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Hodge duality and some properties

  1. Feb 1, 2015 #1
    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.
     
  2. jcsd
  3. Feb 1, 2015 #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:

    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.
     
  4. Feb 1, 2015 #3
    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
     
  5. Feb 1, 2015 #4

    Ben Niehoff

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    What is ##\alpha##?
     
  6. Feb 1, 2015 #5
    Any complex function, I mentioned that in the given. @Ben Niehoff
     
  7. Feb 1, 2015 #6

    Ben Niehoff

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    So, a function is a 0-form. What's the Hodge dual of a 0-form?
     
  8. Feb 1, 2015 #7
    $$-1/3 (i \partial _x \alpha) dx \wedge dy \wedge dz?$$ @Ben Niehoff
     
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