# Riemannian Metric and the Hodge-Star in 3D

## Euclidean Space

The flat metric is

$\begin{split}g=\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right]\end{split}$

Given one-forms $$\mathbf{U}=U_{x}dx+U_{y}dy+U_{z}dz$$ and $$\mathbf{V}=V_{x}dx+V_{y}dy+V_{z}dz$$, for brevity we will write them as column vectors

$\begin{split}\mathbf{U}=\left[\begin{array}{c} U_{x}\\ U_{y}\\ U_{z} \end{array}\right]_{1}\qquad\mathbf{V}=\left[\begin{array}{c} V_{x}\\ V_{y}\\ V_{z} \end{array}\right]_{1}\end{split}$

and their inner product is given by

$\mathbf{U}\cdot\mathbf{V}=U_{x}V_{x}+U_{y}V_{y}+U_{z}V_{z}$

Give two-forms $$\mathbf{U}=U_{yz}\,dy\wedge dz+U_{zx}\,dz\wedge dx+U_{xy}\,dx\wedge dy$$ and $$\mathbf{V}=U_{yz}\,dy\wedge dz+U_{zx}\,dz\wedge dx+U_{xy}\,dx\wedge dy$$, again we will express them as column vectors

$\begin{split}\mathbf{U}=\left[\begin{array}{c} U_{yz}\\ U_{zx}\\ U_{xy} \end{array}\right]_{2}\qquad\mathbf{V}=\left[\begin{array}{c} V_{yz}\\ V_{zx}\\ V_{xy} \end{array}\right]_{2}\end{split}$

and their inner product is given by

$\mathbf{U}\cdot\mathbf{V}=U_{yz}V_{yz}+U_{zx}V_{zx}+U_{xy}V_{xy}$

The Hodge star must be such that we must have the following properties satisfied for one-forms

$\mathbf{U}\wedge\star\mathbf{U}=\left(\mathbf{U}\cdot\mathbf{U}\right)\mu$

Clearly, since $$\star dx=dy\wedge dz$$, $$\star dy=dz\wedge dx$$ and $$\star dz=dx\wedge dy$$, then for a one-form $$\mathbf{U}$$

$\begin{split}\mathbf{U}=\left[\begin{array}{c} U_{x}\\ U_{y}\\ U_{z} \end{array}\right]_{1}\qquad\star\mathbf{U}=\left[\begin{array}{c} U_{x}\\ U_{y}\\ U_{z} \end{array}\right]_{2}\end{split}$

and for a two-form

$\begin{split}\mathbf{U}=\left[\begin{array}{c} U_{yz}\\ U_{zx}\\ U_{xy} \end{array}\right]_{2}\qquad\star\mathbf{U}=\left[\begin{array}{c} U_{yz}\\ U_{zx}\\ U_{xy} \end{array}\right]_{1}\end{split}$

## General (Non-Euclidean) Space

In this case the metric (inner product between one-forms) is given by

$\begin{split}g^{ij}=\left[\begin{array}{ccc} g^{xx} & g^{xy} & g^{zx}\\ g^{xy} & g^{yy} & g^{yz}\\ g^{zx} & g^{yz} & g^{zz} \end{array}\right]\end{split}$

The inner product between two one-forms is then given by

$\begin{split}\mathbf{U}\cdot\mathbf{V}=\mathbf{U}^{T}g\mathbf{V}=\left[\begin{array}{ccc} U_{x} & U_{y} & U_{z}\end{array}\right]_{1}\left[\begin{array}{ccc} g^{xx} & g^{xy} & g^{zx}\\ g^{xy} & g^{yy} & g^{yz}\\ g^{zx} & g^{yz} & g^{zz} \end{array}\right]\left[\begin{array}{c} V_{x}\\ V_{y}\\ V_{z} \end{array}\right]_{1}\end{split}$

Expanding the matrix expressions

\begin{split}\begin{aligned} \mathbf{U}\cdot\mathbf{V} & =g^{xx}U_{x}V_{x}+g^{yy}U_{y}V_{y}+g^{zz}U_{z}V_{z}+\\ & \quad+g^{xy}\left(U_{x}V_{y}+U_{y}V_{x}\right)+g^{yz}\left(U_{y}V_{z}+U_{z}V_{y}\right)+g^{zx}\left(U_{x}V_{z}+U_{z}V_{x}\right)\end{aligned}\end{split}

To compute the inner product between two-forms we need to consider terms of the form $$\left(dx^{i}\wedge dx^{j}\right)\cdot\left(dx^{\alpha}\wedge dx^{\beta}\right)$$. The wedge product is an anti-symmetric tensor product, it is given by

$dx^{i}\wedge dx^{j}=dx^{i}\otimes dx^{j}-dx^{j}\otimes dx^{i}$

The inner product between two forms is then given by

\begin{aligned} \left(dx^{i}\wedge dx^{j}\right)\cdot\left(dx^{\alpha}\wedge dx^{\beta}\right) & =\left(dx^{i}\otimes dx^{j}-dx^{j}\otimes dx^{i}\right)\cdot\left(dx^{\alpha}\otimes dx^{\beta}-dx^{\beta}\otimes dx^{\alpha}\right)\end{aligned}

where up to a normalization constant $$N$$

\begin{split}\begin{aligned} \left(dx^{i}\otimes dx^{j}\right)\cdot\left(dx^{\alpha}\otimes dx^{\beta}\right) & =N\left(dx^{i}\cdot dx^{\alpha}\right)\left(dx^{j}\cdot dx^{\beta}\right)\\ & =Ng^{i\alpha}g^{j\beta}\end{aligned}\end{split}

we can deduce that

\begin{split}\begin{aligned} \left(dx^{i}\wedge dx^{j}\right)\cdot\left(dx^{\alpha}\wedge dx^{\beta}\right) & =N\left(g^{i\alpha}g^{j\beta}-g^{i\beta}g^{j\alpha}-g^{j\alpha}g^{i\beta}+g^{j\beta}g^{i\alpha}\right)\\ & =2N\left(g^{i\alpha}g^{j\beta}-g^{i\beta}g^{j\alpha}\right)\end{aligned}\end{split}

The normalization constant will be chosen so that $$dx^{i}\wedge dx^{j}$$ form an orthonormal basis, thefore $$N=1/2$$, and

$\left(dx^{i}\otimes dx^{j}\right)\cdot\left(dx^{\alpha}\otimes dx^{\beta}\right)=\frac{1}{2}g^{i\alpha}g^{j\beta}$

The inner product between two two-forms then will be given by

$\begin{split}\mathbf{U}\cdot\mathbf{V}=\left[\begin{array}{ccc} U_{yz} & U_{zx} & U_{xy}\end{array}\right]_{2}\left[\begin{array}{ccc} g^{yy}g^{zz}-g^{yz\,2} & g^{yz}g^{zx}-g^{xy}g^{zz} & g^{xy}g^{yz}-g^{zx}g^{yy}\\ g^{yz}g^{zx}-g^{xy}g^{zz} & g^{zz}g^{xx}-g^{zx\,2} & g^{zx}g^{xy}-g^{yz}g^{xx}\\ g^{xy}g^{yz}-g^{zx}g^{yy} & g^{zx}g^{xy}-g^{yz}g^{xx} & g^{xx}g^{yy}-g^{xy\,2} \end{array}\right]\left[\begin{array}{c} V_{yz}\\ V_{zx}\\ V_{xy} \end{array}\right]_{2}\end{split}$

where we have used the fact that

\begin{split}\begin{aligned} \left(dx^{i}\wedge dx^{j}\right)\cdot\left(dx^{\alpha}\wedge dx^{\beta}\right) & =\left(dx^{i}\cdot dx^{\alpha}\right)\left(dx^{j}\cdot dx^{\beta}\right)-\left(dx^{i}\cdot dx^{\beta}\right)\left(dx^{j}\cdot dx^{\alpha}\right)\\ & =g^{i\alpha}g^{j\beta}-g^{i\beta}g^{j\alpha}\end{aligned}\end{split}

Expanding the matrix expression

\begin{split}\begin{aligned} \mathbf{U}\cdot\mathbf{V} & =g^{yy}g^{zz}U_{yz}V_{yz}+g^{zz}g^{xx}U_{zx}V_{zx}+g^{xx}g^{yy}U_{xy}V_{xy}+\\ & +g^{yz}g^{zx}\left(U_{yz}V_{zx}+U_{zx}V_{yz}\right)+g^{zx}g^{xy}\left(U_{zx}V_{xy}+U_{xy}V_{zx}\right)\\ & +g^{xy}g^{yz}\left(U_{xy}V_{yz}+U_{yz}V_{xy}\right)\end{aligned}\end{split}

The hodge-star must be such that

$\mathbf{U}\wedge\star\mathbf{U}=(\mathbf{U}\cdot\mathbf{U})\mathbf{\mu}$

where $$\mu$$ is

$\mathbf{\mu}=\frac{1}{\sqrt{\det g^{ij}}}dx\wedge dy\wedge dz$

and

$\det g^{ij}=g^{xx}g^{yy}g^{zz}+2g^{xy}g^{yz}g^{zx}-g^{xx}g^{yz\,2}-g^{yy}g^{zx\,2}-g^{zz}g^{xy\,2}$

The wedge product in 3D between two one-forms is

\begin{split}\begin{aligned} \mathbf{U} & =U_{x}\,dx+U_{y}\,dy+U_{z}\,dz\\ \mathbf{V} & =V_{x}\,dx+V_{y}\,dy+V_{z}\,dz\\ \mathbf{U}\wedge\mathbf{V} & =\left(U_{y}V_{z}-U_{z}V_{y}\right)\,dy\wedge dz+\\ & +\left(U_{z}V_{x}-U_{x}V_{z}\right)\,dz\wedge dx+\\ & +\left(U_{x}V_{y}-U_{y}V_{x}\right)\,dx\wedge dy\end{aligned}\end{split}

and between a one-form and a two form is

\begin{split}\begin{aligned} \mathbf{U} & =U_{x}\,dx+U_{y}\,dy+U_{z}\,dz\\ \mathbf{V} & =V_{yz}\,dy\wedge dz+V_{zx}\,dz\wedge dx+V_{xy}\,dx\wedge dy\\ \mathbf{U}\wedge\mathbf{V} & =\left(U_{x}V_{yz}+U_{y}V_{zx}+U_{z}V_{xy}\right)dx\wedge dy\wedge dz\end{aligned}\end{split}

Now we need to look for a $$\mathbf{V}_{2}=\star\mathbf{U}_{1}$$ such that $$\mathbf{U}_{1}\wedge\mathbf{V}_{2}=(\mathbf{U}_{1}\cdot\mathbf{U}_{1})\mu_{3}$$ (where the subscript in $$\mathbf{U}_{k}$$ indicates its a k-form).

\begin{aligned} U_{x}V_{yz}+U_{y}V_{zx}+U_{z}V_{xy} & =\frac{1}{\sqrt{\det}}\left(g^{xx}U_{x}^{\phantom{x}2}+g^{yy}U_{y}^{\phantom{y}2}+g^{zz}U_{z}^{\phantom{y}2}+2g^{xy}U_{x}U_{y}+2g^{yz}U_{y}U_{z}+2g^{zx}U_{z}U_{x}\right)\end{aligned}

Clearly one way to satisfy that is

\begin{split}\begin{aligned} V_{yz} & =\frac{1}{\sqrt{\det}}\left(g^{xx}U_{x}+g^{xy}U_{y}+g^{zx}U_{z}\right)\\ V_{zx} & =\frac{1}{\sqrt{\det}}\left(g^{xy}U_{x}+g^{yy}U_{y}+g^{yz}U_{z}\right)\\ V_{xy} & =\frac{1}{\sqrt{\det}}\left(g^{zx}U_{x}+g^{yz}U_{y}+g^{zz}U_{z}\right)\end{aligned}\end{split}

Then the expression for the hodge-star is given by

$\begin{split}\star\mathbf{U}_{1}=\frac{1}{\sqrt{\det}}\left[\begin{array}{ccc} g^{xx} & g^{xy} & g^{zx}\\ g^{xy} & g^{yy} & g^{yz}\\ g^{zx} & g^{yz} & g^{zz} \end{array}\right]\mathbf{U}_{1}\end{split}$

Now to compute the hodge-star of a two-form, we need to look for $$\mathbf{V}_{1}=\star\mathbf{U}_{2}$$ such that $$\mathbf{U}_{2}\wedge\mathbf{V}_{1}=(\mathbf{U}_{2}\cdot\mathbf{U}_{2})\mu_{3}$$

$\mathbf{U}\cdot\mathbf{U}=g^{yy}g^{zz}U_{yz}^{\phantom{yz}2}+g^{zz}g^{xx}U_{xz}^{\phantom{yz}2}+g^{xx}g^{yy}U_{xy}^{\phantom{yz}2}+2g^{yz}g^{zx}U_{yz}U_{zx}+2g^{zx}g^{xy}U_{zx}U_{xy}+2g^{xy}g^{yz}U_{xy}U_{yz}$

and the equality that needs to be satisfied becomes

\begin{split}\begin{aligned} g^{yy}g^{zz}U_{yz}^{\phantom{yz}2}+g^{zz}g^{xx}U_{zx}^{\phantom{yz}2}+g^{xx}g^{yy}U_{xy}^{\phantom{yz}2}+2g^{yz}g^{zx}U_{yz}U_{zx}+2g^{zx}g^{xy}U_{zx}U_{xy}+2g^{xy}g^{yz}U_{xy}U_{yz} & =\\ \frac{1}{\sqrt{\det}}\left(V_{x}U_{yz}+V_{y}U_{zx}+V_{z}U_{xy}\right)\end{aligned}\end{split}

Therefore,

\begin{split}\begin{aligned} V_{x} & =\frac{1}{\sqrt{\det}}\left(g^{yy}g^{zz}U_{yz}+g^{yz}g^{zx}U_{zx}+g^{xy}g^{yz}U_{xy}\right)\\ V_{y} & =\frac{1}{\sqrt{\det}}\left(g^{yz}g^{zx}U_{yz}+g^{zz}g^{xx}U_{zx}+g^{zx}g^{xy}U_{xy}\right)\\ V_{z} & =\frac{1}{\sqrt{\det}}\left(g^{xy}g^{yz}U_{yz}+g^{zx}g^{xy}U_{zx}+g^{xx}g^{yy}U_{xy}\right)\end{aligned}\end{split}

Clearly, the following satisfies the above equality

$\begin{split}\star\mathbf{U}_{2}=\frac{1}{\sqrt{\det}}\left[\begin{array}{ccc} g^{yy}g^{zz}-g^{yz\,2} & g^{yz}g^{zx}-g^{xy}g^{zz} & g^{xy}g^{yz}-g^{zx}g^{yy}\\ g^{yz}g^{zx}-g^{xy}g^{zz} & g^{zz}g^{xx}-g^{zx\,2} & g^{zx}g^{xy}-g^{yz}g^{xx}\\ g^{xy}g^{yz}-g^{zx}g^{yy} & g^{zx}g^{xy}-g^{yz}g^{xx} & g^{xx}g^{yy}-g^{xy\,2} \end{array}\right]\mathbf{U}_{1}\end{split}$

## Summary

• zero-form $$\mathbf{U}=U\,1$$

$\star\mathbf{U}=U\,\frac{1}{\sqrt{\det}}\,dx\wedge dy\wedge dz$
• one-form $$\mathbf{U}=U_{x}\,dx+U_{y}\,dy+U_{z}\,dz$$

$\begin{split}\star\mathbf{U}=\frac{1}{\sqrt{\det}}\left[\begin{array}{ccc} g^{xx} & g^{xy} & g^{zx}\\ g^{xy} & g^{yy} & g^{yz}\\ g^{zx} & g^{yz} & g^{zz} \end{array}\right]\mathbf{U}\end{split}$
• two-form $$\mathbf{U}=U_{yz}\,dy\wedge dz+U_{zx}\,dz\wedge dx+U_{xy}\,dx\wedge dy$$

$\begin{split}\star\mathbf{U}=\frac{1}{\sqrt{\det}}\,\left[\begin{array}{ccc} g^{yy}g^{zz}-g^{yz\,2} & g^{yz}g^{zx}-g^{xy}g^{zz} & g^{xy}g^{yz}-g^{zx}g^{yy}\\ g^{yz}g^{zx}-g^{xy}g^{zz} & g^{zz}g^{xx}-g^{zx\,2} & g^{zx}g^{xy}-g^{yz}g^{xx}\\ g^{xy}g^{yz}-g^{zx}g^{yy} & g^{zx}g^{xy}-g^{yz}g^{xx} & g^{xx}g^{yy}-g^{xy\,2} \end{array}\right]\mathbf{U}\end{split}$
• three-form $$\mathbf{U}=U_{xyz}\,dx\wedge dy\wedge dz$$

$\star\mathbf{U}=U_{xyz}\,\sqrt{\det}$