The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Differential Geometry, Geometry

4 Comments »

  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 [...]

    Pingback by Thirteenth Linkfest | October 8, 2011 | Reply

  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 . [...]

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

  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


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