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Let $(X,\Omega)$ be a complex compact Kahler manifold, where $\Omega$ is the fundamental $(1,1)$-form. Moreover let $L$ be a holomorphic line bundle on $X$.

A (smooth) hermitian metric $h$ on $X$ is said admissible with respect to $\Omega$ if $$c_1(L,h)=a\Omega$$ for $a\in\mathbb R$. Here $c_1(L,h)$ is the curvature form or first Chern form of the hermitian line bundle $(L,h)$.

I'd like to visualize geometrically (in some informal way) what is the meaning of the condition $c_1(L,h)=a\Omega$. Admissible hermitian metrics are unique up to a multiplication by a smooth function, so they seem to be very particular hermtian metrics. What is their peculiarity?

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  • $\begingroup$ Are you looking for something more than that they are positive, negative or flat? $\endgroup$ – Gunnar Þór Magnússon Oct 18 '16 at 19:17
  • $\begingroup$ This notion is introduced in Arakelov geometry, and basically I don't know why we need it. $\endgroup$ – Dubious Oct 18 '16 at 19:21
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    $\begingroup$ I think the question is highly non-trivial. See my comments here mathoverflow.net/questions/93522/… . I know for dim=2, but for higher dimension, I don't have any idea. $\endgroup$ – user21574 Nov 22 '17 at 13:57
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I don't know if this answers to your question, and very likely you already know it, but one easy fact is the following.

First of all an obvious necessary condition is that at the level of cohomology classes we have $$ c_1(L)=a[\Omega]. $$ Once you have this, you put any smooth hermitian metric on $L$, say $h_0$, and look at its Chern curvature $c_1(L,h_0)=i/2\pi\,\Theta(L,h_0)$. Since it defines the same cohomology class as $a\Omega$, and since $X$ is Kähler, the $\partial\bar\partial$-lemma tells you that you can find a smooth function $f\colon X\to\mathbb R$ such that $$ c_1(L,h_0)-a\Omega=\frac i{2\pi}\partial\bar\partial f. $$ Now, set $h=e^{f}h_0$. Then, $$ c_1(L,e^fh_0)=-\frac i{2\pi}\partial\bar\partial\log(e^fh_0)=c_1(L,h_0)-\frac i{2\pi}\partial\bar\partial f=a\Omega. $$

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Such hermitian metrics $h$ on the Line bundle are called Kaehler Hermite Einstein metrics. Such metrics are minimal energy, in the sense that their curvature is harmonic (since $2\Delta\omega = \Lambda \partial\bar\partial \omega = 0$ by the Kaehler identities). In addition, however, the orthogonal complement of the Kaehler form (which is negative definite in the "intersection form" $\int \omega^{n-2} c_1(L) c_1(L')$) is zero.

Note that the condition implies that the manifold is projective. In fact the line bundle is ample because the curvature $\Omega$ is a positive form which implies Kodaira vanishing, so the section in $L^N$ for $N \gg 0$ gives an embedding.

IIRC (but I am not actually sure) in the limit $N\to \infty$ the hermitian metric $h$ on $L$ is $1/N$ of the pullback metric of the Fubini-Study metric on $L^N$ under this embedding.

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    $\begingroup$ These metrics are definitely not K-E in general... In particular they cannot be called Kähler-Einstein :) $\endgroup$ – diverietti Oct 18 '16 at 22:27
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    $\begingroup$ Moreover the condition does not always imply that the manifold is projective: when $a=0$ you just can't say. $\endgroup$ – diverietti Oct 18 '16 at 22:49

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