0
$\begingroup$

Let $(M,\omega)$ be a Kaehler manifold and $h$ be its Hermitian form, then in local sense we can write $$\omega=\partial\bar\partial\log h$$ and also if $f$ be the kaehler potential then we can write $$\omega=\partial\bar\partial\log f$$. So, my question is can say $f$ is equal to $h$ up to additional constant? if we have $f$ then how can we find $h$

$\endgroup$
  • $\begingroup$ What's the Hermition (sic?) form of a Kahler metric? $\endgroup$ – Gunnar Þór Magnússon Apr 24 '14 at 19:21
  • $\begingroup$ Gunnar , I mean was hermitian form $\endgroup$ – user21574 Apr 24 '14 at 20:19
  • $\begingroup$ But what is the "Hermitian form" of a Kahler manifold? When I google it, it is defined to be the $(1,1)$-form $\omega$. $\endgroup$ – Deane Yang Apr 24 '14 at 23:03
  • $\begingroup$ Here you can see in equation 1.11. sciencedirect.com/science/article/pii/039304409090019Y $h$ is hermitian structure $\endgroup$ – user21574 Apr 25 '14 at 13:47
  • $\begingroup$ I guess $h$ here is $g=h+i\omega$, $\endgroup$ – user21574 Apr 25 '14 at 13:55
3
$\begingroup$

The function $\psi:=\log(fh^{-1})$ satisfies $\partial\bar\partial \psi=0$, because $ \partial\bar\partial\log f=\partial\bar\partial\log h =\omega$. Such functions are called pluriharmonic. Locally a pluriharmonic function is a real part of a holomorphic function, by Poincare-Dolbeault-Grothendieck lemma. This fact is true globally, because any real-valued holomorphic function vanishes, and therefore the local holomorphic functions can be glued together. Then, $h$ is unique (up to a constant multiplier) if and only if the manifold has no non-constant global holomorphic functions.

$\endgroup$
  • $\begingroup$ The question here is if $h(s_1,s_2)$ (with sections $s_i$)be hermitian structure then how can we find $h$ by using kahler potential $f$? $\endgroup$ – user21574 Apr 25 '14 at 14:31
  • $\begingroup$ If $h$ is the Hermitian form, what is the meaning of $\log h$? $\endgroup$ – Paul Reynolds Apr 26 '14 at 17:41
  • $\begingroup$ It makes no sense, of course; I assumed that the poster confused a metric with its potential $\endgroup$ – Misha Verbitsky Apr 27 '14 at 16:14
  • $\begingroup$ See sciencedirect.com/science/article/pii/039304409090019Y in equation 1.11, IN FACT , $h$ is hermitian metric corresponding to hermitian form, so $\log h$ here must means $\log \hat h$ which $\hat h$ is hermitian metric corresponding to hermitian form in local trivialization $\endgroup$ – user21574 Apr 27 '14 at 18:24
  • $\begingroup$ I don't have access to this article; but anyway, $dd^c$ cannot be applied to metric, or to its logarithm. $\endgroup$ – Misha Verbitsky Apr 27 '14 at 23:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy