# Negative holomorphic sectional curvature

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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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. –  Deane Yang Oct 13 '10 at 20:50
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. –  Deane Yang Oct 14 '10 at 0:51
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? –  Gunnar Magnusson Oct 14 '10 at 5:18
Oh! Hey brother! –  Gunnar Magnusson Oct 14 '10 at 5:21
Hey little brother, thank you very much! I go and check ! –  diverietti Oct 15 '10 at 16:03

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.