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? 



Negative holomorphic sectional curvature
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.


