Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kahler manifolds

  1. Mar 1, 2008 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Hi, everyone:
    I am doing some reading on the Frolicher Spec Seq. and I am trying to
    understand better the Kahler mflds. Specifically:

    What is meant by the fact that the complex structure, symplectic structure
    and Riemannian structure (from being a C^oo mfld.) are "mutually compatible"?

    I read this in both Griffiths and Harris and in the Wiki page.

    I realized I chose a handle that does not have a good abbreviation. Choices
    are:

    What,
    What Would,
    What Would Gauss,
    Do,

    None of them too good.

    Anyway, chronic burnout is getting to me. Thanks for any answer or suggestions
    for refs.
     
  2. jcsd
  3. Mar 2, 2008 #2
    Hi WWGD,

    The compatibility beween the three structures means that

    [tex][\mathbf{u},\mathbf{v}] = (\mathbf{u},i\mathbf{v}),[/tex]

    where [itex]\mathbf{u}[/itex] and [itex]\mathbf{v}[/itex] are tangent vectors, [,] is the symplectic structure, (,) is the riemannian structure and [itex]i[/itex] is the complex structure. So, given two of the structures, the third structure is defined (if it can be defined at all) by this relation.

    This also means that the riemannian and the symplectic structures are the real and imaginary parts of a hermitian structure <,>:

    [tex]<\mathbf{u},\mathbf{v}> = (\mathbf{u},\mathbf{v}) + i [\mathbf{u},\mathbf{v}].[/tex]
     
    Last edited: Mar 2, 2008
  4. Mar 2, 2008 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Thanks, O.O.Things, that was helpful.

    Any chance you could help me with the aspect of the "integrability condition" of the

    Kahler mfld?. (from http://en.wikipedia.org/wiki/Kahler_manifold ).

    Thanks.
     
  5. Mar 2, 2008 #4
    Well, the integrability condition is that the imaginary part of the hermitian metric must be closed, which is required for it to be a symplectic structure.
     
  6. Mar 10, 2008 #5

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Thanks again, O.O.T. I hope it is not too much to ask for a comment on the same
    entry, on the statement that the compatibility between all three structures is
    equivalent to the presentation of the unitary group as:

    U(n)=O(2n)/\Gl(n,C)/\Sp(2n)


    as in the link above. Sorry, I don't see the relation between this presentation
    of U(n) and the compatibility condition. Brother:Can you spare a paradigm?

    Thanks.
     
  7. Mar 10, 2008 #6

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Never mind, I think I got it, thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?