# Hodge Star Operator

1. Jul 21, 2014

Hi, I have real problems with the indices here, can someone give me a step by step explanation how to compute stuff with this formula?

$*\omega = \frac{\sqrt{|g|}}{r!(m-r)} \omega_{\mu_{1}\mu_{2}...\mu_{r}}\epsilon^{\mu_{1}\mu_{2}...\mu_{r}}_{v_{r+1}...v_{m}}dx^{v_{r+1}}\wedge...\wedge dx^{v_m}$

For example

$*(dx^{1} \wedge dx^{2})=dx^{3}$

I know the formula is long and ugly, but I can't make sense out of it :/

Greets

2. Jul 21, 2014

### Terandol

To be honest I would suggest never using any of the explicit formulas to compute the hodge star. They are useful for proving things about the operator sometimes but other than that I avoid them as much as possible. The Hodge star is really a very simple thing to compute intuitively but it's hard to write down explicitly so the formulas are all ugly and over complicate the process.

I would simply use the definition to do the computation. By definition, $*\omega$ satisfies the equation $\mu \wedge *\omega =\langle \mu,\omega \rangle \mathrm{vol}_n$. So in your example, we want $dx^1\wedge dx^2\wedge *(dx^1\wedge dx^2)=dx^1\wedge dx^2\wedge dx^3$. Hence, $*(dx^1\wedge dx^2)=dx^3$. This idea always works if you go term by term. Ie. just take a (simple) form, and wedge the original form and the hodge star of this form (order is important here.) This will give you an n-form so now all you have to do to compute the hodge star is normalize and pick out a sign using the volume form so that the defining equation is satified.

It is even easier to compute if you start with an orthonormal coordinate system for your metric. Then you always have
$$*(dx^{i_1}\wedge\cdots\wedge dx^{i_p})=\pm dx^{i_{p+1}}\cdots dx^{i_n}$$
where you choose the plus sign if $dx^{i_1}\wedge\cdots\wedge dx^{i_n}$ has the same orientation as your volume form and the minus sign otherwise.

So, really all the explicit formula does is pick out all the $dx^{i}$ that do not appear in your form, normalize them using the metric and then picks a plus or minus sign as needed. If you are adamant about using the explicit formula for computations, it can of course be done. To use your example again, we have $r=2$ (this is the degree of the form) and $m=3$ (this is the dimension of the manifold. ) Here $\omega_{u_1,u_2}=0$ always except $\omega_{12}=-\omega_{21}=1$ so we can get rid of all terms except $u_1=1,u_2=2$ and $u_1=2,u_2=1$. So now you can just plug everything into the formula and compute:
$$*\omega=\frac{1}{2(3-2)}\left(\omega_{12}\epsilon_{\nu}^{12} dx^\nu +\omega_{21}\epsilon_{\nu}^{21} dx^\nu\right) =\frac{1}{2}\omega_{12} dx^3-\frac{1}{2}\omega_{21} dx^3 =\frac{1}{2} dx^3+\frac{1}{2}dx^3 =dx^3.$$

3. Jul 21, 2014

In the formula $\mu \wedge *\omega = <\mu, \omega>vol_n$, $vol_{3}= \sqrt{1}dx^1\wedge dx^2 \wedge dx^3$ right? But shouldn't $<\mu, \omega>$ be zero? Because the determinant is zero. And what should I take in general for $\mu$? What about $*(dx^{1}\wedge dx^{3})=-dx^{2}$? How do I get this result?

$(dx^{1} \wedge dx^{3}) \wedge *(dx^{1} \wedge dx^{3})=(dx^{1} \wedge dx^{3})\wedge (-dx^{2})= -dx^{1}\wedge dx^{3} \wedge dx^{2}= dx^{1}\wedge dx^{2} \wedge dx^{3}$

In the complicated formula, what did you do with the $v_{r+1}$ index? Shouldn't it be $v_{3}$ then? And what does that mean? Further there is another definition where one has to contract the epsilon tensor with the metric tensor:

$\epsilon^{\mu_{1}\mu_{2}...\mu_{m}}=g^{\mu_{1}v_1}g^{\mu_{2}v_{2}}...g^{\mu_{m}v_{m}}\epsilon_{v_{1}v_{2}...v_{m}}=g^{-1}\epsilon_{\mu_{1}\mu_{2}...\mu_{m}}$

I know that you can raise and lower indices when you contract with the inverse metrics but why is it necessary in this case?

Greets

Last edited: Jul 21, 2014
4. Jul 21, 2014

### Terandol

Yes, I was using the standard volume form $vol_n=dx^1\wedge dx^2\wedge dx^3$. The determinant is not zero though. These are 2-forms so the relevant determinant using the standard metric is
$$\langle dx^1\wedge dx^2, dx^1\wedge dx^2 \rangle = \mathrm{det} \left( \begin{bmatrix} \langle dx^1, dx^1\rangle & \langle dx^1, dx^2 \rangle \\ \langle dx^2, dx^1\rangle & \langle dx^2, dx^2 \rangle \end{bmatrix} \right)= \mathrm{det} \left( \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \right) =1.$$

The simplest thing to try is to take $\mu=\omega$ if you are trying to compute $*\omega$. This will usually allow you to compute all $*(dx^{i_1} \wedge \cdots \wedge dx^{i_p} )$ which completely computes the hodge star on all forms be linearity. If your coordinates are orthonormal this will always work and will just give the second formula in my previous post. If not, you may have to try a couple different forms to make it work.

Yes, maybe I should have used the notation $\nu=\nu_3$ in my last answer to stick to the notation in the formula. I don't really think it means anything, the formula just splits a collection of m arbitrary indices into a collection of r arbitrary indices labelled by $\mu_i$'s and a collection of m-r arbitrary indices labelled by $\nu_j$. I don't really see any significance to this other than making it easy to see which indices to contract with in the formula but you could just as easily label all $m$ indices as $\mu_i$'s.

Contracting indices is the same thing as using the isomorphism from the tangent space to the cotangent space to transform $(r,s)$-tensors to $(r-i, s+i)$-tensors. So you can contract a (1,0)-tensor and a (0,1)-tensor directly since one is in the tangent space and one is in the cotangent space. However, if you want to contract two (1,0)-tensors you first have to use the metric to convert one of the (1,0)-tensors into a (0,1)-tensor (ie. raise/lower indices of one of the tensors) and then you can contract. My guess is that in the formula they need to contract the levi-civita tensor with some other tensors of the same kind and so they first need to raise the indices to make the tensor contraction possible.

5. Jul 21, 2014

Oh sry, I edited my post while you already answered it. I think I understood it now. Another puzzle stone done. Thank you for your clear explanations !

Greets

Last edited: Jul 21, 2014
6. Aug 12, 2014

### Geometry_dude

In case you need to formalize everything, the correct written out formula for the hodge star on pseudo-Riemannian manifolds with signature $(n^+,n^-)$ is
$$\alpha = \frac{1}{k!} \alpha_{i_1 \dots i_k} \, d x^{i_1}\wedge \dots \wedge d x^{i_k}$$
$$\implies$$
$$\star \alpha = \frac{(-1)^{n^-}}{k! (n-k!)} \, \alpha_{i_1 \dots i_k} \, g^{i_1 j_1} \cdots g^{i_k j_k} \, \varepsilon_{j_1 \dots j_k j_{k+1} \dots j_n} \, \sqrt{\lvert \det g \rvert} \, d x^{j_{k+1}} \wedge \dots \wedge d x^{j_n}$$
and you can use the identities for the Levi Civita symbol to make computations.

Last edited: Aug 12, 2014