Hodge duality and some properties

  • Thread starter PhyAmateur
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  • #1
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So we know that [Hodge duality](http://en.wikipedia.org/wiki/Hodge_dual) works this way

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where p represents the p in p-form and n is the dimensional number.

My question is: How does hodge duality work on the imaginary number "i" on one hand and on partial derivative like $$\partial_x\alpha$$ let us say on the other where $$\alpha$$ is a complex function.
 

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  • #2
Ben Niehoff
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Why is this in GR rather than in the differential geometry forum? The formula you've written is not correct on Lorentzian-signature manifolds.

However, in Euclidean signature, this is correct:

$$⋆(dx^i_1 \wedge ... \wedge dx^i_p)= \frac{1}{(n-p)!} \epsilon^{i_1..i_p}_{i_{p+1}..i_n} dx^{i_{p+1} } \wedge dx^{i_n}$$

where p represents the p in p-form and n is the dimensional number.
And it means exactly the same thing if you make all of your variables complex.

The catch is that on a complex manifold, one is usually more interested in taking the Hodge dual and the complex conjugate simultaneously.
 
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  • #3
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Thanks. I would appreciate if we could move this to the other section or if you could guide me so that I can move it. On the other hand, I want to ask you to please elaborate on your answer, because this answer doesn't make it clear to me if we have for example $$\star{(i\partial _x \alpha)}$$ it is also the partial differential that is worrying me. @Ben Niehoff
 
  • #4
Ben Niehoff
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What is ##\alpha##?
 
  • #6
Ben Niehoff
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So, a function is a 0-form. What's the Hodge dual of a 0-form?
 
  • #7
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$$-1/3 (i \partial _x \alpha) dx \wedge dy \wedge dz?$$ @Ben Niehoff
 

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