The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

2 Comments »

  1. […] it all works out for the same reason parameterization invariance and the change of variables formula do. Passing from the boundary of the singular cube back to the […]

    Pingback by Stokes’ Theorem (proof part 2) « The Unapologetic Mathematician | August 20, 2011 | Reply

  2. […] we use to reparameterize our integral. Of course, this function may not be defined on all of , but it’s defined on , […]

    Pingback by Integrals over Manifolds (part 1) « The Unapologetic Mathematician | September 5, 2011 | Reply


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