Horizontal period maps have bounded metric differential

Pull the flat bundle and its Hodge splitting back to $\mathfrak{H}$. The bundles $E^p=F^p\cap {\bar{F}}^{k-p}$ are smooth: opposition makes the finite matrix of a local splitting invertible, so its projections are smooth functions of holomorphic frames and their conjugates. Holomorphicity, transversality and their conjugates give ${\mathrm{\nabla }}^{1,0}{E}^p\subset E^p\oplus E^{p-1}$ and ${\mathrm{\nabla }}^{0,1}{E}^p\subset E^p\oplus E^{p+1}$: intersect $F^{p-1}$ with ${\bar{F}}^{k-p}$ in the first case, and $F^p$ with ${\bar{F}}^{k-p-1}$ in the second. Write $\mathrm{\nabla }=D+\theta +\eta$ for its diagonal, lowering $(1,0)$ and raising $(0,1)$ parts. The diagonal connection $D$ preserves $h$: on a fixed Hodge block $C$ is a constant scalar and off-diagonal terms pair to zero, while $\mathrm{\nabla }Q=0$. For adjacent blocks put $B_Q(u,v)=Q(u,\bar{v})$, $u\in E^p$, $v\in E^{p-1}$. Differentiating their zero $B_Q$ pairing gives $B_Q({\theta }_{z}u,v)+B_Q(u,{\eta }_{\bar{z}}v)=0$. Since the two Weil scalars satisfy $i^{2p-k}=-i^{2(p-1)-k}$, this is $h({\theta }_{z}u,v)=h(u,{\eta }_{\bar{z}}v)$. Thus $\eta ={\theta }^{†}$ and $\mathrm{\nabla }=D+\theta +{\theta }^{†}$.

In a curve coordinate write $\theta =Adz$. The Hodge-degree minus-one component of flatness ${\mathrm{\nabla }}^2=0$ gives $D_{\bar{z}}A=0$; its degree-zero component gives $F_D({\mathrm{\partial }}_z,{\mathrm{\partial }}_{\bar{z}})=-[A,A^{†}]$. Products of two $(1,0)$ or two $(0,1)$ forms vanish on the curve, so these are exact identities. In a flat marking, differentiating a section of $E^p\subset F^p$ and projecting modulo $F^p$ leaves exactly its $E^{p-1}$ component $A{\mathrm{\mid }}_{E^p}$. Consequently $A$ is $d\mathrm{\Phi }({\mathrm{\partial }}_z)$ under the tangent-space identification in [L1], and the pullback metric coefficient is $e=\mathrm{\parallel }A{\mathrm{\parallel }}^2=\mathrm{tr}(A{A}^{†})\ge 0$.

If the rank $N$ is zero the map is constant. Otherwise let $\mu =N^{-1}{\sum }_{p}p\dim E^p$, let the self-adjoint grading operator $H$ act by $p-\mu$ on $E^p$, and put $M^2=\mathrm{tr}(H^2)={\sum }_p\dim E^p(p-\mu {)}^2$. For several weights the sums run over their Hodge summands; $A$ preserves the weight summands and still satisfies $[H,A]=-A$. If $M=0$, every nonzero summand has the same first index and $A=0$, again proving the claim. Otherwise put $B=[A,A^{†}]$. It is self-adjoint, and cyclicity of the finite trace gives $\mathrm{tr}(BH)=\mathrm{tr}(A^{†}[H,A])=-e$. Cauchy–Schwarz for finite matrix entries implies $\mathrm{\parallel }B{\mathrm{\parallel }}^2{M}^2\ge e^2$. This constant depends only on the fixed Hodge numbers.

On the open set $e>0$, use the connection induced by $D$ on endomorphisms and its Hilbert–Schmidt metric. Metric compatibility and $D_{\bar{z}}A=0$ give $e_z=⟨D_{z}A,A⟩$ and $e_{z\bar{z}}=\mathrm{\parallel }{D}_{z}A{\mathrm{\parallel }}^2+⟨D_{\bar{z}}{D}_{z}A,A⟩$. The curvature equation in step 2.1 gives $[D_z,D_{\bar{z}}]A=-[B,A]$, so $D_{\bar{z}}{D}_{z}A=[B,A]$. Cyclicity of trace yields $⟨[B,A],A⟩=\mathrm{tr}(B(A{A}^{†}-A^{†}A))=\mathrm{\parallel }B{\mathrm{\parallel }}^2$. It follows that ${\mathrm{\partial }}_z{\mathrm{\partial }}_{\bar{z}}\log e=\frac{\mathrm{\parallel }{D}_{z}A{\mathrm{\parallel }}^2+\mathrm{\parallel }B{\mathrm{\parallel }}^2}{e}-\frac{\mathrm{\mid }⟨D_{z}A,A⟩{\mathrm{\mid }}^2}{e^2}\ge \frac{\mathrm{\parallel }B{\mathrm{\parallel }}^2}{e}\ge \frac{e}{M^2}\mathrm{.}$ The first inequality is finite-dimensional Cauchy–Schwarz. This is the required negative horizontal curvewise curvature bound, derived on the actual pulled-back variation without assuming an ambient period-domain curvature formula.

Transfer to the unit disk by a Cayley transform. Under a holomorphic coordinate change $A$ multiplies by its derivative and $e$ by the absolute square, so step 4.1 has the same form. For $0<r<1$ put ${\lambda }_r(\zeta )=4r^2\mathrm{/}(r^2-\mathrm{\mid }\zeta {\mathrm{\mid }}^2{)}^2$; direct differentiation gives ${\mathrm{\partial }}_{\zeta }{\mathrm{\partial }}_{\bar{\zeta }}\log {\lambda }_r={\lambda }_r\mathrm{/}2$. The smooth function $e$ is bounded on the closed radius-$r$ disk, so $e\mathrm{/}{\lambda }_r$ extends by zero on its boundary and attains its maximum. If positive, this maximum occurs where $e>0$, and the second derivative test for its logarithm gives $0\ge \mathrm{\partial }\bar{\mathrm{\partial }}\log (e\mathrm{/}{\lambda }_r)\ge e\mathrm{/}{M}^2-{\lambda }_r\mathrm{/}2$. Thus its maximum is at most $M^2\mathrm{/}2$; the same bound is immediate if its maximum is zero. Let $r\uparrow 1$, obtaining $e(\zeta )\le (M^2\mathrm{/}2)4\mathrm{/}(1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2{)}^2$ throughout the unit disk. No assertion was made about $\log e$ at a zero of $A$.

In the upper-half-plane coordinate this is $e(z)\le M^2\mathrm{/}(2y^2)$. By [L1] and step 2.1, the squared operator norm of $d\mathrm{\Phi }$ relative to $\mathrm{\mid }dz{\mathrm{\mid }}^2\mathrm{/}{y}^2$ is $e(z)y^2$; hence it is at most $M^2\mathrm{/}2$. The constant $C=M\mathrm{/}\sqrt{2}$ therefore works in this normalization, and the appropriate fixed scalar multiple works for any fixed scalar Hodge-metric normalization. The constant-map cases of step 3.1 also satisfy the conclusion. This proves the original bound everywhere for every admitted polarized variation, including ramification points.