\section{Complex differential geometric background and hyperbolicity}

The material in this section is somehow standard, but we take the opportunity here to fix notations and explain some remarkable facts which are not necessarily in everybody's background. We refer to \cite{Dem,Huy05,Zhe00} for an excellent and more systematic treatise of the subject.

Let $X$ be a complex manifold of complex dimension $n$, and let $h$ be a hermitian metric on its tangent space $T_X$, which is considered as a complex vector bundle endowed with the standard complex structure $J$ inherited from the holomorphic coordinates on $X$. Then, the real part $g$ of $h=g-i\omega$ defines a riemannian metric on the underlying real manifold, while its imaginary part $\omega$ defines a $2$-form on $X$. 

Now, one can consider both the riemannian or the hermitian theory on $X$. On the one hand we have the existence of a unique connection $\nabla$ on $T_X^\mathbb R$ ---the Levi--Civita connection--- which is both compatible with the metric $g$ and without torsion. Here the superscript $\mathbb R$ is put on $T_X$ to emphasize that we are looking at the real underlying manifold. We call the square of this connection $R=\nabla^2$, the riemannian curvature of $(T_X^\mathbb R,g)$. It is a $2$-form with values in the endomorphisms of $T_X^\mathbb R$. 

On the other hand, we can complexify $T_X$ and decompose it as a direct sum of the eigenbundles for the complexified complex structure $J\otimes\operatorname{Id}_\mathbb C$ relatives to the eigenvalues $\pm i$:
$$
T_X^\mathbb C=T_X\otimes\mathbb C\simeq T^{1,0}_X\oplus T^{0,1}_X.
$$
We have a natural vector bundle isomorphism
$$
\begin{aligned}
\xi\colon & T_X^\mathbb R\to T^{1,0}_X \\
& v\mapsto \frac 12(v-iJv)
\end{aligned}
$$
which is moreover $\mathbb C$-linear: $\xi\circ J=i\xi$. There is a natural way to define a hermitian metric on $T_X^\mathbb C$, as follows. We first consider the $\mathbb C$-bilinear extension $g^\mathbb C$ of $g$, and then its sesquilinear form $\tilde h$ made up using complex conjugation in $T_X^\mathbb C$:
$$
\tilde h(\bullet,\bullet):=g^\mathbb C (\bullet,\bar\bullet).
$$
Such a hermitian metric realize the direct sum decomposition above as an orthogonal decomposition. The complexification of $\omega$, which we still call $\omega$ by an abuse of notation, is then a real positive $(1,1)$-form. These three notions, namely a hermitian metric on $T_X$, a hermitian metric on $T^{1,0}_X$, and a real positive $(1,1)$-form are essentially the same, since there is a canonical way to pass from one to the other.

Now, we know that there exists a unique connection $D$ on $T_X^{1,0}$ which is both compatible with $\tilde h$ and the complex structure: the Chern connection. We call the square of this connection $\Theta=D^2$, the Chern curvature of $(T_X^{1,0},\tilde h)$. It is a $(1,1)$-form with values in the anti-hermitian endomorphisms of $(T_X^{1,0},\tilde h)$.

A basic question is then: can we compare these two theories \textsl{via} $\xi$? The answer is classical and surprisingly simple. The riemannian theory and the hermitian one are the same if and only if the metric $h$ is KŠhler, \textsl{i.e.} if and only if $d\omega=0$. In other words, the metric is KŠhler if and only if
$$
D=\xi\circ\nabla\circ\xi^{-1},
$$
and of course, in this case, $\Theta=\xi\circ R\circ\xi^{-1}$.

\subsection{Notions of curvature in riemannian and hermitian geometry and their correlation in the KŠhler case.}

We know give a brief overview of the different notions of curvature in the setting respectively of riemannian geometry and hermitian geometry. Then, we shall compare them in the case of KŠhler metrics, with particular attention to the \lq\lq propagation\rq\rq{} of signs.

\subsubsection{The riemannian case}

Let $(M,g)$ be a riemannian manifold, $\nabla$ its Levi--Civita connection and $R$ its riemannian curvature. To this data it is attached the classical notion of \emph{sectional curvature} $\mathcal K_g$ of $g$. It is a function which assigns to each $2$-plane $\pi=\operatorname{Span}(v,w)$ in $T_M$ the real number
$$
\mathcal K_g(\pi)=-\frac{\langle R(v,w)\cdot v,w\rangle_g}{||v||^2_g||w||^2_g-|\langle v,w\rangle_g|^2}.
$$
One can verify that this function completely determines the riemannian curvature tensor. One usually also considers other \lq\lq easier\rq\rq{} tensors obtained by performing some type of contractions on $R$, for instance the \emph{Ricci curvature} $r_g$ and the \emph{scalar curvature} $s_g$. The former is a symmetric $2$-tensor defined by
$$
r_g(u,v)=\operatorname{tr}_{T_M}\bigl(w\mapsto R(w,u)\cdot v\bigr).
$$
The latter is the real function on $M$ obtained by taking the trace of the Ricci curvature with respect to $g$:
$$
s_g=\operatorname{tr}_{g}r_g.
$$
One can straightforwardly show that, up to a positive factor which depends only on the dimension of $M$, the Ricci curvature can be obtained as an average of sectional curvatures and the scalar curvature as an average of Ricci curvatures. In particular the sign of the sectional curvature \lq\lq dominates\rq\rq{} the sign of the Ricci curvature which, in turn, \lq\lq dominates\rq\rq{} the sign of the scalar curvature. 

\subsubsection{The hermitian case}

We now look at the complex case. We start more generally with the notion of \emph{Griffiths curvature} for a holomorphic hermitian vector bundle $(E,h)$ over a complex manifold $X$. In this situation, we also have a unique connection both compatible with the metric and the holomorphic structure on $E$, whose curvature we still call $\Theta$. 

It is a $(1,1)$-form with values in the anti-hermitian endomorphisms of $(E,h)$. The Griffiths curvature assigns to each pair $(v,\zeta)\in T_{X,x}^{1,0}\times E_x$, $x\in X$, the real number given by
$$
\theta_{E,h}(v,\zeta)=\langle\Theta(v,\bar v)\cdot\zeta,\zeta\rangle_h.
$$
It has the remarkable property (which is a special case of what is called more generally Griffiths formulae) that it decreases when passing to holomorphic subbundles. Suppose that $S\subseteq E$ is a holomorphic subbundle of $E$ and endow it with the restriction metric $h|_S$. Then, given $x\in X$, $\zeta\in S_x\subseteq E_x$, and $v\in T_{X,x}^{1,0}$, we always have
$$
\theta_{S,h|_S}(v,\zeta)\le\theta_{E,h}(v,\zeta).
$$
Now, we look more closely at the special case where $E=T^{1,0}_X$, and the hermitian metric is $\tilde h$ as above. In this case, the Griffiths curvature is nothing but (up to normalization) what is classically called \emph{holomorphic bisectional curvature}. To be more precise, given a point $x\in X$ and two non-zero holomorphic tangent vectors $v,w\in T^{1,0}_{X,x}\setminus\{0\}$ we define the holomorphic bisectional curvature in the directions given by $v,w$ as 
$$
\operatorname{HBC}_h(x,[v],[w])=\frac{\theta_{T_X^{1,0},\tilde h}(v,w)}{||v||^2_{\tilde h}||w||^2_{\tilde h}}.
$$
By a slight abuse of notation, we may possibly confuse and interchange $h$, $\tilde h$ and $\omega$ in what follows.

The \emph{holomorphic sectional curvature} is defined to be the restriction of the holomorphic bisectional curvature to the diagonal:
$$
\operatorname{HSC}_h(x,[v])=\operatorname{HBC}_h(x,[v],[v])=\frac{\theta_{T_X^{1,0},\tilde h}(v,v)}{||v||^4_{\tilde h}}.
$$

In the spirit of the riemannian case, we can construct a closed real $(1,1)$-form, the \emph{Chern--Ricci form}, by taking the trace with respect to the endomorphism part of the Chern curvature. We also normalize it in such a way that its cohomology class coincides with the first Chern class of the manifold, namely:
$$
\operatorname{Ric}_\omega=\frac i{2\pi}\operatorname{tr}_{T^{1,0}_X}\Theta.
$$
The new feature here is that the Chern--Ricci tensor is a $2$-form always belonging to a fixed cohomology class, the first Chern class of $X$, independently of the choice of the metric. This is because, in general, the trace of the curvature of a vector bundle is the curvature of the induced connection on the determinant bundle, which in this case is the dual of the canonical bundle of $X$. By taking again the trace, but this time with respect to $\omega$, we get what is called the \emph{Chern scalar curvature}. Thus, it is by definition the unique real function $\operatorname{scal}_\omega\colon X\to\mathbb R$ such that
$$
\operatorname{Ric}_\omega\wedge\frac{\omega^{n-1}}{(n-1)!}=\operatorname{scal}_\omega\,\frac{\omega^{n}}{n!}.
$$

\subsubsection{Negativity of the holomorphic sectional curvature and hyperbolicity}

Before going further and explore ---when the metric is KŠhler--- the links among the different notions of curvature we introduced in the riemannian and hermitian setting, we would like to explain here how the negativity of the holomorphic sectional curvature implies the Kobayashi hyperbolicity of the manifold. We want to do it here before entering the KŠhler world, since this is a purely hermitian fact. For our purposes, it is sufficient to deal with the smooth compact case even if more general statements can be established.

\begin{theorem}
Let $(X,\omega)$ be a compact hermitian manifold such that we have $\operatorname{HSC}_\omega<0$. Then, $X$ is Kobayashi hyperbolic.
\end{theorem}

In the next section we will see that the negativity of the holomorphic sectional curvature is a sufficient but not necessary condition for Kobayashi hyperbolicity.

\begin{proof}
By Brody's theorem \cite{Bro78} (see \cite[(3.6.3) Theorem]{Kob98} for a more modern account), it is sufficient to show that every holomorphic map $f\colon\mathbb C\to X$ whose derivative is $\omega$-bounded is constant. So let $f$ be such a map a consider the function
$$
\begin{aligned}
F\colon & \mathbb C\to\mathbb R\cup\{-\infty\} \\
& t\mapsto \log||f'(t)||^2_\omega,
\end{aligned}
$$
which is clearly upper semi-continuous and bounded from above. Suppose by contradiction that $F$ is not identically $-\infty$, which corresponds to the fact that $f$ is not constant. Then, of course, the locus where $\log||f'(t)||^2_\omega$ is $-\infty$ is a discrete set. We now check that $\log||f'(t)||^2_\omega$ is a subharmonic function on the whole $\mathbb C$, which is moreover strictly subharmonic over $\{f'\ne 0\}$. Since any bounded subharmonic function on $\mathbb C$ is constant \cite[Proposition 2.7.3]{Kli91}, this gives a contradiction, because a constant function cannot be strictly subharmonic somewhere.

First of all we show the subharmonicity of $\log||f'||^2_\omega$, by showing that for all positive integer $k$ the smooth  functions defined on the whole complex plane $\psi_\varepsilon=\log(||f'||^2_\omega+\varepsilon)$ are subharmonic, \textsl{i.e.} $i\partial\bar\partial\psi_\varepsilon\ge 0$. For, since
$$
\log||f'||^2_\omega=\lim_{\varepsilon\to 0}\psi_\varepsilon
$$
pointwise, and the sequence $\{\psi_\varepsilon\}$ is decreasing, then the subharmonicity of $\log||f'||^2_\omega$ follows from \cite[Theorem 2.6.1, (ii)]{Kli91}. 

So, fix $\varepsilon>0$ and consider a point $t_0\in\mathbb C$. Call $x_0=f(t_0)\in X$ and choose holomorphic coordinates $(z_1,\dots,z_n)$ for $X$ centered at $x_0$ so that $x_0$ corresponds to $z=0$, and write $f=(f_1,\dots,f_n)$ for $f$ in these coordinates. Moreover, chose a normal coordinate frame $\{e_1,\dots,e_n\}$ for $(T_X,\omega)$ at $x_0$ \cite[(12.10) Proposition]{Dem}. With this choice we have that
$$
\langle e_l(z),e_m(z)\rangle_\omega=\delta_{lm}-\sum_{j,k=1}^nc_{jklm}\,z_j\bar z_k +O(|z|^3),
$$
where the $c_{jklm}$'s are the coefficients of the Chern curvature of $\omega$. Observe that, since the metric is not supposed to be KŠhler, we can not chose holomorphic coordinates around $x_0$ such that the $e_l$'s can be taken simply to be $\partial/\partial z_l$; nevertheless, by a constant change of coordinates, we can suppose that $e_l(x_0)$ equals $\partial/\partial z_l(x_0)$, at least at $x_0$. Now, of course there exists holomorphic functions $\varphi_j$, $j=1,\dots,n$, defined on a neighborhood of $t_0$ such that 
$$
f'(t)=\sum_{j=1}^n \varphi_j(t)\,e_j(t),
$$
so that around $t_0$ we have
$$
||f'(t)||^2_\omega=|\varphi(t)|^2-\sum_{j,k,l,m=1}^nc_{jklm}\, f_j(t)\overline{f_k(t)}\varphi_l(t)\overline{\varphi_m(t)}+O(|f|^3).
$$
Moreover, we have $f'_j(t_0)=\varphi_j(t_0)$ for all $j$, since the $e_l$'s and the $\partial/\partial z_l$'s agree at $t_0$.

Remark that, since $X$ is compact, there exists a positive constant $\kappa$ such that $\HSC_\omega<-\kappa$. This condition reads in our coordinates
$$
\sum_{j,k,l,m=1}^n c_{jklm}\, v_j\bar v_k v_l\bar v_m< -\kappa |v|^4, \quad\forall v=(v_1,\dots, v_n)\in \mathbb C^n.
$$
Now, we have to compute $\partial||f'||^2_\omega$, $\bar\partial||f'||^2_\omega$ and $\partial\bar\partial||f'||^2_\omega$ at $t_0$, since
$$
\partial\bar\partial\psi_\varepsilon=\frac{-1}{(||f'||^2_\omega+\varepsilon)^2}\partial||f'||^2_\omega\wedge\bar\partial||f'||^2_\omega+\frac 1{||f'||^2_\omega+\varepsilon}\partial\bar\partial||f'||^2_\omega.
$$
We find
$$
\begin{aligned}
&\partial||f'||^2_\omega|_{t_0}=\langle \varphi'(t_0),f'(t_0)\rangle \,dt, \\
&\bar\partial||f'||^2_\omega|_{t_0}=\langle f'(t_0),\varphi'(t_0)\rangle\,d\bar t, \\
&i\partial\bar\partial||f'||^2_\omega|_{t_0}  =i\biggl(|\varphi'(t_0)|^2-\sum_{j,k,l,m=1}^nc_{jklm}\,f'_j(t_0)\overline{f'_k(t_0)}f'_l(t_0)\overline{f'_m(t_0)}\biggr)\,dt\wedge d\bar t \\
&\qquad\qquad\qquad >i\bigl(|\varphi'(t_0)|^2+\kappa |f'(t_0)|^4\bigr)\,dt\wedge d\bar t,
\end{aligned}
$$
where the brackets just mean the standard hermitian product in $\mathbb C^n$. Putting all this together we obtain
$$
\begin{aligned}
i\partial\bar\partial\psi_\varepsilon|_{t_0} &>i\biggl( \frac{-|\langle f'(t_0),\varphi'(t_0) \rangle|^2}{(|f'(t_0)|^2+\varepsilon)^2}+\frac{|\varphi'(t_0)|^2+\kappa |f'(t_0)|^4}{|f'(t_0)|^2+\varepsilon}\biggr)\,dt\wedge d\bar t \\
&\ge i\biggl( \frac{-|f'(t_0)|^2|\varphi'(t_0)|^2}{(|f'(t_0)|^2+\varepsilon)^2}+\frac{|\varphi'(t_0)|^2+\kappa |f'(t_0)|^4}{|f'(t_0)|^2+\varepsilon}\biggr)\,dt\wedge d\bar t \\
&=i\,\frac{\kappa|f'(t_0)|^6+\varepsilon\bigl(|\varphi'(t_0)|^2+\kappa|f'(t_0)|^4\big)}{(|f'(t_0)|^2+\varepsilon)^2}\,dt\wedge d\bar t \ge 0,
\end{aligned}
$$
where we used  Cauchy--Schwarz for the second inequality, and so $\psi_\varepsilon$ is subharmonic at each point.

To conclude, observe that the very same computation with $\varepsilon=0$ makes sense away from points where $f'=0$, and give moreover strict positivity for $i\partial\bar\partial\log ||f'||^2_\omega$ at these points. Which means that, away from $\{f'=0\}$, $\log ||f'||^2_\omega$ is strictly subharmonic, as desired.
\end{proof}

\subsubsection{The KŠhler case}

Suppose now that $(X,\omega)$ is a KŠhler manifold, with $\omega=-\Im h$. Let $(X,g=\Re h)$ be the underlying riemannian manifold. We saw that this is precisely the case when $\Theta$ and $R$ correspond to each other via $\xi$. In this setting, using $\xi$, one can show easily the following useful relation between the holomorphic bisectional curvature and the riemannian sectional curvature. Let $v,w\in T_{X,x}$ be two independent (real) tangent vectors. Then,
\begin{multline*}
\operatorname{HBC}_\omega(x,[\xi(v)],[\xi(w)])=\frac{||Jv||^2_g||w||^2_g-|\langle Jv,w\rangle_g|^2}{||v||^2_g||w||^2}\,\mathcal K_g\bigl(\operatorname{Span}\{Jv,w\}\bigr)\\
+\frac{||v||^2_g||w||^2_g-|\langle v,w\rangle_g|^2}{||v||^2_g||w||^2}\,\mathcal K_g\bigl(\operatorname{Span}\{v,w\}\bigr).
\end{multline*}
In particular, if $\mathcal K_g$ has a sign at a certain point, so does $\operatorname{HBC}_\omega$, and the sign is the same. Moreover, by specializing at the diagonal, we get
$$
\operatorname{HSC}_\omega(x,[\xi(v)])=\mathcal K_g\bigl(\operatorname{Span}\{Jv,v\}\bigr),
$$
that is, the holomorphic sectional curvature is nothing but the riemannian sectional curvature computed on complex $2$-planes.

In the KŠhler setting, not surprisingly, also the riemannian Ricci curvature and the Chern--Ricci forms as well as the corresponding scalar curvatures are related to each other. The precise relation is as follows, for $v,w\in T_{X,x}$:
$$
\operatorname{Ric}_\omega\bigl(\xi(v),\overline{\xi(w)}\bigr)=\frac i{4\pi}\bigl(r_g(v,w)-ir_g(Jv,w)\bigr).
$$
In particular, since for $g$ the real part of a KŠhler metric the Ricci tensor is $J$-invariant, we get that
$$
-i\operatorname{Ric}_\omega\bigl(\xi(v),\overline{\xi(v)}\bigr)=\frac 1{4\pi}r_g(v,v),
$$
that is $\operatorname{Ric}_\omega$ is a positive (resp. semi-positive, negative, semi-negative) $(1,1)$-form if and only if $r_g$ is a positive (resp. semi-positive, negative, semi-negative) symmetric bilinear form (observe that $-i\operatorname{Ric}_\omega\bigl(\bullet,\bar\bullet\bigr)$ is nothing but the real quadratic form associated to the real $(1,1)$-form $\operatorname{Ric}_\omega$). From this we also infer that
$$
\operatorname{scal}_\omega=\frac 1{4\pi}\,s_g.
$$
\begin{remark}\label{totscalcurv}
In the compact KŠhler case, since both $\omega$ and $\operatorname{Ric}_\omega$ are closed forms, the total scalar curvature becomes a cohomological invariant, namely
$$
\begin{aligned}
\int_X s_g\,dV_g &=4\pi\int_X\operatorname{scal}_\omega\frac{\omega^n}{n!}\\
&=4\pi\int_X\operatorname{Ric}_\omega\wedge\frac{\omega^{n-1}}{(n-1)!}\\
&=\frac{4\pi}{(n-1)!}\,c_1(X)\cdot[\omega]^{n-1}.
\end{aligned}
$$
In particular, the total scalar curvature of a compact KŠhler manifold with vanishing first Chern class must always be zero. This observation will be useful later.
\end{remark}

Now we explain, still in the KŠhler case,  the link between the signs respectively of the holomorphic bisectional curvature and Ricci curvature, and of the holomorphic sectional curvature with the scalar curvature.

\begin{proposition}\label{average}
Let $(X,\omega)$ be compact KŠhler manifold of complex dimension $n$, $x_0\in X$, and $v\in T^{1,0}_{X,x_0}\setminus\{0\}$. Then, we have
$$
-\frac{2\pi i}{n||v||^2_{\tilde h}}\operatorname{Ric}_\omega(v,\bar v)=\dashint_{S^{2n-1}} \HBC_\omega([v],[w])\,d\sigma(w),
$$
and
$$
\frac{4\pi}{n(n+1)}\scal_\omega(x_0)=\dashint_{S^{2n-1}}\HSC_\omega([v])\,d\sigma(v),
$$
where $d\sigma$ is the Lebesgue measure on the $\tilde h$-unit sphere $S^{2n-1}$ in $T^{1,0}_{X,x_0}$, and by $\dashint$ we mean taking the average, \textsl{i.e.}
$$
\dashint_{S^{2n-1}}=\frac{(n-1)!}{2\pi^n}\int_{S^{2n-1}}.
$$
\end{proposition}

In particular, we find that the sign of the holomorphic bisectional curvature of a KŠhler metric dominates the sign of the Ricci curvature, while the sign of the holomorphic sectional curvature dominates the sign of the scalar curvature. 

On the other hand, recall that the holomorphic sectional curvature of a KŠhler metric \emph{completely} determines the curvature tensor \cite[Lemma 7.19]{Zhe00}, so that the point is really whether and how it does spread the sign.


\begin{proof}
Fix holomorphic coordinates around $x_0$ such that 
$$
\omega(x_0)=i\sum_{j=1}^ndz_j\wedge d\bar z_j,
$$ 
\textsl{i.e.} such that $\{\partial/\partial z_j\}_{j=1,\dots,n}$ is a $\tilde h$-unitary basis at $x_0$. Now write the Chern curvature tensor at $x_0$ in these coordinates, to get
$$
\Theta\bigl(T^{1,0}_X,\tilde h\bigr)_{x_0}=\sum_{j,k,l,m=1}^n c_{jklm}\,dz_j\wedge dz_k\otimes\biggl(\frac{\partial}{\partial z_l}\biggr)^*\otimes\frac{\partial}{\partial z_m}.
$$
Since $\omega$ is KŠhler, and we have chosen a unitary basis at $x_0$, we have the well known KŠhler symmetries for the coefficients $c_{jklm}$'s of the curvature:
$$
c_{jklm}=c_{lmjk}=c_{lkjm}=c_{jmlk}=\overline{c_{kjml}}.
$$
Now we express the holomorphic bisectional curvature in coordinates, to get
$$
\HBC_\omega([v],[w])=\frac 1{|v|^2|w|^2}\sum_{j,k,l,m=1}^nc_{jklm}\,v_j\bar v_k w_l\bar w_m,
$$
where $v=\sum_j v_j\,\partial/\partial z_j$ and $w=\sum_j w_j\,\partial/\partial z_j$.
In particular, 
$$
\HSC_\omega([v])=\frac 1{|v|^4}\sum_{j,k,l,m=1}^nc_{jklm}\,v_j\bar v_k v_l\bar v_m.
$$
Moreover, we get for the Chern--Ricci curvature and for the Chern--scalar curvature the following expressions:
$$
\Ric_\omega=\frac i{2\pi}\sum_{j,k,l=1}^n c_{jkll}\,dz_j\wedge dz_k,
$$
and
$$
\scal_\omega=\frac 1{2\pi}\sum_{j,l}c_{jjll}.
$$
Now, we compute
$$
\dashint_{S^{2n-1}} \HBC_\omega([v],[w])\,d\sigma(w) = \frac 1{|v|^2}\sum_{j,k,l,m=1}^nc_{jklm}\,v_j\bar v_k \dashint_{S^{2n-1}} w_l\bar w_m\,d\sigma(w).
$$
The integral on the right hand side is easily seen to be zero unless $l=m$, in which case gives $1/n$. Thus, we obtain
$$
\begin{aligned}
\dashint_{S^{2n-1}} \HBC_\omega([v],[w])\,d\sigma(w) &=\frac 1{n|v|^2}\sum_{j,k,l=1}^nc_{jkll}\,v_j\bar v_k \\
&=-\frac{2\pi i}{n||v||^2_{\tilde h}}\operatorname{Ric}_\omega(v,\bar v).
\end{aligned}
$$
For the holomorphic sectional curvature we have instead
$$
\dashint_{S^{2n-1}} \HSC_\omega([v])\,d\sigma(v) = \sum_{j,k,l,m=1}^nc_{jklm}\dashint_{S^{2n-1}}v_j\bar v_k  v_l\bar v_m\,d\sigma(v).
$$
Once again, it is easy to see that the integrals on the right hand side must be zero unless $j=k$ and $l=m$, or $j=m$ and $k=l$. We have (see for instance \cite{Ber66} or \cite[Lemma 2.2]{Div16} for a detailed and more general computation):
$$
\dashint_{S^{2n-1}}|v_j|^2|v_k|^2\,d\sigma(v)=
\begin{cases}
\frac 2{n(n+1)} & \textrm{if $j=k$} \\
\frac 1{n(n+1)} & \textrm{otherwise},
\end{cases}
$$
so that
$$
\begin{aligned}
\sum_{j,k,l,m=1}^nc_{jklm}\dashint_{S^{2n-1}}v_j\bar v_k  v_l\bar v_m\,d\sigma(v) &=
\sum_{j,l=1}^n c_{jjll}\dashint_{S^{2n-1}}|v_j|^2|v_l|^2\,d\sigma(v) \\
&\qquad+\sum_{j,k=1}^n c_{jkkj}\dashint_{S^{2n-1}}|v_j|^2|v_k|^2\,d\sigma(v) \\
&\qquad\qquad\qquad-\sum_{j=1}^n c_{jjjj}\dashint_{S^{2n-1}}|v_j|^4\,d\sigma(v)\\
&= 2\sum_{j,l=1}^n c_{jjll}\dashint_{S^{2n-1}}|v_j|^2|v_l|^2\,d\sigma(v) \\
&\qquad-\sum_{j=1}^n c_{jjjj}\dashint_{S^{2n-1}}|v_j|^4\,d\sigma(v)\\
&=2\sum_{j\ne l} c_{jjll}\dashint_{S^{2n-1}}|v_j|^2|v_l|^2\,d\sigma(v)\\
&\qquad+\sum_{j=1}^n c_{jjjj}\dashint_{S^{2n-1}}|v_j|^4\,d\sigma(v)\\
&= \frac 2{n(n+1)}\sum_{j,l=1}^n c_{jjll} \\
&=\frac{4\pi}{n(n+1)}\,\scal_\omega(x_0),
\end{aligned}
$$
where, for the second equality, we have used the KŠhler symmetry $c_{jkkj}=c_{jjkk}$.
\end{proof}

The situation can be therefore summarized as follows:
$$
\xymatrix{
 & \mathcal K_{g} \ar@{=>}[r]\ar@{==>}[ddl] & r_g \ar@{=>}[r] & s_g  \\
  & & \Ric_\omega  \ar@{<==>}[u] \ar@{=>}[dr] &  \\
\HBC_\omega \ar@{=>}[r] \ar@{=>}[urr] & \HSC_\omega \ar@{=>}[rr] \ar@{<->}[ur]_{?} & & \scal_\omega \ar@{<==>}[uu]}
$$
The arrows $\Rightarrow$ in the diagram mean that the positivity (resp. semi-positivity, negativity, semi-negativity) of the source curvature implies the positivity (resp. semi-positivity, negativity, semi-negativity) of the target curvature. These arrows are always valid, even in the non KŠhler setting. On the other hand, the dashed arrows are valid in the KŠhler case only. It is however \textsl{a priori} unclear if and how the sign of the holomorphic sectional curvature propagates and determines the signs of the Ricci curvature. 

So, Yau's conjecture deals exactly with this issue, at least in the case of negativity: if $(X,\omega)$ is a compact KŠhler manifold such that $\HSC_\omega<0$, then there exists a (possibly different) KŠhler metric $\omega'$ on $X$ such that $\Ric_{\omega'}<0$. Note that, in particular, this implies that $K_X$ is ample and therefore, by Kodaira's embedding theorem, that $X$ is a projective algebraic manifold.

\begin{remark}
It is somehow embarrassing, but we don't dispose ---at our best knowledge--- any example of a (compact) KŠhler manifold $(X,\omega)$ such that $\HSC_\omega$ is negative but $\HBC_\omega$ or $\Ric_\omega$ do not have a sing. It would be of course highly desirable to have such an example, if any.
\end{remark}