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Let X be a complex hermitian manifold with hermitian form $\omega$. How can you prove that if $\omega$ has negative holomorphic sectional curvature, then its scalar curvature is negative, too?

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    $\begingroup$ This appears to me to be a straightforward consequence of the definitions of scalar and holomorphic sectional curvature and of a metric compatible with a complex structure. $\endgroup$
    – Deane Yang
    Oct 13, 2010 at 20:50
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    $\begingroup$ On second thought I was careless. I do not see how to prove this and am not sure it is true. You can define something called holomorphic scalar curvature and that is negative. $\endgroup$
    – Deane Yang
    Oct 14, 2010 at 0:51
  • $\begingroup$ For the Kahler case you can see this on pages 177-178 of "Complex differential geometry" by Fangyang Zheng. For the non-Kahler case... do you like long calculations with curvature tensors in local coordinates? $\endgroup$ Oct 14, 2010 at 5:18
  • $\begingroup$ Oh! Hey brother! $\endgroup$ Oct 14, 2010 at 5:21
  • $\begingroup$ Hey little brother, thank you very much! I go and check ! $\endgroup$
    – diverietti
    Oct 15, 2010 at 16:03

2 Answers 2

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Here is the answer.

Let $(X,\omega)$ be a Kähler $n$-dimensional manifold. Fix a point $x_0\in X$ an choose local holomorphic coordinates $(z_1,\dots,z_n)$ centered at $x_0$ and such that $(\partial/\partial z_1,\dots,\partial/\partial z_n)$ is unitary at $x_0$. Let $$ \Theta_{x_0}(T_X,\omega)=\sum_{j,k,l,m=1}^nc_{jklm}\hspace{0.3mm}dz_j\wedge d\bar z_k\otimes\left(\frac\partial{\partial z_l}\right)^*\otimes\frac\partial{\partial z_m} $$ be the Chern curvature at the point $x_0$. Consider the induced hermitian form on rank one tensors of $T_X\otimes T_X$ given by $$ \theta_{T_{X,x_o}}(v\otimes w)=\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}v_j\bar v_k w_l\bar w_m, $$ where $$ v,w\in T_{X,x_0},\quad v=\sum v_j\hspace{0.3mm}\frac\partial{\partial z_j},\quad w=\sum w_j\hspace{0.3mm}\frac\partial{\partial z_j}. $$ With this notation, the holomorphic sectional curvature in the direction of $v\in T_{X,x_0}\setminus\{0\}$ is given by $$ \frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v). $$ The idea now is to take the average on the $\omega$-unit sphere $S^{2n-1}$ and try to deduce something on the scalar curvature at the point $x_0$ which is given by $$ s(x_0)=2\sum_{j,k=1}^nc_{jjkk}. $$ So, let's compute the integral $$ \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi), $$ where $d\sigma(\xi)$ is the probability Haar measure on $S^{2n-1}$. It is not hard to see that the integral $$ \int_{S^{2n-1}}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) $$ vanishes unless $j=k$ and $l=m$ or $j=m$ and $k=l$. Thus, we have to compute $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi),\quad j,k=1,\dots,n. $$ It is classically known that $$ \int_{S^{2n-1}}|\xi_j|^4\hspace{0.3mm}d\sigma(\xi)=\frac 2{n(n+1)},\quad j=1,\dots,n, $$ and $$ \int_{S^{2n-1}}|\xi_j|^2|\xi_k|^2\hspace{0.3mm}d\sigma(\xi)=\frac 1{n(n+1)},\quad 1\le j\ne k\le n. $$ Then, we get $$ \begin{aligned} \int_{S^{2n-1}}\sum_{j,k,l,m}^nc_{jklm}\hspace{0.3mm}\xi_j\bar \xi_k \xi_l\bar \xi_m\hspace{0.3mm}d\sigma(\xi) & =\sum_{j,k=1}^nc_{jjkk}\left(\delta_{jk}\frac 2{n(n+1)}+(1-\delta_{jk})\frac 2{n(n+1)}\right) \\ & = \frac 2{n(n+1)}\sum_{j,k=1}^nc_{jjkk}=\frac 1{n(n+1)}s(x_0), \end{aligned} $$ where we have used the Kähler identity $c_{jklm}=c_{jmlk}$.

Thus, if $\frac{1}{||v||_\omega^4}\theta_{T_{X,x_o}}(v\otimes v)$ is negative, so is its average and we are done.

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  • $\begingroup$ Dear diverietti - The volume containing the article you request can be viewed in 4 libraries near you, the closest being the one at Institut Henri Poincaré. If Liliane is as efficient as what she used to be, she should be able to have the paper brought to you in a few days. $\endgroup$
    – ACL
    Dec 1, 2010 at 7:08
  • $\begingroup$ Thanks a lot ACL... It's nice to hear you calling me "Diverietti"... :) $\endgroup$
    – diverietti
    Dec 1, 2010 at 22:37
  • $\begingroup$ @diverietti If you wanted a reference, I believe Berger proved this back in '66: Berger, M., Sur les variétés d'Einstein compactes. In Comptes Rendus de la IIIe Réunion du Groupement des Mathématiciens d'Expression Latine (Namur, 1965), 35--55, Librairie Universitaire, Louvain, 1966. $\endgroup$
    – user105074
    Dec 1, 2020 at 1:36
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Although I'm sure you know this, but as stated, the original question can be misleading. As you have shown, if $\omega$ is a Kähler metric with negative holomorphic sectional curvature, then $\omega$ has negative scalar curvature. If the metric is not Kähler, however, then the (Chern) holomorphic sectional curvature does not control the (Chern) scalar curvature.

If we write $R_{i \bar{j} k \bar{\ell}}$ for the components of the Chern connection in a local frame, then tracing the curvature with the metric yields four Ricci curvatures $\text{Ric}^{(1)} = g^{k \bar{\ell}} R_{i \bar{j} k \bar{\ell}}$, $\text{Ric}^{(2)} = g^{i \bar{j}} R_{i \bar{j} k \bar{\ell}}$, $\text{Ric}^{(3)}= g^{k \bar{j}} R_{i \bar{j} k \bar{\ell}}$, and $\text{Ric}^{(4)} = g^{i \bar{\ell}} R_{i \bar{j} k \bar{\ell}}$. Using the metric to trace again, we see that $\text{Ric}^{(1)}$ and $\text{Ric}^{(2)}$ trace to the same scalar curvature $s$, while $\text{Ric}^{(3)}$ and $\text{Ric}^{(4)}$ trace to $\hat{s}$. In general, $s \neq \hat{s}$, and the holomorphic sectional curvature only controls $s+\hat{s}$. This is essentially what your argument in the Kähler case shows if you don't use the Kähler symmetries :-) Of course, if the metric is balanced $d\omega^{n-1}=0$ and the manifold is compact, then $s = \hat{s}$.

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  • $\begingroup$ Thanks for the answer! $\endgroup$
    – diverietti
    Jan 8 at 16:17

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