# 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 $$\mathbf{U}=U_{xyz}\,dx\wedge dy\wedge dz$$

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

# Riemanian Metric and the Hodge-Star

In elementary geometry the inner product between two vectors $$\mathbf{U}=U^{i}\frac{\partial}{\partial x^{i}}$$ and $$\mathbf{V}=V^{i}\frac{\partial}{\partial x^{i}}$$ is given by

$\mathbf{U}\cdot\mathbf{V}=\sum_{i}U^{i}V^{i}$

In the general case, the Riemanian metric is a type $$(0,2)$$ tensor field which at each point of the manifold $$m\in M$$ satisfies

\begin{split}\begin{aligned} g_{m}(\mathbf{U},\mathbf{V}) & =g_{m}(\mathbf{V},\mathbf{U}) & \text{symmetry}\\ g_{m}(\mathbf{U},\mathbf{U}) & \geq0 & \text{postive-definiteness}\end{aligned}\end{split}

The metric tensor can be expressed as

$g_{m}=g_{ij}(m)\,dx^{i}\otimes dx^{j}$

where the components are given by

$g_{ij}(m)=g_{m}\left(\frac{\partial}{\partial x^{i}},\frac{\partial}{\partial x^{j}}\right)$

There exists an isomorphism between $$T_{m}M$$ and $$T_{m}^{*}M$$. Let $$U$$ be a vector field ($$(1,0)$$ tensor field), and let $$\alpha$$ be a one-form ($$(0,1)$$ tensor field). Then the isomorphism is expressed as

$\alpha_{i}=g_{ij}U^{j}\quad U^{i}=g^{ij}\alpha_{j}$

where $$g$$ with raised indices is the inverse of $$g$$ with lowered indices, namely

$g^{ij}g_{jk}=g_{ij}g^{jk}=\delta_{i}^{k}$

Thus the flat $$(\flat)$$ and sharp $$(\sharp)$$ oprators are easily expressible in terms of the metric

$\left(U^{\flat}\right)_{i}=g_{ij}U^{j}\qquad U^{i}=g^{ij}\left(U^{\flat}\right)_{j}$
$\left(\alpha^{\sharp}\right)^{i}=g^{ij}\alpha_{j}\qquad\alpha_{i}=g_{ij}\left(\alpha^{\sharp}\right)^{j}$

From now on the convention that we will follow will be that components with raised indices will be the components of vector fields, and the components with lowered indices will be the components of one-forms. Thus $$V^{i}$$ will denote the components of a vector field, and $$V_{i}$$ will denote the components of a one-form.

## Eucledian Space

The flat metric is

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

Given two one-forms

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

their inner product is given by

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

The hodge-star must be such that we must have the following propreties satisfied for one-forms

\begin{split}\begin{aligned} \mathbf{U}\cdot\star\mathbf{U} & = & 0\\ \mathbf{U}\cdot\mathbf{U} & = & \star\mathbf{U}\cdot\star\mathbf{U}\end{aligned}\end{split}

Clearly

$\begin{split}\star\mathbf{U}=\left[\begin{array}{c} -U_{y}\\ U_{x} \end{array}\right]=\left[\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array}\right]\mathbf{U}\end{split}$

satisfies these properties.

## General (Non-Eucledian) Space

In this case the symmetric bilinear form defining the inner product between two vectors

$\begin{split}g_{ij}=\left[\begin{array}{cc} g_{xx} & g_{xy}\\ g_{xy} & g_{yy} \end{array}\right]\end{split}$

The correspond inner product between two one-forms on the other hand will be given by

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

and the two are inverses of each other

$\begin{split}\left[\begin{array}{cc} g_{xx} & g_{xy}\\ g_{xy} & g_{yy} \end{array}\right]=\left[\begin{array}{cc} g^{xx} & g^{xy}\\ g^{xy} & g^{yy} \end{array}\right]^{-1}\end{split}$

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

$\begin{split}\mathbf{U}\cdot\mathbf{V}=\mathbf{U}^{T}g\mathbf{V}=\left[\begin{array}{cc} U_{x} & U_{y}\end{array}\right]\left[\begin{array}{cc} g^{xx} & g^{xy}\\ g^{xy} & g^{yy} \end{array}\right]\left[\begin{array}{c} V_{x}\\ V_{y} \end{array}\right]\end{split}$

Expanding the matrix expressions

$\mathbf{U}\cdot\mathbf{V}=g^{xx}U_{x}V_{x}+g^{xy}\left(U_{x}V_{y}+U_{y}V_{x}\right)+g^{yy}U_{y}V_{y}$

The hodge-star must be such that

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

where $$\mu$$ is the normalized volume form, and in 2D it is given by the following two equivalent expressions

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

Since we are working with one-forms the relevant inner product will be the one between one-forms, and we will use the second expression.

The wedge product in 2D is given in components as

$\mathbf{U}\wedge\mathbf{V}=U_{i}V_{j}\,dx^{i}\wedge dx^{j}$
$\mathbf{U}\wedge\mathbf{V}=\left(U_{x}V_{y}-U_{y}V_{x}\right)dx\wedge dy$

Now we need to look for a $$\mathbf{V}=\star\mathbf{U}$$ such that $$\mathbf{U}\wedge\mathbf{V}=(\mathbf{U}\cdot\mathbf{U})\mu$$, or

$U_{x}V_{y}-U_{y}V_{x}=\frac{1}{\sqrt{\det g^{ij}}}\left(g^{xx}U_{x}^{\phantom{x}2}+2g^{xy}U_{x}U_{y}+g^{yy}U_{y}^{\phantom{y}2}\right)$

Clearly one way to satify that is

\begin{split}\begin{aligned} V_{x} & = & -g^{yy}U_{y}-g^{xy}U_{x}\\ V_{y} & = & g^{xx}U_{x}+g^{xy}U_{y}\end{aligned}\end{split}

and noting that $$\det g^{ij}=g^{xx}g^{yy}-g^{xy\,2}$$, then the expression for the hodge-star is given by

$\begin{split}\star\mathbf{U}=\mathbf{V}=\frac{1}{\sqrt{g^{xx}g^{yy}-g^{xy\,2}}}\left[\begin{array}{c} -g^{yy}U_{y}-g^{xy}U_{x}\\ g^{xx}U_{x}+g^{xy}U_{y} \end{array}\right]\end{split}$

which for the flat metric ($$g^{xx}=g^{yy}=1$$ and $$g^{xy}=0$$) reverts to our expression in Eucledian space.

To check that this expression of the hodge star is indeed indempotent (up to a sign), we apply it twice

\begin{split}\begin{aligned} \star\star\mathbf{U} & =\frac{1}{\left|\det g^{ij}\right|}\left[\begin{array}{c} -g^{yy}\left(g^{xx}U_{x}+\cancel{g^{xy}U_{y}}\right)-g^{xy}\left(-\cancel{g^{yy}U_{y}}-g^{xy}U_{x}\right)\\ g^{xx}\left(-g^{yy}U_{y}-\bcancel{g^{xy}U_{x}}\right)+g^{xy}\left(\bcancel{g^{xx}U_{x}}+g^{xy}U_{y}\right) \end{array}\right]\\ & =\frac{1}{\left|\det g^{ij}\right|}\left[\begin{array}{c} -g^{yy}g^{xx}U_{x}+g^{xy\,2}U_{x}\\ -g^{xx}g^{yy}U_{y}+g^{xy\,2}U_{y} \end{array}\right]\\ & =-\frac{g^{xx}g^{yy}-g^{xy\,2}}{\left|\det g^{ij}\right|}\left[\begin{array}{c} U_{x}\\ U_{y} \end{array}\right]\\ & =-\left[\begin{array}{c} U_{x}\\ U_{y} \end{array}\right]=-\mathbf{U}\end{aligned}\end{split}

## Summary

• zero-form $$\mathbf{U}=U\,1$$

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

$\begin{split}\star\mathbf{U}=\frac{1}{\sqrt{\det g^{ij}}}\left[\begin{array}{c} -g^{yy}U_{y}-g^{xy}U_{x}\\ g^{xx}U_{x}+g^{xy}U_{y} \end{array}\right]\end{split}$
$\begin{split}\star\mathbf{U}=\frac{1}{\sqrt{\det g^{ij}}}\left[\begin{array}{cc} -g^{xy} & -g^{yy}\\ g^{xx} & g^{xy} \end{array}\right]\mathbf{U}\end{split}$
• two-form $$\mathbf{U}=U\,dx\wedge dy$$

$\star\mathbf{U}=U\,\sqrt{\det g^{ij}}$