# The Unapologetic Mathematician

## The Hodge Star in Coordinates

It 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 $(U,x)$ of an oriented Riemannian manifold $M$, which means that we have a canonical volume form that locally looks like

$\displaystyle\omega=\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n$

Now, we know that any $k$-form on $U$ can be written out as a sum of functions times $k$-fold wedges:

$\displaystyle\eta=\sum\limits_{1\leq i_1<\dots

Since the star operation is linear, we just need to figure out what its value is on the $k$-fold wedges. And for these the key condition is that for every $k$-form $\zeta$ we have

$\displaystyle\zeta\wedge*(dx^{i_1}\wedge\dots\wedge dx^{i_k})=\langle\zeta,dx^{i_1}\wedge\dots\wedge dx^{i_k}\rangle\omega$

Since both sides of this condition are linear in $\zeta$, we also only need to consider values of $\zeta$ which are $k$-fold wedges. If $\zeta$ is not the same wedge as $\eta$, then the inner product is zero, while if $\zeta=\eta$ then

\displaystyle\begin{aligned}(dx^{i_1}\wedge\dots\wedge dx^{i_k})\wedge*(dx^{i_1}\wedge\dots\wedge dx^{i_k})&=\langle dx^{i_1}\wedge\dots\wedge dx^{i_k},dx^{i_1}\wedge\dots\wedge dx^{i_k}\rangle\omega\\&=\det\left(\langle dx^{i_j},dx^{i_k}\rangle\right)\omega\\&=\det\left(\delta^{jk}\right)\omega\\&=\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\end{aligned}

And so $*(dx^{i_1}\wedge\dots\wedge dx^{i_k})$ must be $\pm\sqrt{\lvert g_{ij}\rvert}$ times the $n-k$-fold wedge made up of all the $dx^i$ that do not show up in $\eta$. The positive or negative sign is decided by which order gives us an even permutation of all the $dx^i$ on the left-hand side of the above equation.

October 8, 2011 - Posted by | Differential Geometry, Geometry

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

2. I have a question for you, if I wanted to show the rate of change in a person personality due to a control stimuli, how do you think the formula should llook like.

Comment by Marcus Cox | October 9, 2011 | Reply

3. […] easiest to work this out in coordinates. If is some -fold wedge then is times the wedge of all the indices that don’t show up in . […]

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4. […] implications does this have on the coordinate expression of the Hodge star? It’s pretty much the same, except for the determinant part. You can think about it yourself, […]

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

5. Hi John, just an observation. When you pick local coordinates, the frame for the cotangent bundle dx_{i} i=1,,,,,n induced by the local coordinates is not orthonormal in general, so the inner product of two different wedges is not zero. In the same way is not delta_{i,j}. I think this only works if you pick Riemannian normal coordinates centered at some point “p”, even in this case this expression for the Hodge star operator would be valid only for the center point “p” and not for other points in this chart.

Comment by Josh | August 12, 2014 | Reply