# The Unapologetic Mathematician

## (Hyper-)Surface Integrals

The flip side of the line integral is the surface integral.

Given an $n$-manifold $M$ we let $S$ be an oriented $n-1$-dimensional “hypersurface”, which term we use because in the usual case of $M=\mathbb{R}^3$ a hypersurface has two dimensions — it’s a regular surface. The orientation on a hypersurface consists of $n-1$ tangent vectors which are all in the image of the derivative of the local parameterization map, which is a singular cube.

Now we want another way of viewing this orientation. Given a metric on $M$ we can use the inverse of the Hodge star from $M$ on the orientation $n-1$-form of $S$, which gives us a covector $\nu(p)\in\mathcal{T}^*_pM$ defined at each point $p$ of $S$. Roughly speaking, the orientation form of $S$ defines an $n-1$-dimensional subspace of $\mathcal{T}^*_pM$: the covectors “tangent to $S$“. The covector $\nu(p)$ is “perpendicular to $S"$, and since $S$ has only one fewer dimension than $M$ there are only two choices for the direction of such a vector. The choice of one or the other defines a distinction between the two sides of $S$.

If we flip around our covectors into vectors — again using the metric — we get something like a vector field defined on $S$. I say “something like” because it’s only defined on $S$, which is not an open subspace of $M$, and strictly speaking a vector field must be defined on an open subspace. It’s not hard to see that we could “thicken up” $S$ into an open subspace and extend our vector field smoothly there, so I’m not really going to make a big deal of it, but I want to be careful with my language; it’s also why I didn’t say we get a $1$-form from the Hodge star. Anyway, we will call this vector-valued function $dS$, for reasons that will become apparent shortly.

Now what if we have another vector-valued function $F$ defined on $S$ — for example, it could be a vector field defined on an open subspace containing $S$. We can define the “surface integral of $F$ through $S$“, which measures how much the vector field flows through the surface in the direction our orientation picks out as “positive”. And we measure this amount at any given point $p$ by taking the covector at $p$ provided by the Hodge star and evaluating it at the vector $F(p)$. This gives us a value that we can integrate over the surface. This evaluation can be flipped around into our vector field notation and language, allowing us to write down the integral as

$\displaystyle\int\limits_SF\cdot dS$

because the “dot product” $\left[F\cdot dS\right](p)$ is exactly what it means to evaluate the covector dual to $dS(p)$ at the vector $F(p)$. This should look familiar from multivariable calculus, but I’ve been saying that a lot lately, haven’t I?

We can also go the other direction and make things look more abstract. We could write our vector field as a $1$-form $\alpha$, which lets us write our surface integral as

$\displaystyle\int\limits_S\langle\alpha,\nu\rangle\omega_M=\int\limits_S\alpha\wedge*\nu$

where $\omega_M$ is the volume form on $M$. But then we know that $*\nu=\omega_S$ the volume form on $S$. That is, we can define a hypersurface integral of any $1$-form $\alpha$ across any hypersurface $S$ equipped with a volume form $\omega_S$ by

$\displaystyle\int\limits_S\alpha\wedge\omega_S$

whether or not we have a metric on $M$.

October 27, 2011

## The Fundamental Theorem of Line Integrals

At last we can state one of our special cases of Stokes’ theorem. We don’t need to prove it, since we already did in far more generality, but it should help put our feet on some familiar ground.

So, we take an oriented curve $c$ in a manifold $M$. The image of $c$ is an oriented submanifold of $M$, where the “orientation” means picking one of the two possible tangent directions at each point along the image of the curve. As a high-level view, we can characterize the orientation as the direction that we traverse the curve, from one “starting” endpoint to the other “ending” endpoint.

Given any $1$-form $\alpha$ on the image of $c$ — in particular, given an $\alpha$ defined on $M$ — we can define the line integral of $\alpha$ over $c$. We already have a way of evaluating line integrals: pull the $1$-form back to the parameter interval of $c$ and integrate there as usual. But now we want to use Stokes’ theorem to come up with another way. Let’s write down what it will look like:

$\displaystyle\int\limits_cd\omega=\int\limits_{\partial c}\omega$

where $\omega$ is some $0$-form. That is: a function. This tells us that we can only make this work for “exact” $1$-forms $\alpha$, which can be written in the form $\alpha=df$ for some function $f$.

But if this is the case, then life is beautiful. The (oriented) boundary of $c$ is easy: it consists of two $0$-faces corresponding to the two endpoints. The starting point gets a negative orientation while the ending point gets a positive orientation. And so we write

$\displaystyle\int\limits_cdf=\int\limits_{\partial c}f=\sum\limits_{j=0,1}(-1)^{1+j}f(c_{1,j})=f(c(1))-f(c(0))$

That is, we just evaluate $f$ at the two endpoints and subtract the value at the start from the value at the end!

What does this look like when we have a metric and we can rewrite the $1$-form $\alpha$ as a vector field $F$? In this case, $\alpha$ is exact if and only if $F$ is “conservative”, which just means that $F=\nabla f$ for some function $f$. Then we can write

$\displaystyle\int\limits_c\nabla f\cdot ds=f(c(1))-f(c(0))$

which should look very familiar from multivariable calculus.

We call this the fundamental theorem of line integrals by analogy with the fundamental theorem of calculus. Indeed, if we set this up in the manifold $\mathbb{R}$, we get back exactly the second part of the fundamental theorem of calculus back again.

October 24, 2011

## Line Integrals

We now define some particular kinds of integrals as special cases of our theory of integrals over manifolds. And the first such special case is that of a line integral.

Consider an oriented curve $c$ in the manifold $M$. We know that this is a singular $1$-cube, and so we can pair it off with a $1$-form $\alpha$. Specifically, we pull back $\alpha$ to $c^*\alpha$ on the interval $[0,1]$ and integrate.

More explicitly, the pullback $c^*\alpha$ is evaluated as

$\displaystyle\left[c^*\alpha\left(\frac{d}{dt}\right)\right](t_0)=\left[\alpha_{c(t_0)}\right]\left(c_{*t_0}\frac{d}{dt}\bigg\vert_{t_0}\right)$

That is, for a $t_0\in[0,1]$, we take the value $\alpha_{c(t_0)}\in\mathcal{T}^*_{c(t_0)}M$ of the $1$-form $\alpha$ at the point $c(t_0)\in M$ and the tangent vector $c'(t_0)\in\mathcal{T}_{c(t_0)}M$ and pair them off. This gives us a real-valued function which we can integrate over the interval.

So, why do we care about this particularly? In the presence of a metric, we have an equivalence between $1$-forms $\alpha$ and vector fields $F$. And specifically we know that the pairing $\alpha_{c(t)}\left(c'(t)\right)$ is equal to the inner product $\langle F(c(t)),c'(t)\rangle$ — this is how the equivalence is defined, after all. And thus the line integral looks like

$\displaystyle\int\limits_c\alpha=\int\limits_{[0,1]}\langle F(c(t)),c'(t)\rangle\,dt$

Often the inner product is written with a dot — usually called the “dot product” of vectors — in which case this takes the form

$\displaystyle\int\limits_{[0,1]}F(c(t))\cdot c'(t)\,dt$

We also often write $ds=c'(t)\,dt$ as a “vector differential-valued function”, in which case we can write

$\displaystyle\int\limits_{[0,1]}F\cdot ds$

Of course, we often parameterize a curve by a more general interval $I$ than $[0,1]$, in which case we write

$\displaystyle\int\limits_IF\cdot ds$

This expression may look familiar from multivariable calculus, where we first defined line integrals. We can now see how this definition is a special case of a much more general construction.

October 21, 2011

## The Codifferential

From our calculation of the square of the Hodge star we can tell that the star operation is invertible. Indeed, since $*^2=(-1)^{k(n-k)}\lvert g_{ij}\rvert$ — applying the star twice to a $k$-form in an $n$-manifold with metric $g$ is the same as multiplying it by $(-1)^{k(n-k)}$ and the determinant of the matrix of $g$ — we conclude that $*^{-1}=(-1)^{k(n-k)}\lvert g^{ij}\rvert*$.

With this inverse in hand, we will define the “codifferential”

$\displaystyle\delta=(-1)^k*^{-1}d*$

The first star sends a $k$-form to an $n-k$-form; the exterior derivative sends it to an $n-k+1$-form; and the inverse star sends it to a $k-1$-form. Thus the codifferential goes in the opposite direction from the differential — the exterior derivative.

Unfortunately, it’s not quite as algebraically nice. In particular, it’s not a derivation of the algebra. Indeed, we can consider $fdx$ and $gdy$ in $\mathbb{R}^3$ and calculate

\displaystyle\begin{aligned}\delta(fdx)&=-*d*(fdx)=-*d(fdy\wedge dz)=-*\frac{\partial f}{\partial x}dx\wedge dy\wedge dz=-\frac{\partial f}{\partial x}\\\delta(gdy)&=-*d*(gdy)=-*d(gdz\wedge dx)=-*\frac{\partial g}{\partial y}dy\wedge dz\wedge dx=-\frac{\partial g}{\partial y}\end{aligned}

while

\displaystyle\begin{aligned}\delta(fgdx\wedge dy)&=-*d*(fgdx\wedge dy)\\&=-*d(fgdz)\\&=-*\left(\left(\frac{\partial f}{\partial x}g+f\frac{\partial g}{\partial x}\right)dx\wedge dz+\left(\frac{\partial f}{\partial y}g+f\frac{\partial g}{\partial y}\right)dy\wedge dz\right)\\&=\left(\frac{\partial f}{\partial x}g+f\frac{\partial g}{\partial x}\right)dy-\left(\frac{\partial f}{\partial y}g+f\frac{\partial g}{\partial y}\right)dx\end{aligned}

but there is no version of the Leibniz rule that can account for the second and third terms in this latter expansion. Oh well.

On the other hand, the codifferential $\delta$ is (sort of) the adjoint to the differential. Adjointness would mean that if $\eta$ is a $k$-form and $\zeta$ is a $k+1$-form, then

$\displaystyle\langle d\eta,\zeta\rangle=\langle\eta,\delta\zeta\rangle$

where these inner products are those induced on differential forms from the metric. This doesn’t quite hold, but we can show that it does hold “up to homology”. We can calculate their difference times the canonical volume form

\displaystyle\begin{aligned}\left(\langle d\eta,\zeta\rangle-\langle\eta,\delta\zeta\rangle\right)\omega&=d\eta\wedge*\zeta-\eta\wedge*\delta\zeta\\&=d\eta\wedge*\zeta-(-1)^k\eta\wedge**^{-1}d*\zeta\\&=d\eta\wedge*\zeta-(-1)^k\eta\wedge d*\zeta\\&=d\left(\eta\wedge*\zeta\right)\end{aligned}

which is an exact $n$-form. It’s not quite as nice as equality, but if we pass to De Rham cohomology it’s just as good.

October 21, 2011

## The Hodge Star, Squared

It’s interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of $\mathbb{R}^3$ we always get back exactly what we started with. That is, in this case, $*^2=1$ — the identity operation.

It’s easiest to work this out in coordinates. If $\eta=dx^{i_1}\wedge\dots\wedge dx^{i_k}$ is some $k$-fold wedge then $*\eta$ is $\pm\sqrt{\lvert g_{ij}\rvert}$ times the wedge of all the indices that don’t show up in $\eta$. But then $**\eta$ is $\pm\sqrt{\lvert g_{ij}\rvert}$ times the wedge of all the indices that don’t show up in $*\eta$, which is exactly all the indices that were in $\eta$ in the first place! That is, $**\eta=\pm\lvert g_{ij}\rvert\eta$, where the sign is still a bit of a mystery.

But some examples will quickly shed some light on this. We can even extend to the pseudo-Riemannian case and pick a coordinate system so that $\langle dx^i,dx^j\rangle=\epsilon_i\delta^{ij}$, where $\epsilon_i=\pm1$. That is, any two $dx^i$ are orthogonal, and each $dx^i$ either has “squared-length” $1$ or $-1$. The determinant $\lvert g_{ij}\rvert$ is simply the product of all these signs. In the Riemannian case this is simply $1$.

Now, let’s start with an easy example: let $\eta$ be the wedge of the first $k$ indices:

$\displaystyle\eta=dx^1\wedge\dots\wedge dx^k$

Then $*\eta$ is (basically) the wedge of the other indices:

$\displaystyle*\eta=\sqrt{\lvert g_{ij}\rvert}dx^{k+1}\wedge\dots\wedge dx^n$

The sign is positive since $\eta\wedge*\eta$ already has the right order. But now we flip this around the other way:

$\displaystyle**\eta=\pm\lvert g_{ij}\rvert dx^1\wedge\dots\wedge dx^k$

but this should obey the same rule as ever:

\displaystyle\begin{aligned}*\eta\wedge**\eta&=\pm\lvert g_{ij}\rvert\sqrt{\lvert g_{ij}\rvert}dx^{k+1}\wedge\dots\wedge dx^n\wedge dx^1\wedge\dots\wedge dx^k\\&=\pm\lvert g_{ij}\rvert(-1)^{k(n-k)}\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\\&=\pm\lvert g_{ij}\rvert(-1)^{k(n-k)}\omega\end{aligned}

where we pick up a factor of $-1$ each time we pull one of the last $k$ $1$-forms leftwards past the first $n-k$. We conclude that the actual sign must be $\lvert g_{ij}\rvert(-1)^{k(n-k)}$ so that this result is exactly $\omega$. Similar juggling for other selections of $\eta$ will give the same result.

In our special Riemannian case with $n=3$, then no matter what $k$ is we find the sign is always positive, as we expected. The same holds true in any odd dimension for Riemannian manifolds. In even dimensions, when $k$ is odd then so is $n-k$, and so $*^2=-1$ — the negative of the identity transformation. And the whole situation gets more complicated in the pseudo-Riemannian version depending on the number of $-1$s in the diagonalized metric tensor.

October 18, 2011

## The Divergence Operator

One fact that I didn’t mention when discussing the curl operator is that the curl of a gradient is zero: $\nabla\times(\nabla f)=0$. In our terms, this is a simple consequence of the nilpotence of the exterior derivative. Indeed, when we work in terms of $1$-forms instead of vector fields, the composition of the two operators is $*d(df)$, and $d^2$ is always zero.

So why do we bring this up now? Because one of the important things to remember from multivariable calculus is that the divergence of a curl is also automatically zero, and this will help us figure out what a divergence is in terms of differential forms. See, if we take our vector field and consider it as a $1$-form, the exterior derivative is already known to be (essentially) the curl. So what else can we do?

We use the Hodge star again to flip the $1$-form back to a $2$-form, so we can apply the exterior derivative to that. We can check that this will be automatically zero if we start with an image of the curl operator; our earlier calculations show that $*^2$ is always the identity mapping — at least on $\mathbb{R}^3$ with this metric — so if we first apply the curl $*d$ and then the steps we’ve just suggested, the result is like applying the operator $d**d=dd=0$.

There’s just one catch: as we’ve written it this gives us a $3$-form, not a function like the divergence operator should! No matter; we can break out the Hodge star once more to flip it back to a $0$-form — a function — just like we want. That is, the divergence operator on $1$-forms is $*d*$.

Let’s calculate this in our canonical basis. If we start with a $1$-form $\alpha=Pdx+Qdy+Rdz$ then we first hit it with the Hodge star:

$\displaystyle*\alpha=Pdy\wedge dz+Qdz\wedge dx+Rdx\wedge dy$

Next comes the exterior derivative:

$\displaystyle d*\alpha=\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right)dx\wedge dy\wedge dz$

and then the Hodge star again:

$\displaystyle*d*\alpha=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

which is exactly the definition (in coordinates) of the usual divergence $\nabla\cdot F$ of a vector field $F$ on $\mathbb{R}^3$.

October 13, 2011

## The Curl Operator

Let’s continue our example considering the special case of $\mathbb{R}^3$ as an oriented, Riemannian manifold, with the coordinate $1$-forms $\{dx, dy, dz\}$ forming an oriented, orthonormal basis at each point.

We’ve already see the gradient vector $\nabla f$, which has the same components as the differential $df$:

\displaystyle\begin{aligned}df&=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz\\\nabla f&=\begin{pmatrix}\displaystyle\frac{\partial f}{\partial x}\\\displaystyle\frac{\partial f}{\partial y}\\\displaystyle\frac{\partial f}{\partial z}\end{pmatrix}\end{aligned}

This is because we use the metric to convert from vector fields to $1$-forms, and with respect to our usual bases the matrix of the metric is the Kronecker delta.

We will proceed to define analogues of the other classical differential operators you may remember from multivariable calculus. We will actually be defining operators on differential forms, but we will use this same trick to identify vector fields and $1$-forms. We will thus not usually distinguish our operators from the classical ones, but in practice we will use the classical notations when acting on vector fields and our new notations when acting on $1$-forms.

Anyway, the next operator we come to is the curl of a vector field: $F\mapsto\nabla\times F$. Of course we’ll really start with a $1$-form instead of a vector field, and we already know a differential operator to use on forms. Given a $1$-form $\alpha$ we can send it to $d\alpha$.

The only hangup is that this is a $2$-form, while we want the curl of a vector field to be another vector field. But we do have a Hodge star, which we can use to flip a $2$-form back into a $1$-form, which is “really” a vector field again. That is, the curl operator corresponds to the differential operator $*d$ that takes $1$-forms back to $1$-forms.

Let’s calculate this in our canonical basis, to see that it really does look like the familiar curl. We start with a $1$-form $\alpha=Pdx+Qdy+Rdz$. The first step is to hit it with the exterior derivative, which gives

\displaystyle\begin{aligned}d\alpha=&dP\wedge dx+dQ\wedge dy + dR\wedge dz\\=&\left(\frac{\partial P}{\partial x}dx+\frac{\partial P}{\partial y}dy+\frac{\partial P}{\partial z}dz\right)\wedge dx\\&+\left(\frac{\partial Q}{\partial x}dx+\frac{\partial Q}{\partial y}dy+\frac{\partial Q}{\partial z}dz\right)\wedge dy\\&+\left(\frac{\partial R}{\partial x}dx+\frac{\partial R}{\partial y}dy+\frac{\partial R}{\partial z}dz\right)\wedge dz\\=&\frac{\partial P}{\partial y}dy\wedge dx+\frac{\partial P}{\partial z}dz\wedge dx\\&+\frac{\partial Q}{\partial x}dx\wedge dy+\frac{\partial Q}{\partial z}dz\wedge dy\\&+\frac{\partial R}{\partial x}dx\wedge dz+\frac{\partial R}{\partial y}dy\wedge dz\\=&\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dy\wedge dz\\&+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dz\wedge dx\\&+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\wedge dy\end{aligned}

Next we hit this with the Hodge star. We’ve already calculated how the Hodge star affects the canonical basis of $2$-forms, so this is just a simple lookup to find:

$\displaystyle*d\alpha=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)dx+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)dy+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dz$

which are indeed the usual components of the curl. That is, if $\alpha$ is the $1$-form corresponding to the vector field $F$, then $*d\alpha$ is the $1$-form corresponding to the vector field $\nabla\times F$.

October 12, 2011

## A Hodge Star Example

I want to start getting into a nice, simple, concrete example of the Hodge star. We need an oriented, Riemannian manifold to work with, and for this example we take $\mathbb{R}^3$, which we cover with the usual coordinate patch with coordinates we call $\{x,y,z\}$.

To get a metric, we declare the coordinate covector basis $\{dx,dy,dz\}$ to be orthonormal, which means that we have the matrix

$\displaystyle g^{ij}=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}$

and also the inner product matrix

$\displaystyle g_{ij}=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}$

since we know that $g_{ij}$ and $g^{ij}$ are inverse matrices. And so we get the canonical volume form

$\displaystyle\omega=\sqrt{\det\left(g_{ij}\right)}dx\wedge dy\wedge dz=dx\wedge dy\wedge dz$

We declare our orientation of $\mathbb{R}^3$ to be the one corresponding to this top form.

Okay, so now we can write down the Hodge star in its entirety. And in fact we’ve basically done this way back when we were talking about the Hodge star on a single vector space:

\displaystyle\begin{aligned}*1&=dx\wedge dy\wedge dz\\ *dx&=dy\wedge dz\\ *dy&=-dx\wedge dz=dz\wedge dx\\ *dz&=dx\wedge dy\\ *(dx\wedge dy)&=dz\\ *(dx\wedge dz)&=-dy\\ *(dy\wedge dz)&=dx\\ *(dx\wedge dy\wedge dz)&=1\end{aligned}

So, what does this buy us? Something else that we’ve seen before in the context of a single vector space. Let’s say that $v$ and $w$ are two vector fields defined on an open subset $U\subseteq\mathbb{R}^3$. We can write these out in our coordinate basis:

\displaystyle\begin{aligned}v&=v_x\frac{\partial}{\partial x}+v_y\frac{\partial}{\partial y}+v_z\frac{\partial}{\partial z}\\w&=w_x\frac{\partial}{\partial x}+w_y\frac{\partial}{\partial y}+w_z\frac{\partial}{\partial z}\end{aligned}

Now, we can use our metric to convert these vectors to covectors — vector fields to $1$-forms. We use the matrix $g_{ij}$ to get

\displaystyle\begin{aligned}g(v,\underline{\hphantom{X}})&=v_xdx+v_ydy+v_zdz\\g(w,\underline{\hphantom{X}})&=w_xdx+w_ydy+w_zdz\end{aligned}

Next we can wedge these together

\displaystyle\begin{aligned}g(v,\underline{\hphantom{X}})\wedge g(w,\underline{\hphantom{X}})=&(v_yw_z-v_zw_y)dy\wedge dz\\&+(v_zw_x-v_xw_z)dz\wedge dx\\&+(v_xw_y-v_yw_x)dx\wedge dy\end{aligned}

Now we come to the Hodge star!

\displaystyle\begin{aligned}*(g(v,\underline{\hphantom{X}})\wedge g(w,\underline{\hphantom{X}}))=&(v_yw_z-v_zw_y)dx\\&+(v_zw_x-v_xw_z)dy\\&+(v_xw_y-v_yw_x)dz\end{aligned}

and now we’re back to a $1$-form, so we can use the metric to flip it back to a vector field:

\displaystyle\begin{aligned}g\left(*(g(v,\underline{\hphantom{X}})\wedge g(w,\underline{\hphantom{X}})),\underline{\hphantom{X}}\right)=&(v_yw_z-v_zw_y)\frac{\partial}{\partial x}\\&+(v_zw_x-v_xw_z)\frac{\partial}{\partial y}\\&+(v_xw_y-v_yw_x)\frac{\partial}{\partial z}\end{aligned}

Here, the outermost $g(\underline{\hphantom{X}},\underline{\hphantom{X}})$ is the inner product on $1$-forms, while the inner ones are the inner product on vector fields. This is exactly the cross product of vector fields on $\mathbb{R}^3$.

October 11, 2011

## The Hodge Star in Coordinates

It will be useful to be able to write down the Hodge star in a local coordinate system. So let’s say that we’re in an oriented coordinate patch $(U,x)$ of an oriented Riemannian manifold $M$, which means that we have a canonical volume form that locally looks like

$\displaystyle\omega=\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n$

Now, we know that any $k$-form on $U$ can be written out as a sum of functions times $k$-fold wedges:

$\displaystyle\eta=\sum\limits_{1\leq i_1<\dots

Since the star operation is linear, we just need to figure out what its value is on the $k$-fold wedges. And for these the key condition is that for every $k$-form $\zeta$ we have

$\displaystyle\zeta\wedge*(dx^{i_1}\wedge\dots\wedge dx^{i_k})=\langle\zeta,dx^{i_1}\wedge\dots\wedge dx^{i_k}\rangle\omega$

Since both sides of this condition are linear in $\zeta$, we also only need to consider values of $\zeta$ which are $k$-fold wedges. If $\zeta$ is not the same wedge as $\eta$, then the inner product is zero, while if $\zeta=\eta$ then

\displaystyle\begin{aligned}(dx^{i_1}\wedge\dots\wedge dx^{i_k})\wedge*(dx^{i_1}\wedge\dots\wedge dx^{i_k})&=\langle dx^{i_1}\wedge\dots\wedge dx^{i_k},dx^{i_1}\wedge\dots\wedge dx^{i_k}\rangle\omega\\&=\det\left(\langle dx^{i_j},dx^{i_k}\rangle\right)\omega\\&=\det\left(\delta^{jk}\right)\omega\\&=\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\end{aligned}

And so $*(dx^{i_1}\wedge\dots\wedge dx^{i_k})$ must be $\pm\sqrt{\lvert g_{ij}\rvert}$ times the $n-k$-fold wedge made up of all the $dx^i$ that do not show up in $\eta$. The positive or negative sign is decided by which order gives us an even permutation of all the $dx^i$ on the left-hand side of the above equation.

October 8, 2011

## The Hodge Star on Differential Forms

Let’s say that $M$ is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on differential forms, and of course we have a wedge product of differential forms. We have almost everything we need to define an analogue of the Hodge star on differential forms; we just need a particular top — or “volume” — form at each point.

To this end, pick one or the other orientation, and let $(U,x)$ be a coordinate patch such that the form $dx^1\wedge\dots\wedge dx^n$ is compatible with the chosen orientation. We’d like to use this form as our top form, but it’s heavily dependent on our choice of coordinates, so it’s very much not a geometric object — our ideal choice of a volume form will be independent of particular coordinates.

So let’s see how this form changes; if $(V,y)$ is another coordinate patch, we can assume that $U=V$ by restricting each patch to their common intersection. We’ve already determined that the forms differ by a factor of the Jacobian determinant:

$\displaystyle dx^1\wedge\dots\wedge dx^n=\det\left(\frac{\partial x^i}{\partial x^j}\right)dy^1\wedge\dots\wedge dy^n$

What we want to do is multiply our form by some function that transforms the other way, so that when we put them together the product will be invariant.

Now, we already have something else floating around in our discussion: the metric tensor $g$. When we pick coordinates $x^i$ we get a matrix-valued function:

$\displaystyle g^x_{ij}=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$

and similarly with respect to the alternative coordinates $y^i$:

$\displaystyle g^y_{ij}=g\left(\frac{\partial}{\partial y^i},\frac{\partial}{\partial y^j}\right)$

So, what’s the difference between these two matrix-valued functions? We can calculate two ways:

\displaystyle\begin{aligned}g^x_{ij}&=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)\\&=g\left(\sum\limits_{k=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial}{\partial y^k},\sum\limits_{l=1}^n\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j}g\left(\frac{\partial}{\partial y^k},\frac{\partial}{\partial y^l}\right)\\&=\sum\limits_{k,l=1}^n\frac{\partial y^k}{\partial x^i}\frac{\partial y^l}{\partial x^j}g^y_{kl}\end{aligned}

That is, we transform the metric tensor with two copies of the inverse Jacobian matrix. Indeed, we could have come up with this on general principles, since $g$ has type $(0,2)$ — a tensor of type $(m,n)$ transforms with $m$ copies of the Jacobian and $n$ copies of the inverse Jacobian.

Anyway, now we can take the determinant of each side:

$\displaystyle\lvert g^x_{ij}\rvert=\left\lvert\frac{\partial y^i}{x^j}\right\rvert^2\lvert g^y_{ij}\rvert$

and taking square roots we find:

$\displaystyle\sqrt{\lvert g^x_{ij}\rvert}=\left\lvert\frac{\partial y^i}{x^j}\right\rvert\sqrt{\lvert g^y_{ij}\rvert}$

Thus the square root of the metric determinant is a function that transforms from one coordinate patch to the other by the inverse Jacobian determinant. And so we can define:

$\displaystyle\omega_U=\sqrt{\lvert g^x_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\in\Omega^n_M(U)$

which does depend on the coordinate system to write down, but which is actually invariant under a change of coordinates! That is, $\omega_U=\omega_V$ on the intersection $U\cap V$. Since the algebras of differential forms form a sheaf $\Omega^n_M$, we know that we can patch these $\omega_U$ together into a unique $\omega\in\Omega^n_M(M)$, and this is our volume form.

And now we can form the Hodge star, point by point. Given any $k$-form $\eta$ we define the dual form $*\eta$ to be the unique $n-k$-form such that

$\displaystyle\zeta\wedge*\eta=\langle\zeta,\eta\rangle\omega$

for all $k$-forms $\zeta\in\Omega^k(M)$. Since at every point $p\in M$ we have an inner product and a wedge $\omega(p)\in A^n(\mathcal{T}^*_pM)$, we can find a $*\eta(p)\in A^{n-k}(\mathcal{T}^*_pM)$. Some general handwaving will suffice to show that $*\eta$ varies smoothly from point to point.

October 6, 2011