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Metric on a Riemannian manifold $(M,g)$ is Einstein, if for some function $\lambda\colon M\to \mathbb R$ $$ Ric(g)=\lambda g. $$ It is well know, that such $\lambda$ is, in fact, a constant.

The notion of Einstein metric fits perfectly into the world of K\"ahler manifolds, in this case such a metric is called K\"ahler-Einstein. Existence of K\"ahler-Einstein metrics is related to many deep results in differential and algebraic geometry.

I wonder whether there is something interesting going on for an arbitrary complex Hermitian manifold $(M, g, J)$. More specifically, let $\nabla$ be the Chern connection on a Hermitian manifold $(M, g, J)$ and let $\Omega\in \Lambda^{1,1}T^*M\otimes\Lambda^{1,1}T^*M$ be the curvature form of this connection $$ \Omega(\xi,\bar\eta,\zeta,\bar\nu):=g\Bigl( ( \nabla_\xi\nabla_{\bar\eta}-\nabla_{\bar\eta}\nabla_\xi- \nabla_{[\xi,\bar\eta]} )\zeta, \bar\nu \Bigr). $$ Note that unlike the K\"ahler situation, $\nabla$ has torsion and $\Omega$ has less symmetries, then usually. In particular, we can define two different Chern-Ricci forms: $$ \Theta^{(1)}_{i\bar j}=g^{k\bar l}\Omega_{i\bar jk\bar l},\quad \Theta^{(2)}_{k\bar l}=g^{i\bar j}\Omega_{i\bar jk\bar l}. $$

One might play the same game as in Riemannian case, and introduce two versions of Chern-Einstein metrics (is there a common name for these phenomena?) $$ g_{i\bar j}=\lambda_1 \Theta^{(1)}_{i\bar j}, \mbox{ and } g_{i\bar j}=\lambda_2 \Theta^{(2)}_{i\bar j}. $$

Remark. In the second case the metric $g$ on $T^{1,0}M$ is Hermitian-Einstein. Hermitian-Einstein is not necessarily Chern-Einstein for $\Theta^{(2)}$.

Very vaguely, I would like to know

Q1 What is know about the existence of Chern-Einstein metrics on Hermitian manifolds (I am mostly interested in dimensions >2)?

Q2 What are topological/geometrical obstructions to the existence of such metrics?

A more direct question is

Q3 Is it true that $\lambda_1$ and $\lambda_2$ (whic are apriori functions on $M$) are necessarily constant?

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1 Answer 1

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Let me write your equations instead as $$\Theta^{(1)}_{i\bar{j}}=\lambda_1 g_{i\bar{j}},$$ $$\Theta^{(2)}_{i\bar{j}}=\lambda_2 g_{i\bar{j}},$$ where $\lambda_1,\lambda_2$ are real-valued functions. The Chern-Ricci form $\Theta^{(1)}=\sqrt{-1}\Theta^{(1)}_{i\bar{j}}dz^i\wedge d\bar{z}^j$ is closed.

The case $\lambda_1\equiv 0$ is easy to deal with. Such a Hermitian metric exists if and only if the Chern-Ricci form defines the zero class in the Bott-Chern cohomology $H^{1,1}_{\rm BC}(X,\mathbb{R})$, i.e. if and only if for some (and hence any) Hermitian metric $g$ we have $\Theta^{(1)}_g=\sqrt{-1}\partial\bar{\partial}F$ for some smooth function $F$ on $X$. In this case the conformal metric $e^{F/n}g$ has vanishing Chern-Ricci form. The vanishing of this cohomology class implies the vanishing of $c_1(X)$ in $H^2(X,\mathbb{R})$, but not conversely.

When $\lambda_1$ is not identically zero, then $\lambda_1$ is constant if and only if the torsion $1$-form $\tau$ of $g$ vanishes identically, where $\tau=T_{ij}^j dz^i,$ and $T_{ij}^k$ is the torsion of $g$. This is a result of Goldberg. In this case, we must have that $g$ is in fact Kähler, since $d\Theta^{(1)}=0$.

In general, the equation $\Theta^{(2)}_{i\bar{j}}=\lambda_2 g_{i\bar{j}}$ does not imply that $\lambda_2$ is constant. Also, even if $\lambda_2$ is constant and nonzero this does not imply that $g$ is Kähler. A simple example is the standard Hermitian metric $$g_{i\bar{j}}=\frac{\delta_{ij}}{|z|^2+|w|^2},$$ on the Hopf surface $\mathbb{C}^2/((z,w)\sim (2z,2w))$, which satisfies $\Theta^{(2)}_{i\bar{j}}=g_{i\bar{j}}$.

In general since such metrics are Hermitian-Einstein, you get topological obstructions from the Chern number inequality.

Lastly, you could also define a third Ricci curvature $$\Theta^{(3)}_{i\bar{j}}=g^{k\bar{\ell}}\Omega_{i\bar{\ell}k\bar{j}}.$$ Then the result of Goldberg also holds in this case.

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  • $\begingroup$ Thank you for the answer. Of course, I've mixed up the constants the definition of Chern-Einstein metrics. Do you know any examples when $\lambda_2$ is not zero? $\endgroup$ Commented Apr 15, 2016 at 14:10
  • $\begingroup$ The Hopf surface example above has $\lambda_2=1$, is that what you want? $\endgroup$
    – YangMills
    Commented Apr 15, 2016 at 14:16
  • $\begingroup$ Sorry, I meant non-constant, of course. $\endgroup$ Commented Apr 15, 2016 at 14:33
  • $\begingroup$ @YangMills What about $\Theta^{(4)}_{i \overline{j}} = g^{k \overline{\ell}} \Omega_{k \overline{j} i \overline{\ell}}$? $\endgroup$
    – AmorFati
    Commented Aug 8, 2021 at 21:35
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    $\begingroup$ @VeryConfused that is equal to the complex conjugate of $\Theta^{(3)}$ $\endgroup$
    – YangMills
    Commented Aug 10, 2021 at 14:33

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