int\limits_{f([0,1]^k)}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_{[0,1]^k}\left[c^*\omega\right]\left(\frac{\partial}{\partial u^1},\dots,\frac{\partial}{\partial u^k}\right)\,d(u^1,\dots,u^k)\\&=\int\limits_c\omega\end{aligned}

where we use our expression for the integral of $\omega$ over the image $f([0,1]^k)$ in passing from the third to the fourth line. Thus the integral is a geometric quantity, depending only on the image $c([0,1]^k)$ and the $k$-form $\omega$ rather than on any detail of the parameterization itself.

August 4, 2011 - Posted by | Differential Topology, Topology