# The Unapologetic Mathematician

## Switching Orientations

If we have an oriented manifold $M$, then we know that the underlying manifold has another orientation available; if $\alpha$ is a top form that gives $M$ its orientation, then $-\alpha$ gives it the opposite orientation. We will write $-M$ for the same underlying manifold equipped with this opposite orientation.

Now it turns out that the integrals over the same manifold with the two different orientations are closely related. Indeed, if $\omega$ is any $n$-form on the oriented $n$-manifold $M$, then we find

$\displaystyle\int\limits_{-M}\omega=-\int\limits_M\omega$

Without loss of generality, we may assume that $\omega$ is supported within the image of a singular cube $c$. If not, we break it apart with a partition of unity as usual.

Now, if $c:[0,1]^n\to M$ is orientation-preserving, then we can come up with another singular cube that reverses the orientation. Indeed, let $f(u^1,u^2,\dots,u^n)=(-u^1,u^2,\dots,u^n)$. It’s easy to see that $f^*$ sends $du^1$ to $-du^1$ and preserves all the other $du^i$. Thus it sends $du^1\wedge\dots\wedge du^n$ to its negative, which shows that it’s an orientation-reversing mapping from the standard $n$-cube to itself. Thus we conclude that the composite $\tilde{c}=c\circ f$ is an orientation-reversing singular cube with the same image as $c$.

But then $\tilde{c}:[0,1]\to-M$ is an orientation-preserving singular cube containing the support of $\omega$, and so we can use it to calculate integrals over $-M$. Working in from each side of our proposed equality we find

\displaystyle\begin{aligned}\int\limits_M\omega&=\int\limits_c\omega=\int\limits_{[0,1]^n}c^*\omega\\\int\limits_{-M}\omega&=\int\limits_{\tilde{c}}\omega=\int\limits_{[0,1]^n}\tilde{c}^*\omega=\int\limits_{[0,1]^n}f^*c^*\omega\end{aligned}

We know that we can write

$\displaystyle c^*\omega=g(u^1,\dots,u^n)du^1\wedge\dots\wedge du^n$

for some function $g$. And as we saw above, $f^*$ sends $du^1\wedge\dots\wedge du^n$ to its negative. Thus we conclude that

$\displaystyle\tilde{c}^*\omega=-=g(u^1,\dots,u^n)du^1\wedge\dots\wedge du^n$

meaning that when we calculate the integral over $-M$ we’re using the negative of the form on $[0,1]^n$ that we use when calculating the integral over $M$.

This makes it even more sensible to identify an orientation-preserving singular cube with its image. When we write out a chain, a positive multiplier has the sense of counting a point in the domain more than once, while a negative multiplier has the sense of counting a point with the opposite orientation. In this sense, integration is “additive” in the domain of integration, as well as linear in the integrand.

The catch is that this only works when $M$ is orientable. When this condition fails we still know how to integrate over chains, but we lose the sense of orientation.

About these ads

September 8, 2011 - Posted by | Differential Topology, Topology

## 2 Comments »

1. […] if is orientation-reversing from to then it’s orientation-preserving from to , and we already know that integrals over are the negatives of those over . Further, we can assume that the support of […]

Pingback by Integrals and Diffeomorphisms « The Unapologetic Mathematician | September 12, 2011 | Reply

2. […] we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the face of a singular -cube to cancel each other […]

Pingback by Stokes’ Theorem on Manifolds « The Unapologetic Mathematician | September 16, 2011 | Reply