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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?

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    $\begingroup$ 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. $\endgroup$ – ACuriousMind Jul 21 '14 at 0:56
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    $\begingroup$ @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. $\endgroup$ – Wouter Jul 21 '14 at 3:18
  • $\begingroup$ The dual of a $2$-form is a $1$-form in this case. The Hodge dual associates $k$ forms with $n-k$ forms uniquely. $\endgroup$ – Cameron Williams Jul 21 '14 at 3:55
  • $\begingroup$ 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. $\endgroup$ – Semiclassical Jul 21 '14 at 3:58
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    $\begingroup$ The first example is explicitely explained here. $\endgroup$ – Dilaton Jul 22 '14 at 12:39
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You have the formulae $$\omega = \frac{1}{r!}\omega_{\mu_{1}\mu_{2}...\mu_{r}}dx^{\mu_{1}} \wedge dx^{\mu_{2}} \wedge...\wedge dx^{\mu_{r}} \in \Omega^{r}(M)$$ $$\star\omega = \frac{\sqrt{|g|}}{r!(m-r)} \omega_{\mu_{1}\mu_{2}...\mu_{r}}\epsilon^{\mu_{1}\mu_{2}...\mu_{r}}_{\nu_{r+1}...\nu_{m}}dx^{\nu_{r+1}}\wedge...\wedge dx^{\nu_m}\in \Omega^{m-r}(M)$$ in general. Let's consider your $(r,m)=(2,3)$ case. Then these become $$\omega = \frac{1}{2!}\omega_{\mu_{1}\mu_{2}}dx^{\mu_{1}} \wedge dx^{\mu_{2}}\in\Omega^{2}(M)\implies \star\omega = \frac{\sqrt{|g|}}{2} \omega_{\mu_{1}\mu_{2}}\epsilon^{\mu_{1}\mu_{2}}_{\nu_{3}}dx^{\nu_{3}}\in \Omega^{1}(M)$$ Note that only $\nu_3$ shows up once since $r+1=m=3$. So $\star\omega$ is indeed a 1-form. Concretely, take the metric to be orthonormal i.e. $\sqrt{|g|}=1$, and suppose $\omega_{12}=-\omega_{21}=1$ and $\omega_{ij}=0$ otherwise. Then from the antisymmetry of the wedge product and the $\epsilon$ symbol, we have $\omega=dx_1\wedge dx_2$ and $$\star \omega =\frac{1}{2}\left(\epsilon^{12}_\nu dx^\nu-\epsilon^{21}_\nu dx^\nu\right)=\epsilon^{12}_\nu dx^\nu=\epsilon^{12\nu}dx_\nu=dx_3$$ as expected. Note that in the last two steps we have implicitly used $g_{\mu\nu}$ to raise/lower the index $\nu$ and $\epsilon^{12k}=\delta_{3k}.$

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  • $\begingroup$ Hey thank you, that helped ! I was confused because of this $v_{3}$ index, but it has no meaning other then summation I guess. $\endgroup$ – user117640 Jul 25 '14 at 20:29
  • $\begingroup$ That's the game that's always played, yeah. Each repeated index is just a dummy. (That's why, you'll notice, I called it as $\nu$ in the last line---after all, it doesn't matter what it's name is as long as I'm consistent with it.) $\endgroup$ – Semiclassical Jul 25 '14 at 20:30
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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.

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  • $\begingroup$ Hi, I'm sorry, but I know nothing about Clifford Algebra. But thx for the afford. $\endgroup$ – user117640 Jul 22 '14 at 20:38

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