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

Proving Kähler, finding the Kähler form

  1. Nov 4, 2013 #1
    Hi!

    I have a question about Kähler manifolds. Of course there are many books (I prefer Nakahara) and lecture notes on this topic, but as a physicist I need a very hands-on way of dealing with metrics, etc.

    Given a metric, what is the simplest way to find the Kähler form and to prove the Kähler property? I mean, e.g. Nakahara describes Kähler, but is there something simpler one can do (perhaps just in some cases)?

    An example I have encountered is
    \begin{equation}
    ds^2 = (r-u) \left(\frac{\text d r^2}{F(r)} -\frac{\text d u^2}{G(u)} \right) + \frac{1}{r-u} \left(F(r) (\text d t +u \text d z)^2 - G(u) (\text d t+ r \text d z)^2 \right) ~,
    \end{equation}
    where $F$ and $G$ may be any functions of one variable. This metric is claimed to be Kähler with Kähler form
    \begin{equation}
    \Omega = \text d (r+ u) \wedge \text d t + \text d (r u) \wedge \text d z ~,
    \end{equation}
    which is obviously closed.

    Following Nakahara, one needs to change coordinates to make the metric Hermitian, but can the Kähler form be read off more directly?

    Cheers!
     
  2. jcsd
  3. Nov 4, 2013 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Given a Riemannian metric ##g(X,Y)## and an almost complex structure ##J##, the Kahler form is defined as ##\Omega(X,Y) = g(JX,Y)## (eq (8.55) in Nakahara). The Kahler form will be closed if the almost complex structure is covariantly constant in the Riemannian metric. I'm not sure what complex structure is being used in the example you gave, but it's probably straightforward, if tedious, to work it out.
     
  4. Nov 4, 2013 #3

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Adding to what fzero said: You first need to find the complex structure and show that it is integrable. For a generic metric, this might be rather non-obvious. But in most actual applications, the metric is written in such a way that the complex structure is easy to see.

    What one does, generally, is to write the metric in terms of orthonormal frames. Then you know that the almost-complex structure must simply rotate these frames. In your example, the metric is already a sum of four squares, so the hard work is already done.

    Looking at your metric, it can have signature (++++), (++--), or (----), depending on the signs of F and G. In order to apply complex geometry, you must group things in pairs that have the same signature. So the two terms with ##F(r)## in front go together, and the two terms with ##G(u)## go together. From here it should be easy to postulate an almost-complex structure J. Then you apply standard techniques to show that J is integrable.

    Finally, given J, the Kahler form is well-defined.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving Kähler, finding the Kähler form
  1. Differential forms (Replies: 8)

  2. Connections and forms (Replies: 12)

Loading...