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A technical question that has to do with Hodge Star

  1. Apr 19, 2015 #1
    If we have an equation that looks like $$H=Y$$ and we want to multiply H by either $$ReM_{IJ}$$ or $$ImM_{IJ}$$ where $$M_{IJ}$$ is a complex matrix. But the thing is that $$Y=\star(...)$$ where $$\star$$ is hodge star and (...) is set of complex functions and other numerical stuff, my question is technical here, say we decide to multiply H by $$ReM_{IJ}$$ can we move $$ReM_{IJ}$$ into the parenthesis and jump over the Hodge star? That is to say $$ReM_{IJ}H=\star(ReM_{IJ} ...)$$ or this is absolutely wrong and we should keep $$ReN_{IJ}$$ outside the Hodge star? That is to say $$ReM_{IJ}H=ReM_{IJ}\star( ...)$$

    EDIT: M_{IJ} is a complex matrix
     
    Last edited: Apr 19, 2015
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  3. Apr 19, 2015 #2

    robphy

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    What's your definition of the Hodge Star in this context?
     
  4. Apr 19, 2015 #3
    Hodge star in R^3.
     
  5. Apr 20, 2015 #4

    Ben Niehoff

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    The Hodge star is a linear map. Careful though, in complex geometry sometimes the Hodge star is a conjugate-linear map. Hence the reason for the question.

    P.S. Your posts would be way easier to read if you use in-line Latex where appropriate.
     
  6. Apr 20, 2015 #5
    What do you mean here by conjugate-linear map? Let me be more specific with my question, do you mean if "c" is a real number or real function then
    $$ \star(c\omega)=(c\star \omega)$$
    If so, then what will happen if c were a complex function instead?

    Note: I am trying to using a single "$" for in-line Latex with no results :(!
     
  7. Apr 20, 2015 #6

    Ben Niehoff

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    Actually, I think the conjugate-linear version is typically written ##\bar \star##. It means the combination of taking the Hodge dual and the complex conjugate. Hence for a complex number ##c##,

    $$\bar \star \; c \omega = \bar c \; \bar \star \; \omega.$$

    As for why you would want to do this: it allows you to define a natural Hermitian product on compact complex manifolds:

    $$\langle \alpha, \, \beta \rangle = \int_M \alpha \wedge \bar \star \, \beta.$$

    On PF, use ## for in-line Latex.
     
  8. Apr 20, 2015 #7
    Oh great, I understand now, then for a linear map ##\star## and if ##c## is a complex function then it will not affect the complex function upon leaping it over the ##\star##

    That is to say, is the following correct if the c is a complex function:

    $$\star(c\omega)=c\star\omega$$

    Thank you for the in-line tip, much neater! And if you may suggest for me a good read about those mappings (linear and complex conjugate) I'd be grateful!
     
  9. Apr 20, 2015 #8

    Ben Niehoff

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    You should read Nakahara's book.
     
  10. Apr 20, 2015 #9
    Ok thank you, but about the last thing I mentioned about the complex function in my previous comment? I would appreciate if you reread my previous comment again? Does the complex function leap over the linear Hodge dual with no problem the same as if it were a real function?
     
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