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Hodge star operator

+ 4 like - 0 dislike
2519 views

Again I have issues with notations. The hodge star operator is defined as :

(m is the dimension of the manifold)

$$\star: \Omega^{r}(M) \rightarrow \Omega^{m-r}(M)$$

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

Where

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

With an r-form

$$\omega = \frac{1}{r!}\omega_{\mu_{1}\mu_{2}...\mu_{r}}dx^{\mu_{1}} \wedge dx^{\mu_{2}} \wedge...\wedge dx^{\mu_{r}} \in \Omega^{r}(M)$$

Gives

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

Now I would like to derive these results

(orthogonal metric and doesn't matter if forms or vectors)

$$\star(e_{2} \wedge e_{3})=e_{1}$$

$$\star(e_{1} \wedge e_{3})=-e_{2}$$

$$\star(e_{1} \wedge e_{2})=e_{3}$$

Another example, let me calculate $r=2$, $m=3$

$$\star(dx \wedge dy)=\sqrt{|g|}\epsilon^{xy}_{\nu_{2}\nu_{3}}dx^{\nu_{3}} \wedge dx^{\nu_3}$$

I have no clue what's going on, is $\nu$ different from $\mu$ ? How does this machinery work?

I know the formula is long and annoying, but can someone give a clear example of how this works?

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user user117640
asked Jul 21, 2014 in Mathematics by user117640 (20 points) [ no revision ]
Hint: If $m = 3$, then your last formula has too many $\nu$ on the right hand side. Also, this looks like a pure math question to me.

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user ACuriousMind
@ACuriousMind Pure math questions that arise in a physical context have recently become on-topic. Arguably, no such context has been explicitly given in this question, but I think it's clear what the physical motivation for it is.

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user Wouter
The dual of a $2$-form is a $1$-form in this case. The Hodge dual associates $k$ forms with $n-k$ forms uniquely.

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user Cameron Williams
Your expression for $*(dx\wedge dy)$ at the end isn't correct: since $r+1=2+1=3$ and $m=3$, the $\nu_2$ shouldn't be there. Similarly, the RHS should only be a one-form.

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user Semiclassical

2 Answers

+ 3 like - 0 dislike

Let us do the two examples you have given. For the first example, I will assume you are working with an Euclidean metric and so:

\(*(e_1 \wedge e_3) = \frac{1}{(3-2)!} \varepsilon^{132} e_2 = - \varepsilon^{123} e_2 = - e_2\)

For the second example, I believe your expression on the right-hand side is wrong for \(r=2\) and \(m=3\). I think it should be:

\(*(\mathrm{d} x \wedge \mathrm{d} y) = \frac{ \sqrt{|g|}}{(3-2)!} \varepsilon^{xyz} \mathrm{d} z = \sqrt{|g|} \mathrm{d} z\)

answered Jul 21, 2014 by Hunter (510 points) [ no revision ]

Hi, I'm the guy who asked that question from math exchange. Thx for the answer, that helped. I might also import my account to physicsoverflow. I think this is the best answer.

@user117640 No problem, happy to help.

+ 2 like - 0 dislike

It may sometimes be easier to compute Hodge duals using an orthogonal basis and clifford algebra.

Clifford algebra allows you to write a lot of complicated tensor expressions as simple products of vectors, using the "geometric product". If $a, b$ are orthogonal vectors with respect to some metric, and $u, v, w$ are arbitrary vectors, then under the geometric product,

$$aa \equiv g(a,a) = |a|^2, \quad ab = -ba, \quad (uv)w = u(vw)$$

So this captures the inner product, encoded in the metric; the need for antisymmetric tensors in $ab = - ba$, and associativity allows us to drop unneeded parentheses and talk about long products of vectors as entities in their own right.

Now then, suppose you have orthogonal coordinate one-forms $e^t, e^x, e^y, e^z$. Then there is a well-defined volume-form $i$:

$$i \equiv e^t e^x e^y e^z$$

Again, note that I said orthogonal. What happens if you take a two-form like $e^x e^y$ and multiply it by $i$?

$$i e^x e^y = e^t e^x e^y e^z e^x e^y =e^t (e^x e^y)(e^x e^y) e^z = - e^t e^z g^{xx} g^{yy} $$

where $g^{xx}$ and $g^{yy}$ arose from the inner products of forms. The sign might be "wrong" for a given convention, but this is doing exactly the same sort of thing the Hodge dual does.

I'll illustrate that, in the case of 3d with a Cartesian metric, you get the same results this way as you would using the index definition. In 3d, $i = e_1 e_2 e_3$, so we get

$$\star e_1 = i e_1 = e_1 e_2 e_3 e_1 = -e_1 e_2 e_1 e_3 = + e_1 e_1 e_2 e_3 = e_2 e_3$$

The other equalities follow similarly.

This post imported from StackExchange Mathematics at 2014-07-21 11:20 (UCT), posted by SE-user Muphrid
answered Jul 21, 2014 by Muphrid (20 points) [ no revision ]

Hi, I'm the guy from math exchange who asked the question. Thx for the afford, but I know nothing about Clifford Algebra.

Hi @user117640 welcome to PhysicsOverflow.

If you would like to get access to your imported account, you can write a mail to me or to admin@physicsoverflow.org to obtain temporary login data. I imported the question because I disapproved the migration and think it is interesting for physicists too.

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