# The Unapologetic Mathematician

## Oriented Manifolds with Boundary

Let’s take a manifold with boundary $M$ and give it an orientation. In particular, for each $p\in M$ we can classify any ordered basis as either positive or negative with respect to our orientation. It turns out that this gives rise to an orientation on the boundary $\partial M$.

Now, if $p\in\partial M$ is a boundary point, we’ve seen that we can define the tangent space $\mathcal{T}_pM$, which contains — as an $n-1$-dimensional subspace — $\mathcal{T}_p(\partial M)$. This subspace cuts the tangent space into two pieces, which we can distinguish as follows: if $(U,x)$ is a coordinate patch around $p$ with $x(p)=0$, then the image of $\partial M$ near $p$ is a chunk of the hyperplane $x^n=0$. The inside of $M$ corresponds to the area where $x^n>0$, while the outside corresponds to $x^n>0$.

And so the map $x_{*p}$ sends a vector $v\in\mathcal{T}_pM$ to a vector in $\mathcal{T}_0\mathbb{R}^n$, which either points up into the positive half-space, along the hyperplane, or down into the negative half-space. Accordingly, we say that $v$ is “inward-pointing” if $x_{*p}(v)$ lands in the first category, and “outward-pointing” if if lands in the last. We can tell the difference by measuring the $n$th component — the value $\left[x_{*p}(v)\right](u^n)=v(u^n\circ x)=v(x^n)$. If this value is positive the vector is inward-pointing, while if it’s positive the vector is outward-pointing.

This definition may seem to depend on our choice of coordinate patch, but the division of $\mathcal{T}_pM$ into halves is entirely geometric. The only role the coordinate map plays is in giving a handy test to see which half a vector lies in.

Now we are in a position to give an orientation to the boundary $\partial M$, which we do by specifying which bases of $\partial M$ are “positively oriented” and which are “negatively oriented”. Specifically, if $v_1,\dots,v_{n-1}$ is a basis of $\mathcal{T}_p(\partial M)\subseteq\mathcal{T}_pM$\$ then we say it’s positively oriented if for any outward-pointing $v\in\mathcal{T}_pM$ the basis $v,v_1,\dots,v_{n-1}$ is positively oriented as a basis of $\mathcal{T}_pM$, and similarly for negatively oriented bases.

We must check that this choice does define an orientation on $\partial M$. Specifically, if $(V,y)$ is another coordinate patch with $y(p)=0$, then we can set up the same definitions and come up with an orientation on each point of $V\cap\partial M$. If $U$ and $V$ are compatibly oriented, then $U\cap\partial M$ and $V\cap\partial M$ must be compatible as well.

So we assume that the Jacobian of $y\circ x^{-1}$ is everywhere positive on $U\cap V$. That is

$\displaystyle\det\left(\frac{\partial(y^i\circ x^{-1})}{\partial u^j}\right)>0$

We can break down $x$ and $y$ to strip off their last components. That is, we write $x(q)=(\tilde{x}(q),x^n(q))$, and similarly for $y$. The important thing here is that when we restrict to the boundary $U\cap\partial M$ the $\tilde{x}$ work as a coordinate map, as do the $\tilde{y}$. So if we set $u^n=0$ and vary any of the other $u^j$, the result of $y^n(x^{-1}(u))$ remains at zero. And thus we can expand the determinant above:

\displaystyle\begin{aligned}0&<\det\left(\frac{\partial(y^i\circ x^{-1})}{\partial u^j}\bigg\vert_{u^n=0}\right)\\&=\det\begin{pmatrix}\displaystyle\frac{\partial(y^1\circ x^{-1})}{\partial u^1}\bigg\vert_{u^n=0}&\cdots&\displaystyle\frac{\partial(y^1\circ x^{-1})}{\partial u^{n-1}}\bigg\vert_{u^n=0}&\displaystyle\frac{\partial(y^1\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}\\\vdots&\ddots&\vdots&\vdots\\\displaystyle\frac{\partial(y^{n-1}\circ x^{-1})}{\partial u^1}\bigg\vert_{u^n=0}&\cdots&\displaystyle\frac{\partial(y^{n-1}\circ x^{-1})}{\partial u^{n-1}}\bigg\vert_{u^n=0}&\displaystyle\frac{\partial(y^{n-1}\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}\\{0}&\cdots&0&\displaystyle\frac{\partial(y^n\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}\end{pmatrix}\end{aligned}

The determinant is therefore the determinant of the upper-left $n\times n$ submatrix — which is the Jacobian determinant of the transition function $\tilde{y}\circ\tilde{x}^{-1}$ on the intersection $(U\cap\partial M)\cap(V\cap\partial M$ — times the value in the lower right.

If the orientations induced by those on $U$ and $V$ are to be compatible, then this Jacobian determinant on the boundary must be everywhere positive. Since the overall determinant is everywhere positive, this is equivalent to the lower-right component being everywhere positive on the boundary. That is:

$\displaystyle\frac{\partial(y^n\circ x^{-1})}{\partial u^n}\bigg\vert_{u^n=0}$

But this asks how the $n$th component of $y$ changes as the $n$th component of $x$ increases; as we move away from the boundary. But, at least where we start on the boundary, $y^n$ can do nothing but increase! And thus this partial derivative must be positive, which proves our assertion.

September 16, 2011

## The Tangent Space at the Boundary

If we have a manifold with boundary $M$, then at all the interior points $M\setminus\partial M$ it looks just like a regular manifold, and so the tangent space is just the same as ever. But what happens when we consider a point $p\in\partial M$?

Well, if $(U,x)$ is a chart around $p$ with $x(p)=0\in H^n$, then we see that the part of the boundary within $U$$U\cap\partial M$ — is the surface $\{q\in U\vert x^n(q)=0\}$. The point $0\in H^n\subseteq\mathbb{R}^n$ has a perfectly good tangent space as a point in $\mathbb{R}^n$: $\mathcal{T}_0\mathbb{R}^n$. We will consider this to be the tangent space of $H^n$ at zero, even though half of its vectors “point outside” the space itself.

We can use this to define the tangent space $\mathcal{T}_pM$. Indeed, the function $x^{-1}$ goes from $H^n$ to $M$ and takes the point $0$ to $p$; it only makes sense to define $\mathcal{T}_pM$ as $(x^{-1})_{*0}\left(\mathcal{T}_0\mathbb{R}^n\right)$.

This is all well and good algebraically, but geometrically it seems that we’re letting tangent vectors spill “off the edge” of $M$. But remember our geometric characterization of tangent vectors as equivalence classes of curves — of “directions” that curves can go through $p$. Indeed, a curve could well run up to the edge of $M$ at the point $p$ in any direction that — if continued — would leave the manifold through its boundary. The geometric definition makes it clear that this is indeed the proper notion of the tangent space at a boundary point.

Now, let $y$ be the function we get by restricting $x$ to the boundary $\partial M$. The function $y^{-1}$ sends the boundary $\partial H^n\cong\mathbb{R}^{n-1}\times\{0\}$ to the boundary $\partial M$ — at least locally — and there is an inclusion $i:\partial M\to M$. On the other hand, there is an inclusion $j:\mathbb{R}^{n-1}\times\{0\}\to\mathbb{R}^n$, which $x^{-1}$ then sends to $U$ — again, at least locally. That is, we have the equation

$\displaystyle i\circ y^{-1}=x^{-1}\circ j$

Taking the derivative, we see that

$\displaystyle i_*\left((y^{-1})_*\left(\mathcal{T}_0\mathbb{R}^{n-1}\right)\right)=(x^{-1})_*\left(j_*\left(\mathcal{T}_0\mathbb{R}^{n-1}\right)\right)$

But $i_*$ must be the inclusion of the subspace $\mathcal{T}_p(\partial M)$ into the tangent space $\mathcal{T}_pM$. That is, the tangent vectors to the boundary manifold are exactly those tangent vectors on the boundary that $x_*$ sends to tangent vectors in $\mathcal{T}_0\mathbb{R}^n$ whose $n$th component is zero.

September 15, 2011

## Manifolds with Boundary

Ever since we started talking about manifolds, we’ve said that they locally “look like” the Euclidean space $\mathbb{R}^n$. We now need to be a little more flexible and let them “look like” the half-space $H^n=\{x\in\mathbb{R}^n\vert x^n\geq0\}$.

Away from the subspace $x^n=0$, $H^n$ is a regular $n$-dimensional manifold — we can always find a small enough ball that stays away from the edge — but on the boundary subspace it’s a different story. Just like we wrote the boundary of a singular cubic chain, we write $\partial H^n=\{x\in\mathbb{R}^n\vert x^n=0\}$ for this boundary. Any point $p\in M$ that gets sent to $\partial H^n$ by a coordinate map $x$ must be sent to $\partial H^n$ by every coordinate map. Indeed, if $y$ is another coordinate map on the same patch $U$ around $p$, then the transition function $y\circ x^{-1}:H^n\to H^n$ must be a homeomorphism from $x(U)$ onto $y(U)$, and so it must send boundary points to boundary points. Thus we can define the boundary $\partial M$ to be the collection of all these points.

Locally, $\partial M$ is an $n-1$-dimensional manifold. Indeed, if $(U,x)$ is a coordinate patch around a point $p\in\partial M$ then $x(p)\in\partial H^n\cap x(U)$, and thus the preimage $x^{-1}(\partial H^n)$ is an $n-1$-dimensional coordinate patch around $p$. Since every point is contained in such a patch, $\partial M$ is indeed an $n-1$-dimensional manifold.

As for smooth structures on $M$ and $\partial M$, we define them exactly as usual; real-valued functions on a patch $(U,x)$ of $M$ containing some boundary points are considered smooth if and only if the composition $f\circ x^{-1}$ is smooth as a map from (a portion of) the half-space to $\mathbb{R}$. And such a function is smooth at a boundary point of the half-space if and only if it’s smooth in some neighborhood of the point, which extends — slightly — across the boundary.

September 13, 2011

## Integrals and Diffeomorphisms

Let’s say we have a diffeomorphism $f:M^n\to N^n$ from one $n$-dimensional manifold to another. Since $f$ is both smooth and has a smooth inverse, we must find that the Jacobian is always invertible; the inverse of $J_f$ at $p\in M$ is $J_{f^{-1}}$ at $f(p)\in N$. And so — assuming $M$ is connected — the sign of the determinant must be constant. That is, $f$ is either orientation preserving or orientation-reversing.

Remembering that diffeomorphism is meant to be our idea of what it means for two smooth manifolds to be “equivalent”, this means that $N$ is either equivalent to $M$ or to $-M$. And I say that this equivalence comes out in integrals.

So further, let’s say we have a compactly-supported $n$-form $\omega$ on $N$. We can use $f$ to pull back $\omega$ from $N$ to $M$. Then I say that

$\displaystyle\int\limits_Mf^*\omega=\pm\int\limits_N\omega$

where the positive sign holds if $f$ is orientation-preserving and the negative if $f$ is orientation-reversing.

In fact, we just have to show the orientation-preserving side, since if $f$ is orientation-reversing from $M$ to $N$ then it’s orientation-preserving from $-M$ to $N$, and we already know that integrals over $-M$ are the negatives of those over $M$. Further, we can assume that the support of $f^*\omega$ fits within some singular cube $c:[0,1]^n\to M$, for if it doesn’t we can chop it up into pieces that do fit into cubes $c_i$, and similarly chop up $N$ into pieces that fit within corresponding singular cubes $f\circ c_i$.

But now it’s easy! If $f^*\omega$ is supported within the image of an orientation-preserving singular cube $c$, then $\omega$ must be supported within $f\circ c$, which is also orientation-preserving since both $f$ and $c$ are, by assumption. Then we find

\displaystyle\begin{aligned}\int\limits_N\omega&=\int\limits_{f\circ c}\\&=\int\limits_{f(c([0,1]^n))}\omega\\&=\int\limits_{c([0,1]^n)}f^*\omega\\&=\int\limits_cf^*\omega\\&=\int\limits_Mf^*\omega\end{aligned}

In this sense we say that integrals are preserved by (orientation-preserving) diffeomorphisms.

September 12, 2011

## 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.

September 8, 2011

## Integrals over Manifolds (part 2)

Okay, so we can now integrate forms as long as they’re supported within the image of an orientation-preserving singular cube. But what if the form $\omega$ is bigger than that?

Well, paradoxically, we start by getting smaller. Specifically, I say that we can always find an orientable open cover of $M$ such that each set in the cover is contained within the image of a singular cube.

We start with any orientable atlas, which gives us a coordinate patch $(U,x)$ around any point $p$ we choose. Without loss of generality we can pick the coordinates such that $x(p)=0$. There must be some open ball around $0$ whose closure is completely contained within $x(U)$; this closure is itself the image of a singular cube, and the ball obviously contained in its closure. Hitting everything with $x^{-1}$ we get an open set — the inverse image of the ball — contained in the image of a singular cube, all of which contains $p$. Since we can find such a set around any point $p\in M$ we can throw them together to get an open cover of $M$.

So, what does this buy us? If $\omega$ is any compactly-supported $n$ form on an $n$-dimensional manifold $M$, we can cover its support with some open subsets of $M$, each of which is contained in the image of a singular $n$-cube. In fact, since the support is compact, we only need a finite number of the open sets to do the job, and throw in however many others we need to cover the rest of $M$.

We can then find a partition of unity $\Phi=\{\phi\}$ subordinate to this cover of $M$. We can decompose $\omega$ into a (finite) sum:

$\displaystyle\omega=\sum\limits_{\phi\in\Phi}\phi\omega$

which is great because now we can define

$\displaystyle\int\limits_M\omega=\sum\limits_{\phi\in\Phi}\int\limits_M\phi\omega$

But now we must be careful! What if this definition depends on our choice of a suitable partition of unity? Well, say that $\Psi=\{\psi\}$ is another such partition. Then we can write

$\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_M\phi\omega=\sum\limits_{\phi\in\Phi}\int\limits_M\sum\limits_{\psi\in\Psi}\psi\phi\omega=\sum\limits_{\psi\in\Psi}\int\limits_M\sum\limits_{\phi\in\Phi}\phi\psi\omega=\sum\limits_{\psi\in\Psi}\int\limits_M\psi\omega$

so we get the same answer no matter which partition we use.

September 7, 2011

## 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

## Orientable Atlases

If we orient a manifold $M$ by picking an everywhere-nonzero top form $\omega$, then it induces an orientation on each coordinate patch $(U,x)$. Since each one also comes with its own orientation form, we can ask whether they’re compatible or not.

And it’s easy to answer; just calculate

$\displaystyle\omega\left(\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}\right)$

and if the answer is positive then the two are compatible, while if it’s negative then they’re incompatible. But no matter; just swap two of the coordinates and we have a new coordinate map on $U$ whose own orientation is compatible with $\omega$.

This shows that we can find an atlas on $M$ all of whose patches have compatible orientations. Given any atlas at all for $M$, either use a coordinate patch as is or swap two of its coordinates depending on whether its native orientation agrees with $\omega$ or not. In fact, if we’re already using a differentiable structure — containing all possible coordinate patches which are (smoothly) compatible with each other — then we just have to throw out all the patches which are (orientably) incompatible with $\omega$.

The converse, as it happens, is also true: if we can find an atlas for $M$ such that for any two patches $(U,x)$ and $(V,y)$ the Jacobian of the transition function is everywhere positive on the intersection $U\cap V$, then we can find an everywhere-nonzero top form to orient the whole manifold.

Basically, what we want is to patch together enough of the patches’ native orientations to cover the whole manifold. And as usual for this sort of thing, we pick a partition of unity subordinate to our atlas. That is, we have a countable, locally finite collection of functions $\{\phi_i\}$ so that $\phi_i$ is supported within the patch $(U_i,x_i)$. Then we define the $n$-form $\omega_i$ on $U_i$ by

$\displaystyle\omega_i(p)=\phi_i(p)dx_i^1\wedge\dots\wedge dx_i^n$

and by $0$ outside of $U_i$. Adding up all the $\omega_i$ gives us our top form; the sum makes sense because it’s locally finite, and at each point we don’t have to worry about things canceling off because each orientation form $\omega_i$ is a positive multiple of each other one wherever they’re both nonzero.

August 31, 2011

## Compatible Orientations

Any coordinate patch $(U,x)$ in a manifold $M$ is orientable. Indeed, the image $x(U)\subseteq\mathbb{R}^n$ is orientable — we can just use $du^1\wedge\dots\wedge du^n$ to orient $\mathbb{R}^n$ — and given a choice of top form on $x(U)$ we can pull it back along $x$ to give an orientation of $U$ itself.

But what happens when we bring two patches $U$ and $V$ together? They may each have orientations given by top forms $\omega_U$ and $\omega_V$. We must ask whether they are “compatible” on their overlap. And compatibility means each one picks out the same end of $\Lambda*_k(U\cap V)$ at each point. But this just means that — when restricted to the intersection $U\cap V$$\omega_U=f\omega_V$ for some everywhere-positive smooth function $f$.

Another way to look at the same thing is to let $\omega_U$ be the pullback $x^*(du^1\wedge\dots\wedge du^n)=dx^1\wedge\dots\wedge dx^n$, and $\omega_V=dy^1\wedge\dots\wedge dy^n$. Then we must ask what this function $f$ is. It must exist even if the orientations are incompatible, since $\omega_V$ is never zero, but what is it?

A little thought gives us our answer: $f$ is the Jacobian determinant of the coordinate transformation from one patch to the other. Indeed, we use the Jacobian to change bases on the cotangent bundle, and transforming between these top forms amounts to taking the determinant of the transformation between the $1$-forms $du^i$ and $dv^j$.

So what does this mean? It tells us that if the Jacobian of the coordinate transformation relating two coordinate patches is everywhere positive, then the coordinates have compatible orientations. On the other hand, if the coordinate transformation’s Jacobian is everywhere negative, then the coordinates also have compatible orientations. Why? because even though the sample orientations differ, we can just use $\omega_U$ and $-\omega_V$, which do give the same orientation everywhere.

The problem comes up when the Jacobian is sometimes positive and sometimes negative. Now, it can never be zero, but if the intersection has more than one component it may be positive on one and negative on the other. Then if you pick orientations which coincide on one part of the overlap, they must necessarily disagree on the other part, and no coherent orientation can be chosen for the whole manifold.

I won’t go into this example in full detail yet, but this is essentially what happens with the famous Möbius strip: glue two strips of paper together at one end and we can coherently orient their union. But if we give a half-twist to the other ends before gluing them, we cannot coherently orient the result. The Jacobian is positive on the one overlap and negative on the other.

August 29, 2011