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.

October 8, 2011 Posted by | Differential Geometry, Geometry | 5 Comments