# The Unapologetic Mathematician

## Inner Products of Vector Fields

Now that we can define the inner product of two vectors using a metric $g$, we want to generalize this to apply to vector fields.

This should be pretty straightforward: if $v$ and $w$ are vector fields on an open region $U$ it gives us vectors $v_p$ and $w_p$ at each $p\in U$. We can hit these pairs with $g_p$ to get $g_p(v_p,w_p)$, which is a real number. Since we get such a number at each point $p$, this gives us a function $g(v,w):U\to\mathbb{R}$.

That this $g$ is a bilinear function is clear. In fact we’ve already implied this fact when saying that $g$ is a tensor field. But in what sense is it an inner product? It’s symmetric, since each $g_p$ is, and positive definite as well. To be more explicit: $g_p(v_p,v_p)\geq0$ with equality if and only if $v_p$ is the zero vector in $\mathcal{T}_pM$. Thus the function $g(v,v)$ always takes on nonneative values, is zero exactly where $v$ is, and is the zero function if and only if $v$ is the zero vector field.

What about nondegeneracy? This is a little trickier. Given a nonzero vector field, we can find some point $p$ where $v_p$ is nonzero, and we know that there is some $w_p$ such that $g_p(v_p,w_p)\neq0$. In fact, we can find some region $U$ around $p$ where $v$ is everywhere nonzero, and for each point $q\in U$ we can find a $w_q$ such that $g_q(v_q,w_q)\neq0$. The question is: can we do this in such a way that $w_q$ is a smooth vector field?

The trick is to pick some coordinate map $x$ on $U$, shrinking the region if necessary. Then there must be some $i$ such that

$\displaystyle g_p\left(v_p,\frac{\partial}{\partial x^i}\bigg\vert_p\right)\neq0$

because otherwise $g_p$ would be degenerate. Now we get a smooth function near $p$:

$\displaystyle g\left(v,\frac{\partial}{\partial x^i}\right)$

which is nonzero at $p$, and so must be nonzero in some neighborhood of $p$. Letting $w$ be this coordinate vector field gives us a vector field that when paired with $v$ using $g$ gives a smooth function that is not identically zero. Thus $g$ is also nonzero, and is worthy of the title “inner product” on the module of vector fields $\mathfrak{X}(U)$ over the ring of smooth functions $\mathcal{O}(U)$.

Notice that we haven’t used the fact that the $g_p$ are positive-definite except in the proof that $g$ is, which means that if $g$ is merely pseudo-Riemannian then $g$ is still symmetric and nondegenerate, so it’s still sort of like an inner product, like an symmetric, nondegenerate, but indefinite form is still sort of like an inner product.

September 30, 2011

## Isometries

Sorry for the delay but it’s been sort of hectic with work, my non-math hobbies, and my latest trip up to DC.

Anyway, now that we’ve introduced the idea of a metric on a manifold, it’s natural to talk about mappings that preserve them. We call such maps “isometries”, since they give the same measurements on tangent vectors before and after their application.

Now, normally there’s no canonical way to translate tensor fields from one manifold to another so that we can compare them, but we’ve seen one case where we can do it: pulling back differential forms. This works because differential forms are entirely made from contravariant vector fields, so we can pull back by using the derivative to push forward vectors and then evaluate.

So let’s get explicit: say we have a metric $g_N$ on the manifold $N$, which gives us an inner product $g_{N,q}$ on each tangent space $\mathcal{T}_qN$. If we have a smooth map $f:M\to N$, we want to use it to define a metric $f^*g_N$ on $M$. That is, we want an inner product $(f^*g_N)_p$ on each tangent space $\mathcal{T}_pM$.

Now, given vectors $v$ and $w$ in $\mathcal{T}_pM$, we can use the derivative $f_*$ to push them forward to $f_*v$ and $f_*w$ in $\mathcal{T}_{f(p)}N$. We can hit these with the inner product $g_{N,f(p)}$, defining

$\displaystyle\left[(f^*g_N)_p\right]\left(v,w\right)=g_{N,f(p)}\left(f_*v,f_*w\right)$

It should be straightforward to check that this is indeed an inner product. To be thorough, we must also check that $f^*g_N$ is actually a tensor field. That is, as we move $p$ continuously around $M$ the inner product $(f^*g_N)_p$ varies smoothly.

To check this, we will use our trick: let $(U,x)$ be a coordinate patch around $p$, giving us the basic coordinate vector fields $\frac{\partial}{\partial x^i}$ in the patch. If $(V,y)$ is a coordinate patch around $f(p)$, then we know how to calculate the derivative $f_*$ applied to these vectors:

$\displaystyle f_*\frac{\partial}{\partial x^i}=\sum\limits_{k=1}^n\frac{\partial(y^k\circ f)}{\partial x^i}\frac{\partial}{\partial y^k}$

so we can stick this into the above calculation:

\displaystyle\begin{aligned}\left[(f^*g_N)_p\right]\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)&=g_{N,f(p)}\left(f_*\frac{\partial}{\partial x^i},f_*\frac{\partial}{\partial x^j}\right)\\&=g_{N,f(p)}\left(\sum\limits_{k=1}^n\frac{\partial(y^k\circ f)}{\partial x^i}\frac{\partial}{\partial y^k},\sum\limits_{l=1}^n\frac{\partial(y^l\circ f)}{\partial x^j}\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial(y^k\circ f)}{\partial x^i}\frac{\partial(y^l\circ f)}{\partial x^j}g_{N,f(p)}\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^l}\right)\end{aligned}

Since we assumed that $g_N$ is a metric, the evaluation on the right side is a smooth function, and thus the left side is as well. So we conclude that $f^*g_N$ is a smooth tensor field, and thus a metric.

Now if $M$ comes with its own metric $g_M$, we can ask if the pull-back $f^*g_N$ is equal to $g_M$ at each point. If it is, then we call $f$ an isometry. It’s also common to say that $f$ “preserves the metric”, even though the metric gets pulled back not pushed forward.

September 27, 2011

## (Pseudo-)Riemannian Metrics

Ironically, in order to tie what we’ve been doing back to more familiar material, we actually have to introduce more structure. It’s sort of astonishing in retrospect how much structure comes along with the most basic, intuitive cases, or how much we can do before even using that structure.

In particular, we need to introduce something called a “Riemannian metric”, which will move us into the realm of differential geometry instead of just topology. Everything up until this point has been concerned with manifolds as “shapes”, but we haven’t really had any sense of “size” or “angle” or anything else we could measure. Having these notions — and asking that they be preserved — is the difference between geometry and topology.

Anyway, a Riemannian metric on a manifold $M$ is nothing more than a certain kind of tensor field $g$ of type $(0,2)$ on $M$. At each point $p\in M$, the field $g$ gives us a tensor:

$\displaystyle g_p\in\mathcal{T}_p^*M\otimes\mathcal{T}_p^*M\cong\left(\mathcal{T}_pM\otimes\mathcal{T}_pM\right)^*$

We can interpret this as a bilinear function which takes in two vectors $v_p,w_p\in\mathcal{T}_pM$ and spits out a number $g_p(v_p,w_p)$. That is, $g_p$ is a bilinear form on the space $\mathcal{T}_pM$ of tangent vectors at $p$.

So, what makes $g$ into a Riemannian metric? We now add the assumption that $g_p$ is not just a bilinear form, but that it’s an inner product. That is, $g_p$ is symmetric, nondegenerate, and positive-definite. We can let the last condition slip a bit, in which case we call $g$ a “pseudo-Riemannian metric”. When equipped with a metric, we call $M$ a “(pseudo-)Riemannian manifold”.

It’s common to also say “Riemannian” in the case of negative-definite metrics, since there’s little difference between the cases of signature $(n,0)$ and $(0,n)$. Another common special case is that of a “Lorentzian” metric, which is signature $(n-1,1)$ or $(1,n-1)$.

As we might expect, $g$ is called a metric because it lets us measure things. Specifically, since $g_p$ is an inner product it gives us notions of the length and angle for tangent vectors at $p$. We must be careful here; we do not yet have a way of measuring distances between points on the manifold $M$ itself. The metric only tells us about the lengths of tangent vectors; it is not a metric in the sense of metric spaces. However, if two curves cross at a point $p$ we can use their tangent vectors to define the angle between the curves, so that’s something.

September 20, 2011

## Stokes’ Theorem on Manifolds

Now we come back to Stokes’ theorem, but in the context of manifolds with boundary.

If $M$ is such a manifold of dimension $n$, and if $\omega$ is a compactly-supported $n$-form, then as usual we can use a partition of unity to break up the form into pieces, each of which is supported within the image of an orientation-preserving singular $n$-cube. For each singular cube $c$, either the image $c([0,1]^n)$ is contained totally within the interior of $M$, or it runs up against the boundary. In the latter case, without loss of generality, we can assume that $c([0,1]^n)\cap M$ is exactly the face $c_{n,0}([0,1]^{n-1})$ of $c$ where the $n$th coordinate is zero.

In the first case, our work is easy:

$\displaystyle\int\limits_Md\omega=\int\limits_cd\omega=\int\limits_{\partial d}\omega=\int\limits_{\partial M}\omega$

since $\omega$ is zero everywhere along the image of $\partial c$, and along $\partial M$.

In the other case, the vector fields $\frac{\partial}{\partial u^i}$ — in order — give positively-oriented basss of the tangent spaces of the standard $n$-cube. As $c$ is orientation, preserving, the ordered collection $\left(c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^n}\right)$ gives positively-oriented bases of the tangent spaces of the image of $c$. The basis $\left(c_*\left(-\frac{\partial}{\partial u^n}\right),c_*\frac{\partial}{\partial u^1},\dots,c_*\frac{\partial}{\partial u^{n-1}}\right)$ is positively-oriented if and only if $n$ is even, since we have to pull the $n$th vector past $n-1$ others, picking up a negative sign for each one. But for a point $(a,0)$ with $a\in[0,1]^{-1}$, we see that

$\displaystyle c_{*(a,0)}\left(\frac{\partial}{\partial u^i}\right)=(c_{n,0})_{*a}\left(\frac{\partial}{\partial u^i}\right)$

for all $1\leq i\leq n-1$. That is, these image vectors are all within the tangent space of the boundary, and in this order. And since $c_*\left(-\frac{\partial}{\partial u^n}\right)$ is outward-pointing, this means that $c_{n,0}:[0,1]^{n-1}\to\partial M$ is orientation-preserving if and only if $n$ is even.

Now we can calculate

\displaystyle\begin{aligned}\int\limits_Md\omega&=\int\limits_cd\omega\\&=\int\limits_{\partial c}\omega\\&=\int\limits_{(-1)^nc_{n,0}}\omega\\&=(-1)^n\int\limits_{c_{n,0}}\omega\\&=(-1)^n(-1)^n\int\limits_{\partial M}\omega\\&=\int\limits_{\partial M}\omega\end{aligned}

where we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the $(n,0)$ face of a singular $n$-cube to cancel each other off.

So in general we find

\displaystyle\begin{aligned}\int\limits_{\partial M}\omega&=\sum\limits_{\phi\in\Phi}\int\limits_{\partial M}\phi\omega\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md(\phi\omega)\\&=\sum\limits_{\phi\in\Phi}\int\limits_M\left(d\phi\wedge\omega+\phi d\omega\right)\\&=\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega+\int\limits_Md\omega\end{aligned}

The last sum is finite, since on of the support of $\omega$ all but finitely many of the $\phi$ are constantly zero, meaning that their differentials are zero as well. Since the sum is (locally) finite, we have no problem pulling it all the way inside:

$\displaystyle\sum\limits_{\phi\in\Phi}\int\limits_Md\phi\wedge\omega=\int\limits_Md\left(\sum\limits_{\phi\in\Phi}\phi\right)\wedge\omega=\int\limits_Md\left(1\right)\wedge\omega=0$

so the sum cancels off, leaving just the integral, as we’d expect. That is, under these circumstances,

$\displaystyle\int\limits_Md\omega=\int\limits_{\partial M}\omega$

which is Stokes’ theorem on manifolds.

September 16, 2011

## 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