# 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

## Orientation-Preserving Mappings

Of course now that we have more structure, we have more structured maps. But this time it’s not going to be quite so general; we will only extend our notion of an embedding, and particularly of an embedding in codimension zero.

That is, let $f:M\to N$ be an embedding of manifolds where each of $M$ and $N$ has dimension $n$. Since their dimensions are the same, the codimension of this embedding — the difference between the dimension of $N$ and that of $M$ — is $0$. If $M$ and $N$ are both oriented, then we say that $f$ preserves the orientation if the pullback of any $n$-form on $N$ which gives the chosen orientation gives us an $n$-form on $M$ which gives its chosen orientation. We easily see that this concept wouldn’t even make sense if $M$ and $N$ didn’t have the same dimension.

More specifically, let $M$ and $N$ be oriented by $n$-forms $\omega_M$ and $\omega_N$, respectively. If $f^*\omega_N=\lambda\omega_M$ for some smooth, everywhere-positive $\lambda\in\mathcal{O}(M)$, we say that $f$ is orientation-preserving. The specific choices of $\omega_M$ and $\omega_N$ don’t matter; if $\omega_M'$ gives the same orientation on $M$ then we must have $\omega_M=\phi\omega_M'$ for some smooth, everywhere-positive $\phi$, and $f^*\omega_N=\lambda\phi\omega_M'$; if $\omega_N'$ gives the same orientation on $N$ then we must have $\omega_N'=\phi\omega_N$ for some smooth, everywhere-positive $\phi$, and $f^*\omega_N'=\phi f^*\omega_N=\phi\lambda\omega_M$.

In fact, we have a convenient way of coming up with test forms. Let $(U,x)$ be a coordinate patch on $M$ around $p$ whose native orientation agrees with that of $M$, and let $(V,y)$ be a similar coordinate patch on $N$ around $f(p)$. Now we have neighborhoods of $p$ and $f(p)$ between which $f$ is a diffeomorphism, and we have top forms $dx^1\wedge\dots\wedge dx^n$ and $dy^1\wedge\dots\wedge dy^n$ in $U$ and $V$, respectively. Pulling back the latter form we find

\displaystyle\begin{aligned}f^*(dy^1\wedge\dots\wedge dy^n)&=d(y^1\circ f)\wedge\dots\wedge d(y^n\circ f)\\&=\left(\sum\limits_{i_1=1}^n\frac{\partial(y^1\circ f)}{\partial x^{i_1}}dx^{i_1}\right)\wedge\dots\wedge\left(\sum\limits_{i_n=1}^n\frac{\partial(y^n\circ f)}{\partial x^{i_n}}dx^{i_n}\right)\\&=\det\left(\frac{\partial(y^j\circ f)}{\partial x^{i}}\right)dx^1\wedge\dots\wedge dx^n\end{aligned}

That is, the pullback of the (local) orientation form on $N$ differs from the (local) orientation form on $M$ by a factor of the Jacobian determinant of the function $f$ with respect to these coordinate maps. This repeats what we saw in the case of transition functions between coordinates. And so if whenever we pick local coordinates on $M$ and $N$ we find an everywhere-positive Jacobian determinant of $f$, then $f$ preserves orientation.

September 1, 2011