# The Unapologetic Mathematician

## 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 - Posted by | Differential Geometry, Geometry

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