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 $-1$ s in the diagonalized metric tensor.

5 Comments »

[…] 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
[…] 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
[…] 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
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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