# 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.

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October 6, 2011 - Posted by | Differential Geometry, Geometry

## 12 Comments »

1. […] 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 […]

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2. […] (Pseudo)-Riemannian Metrics, Isometries, Inner Products on 1-Forms, The Hodge Star in Coordinates, The Hodge Star on Differential Forms, Inner Products on Differential […]

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3. […] 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 , which we […]

Pingback by A Hodge Star Example « The Unapologetic Mathematician | October 11, 2011 | Reply

4. […] is a -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 -form back into a -form, which is “really” a vector field […]

Pingback by The Curl Operator « The Unapologetic Mathematician | October 12, 2011 | Reply

5. […] use the Hodge star again to flip the -form back to a -form, so we can apply the exterior derivative to that. We can […]

Pingback by The Divergence Operator « The Unapologetic Mathematician | October 13, 2011 | Reply

6. […] interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of we always get back exactly what we […]

Pingback by The Hodge Star, Squared « The Unapologetic Mathematician | October 18, 2011 | Reply

7. […] we want another way of viewing this orientation. Given a metric on we can use the inverse of the Hodge star from on the orientation -form of , which gives us a covector defined at each point of . Roughly […]

Pingback by (Hyper-)Surface Integrals « The Unapologetic Mathematician | October 27, 2011 | Reply

8. […] do not arise from taking the exterior derivatives of -forms. If is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on which are not the divergence of […]

Pingback by Compact Oriented Manifolds without Boundary have Nontrivial Homology « The Unapologetic Mathematician | November 24, 2011 | Reply

9. […] remember that we’re working in our standard with the standard metric, which lets us use the Hodge star to flip a -form into a -form, and a -form into a vector field! The result is exactly the field […]

Pingback by Gauss’ Law « The Unapologetic Mathematician | January 11, 2012 | Reply

10. […] Now I want to juggle around some of these Hodge stars: […]

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11. […] we push ahead with the Faraday field in hand, we need to properly define the Hodge star in our four-dimensional space, and we need a pseudo-Riemannian metric to do this. Before we were […]

Pingback by Minkowski Space « The Unapologetic Mathematician | March 7, 2012 | Reply

12. […] Hodge Star – which defines an operation where if one has a k-dimensional vector representation of a larger n-dimensional space, one can find a dual representation in the n-k dimensional space – e.g. it is a transformation operation of the basis for some function […]

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