[proofplan]
The proof uses the standard Lie-theoretic description of the tangent space to a polarized period domain and then applies the Griffiths-Schmid curvature formula in a horizontal direction. The formula gives the holomorphic sectional curvature as the negative quotient of the squared Hodge norm of $[\xi,\xi^*]$ by $\|\xi\|_F^4$. It remains only to show that this commutator is nonzero for $0\ne \xi\in \mathfrak g_F^{-1,1}$; this follows because such a $\xi$ lowers the finite Hodge grading, hence is nilpotent, while a normal nilpotent operator on a positive definite Hermitian space must vanish.
[/proofplan]
[step:Identify horizontal tangent vectors with endomorphisms of Hodge type $(-1,1)$]
By the standard tangent-space description of the classifying space of polarized Hodge structures, the holomorphic tangent space at $F$ is identified with
\begin{align*}
T_FD\cong \bigoplus_{a<0}\mathfrak g_F^{a,-a}.
\end{align*}
Under this identification, [Griffiths transversality](/theorems/9129) singles out the horizontal subspace
\begin{align*}
T_F^{\mathrm{hor}}D\cong \mathfrak g_F^{-1,1}.
\end{align*}
Thus the given vector $\xi$ is simultaneously a tangent vector to $D$ at $F$ and a complex-linear endomorphism
\begin{align*}
\xi:H_{\mathbb C}\to H_{\mathbb C}
\end{align*}
satisfying
\begin{align*}
\xi\left(H_F^{p,k-p}\right)\subset H_F^{p-1,k-p+1}
\end{align*}
for every index $p$.
[/step]
[step:Apply the Griffiths-Schmid curvature formula in the horizontal direction]
We use the Griffiths-Schmid curvature formula for the invariant Hodge metric on a polarized period domain in a horizontal tangent direction (citing a result not yet in the wiki: Griffiths-Schmid curvature formula for period domains). Its hypotheses apply here because $D$ is the polarized period domain for $(H_{\mathbb Z},Q)$, the metric is the invariant Hodge metric, and $\xi\in T_F^{\mathrm{hor}}D\cong \mathfrak g_F^{-1,1}$ is horizontal. With the holomorphic sectional curvature normalization fixed in the statement, the formula gives
\begin{align*}
K_D(F,\xi)=-\frac{\|[\xi,\xi^*]\|_F^2}{\|\xi\|_F^4}.
\end{align*}
Since $\xi\ne 0$, the denominator satisfies
\begin{align*}
\|\xi\|_F^4>0.
\end{align*}
It remains to prove
\begin{align*}
[\xi,\xi^*]\ne 0.
\end{align*}
[guided]
We now invoke the exact curvature input used in this theorem. The relevant result is the Griffiths-Schmid curvature formula for period domains in horizontal directions (citing a result not yet in the wiki: Griffiths-Schmid curvature formula for period domains). It applies because the ambient space $D$ is precisely a classifying space of polarized Hodge structures, the metric in the theorem is the invariant Hodge metric, and the vector $\xi$ lies in the horizontal subspace
\begin{align*}
T_F^{\mathrm{hor}}D\cong \mathfrak g_F^{-1,1}.
\end{align*}
Under the Griffiths-Schmid sign and normalization convention for holomorphic sectional curvature, the formula states that the curvature of the complex line $\mathbb C\xi$ is
\begin{align*}
K_D(F,\xi)=-\frac{\|[\xi,\xi^*]\|_F^2}{\|\xi\|_F^4}.
\end{align*}
Here $\xi^*$ is the adjoint for the positive definite Hodge metric on $H_{\mathbb C}$, not the adjoint with respect to the polarization form $Q$ itself. This distinction matters because the positivity argument below uses the Hilbert-space identities associated to a positive definite Hermitian [inner product](/page/Inner%20Product).
Since $\xi\ne 0$, its Hodge norm is positive:
\begin{align*}
\|\xi\|_F>0.
\end{align*}
Therefore
\begin{align*}
\|\xi\|_F^4>0.
\end{align*}
The displayed curvature formula already proves nonpositivity. To obtain strict negativity, the only remaining issue is to rule out the case
\begin{align*}
[\xi,\xi^*]=0.
\end{align*}
That is the role of the nilpotent-normal argument in the next step.
[/guided]
[/step]
[step:Show that a nonzero horizontal endomorphism has nonzero commutator with its Hodge adjoint]
Because only finitely many Hodge summands $H_F^{p,k-p}$ are nonzero and $\xi$ lowers the first Hodge index by one, there exists an integer $N\ge 1$ such that
\begin{align*}
\xi^N=0.
\end{align*}
Thus $\xi$ is nilpotent as an endomorphism of the finite-dimensional Hermitian [vector space](/page/Vector%20Space) $H_{\mathbb C}$.
Suppose, for contradiction, that
\begin{align*}
[\xi,\xi^*]=0.
\end{align*}
Then $\xi\xi^*=\xi^*\xi$, so $\xi$ is normal with respect to the Hodge metric. We prove that a normal nilpotent endomorphism of a finite-dimensional positive definite Hermitian vector space is zero.
Choose the smallest integer $m\ge 1$ such that $\xi^m=0$. If $\xi\ne 0$, then $m\ge 2$, and there exists $v\in H_{\mathbb C}$ such that
\begin{align*}
w:=\xi^{m-1}v
\end{align*}
is nonzero. Since $\xi^m=0$, this vector satisfies
\begin{align*}
\xi w=0.
\end{align*}
For a normal operator, $\ker \xi=\ker \xi^*$. Indeed, if $u\in \ker \xi$, then
\begin{align*}
\|\xi^*u\|_F^2=(\xi\xi^*u,u)_F=(\xi^*\xi u,u)_F=0,
\end{align*}
so $\xi^*u=0$. The reverse implication is identical with $\xi$ and $\xi^*$ interchanged. Therefore
\begin{align*}
w\in \ker \xi^*.
\end{align*}
On the other hand, $w=\xi(\xi^{m-2}v)$ lies in $\operatorname{im}\xi$. Since every vector in $\operatorname{im}\xi$ is orthogonal to every vector in $\ker \xi^*$, the vector $w$ is orthogonal to itself. Hence
\begin{align*}
\|w\|_F^2=0.
\end{align*}
The Hodge metric is positive definite, so $w=0$, contradicting the choice of $w$. Therefore the assumption $[\xi,\xi^*]=0$ is impossible, and
\begin{align*}
[\xi,\xi^*]\ne 0.
\end{align*}
[/step]
[step:Conclude strict negativity of the horizontal holomorphic sectional curvature]
The Hodge-metric norm on $\mathfrak g_{\mathbb C}$ is positive definite, and the previous step gives
\begin{align*}
[\xi,\xi^*]\ne 0.
\end{align*}
Therefore
\begin{align*}
\|[\xi,\xi^*]\|_F^2>0.
\end{align*}
Combining this with the Griffiths-Schmid formula
\begin{align*}
K_D(F,\xi)=-\frac{\|[\xi,\xi^*]\|_F^2}{\|\xi\|_F^4}
\end{align*}
and with $\|\xi\|_F^4>0$, we obtain
\begin{align*}
K_D(F,\xi)<0.
\end{align*}
This proves both the displayed curvature formula and the strict negativity for every nonzero horizontal tangent vector $\xi\in T_F^{\mathrm{hor}}D$.
[/step]
