The Unapologetic Mathematician

Mathematics for the interested outsider

The Hodge Star, Squared

It’s interesting to look at what happens when we apply the Hodge star twice. We just used the fact that in our special case of \mathbb{R}^3 we always get back exactly what we started with. That is, in this case, *^2=1 — the identity operation.

It’s easiest to work this out in coordinates. If \eta=dx^{i_1}\wedge\dots\wedge dx^{i_k} is some k-fold wedge then *\eta is \pm\sqrt{\lvert g_{ij}\rvert} times the wedge of all the indices that don’t show up in \eta. But then **\eta is \pm\sqrt{\lvert g_{ij}\rvert} times the wedge of all the indices that don’t show up in *\eta, which is exactly all the indices that were in \eta in the first place! That is, **\eta=\pm\lvert g_{ij}\rvert\eta, where the sign is still a bit of a mystery.

But some examples will quickly shed some light on this. We can even extend to the pseudo-Riemannian case and pick a coordinate system so that \langle dx^i,dx^j\rangle=\epsilon_i\delta^{ij}, where \epsilon_i=\pm1. That is, any two dx^i are orthogonal, and each dx^i either has “squared-length” 1 or -1. The determinant \lvert g_{ij}\rvert is simply the product of all these signs. In the Riemannian case this is simply 1.

Now, let’s start with an easy example: let \eta be the wedge of the first k indices:

\displaystyle\eta=dx^1\wedge\dots\wedge dx^k

Then *\eta is (basically) the wedge of the other indices:

\displaystyle*\eta=\sqrt{\lvert g_{ij}\rvert}dx^{k+1}\wedge\dots\wedge dx^n

The sign is positive since \eta\wedge*\eta already has the right order. But now we flip this around the other way:

\displaystyle**\eta=\pm\lvert g_{ij}\rvert dx^1\wedge\dots\wedge dx^k

but this should obey the same rule as ever:

\displaystyle\begin{aligned}*\eta\wedge**\eta&=\pm\lvert g_{ij}\rvert\sqrt{\lvert g_{ij}\rvert}dx^{k+1}\wedge\dots\wedge dx^n\wedge dx^1\wedge\dots\wedge dx^k\\&=\pm\lvert g_{ij}\rvert(-1)^{k(n-k)}\sqrt{\lvert g_{ij}\rvert}dx^1\wedge\dots\wedge dx^n\\&=\pm\lvert g_{ij}\rvert(-1)^{k(n-k)}\omega\end{aligned}

where we pick up a factor of -1 each time we pull one of the last k 1-forms leftwards past the first n-k. We conclude that the actual sign must be \lvert g_{ij}\rvert(-1)^{k(n-k)} so that this result is exactly \omega. Similar juggling for other selections of \eta will give the same result.

In our special Riemannian case with n=3, then no matter what k is we find the sign is always positive, as we expected. The same holds true in any odd dimension for Riemannian manifolds. In even dimensions, when k is odd then so is n-k, and so *^2=-1 — the negative of the identity transformation. And the whole situation gets more complicated in the pseudo-Riemannian version depending on the number of -1s in the diagonalized metric tensor.

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October 18, 2011 - Posted by | Differential Geometry, Geometry

5 Comments »

  1. […] our calculation of the square of the Hodge star we can tell that the star operation is invertible. Indeed, since — applying the star twice […]

    Pingback by The Codifferential « The Unapologetic Mathematician | October 21, 2011 | Reply

  2. […] if we define as another -form then we know it corresponds to the curl . But on the other hand we know that in dimension we have , and so we find as well. Thus we […]

    Pingback by The Classical Stokes Theorem « The Unapologetic Mathematician | November 23, 2011 | Reply

  3. […] that the square of the Hodge star has the opposite sign from the Riemannian case; when is odd the double Hodge dual of a -form is […]

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

  4. What does the g in sqrt mean?

    Comment by Jojo | March 5, 2014 | Reply

    • I explain this in the third paragraph. Read better.

      Comment by John Armstrong | March 5, 2014 | Reply


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