Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Hodge star operator

+ 4 like - 0 dislike
4760 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 20, 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 (520 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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...