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:

$\displaystyle\eta=\sum\limits_{1\leq i_1<\dots<i_k\leq n}\eta_{i_1\dots i_k}dx^{i_1}\wedge\dots\wedge dx^{i_k}$

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.

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

4 Comments »

[...] 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 [...]

Pingback by Thirteenth Linkfest | October 8, 2011 | Reply
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
[...] 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 . [...]

Pingback by The Hodge Star, Squared « The Unapologetic Mathematician | October 18, 2011 | Reply
[...] 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

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