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

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