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 $\mathbb{R}^3$ .

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October 11, 2011 - Posted by John Armstrong | Differential Geometry, Geometry

2 Comments »

[...] continue our example considering the special case of as an oriented, Riemannian manifold, with the coordinate -forms [...]

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[...] can check that this will be automatically zero if we start with an image of the curl operator; our earlier calculations show that is always the identity mapping — at least on with this metric — so if we [...]

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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