On a compact Kahler manifold, let $g$ be the unique Kahler-Einstein metric with $\mathrm{Ricc}(g)=-g$, proved to exist by Yau and Aubin when the first Chern class $C_1(M)<0$.

Question: Does $g$ satisfy uniqueness up to homothety in the following sense: if $h$ is another metric satisfying $\mathrm{Ricc}(h)=-g$ then $h=cg$ for some $c>0$?


2 Answers 2


so here I guess $M$ is a compact Kähler manifold. Thanks to Yau theorem, we know that there exists a unique Kähler metric $h$ in each Kähler cohomology class such that $\mathrm{Ric}(h)=-g$ (more generally you can prescribed the Ricci form to be any form in the cohomology class of $c_1(M)$).

So to answer your question, there are a lot of such metrics $h$, but there is only one such in each Kähler cohomology class of $M$. In particular, if $h$ lies in a (positive) multiple of $-c_1(M)$, then $h$ is proportional to $g$.

  • $\begingroup$ If $\mathrm{Ricc}(g)=-g$ then we know that $\mathrm{Ricc}(cg)=-g$ for any $c>0$. So my question is: are these metrics $h=cg$ for $c>0$ the only others that satisfy $\mathrm{Ricc}(h)=-g$? $\endgroup$
    – mdg
    Commented Oct 21, 2015 at 20:06
  • $\begingroup$ Addition to above comment: Are the metrics $h=cg$ the only ones on the manifold satisfying $\mathrm{Ricc}(h)=-g$, regardless cohomology classes and even whether the metrics are Kahler? $\endgroup$
    – mdg
    Commented Oct 21, 2015 at 20:17
  • 2
    $\begingroup$ If the dimension of $H^{1,1}$ is strictly larger than $1$, then as Henri explained there are other Kahler metrics (not of the form $cg$ with $c$ a constant) with the same Ricci curvature as $g$. $\endgroup$
    – YangMills
    Commented Oct 22, 2015 at 4:56
  • $\begingroup$ @ YangMills Given $C_1(M)<0$, are there instances in which $\mathrm{dim}\;H^{1,1}\leq1$? $\endgroup$
    – mdg
    Commented Oct 22, 2015 at 20:10
  • $\begingroup$ Yes, plenty. For instance, every hypersurface in $\mathbb P^n$ of degree $d>n+1$ (n>3) has the same Picard group as $\mathbb P^n$ by Grothendieck-Lefschetz theorem, so $h^{1,1}=1$. However, if you take the product of two curves of genus $g \ge 2$, you get $h^{1,1}\ge 2$. $\endgroup$
    – Henri
    Commented Oct 22, 2015 at 21:44

In general with your assumption

$$Ric(\omega)=\sqrt{-1}\partial_i\bar\partial_j\log\det g_{i\bar j}$$

so by this formula we have always $Ric(cg)=Ric(g)$, and if $h,h'$ satisfies in $Ric(h)=−g$, $Ric c(h')=−g$ then in general it is sufficient to use of this formula and remove $\partial\bar\partial$ of bothsides and we get $\omega^n=e^c\omega'^n$ or $\det h_{i\bar j}=e^c\det h'_{i\bar j}$ for a constant $c$, so a lot of such metrics exists

and by Aubin's proof (he was a first French Mathematician who proved it and later Yau gave different proof) and Calabi himself also gave a proof for unicity of such metrics when the first Chern class is negative

if the kahler metric $h'$ lies in a $ c_1(K_X)$,then $h=h'$. See Tian's book

In fact if $\omega'$ be the corresponding metric of $h'$ then by $dd^c$-lemma we can write $\omega'=\omega+\sqrt{-1}\partial\bar\partial \varphi$ and so we will have a non-linear parabolic monge-Ampere equation $$\frac {(\omega+\sqrt{-1}\partial\bar\partial \varphi)^n}{\omega^n}=e^c$$

which its solution is unique. See theorem 19.1 page 129 Lectures on Kahler geometry Andrei Moroianu

  • $\begingroup$ I think you mean $h'\in c_1(K_X)$, and in that case $h'=h$ (no constant needed). $\endgroup$
    – Henri
    Commented Oct 22, 2015 at 14:59

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