Quasi-unipotent monodromy for geometric polarized VHS

Trivialize the pulled-back local system on $\mathfrak{H}$ and write $E^p=F^p\cap {\bar{F}}^{k-p}$. The two filtrations and transversality give ${\mathrm{\nabla }}^{1,0}{E}^p\subset E^p\oplus E^{p-1}$ and ${\mathrm{\nabla }}^{0,1}{E}^p\subset E^p\oplus E^{p+1}$. Let $D$ be the diagonal part of $\mathrm{\nabla }$ and write its lowering part as $Adz$. The diagonal part preserves $h$: differentiate $Q(Cu,\bar{v})$ for two sections of one Hodge summand, use $\mathrm{\nabla }Q=0$, and note that off-diagonal terms are orthogonal. For adjacent summands the derivative of their zero $Q(u,\bar{v})$ pairing, together with the opposite Weil scalars in [F2], makes the raising part exactly $A^{†}d\bar{z}$. Thus $\mathrm{\nabla }=D+Adz+A^{†}d\bar{z}\mathrm{.}$ Differentiating the Hodge filtration in a flat marking identifies $A$ with $d\mathrm{\Phi }({\mathrm{\partial }}_z)$, so [F3] gives $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\le C\mathrm{/}y$. This construction uses the actual polarized variation, not a transitivity or properness assertion about a real Lie group.

Let $v$ be a nonzero flat section along a differentiable curve $z(s)$. Since $D$ preserves $h$ and $\mathrm{\nabla }v=0$, $\frac{d}{ds}\mathrm{\parallel }v{\mathrm{\parallel }}_{h_{z(s)}}^2=-2\mathrm{Re}{h}_{z(s)}((A{z}^{\mathrm{\prime }}(s)+A^{†}{\bar{z}}^{\mathrm{\prime }}(s))v,v)\mathrm{.}$ Finite-dimensional Cauchy–Schwarz and $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}$ give $\mid \frac{d}{ds}\log \mathrm{\parallel }v{\mathrm{\parallel }}_{h_{z(s)}}\mid \le 2\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{H}\mathrm{S}}\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }\le 2C\frac{\mathrm{\mid }{z}^{\mathrm{\prime }}(s)\mathrm{\mid }}{\mathrm{Im}z(s)}\mathrm{.}$ For the horizontal segment from $z$ to $z+1$, integration yields, for every vector in the fixed flat vector space, $e^{-2C\mathrm{/}y}\mathrm{\parallel }v{\mathrm{\parallel }}_{h_z}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{h_{z+1}}\le e^{2C\mathrm{/}y}\mathrm{\parallel }v{\mathrm{\parallel }}_{h_z}\mathrm{.}$ The zero vector satisfies the same inequalities directly.

Let $S$ denote the period-map deck operator, so $\mathrm{\Phi }(z+1)=S\mathrm{\Phi }(z)$. The action on the reference space in a flat marking is the inverse of parallel transport; equivalently some monodromy-action conventions name that inverse itself $T$. Thus $S$ is represented by $T^{-1}$ or $T$ under these conventions. The conclusion about moduli of eigenvalues is the same for both. Since $S$ is real and preserves $Q$, its transport of the Weil operator gives $C_{z+1}=S{C}_z{S}^{-1}$, hence $\mathrm{\parallel }Sv{\mathrm{\parallel }}_{h_{z+1}}=\mathrm{\parallel }v{\mathrm{\parallel }}_{h_z}$. Apply step 2.1 to $Sv$ and use this identity to obtain $\mathrm{\parallel }Sv{\mathrm{\parallel }}_{h_z}\le e^{2C\mathrm{/}y}\mathrm{\parallel }v{\mathrm{\parallel }}_{h_z}$. Applying step 2.1 to $v$ and using $\mathrm{\parallel }v{\mathrm{\parallel }}_{h_{z+1}}=\mathrm{\parallel }{S}^{-1}v{\mathrm{\parallel }}_{h_z}$ gives the identical bound for $S^{-1}$. For a complex eigenvector of the fixed complexification of $S$, positivity of $h_z$ therefore gives $e^{-2C\mathrm{/}y}\le \mathrm{\mid }\lambda \mathrm{\mid }\le e^{2C\mathrm{/}y}\mathrm{.}$ Let $y\to +\mathrm{\infty }$. Every eigenvalue of $S$, and therefore every eigenvalue of $T$, has modulus one. No subsequence, compact-stabilizer lift or homogeneous-space metric comparison is needed.

Put $n=\mathrm{rank}{\mathbb{V}}_{\mathbb{Z}}$. If $n=0$ the assertion holds with exponent one. Otherwise choose an integral basis, and for each positive integer $j$ form the characteristic polynomial $P_j(X)=\det (XI-T^j)$. It is monic of degree $n$ with integer coefficients. Its roots, with multiplicity, are ${\lambda }_1^j,\dots ,{\lambda }_n^j$, where the ${\lambda }_i$ are the eigenvalues of $T$; triangularizing over $\mathbb{C}$ proves this even when $T$ is not diagonalizable. By step 3.1, the coefficient of $X^{n-r}$ has absolute value at most $\left(\genfrac{}{}{0px}{}{n}{r}\right)$. There are only finitely many possible integer coefficient lists, so $P_a=P_b$ for some integers $1\le a<b$. Equality of their finite multisets of roots supplies a permutation $\sigma$ with ${\lambda }_i^a={\lambda }_{\sigma (i)}^b$. Inductively ${\lambda }_i^{a^q}={\lambda }_{{\sigma }^q(i)}^{b^q}$; if $q$ is a positive cycle length of $i$, this becomes ${\lambda }_i^{a^q}={\lambda }_i^{b^q}$. Since $T$ is invertible, ${\lambda }_i\mathrm{\ne }0$, so ${\lambda }_i$ is a root of unity. This finite integer-matrix argument requires no separate theorem about algebraic conjugates.

Choose a positive integer $m$ divisible by the finitely many orders just obtained. Every eigenvalue of $T^m$ is one, so its characteristic polynomial is $(X-1{)}^n$. Cayley–Hamilton gives $(T^m-I{)}^n=0$, proving that $T^m$ is unipotent. Finally a positively oriented loop under the cover $t=s^m$ winds $m$ times around the original puncture; composition of parallel transport therefore gives precisely $T^m$ as the new monodromy. This proves the entire original assertion without a weight-dependent nilpotence claim.