# 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.