# The Unapologetic Mathematician

## 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 - Posted by | Differential Topology, Topology

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