Proof. Clearly $\unicode[STIX]{x1D70C}$ cannot be virtually tensor decomposable because then it should have rank ${\geqslant}4$ . Suppose $\unicode[STIX]{x1D70C}$ is virtually reducible, becoming reducible upon pullback to $p:Z\rightarrow X$ . The image of $\unicode[STIX]{x1D70B}_{1}(Z,z)$ is of finite index in $\unicode[STIX]{x1D70B}_{1}(X,x)$ and the restriction of $\unicode[STIX]{x1D70C}$ to this subgroup is reducible. Therefore we may replace $Z$ by the finite étale covering corresponding to this subgroup, in other words, we may suppose that $p$ is finite étale. We may furthermore suppose it is Galois with group  $G$ . Write the decomposition into isotypical components

$$\begin{eqnarray}V_{\unicode[STIX]{x1D70C}}|_{Z}=\bigoplus V_{i}~\otimes ~W_{i},\end{eqnarray}$$

where $V_{i}$ are irreducible local systems on $Z$ and $W_{i}$ are vector spaces. The Galois group acts on $\{V_{i}\}$ . This action is transitive, otherwise $\unicode[STIX]{x1D70C}$ would be reducible. In particular, all of the $V_{i}$ and all of the $W_{i}$ have the same rank.

Suppose there is more than one isotypical component. Then $V_{i}$ and $W_{i}$ must be of rank $1$ and $G$ acts on the set of isotypical components by permutation. Since the subgroup of $\operatorname{SL}(n,\mathbb{C})$ fixing the decomposition of $V_{\unicode[STIX]{x1D70C}}|_{Z}$ into isotypical components is a maximal torus, the image of $\unicode[STIX]{x1D70C}$ is contained in the normalizer of a maximal torus. Furthermore, assume that $\unicode[STIX]{x1D70C}$ underlies a VHS. The Galois action preserves the Hodge type, and there is a single Hodge type for the rank  $1$ isotypical component (note that the isotypical decomposition is compatible with the Hodge structure). Therefore $\unicode[STIX]{x1D70C}$ has only a single Hodge type and it is unitary. This proves the lemma in the case of several isotypical components.

We may now assume $V_{\unicode[STIX]{x1D70C}}|_{Z}=V_{1}~\otimes ~W_{1}$ . Since the rank is ${\leqslant}3$ , and by hypothesis $V_{\unicode[STIX]{x1D70C}}|_{Z}$ is reducible so the rank of $V_{1}$ is strictly smaller than $n$ , we get that $V_{1}$ has rank $1$ . It means that the restriction and projection to a representation $\unicode[STIX]{x1D70B}_{1}(Z,z)\rightarrow \operatorname{PGL}(n,\mathbb{C})$ is trivial. Since by hypothesis our representation is into $\operatorname{SL}(n,\mathbb{C})$ we get that $\unicode[STIX]{x1D70C}$ has finite image. In particular it is unitary, and a unitary irreducible VHS can have only a single Hodge type. This completes the proof of the lemma.◻

Proof. By the above lemma, $\unicode[STIX]{x1D70C}$ is not tensor decomposable and not virtually reducible.◻

Proof. We claim that the map

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}:L_{1}\oplus L_{2}\rightarrow \unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)\end{eqnarray}$$

induced from $\unicode[STIX]{x1D703}$ has rank $1$ . Indeed, in the local coordinates $\unicode[STIX]{x1D6FC}$ is given by two logarithmic $1$ -forms $\unicode[STIX]{x1D714}_{1}$ and $\unicode[STIX]{x1D714}_{2}$ and the integrability of $\unicode[STIX]{x1D703}$ implies that $\unicode[STIX]{x1D714}_{1}\wedge \unicode[STIX]{x1D714}_{2}=0$ , so they are proportional and $\unicode[STIX]{x1D6FC}$ has rank $1$ . Let $M$ be the saturation of the image of $\unicode[STIX]{x1D6FC}$ in $\unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)$ (i.e., the largest rank $1$ subsheaf of $\unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)$ containing the image of $\unicode[STIX]{x1D6FC}$ ). Then $M$ is a line bundle (as it is a rank $1$ reflexive ${\mathcal{O}}_{X}$ -module on a smooth variety) such that $\unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)/M$ is torsion-free.

By construction we have inclusions $L_{1}\subset M$ and $L_{2}\subset M$ defining effective divisors $B_{1}$ and $B_{2}$ , respectively. Let us set $e^{p,q}=c_{1}(E^{p,q})$ , $l_{i}=c_{1}(L_{i})$ and $b_{i}=c_{1}(B_{i})$ . Then we can write

$$\begin{eqnarray}m=c_{1}(M)=l_{1}+b_{1}=l_{2}+b_{2}.\end{eqnarray}$$

By assumption we have

$$\begin{eqnarray}e^{0,2}+e^{1,1}+e^{2,0}=0\end{eqnarray}$$

in the rational cohomology $H^{2}(\bar{X},\mathbb{Q})$ , so the classes $e^{p,q}$ can be written in terms of $l_{1}$ and $l_{2}$ in the following way:

$$\begin{eqnarray}e^{2,0}={\textstyle \frac{1}{3}}(2l_{1}+l_{2}),\quad e^{1,1}={\textstyle \frac{1}{3}}(l_{2}-l_{1}),\quad e^{0,2}={\textstyle \frac{1}{3}}(-l_{1}-2l_{2}).\end{eqnarray}$$

Therefore

$$\begin{eqnarray}3m=l_{1}+b_{1}+2(l_{2}+b_{2})\geqslant l_{1}+2l_{2}=-3e^{0,2}.\end{eqnarray}$$

In particular, we have $m\geqslant -e^{0,2}$ . Similarly, we have

$$\begin{eqnarray}3m=2(l_{1}+b_{1})+(l_{2}+b_{2})\geqslant 2l_{1}+l_{2}=3e^{2,0},\end{eqnarray}$$

so $m\geqslant e^{2,0}$ .

The following lemma is a corollary of the well-known Bogomolov’s lemma.

Proof. We have a short exact sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{O}}_{\bar{Y}}(-A)^{n-2}\rightarrow \unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)|_{\bar{Y}}\rightarrow \unicode[STIX]{x1D6FA}_{\bar{Y}}(\log D_{Y})\rightarrow 0.\end{eqnarray}$$

Let us consider the composition $N|_{\bar{Y}}\rightarrow \unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)|_{\bar{Y}}\rightarrow \unicode[STIX]{x1D6FA}_{\bar{Y}}(\log D_{Y})$ . By Bogomolov’s lemma (see [Reference BogomolovBog78, Theorem 4] and [Reference Esnault and ViehwegEV92, Corollary 6.9]) $\unicode[STIX]{x1D6FA}_{\bar{Y}}(\log D_{Y})$ does not contain big line bundles. So if this composition is non-zero then $N|_{\bar{Y}}$ is not big.

If the map $N|_{\bar{Y}}\rightarrow \unicode[STIX]{x1D6FA}_{\bar{Y}}(\log D_{Y})$ is zero then by the above exact sequence $N|_{\bar{Y}}$ is contained in ${\mathcal{O}}_{\bar{Y}}(-A)^{n-2}$ , so $\unicode[STIX]{x1D705}(N|_{\bar{Y}})=-\infty$ .◻

Slope $A$ -stability of $(E,\unicode[STIX]{x1D703})$ implies that

$$\begin{eqnarray}e^{0,2}.\,A<0\end{eqnarray}$$

and

$$\begin{eqnarray}-e^{2,0}.\,A=(e^{0,2}+e^{1,1}).\,A<0.\end{eqnarray}$$

If $(e^{0,2})^{2}>0$ then, thanks to Lemma 2.3, the first inequality implies that $-e^{0,2}|_{\bar{Y}}$ is big. But then $M|_{\bar{Y}}$ is big, which contradicts the above lemma. Therefore

$$\begin{eqnarray}(e^{0,2})^{2}\leqslant 0.\end{eqnarray}$$

Similarly, if $(e^{2,0})^{2}>0$ then $e^{2,0}|_{\bar{Y}}$ is big. But then $M|_{\bar{Y}}$ is big, a contradiction. Therefore

$$\begin{eqnarray}(e^{2,0})^{2}\leqslant 0.\end{eqnarray}$$

Let us also recall that

$$\begin{eqnarray}0=c_{2}(E)=e^{2,0}.\,e^{1,1}+e^{1,1}.\,e^{0,2}+e^{0,2}.\,e^{2,0}=-(e^{2,0})^{2}+e^{1,1}.\,e^{0,2}.\end{eqnarray}$$

Therefore

$$\begin{eqnarray}L_{2}^{2}=(e^{1,1}+e^{0,2})^{2}-4e^{1,1}.\,e^{0,2}=(e^{2,0})^{2}-4(e^{2,0})^{2}=-3(e^{2,0})^{2}\geqslant 0.\end{eqnarray}$$

Similarly, we have

$$\begin{eqnarray}L_{1}^{2}=(e^{2,0}-e^{1,1})^{2}=-3(e^{0,2})^{2}\geqslant 0.\end{eqnarray}$$

From stability of $E$ we get $(L_{1}+2L_{2}).A=-3e^{0,2}.A>0$ and hence there exists $i$ with $L_{i}.A>0$ . Lemmas 2.3 and 4.4 imply that $L_{i}|_{\bar{Y}}$ is nef with $L_{i}^{2}=0$ and the degree of $L_{i}|_{\bar{Y}}$ is strictly positive.

Proof. Let us first assume that $i=1$ , i.e., $L_{1}|_{\bar{Y}}$ is nef with $L_{1}^{2}=0$ . Since $B_{2}$ is effective, we have $L_{1}.B_{2}\geqslant 0$ . By the Lefschetz hyperplane theorem for quasi-projective varieties (see [Reference Hamm and LêHL85, Theorem 1.1.3]) we have $\unicode[STIX]{x1D70B}_{1}(Y)\simeq \unicode[STIX]{x1D70B}_{1}(X)$ . So by functoriality of the correspondence between representations and Higgs bundles, the restriction

$$\begin{eqnarray}(E|_{\bar{Y}},\unicode[STIX]{x1D703}_{Y}:E|_{\bar{Y}}\rightarrow ^{\unicode[STIX]{x1D703}|_{Y}}E|_{\bar{Y}}~\otimes ~\unicode[STIX]{x1D6FA}_{\bar{X}}(\log D)|_{\bar{Y}}\rightarrow E|_{\bar{Y}}~\otimes ~\unicode[STIX]{x1D6FA}_{\bar{Y}}(\log D_{Y}))\end{eqnarray}$$

of $(E,\unicode[STIX]{x1D703})$ to $\bar{Y}$ is $A$ -stable (one can also give a direct proof of this fact: see [Reference LangerLan15, Theorem 12] and [Reference LangerLan16]). Since the rational Chern classes of $E|_{\bar{Y}}$ vanish, $(E|_{\bar{Y}},\unicode[STIX]{x1D703}_{Y})$ is stable with respect to every stable polarization and semi-stable with respect to every nef polarization. In particular, $L_{1}|_{\bar{Y}}$ -semi-stability of $(E|_{\bar{Y}},\unicode[STIX]{x1D703}_{Y})$ implies that $L_{1}.(L_{1}+2L_{2})\geqslant 0$ . Therefore we have $L_{1}.L_{2}\geqslant 0$ .

Applying Lemma 2.4 to $M=L_{1}+B_{1}$ we see that $L_{1}^{2}=0$ , $L_{1}.B_{1}=0$ and $L_{1}.M=0$ . Using $M=L_{2}+B_{2}$ we get equalities $L_{1}.L_{2}=L_{1}.B_{2}=0$ . But then

$$\begin{eqnarray}c_{2}(E)={\textstyle \frac{1}{9}}(L_{1}^{2}+L_{1}.L_{2}+L_{2}^{2})=0\end{eqnarray}$$

implies that $L_{2}^{2}=-L_{1}.L_{2}=0$ .

The proof in the case $i=2$ is analogous.◻

This lemma finishes the proof of assertions (i) and (iv) of the proposition. To prove (ii), let us consider $i^{\prime }$ such that $\{i,i^{\prime }\}=\{1,2\}$ . Let us choose a rational $a$ such that $(L_{i^{\prime }}-aL_{i}).A=0$ . Since $(L_{i^{\prime }}-aL_{i})^{2}=0$ this implies that $l_{i^{\prime }}=al_{i}$ in $H^{2}(X,\mathbb{Q})$ .

Assertion (iii) of the proposition follows from the proof of Lemma 4.5. More precisely, if $i=1$ then this proof shows that $L_{1}.M=L_{1}.B_{1}=L_{1}.B_{2}=0$ . So (ii) implies that $L_{2}.M=L_{2}.B_{1}=L_{2}.B_{2}=0$ . The proof in the case $i=2$ is analogous.◻

Proof. We can assume that $X$ has dimension at least $2$ as otherwise there is nothing to be proven. Therefore the complex variation of Hodge structure corresponding to $\unicode[STIX]{x1D70C}$ is of the form described in Proposition 4.3. Let us write $L_{2}\equiv aL_{1}$ . Since the assertion for a representation is equivalent to the assertion for its dual, we can assume that for a general complete intersection surface $S\subset X$ the restriction $L_{1}|_{S}$ is nef with strictly positive degree and $a\leqslant 1$ .

In the notation of Proposition 4.3 we have $M=L_{1}(B_{1})=L_{2}(B_{2})$ . Let us write $B_{1}=B+B_{1}^{\prime }$ and $B_{2}=B+B_{2}^{\prime }$ , where $B$ , $B_{1}^{\prime }$ , $B_{2}^{\prime }$ are effective and $B_{1}^{\prime }$ and $B_{2}^{\prime }$ have no common irreducible components. Recall that $L_{1}.B=L_{1}.B_{1}^{\prime }=0$ and $L_{2}.B=L_{2}.B_{2}^{\prime }=0$ . We set $M^{\prime }:=M(-B)$ . This line bundle has a canonical inclusion into $\unicode[STIX]{x1D6FA}_{X}$ given by composing $M(-B)\subset M$ with $M\rightarrow \unicode[STIX]{x1D6FA}_{X}$ . By definition we also have $M^{\prime }=L_{1}(B_{1}^{\prime })=L_{2}(B_{2}^{\prime })$ . Intersecting both sides of equality $L_{1}+B_{1}^{\prime }=L_{2}+B_{2}^{\prime }$ with $B_{1}^{\prime }$ we get

$$\begin{eqnarray}(B_{1}^{\prime })^{2}=B_{1}^{\prime }.B_{2}^{\prime }\geqslant 0\end{eqnarray}$$

(here we use that $L_{2}.B_{1}^{\prime }=aL_{1}.B_{1}^{\prime }=0$ , and since $B_{1}^{\prime }$ and $B_{2}^{\prime }$ have no common components their intersection is positive). But Lemma 2.4 implies that $(B_{1}^{\prime })^{2}\leqslant 0$ , so we have $(B_{1}^{\prime })^{2}=B_{1}^{\prime }.B_{2}^{\prime }=0$ . Hence, we also get $(B_{2}^{\prime })^{2}=0$ . There exists also some non-negative $\unicode[STIX]{x1D6FC}_{1}$ such that $(B_{1}^{\prime }-\unicode[STIX]{x1D6FC}_{1}L_{1}).A=0$ . Since $(B_{1}^{\prime }-\unicode[STIX]{x1D6FC}_{1}L_{1})^{2}=0$ , we have $B_{1}^{\prime }\equiv \unicode[STIX]{x1D6FC}_{1}L_{1}$ . Similarly, there exists some non-negative $\unicode[STIX]{x1D6FC}_{2}$ such that $B_{2}^{\prime }\equiv \unicode[STIX]{x1D6FC}_{2}L_{1}$ . Note also that $B_{i}^{\prime }=0$ is equivalent to $\unicode[STIX]{x1D6FC}_{i}=0$ , since if $B_{i}^{\prime }$ is a non-zero effective divisor then $B_{i}^{\prime }.A>0$ . Let us recall that $L_{1}+B_{1}^{\prime }=L_{2}+B_{2}^{\prime }$ , so $(1+\unicode[STIX]{x1D6FC}_{1})L_{1}\equiv (a+\unicode[STIX]{x1D6FC}_{2})L_{1}$ . But $L_{1}.A>0$ , so $1+\unicode[STIX]{x1D6FC}_{1}=a+\unicode[STIX]{x1D6FC}_{2}$ . Since $a\leqslant 1$ , this implies that $0\leqslant \unicode[STIX]{x1D6FC}_{1}\leqslant \unicode[STIX]{x1D6FC}_{2}$ .

We consider two cases depending on whether $B_{2}^{\prime }$ is zero or not.

If $B_{2}^{\prime }=0$ then $\unicode[STIX]{x1D6FC}_{2}=\unicode[STIX]{x1D6FC}_{1}=0$ , $B_{1}^{\prime }=0$ and $L:=L_{1}=L_{2}$ with the same map to $M$ and  $\unicode[STIX]{x1D6FA}_{X}$ . In this case we set $N:=E^{2,0}~\otimes ~L_{1}^{-1}$ . Since $\det E=N^{\otimes 3}$ , the line bundle $N$ is $3$ -torsion. Therefore $(E,\unicode[STIX]{x1D703})\simeq (N,0)~\otimes ~\operatorname{Sym}^{2}(F)~\otimes ~\det (F)^{-1}$ , where $F=(F^{1,0}~\oplus ~F^{0,1},\unicode[STIX]{x1D703}_{F})$ is a system of Hodge bundles with $F^{1,0}=L$ , $F^{0,1}={\mathcal{O}}_{X}$ and the Higgs field $\unicode[STIX]{x1D703}_{F}$ given by the inclusion $L\rightarrow \unicode[STIX]{x1D6FA}_{X}$ . Note that $F$ is an $A$ -stable system of Hodge bundles but it does not have vanishing Chern classes so it does not underlie a rank $2$ representation of $\unicode[STIX]{x1D70B}_{1}(X,x)$ . But it comes from a projective representation $\unicode[STIX]{x1D70B}_{1}(X,x)\rightarrow \operatorname{PGL}(2,\mathbb{C})$ . There exists a finite covering $\unicode[STIX]{x1D70B}:Y\rightarrow X$ such that $\unicode[STIX]{x1D70B}^{\ast }L=M^{\otimes 2}$ for some line bundle $M$ on $Y$ . Then $\unicode[STIX]{x1D70B}^{\ast }(E,\unicode[STIX]{x1D703})\simeq \unicode[STIX]{x1D70B}^{\ast }(N,0)~\otimes ~\operatorname{Sym}^{2}(F_{Y})$ , where $F_{Y}=(F_{Y}^{1,0}\oplus F_{Y}^{1,0},\unicode[STIX]{x1D703}_{Y})$ is a system of Hodge bundles with $F_{Y}^{1,0}=M$ , $F_{Y}^{0,1}=M^{-1}$ and the Higgs field $\unicode[STIX]{x1D703}_{Y}$ given by the inclusion $\unicode[STIX]{x1D70B}^{\ast }L\rightarrow \unicode[STIX]{x1D70B}^{\ast }\unicode[STIX]{x1D6FA}_{X}\rightarrow \unicode[STIX]{x1D6FA}_{Y}$ .

This corresponds to case (i) of the theorem.

If $B_{2}^{\prime }\neq 0$ , then by the Hodge index theorem there exist positive integers $b$ and $c$ such that $bM^{\prime }\equiv cB_{2}^{\prime }$ . The idea is now to say that $M^{\prime }$ looks like an effective divisor. Roughly speaking, we try to get a section of $M^{\prime }$ and then argue using this differential form. In practice we need to extract a root, and we only have numerical information so, in the case where the Picard scheme has positive dimension, we need to deal with several cases.

Start by noting that $U:=bM^{\prime }-cB_{2}^{\prime }\in \operatorname{Pic}^{\unicode[STIX]{x1D70F}}(X)$ . Multiplying $b$ and $c$ by the same positive integer we can assume that in fact $U\in \operatorname{Pic}^{0}(X)$ . Therefore we have a non-zero map

$$\begin{eqnarray}U\rightarrow (M^{\prime })^{\otimes b}\end{eqnarray}$$

determined by the divisor $cB_{2}^{\prime }$ . We can write $U=V^{\otimes b}$ for some $V\in \operatorname{Pic}^{0}(X)$ and hence we get a section $\unicode[STIX]{x1D702}\in H^{0}(X,(M^{\prime }~\otimes ~V^{\ast })^{\otimes b})$ (whose divisor is $cB_{2}^{\prime }$ ).

Let $p:Z\rightarrow X$ be a desingularization of the ramified covering defined by taking the $b$ th root of $\unicode[STIX]{x1D702}$ . Over $Z$ we have a tautological section $\unicode[STIX]{x1D6FC}\in H^{0}(Z,p^{\ast }(M^{\prime }~\otimes ~V^{\ast }))$ such that $\unicode[STIX]{x1D6FC}^{\otimes b}=p^{\ast }\unicode[STIX]{x1D702}$ .

In particular, $\unicode[STIX]{x1D6FC}\in H^{0}(Z,\unicode[STIX]{x1D6FA}_{Z}^{1}~\otimes ~p^{\ast }V^{\ast })$ , which means that $p^{\ast }V^{\ast }\in \operatorname{Pic}^{0}(Z)$ is in the jump-locus for twisted sections of $\unicode[STIX]{x1D6FA}_{Z}^{1}$ :

$$\begin{eqnarray}S(Z):=\{N\in \operatorname{Pic}^{0}(Z)\mid \dim H^{0}(Z,\unicode[STIX]{x1D6FA}_{Z}^{1}~\otimes ~N)\geqslant 1\}.\end{eqnarray}$$

The irreducible components of this jump-locus are translates of abelian subvarieties of $\text{Pic}^{0}(Z)$ by torsion points. The fact that the components are translates of abelian varieties was proved in [Reference Green and LazarsfeldGL91] and the assertion about torsion was proved in [Reference SimpsonSim93b]. We distinguish two cases depending on whether $p^{\ast }V^{\ast }$ is a torsion point or not.

Case 1: let us assume that $p^{\ast }V^{\ast }$ is a torsion point of $\text{Pic}^{0}(Z)$ . Then the line bundle $p^{\ast }V^{\ast }$ defines a finite étale covering $q:Z^{\prime }\rightarrow Z$ such that $V^{\ast }$ becomes trivial after pulling back to $Z^{\prime }$ . Let $p^{\prime }$ denote the composition of $q:Z^{\prime }\rightarrow Z$ with $p:Z\rightarrow X$ . Then over $Z^{\prime }$ we get a non-zero section $\unicode[STIX]{x1D709}\in H^{0}(Z^{\prime },\unicode[STIX]{x1D6FA}_{Z^{\prime }}^{1})$ given by the composition of $q^{\ast }\unicode[STIX]{x1D6FC}:{\mathcal{O}}_{Z^{\prime }}\rightarrow q^{\ast }(p^{\ast }(M^{\prime }~\otimes ~V^{\ast }))\simeq q^{\ast }(p^{\ast }(M^{\prime }))=(p^{\prime })^{\ast }M^{\prime }$ with the canonical map $(p^{\prime })^{\ast }M^{\prime }\rightarrow (p^{\prime })^{\ast }\unicode[STIX]{x1D6FA}_{X}\rightarrow \unicode[STIX]{x1D6FA}_{Z^{\prime }}^{1}$ .

The line bundle $(p^{\prime })^{\ast }M^{\prime }\subset \unicode[STIX]{x1D6FA}_{Z^{\prime }}^{1}$ is generically generated by this section $\unicode[STIX]{x1D709}$ .

We can look at the Albanese map defined by $\unicode[STIX]{x1D709}$ (see [Reference SimpsonSim93a, p. 101]). This is the map $\unicode[STIX]{x1D713}:Z^{\prime }\rightarrow A=\operatorname{Alb}(Z^{\prime })/B$ , where $B$ is the sum of all abelian subvarieties of $\operatorname{Alb}(Z^{\prime })$ on which $\unicode[STIX]{x1D709}$ vanishes. By construction $\unicode[STIX]{x1D709}=\unicode[STIX]{x1D713}^{\ast }(\unicode[STIX]{x1D709}_{A})$ for some $1$ -form $\unicode[STIX]{x1D709}_{A}$ on $A$ such that the restriction of $\unicode[STIX]{x1D709}_{A}$ to any non-trivial abelian subvariety of $A$ is non-zero.

We distinguish two subcases depending on the dimension of $\unicode[STIX]{x1D713}(Z^{\prime })$ .

Subcase 1.1: the image of $\unicode[STIX]{x1D713}$ is a curve. In this case taking the Stein factorization of $Z^{\prime }\rightarrow \unicode[STIX]{x1D713}(Z^{\prime })$ , we get a fibration $f:Z^{\prime }\rightarrow C$ over a smooth projective curve and a $1$ -form $\unicode[STIX]{x1D709}_{C}$ on $C$ such that $\unicode[STIX]{x1D709}=f^{\ast }\unicode[STIX]{x1D709}_{C}$ . Considering $\unicode[STIX]{x1D709}$ as a map ${\mathcal{O}}_{Z^{\prime }}\rightarrow \unicode[STIX]{x1D6FA}_{Z^{\prime }}$ , this means that the map factors through $f^{\ast }\unicode[STIX]{x1D709}_{C}:{\mathcal{O}}_{Z^{\prime }}=f^{\ast }{\mathcal{O}}_{C}\rightarrow f^{\ast }\unicode[STIX]{x1D6FA}_{C}$ . Therefore $(p^{\prime })^{\ast }M$ is contained in the saturation of $f^{\ast }\unicode[STIX]{x1D6FA}_{C}$ in $\unicode[STIX]{x1D6FA}_{Z^{\prime }}$ . If $F$ is a smooth fiber of $f$ (or just a multiplicity $1$ irreducible component of a fiber of $f$ ) then this shows that the canonical map $(p^{\prime })^{\ast }M|_{F}\rightarrow \unicode[STIX]{x1D6FA}_{Z^{\prime }/C}|_{F}=\unicode[STIX]{x1D6FA}_{F}$ is zero. This implies that the Kodaira–Spencer maps of our variation of Hodge structures restricted to the fiber $F$ , vanish. Each Hodge subbundle is therefore a flat subbundle. Hence, the restriction of our variation of Hodge structures to $F$ splits into a direct sum of three rank $1$ variations of Hodge structure at the three different Hodge types. Clearly, $\unicode[STIX]{x1D70C}$ is not virtually tensor decomposable. It is also not virtually reducible as the Kodaira–Spencer maps remain non-zero under alterations. So by Theorem 3.3 ( $4\Rightarrow 1$ ), $\unicode[STIX]{x1D70C}$ projectively factors through an orbicurve.

Subcase 1.2: the dimension of the image $\unicode[STIX]{x1D713}(Z^{\prime })$ is at least $2$ . Let $\tilde{A}\rightarrow A$ be the universal covering of $A$ (treated as a complex vector space). This map is only analytic and not algebraic. Let $\tilde{Z}:=Z^{\prime }\times _{A}\tilde{A}$ be the covering of $Z^{\prime }$ defined by the Albanese map $\unicode[STIX]{x1D713}$ and let $\unicode[STIX]{x1D70B}:\tilde{Z}\rightarrow Z^{\prime }$ be the projection on the first factor and $\tilde{\unicode[STIX]{x1D713}}:\tilde{Z}\rightarrow \tilde{A}$ on the second factor (these maps are also only analytic). Let us fix a point $\tilde{z}_{0}\in \tilde{Z}$ . There exists a unique linear function $g_{\tilde{A}}:\tilde{A}\rightarrow \mathbb{C}$ such that $g_{\tilde{A}}(\tilde{\unicode[STIX]{x1D713}}(\tilde{z}_{0}))=0$ and $dg_{\tilde{A}}=\unicode[STIX]{x1D709}_{A}$ . Let $g:\tilde{Z}\rightarrow \mathbb{C}$ be the composition of $g_{\tilde{A}}$ with $\tilde{\unicode[STIX]{x1D713}}$ . Then the Lefschetz theorem (see [Reference SimpsonSim93a, Theorem 1]; since $\dim (\unicode[STIX]{x1D713}(Z^{\prime }))\geqslant 2$ the assumptions of this theorem are satisfied) says that for any fiber $F$ of $g$ (or, in other words, a possibly non-compact leaf of the foliation defined by $\unicode[STIX]{x1D709}$ ), the map $\unicode[STIX]{x1D70B}_{1}(F,\tilde{z}_{0})\rightarrow \unicode[STIX]{x1D70B}_{1}(\tilde{Z},\tilde{z}_{0})$ is surjective.

Note that by our construction the restriction $\unicode[STIX]{x1D70B}^{\ast }\unicode[STIX]{x1D709}|_{F}$ is zero. Choosing $F$ so that it does not lie in the preimage of $B_{2}^{\prime }$ , the inclusion $M^{\prime }\rightarrow \unicode[STIX]{x1D6FA}_{X}$ becomes zero after pulling back to $F$ and composing with the canonical map to $\unicode[STIX]{x1D6FA}_{F}$ . So our variation of Hodge structure splits on $F$ into a direct sum of three line bundles. In this case the image of the representation $\unicode[STIX]{x1D70B}_{1}(F,\tilde{z}_{0})\rightarrow \operatorname{SL}(3,\mathbb{C})$ is contained in a maximal torus. Therefore the monodromy representation is abelian on $\unicode[STIX]{x1D70B}_{1}(\tilde{Z},\tilde{z}_{0})$ . But $\unicode[STIX]{x1D70B}_{1}(Z^{\prime },\tilde{\unicode[STIX]{x1D713}}(\tilde{z}_{0}))/\unicode[STIX]{x1D70B}_{1}(\tilde{Z},\tilde{z}_{0})$ is an abelian group, so the monodromy of our original VHS would be solvable. Then the Zariski closure $G\subset \operatorname{SL}(3,\mathbb{C})$ of $\text{im}\,\unicode[STIX]{x1D70C}$ is also solvable (this follows from [Reference BorelBor91, I.2.4 Proposition]). On the other hand, since $\unicode[STIX]{x1D70C}$ is irreducible, the connected component $G^{0}$ is reductive. Hence, $G^{0}$ is contained in a maximal torus of $\operatorname{SL}(3,\mathbb{C})$ . Taking the étale covering $q:\tilde{X}\rightarrow X$ corresponding to the image of $\unicode[STIX]{x1D70B}_{1}(X,x)\rightarrow G/G^{0}$ , we see that the monodromy of $q^{\ast }V_{\unicode[STIX]{x1D70C}}$ is contained in the maximal torus of $\operatorname{SL}(3,\mathbb{C})$ . But then the corresponding pullback of the system of Hodge bundles is a direct sum of line bundles with trivial maps between the components, contradicting the fact that the original VHS was of type $(1,1,1)$ .

Case 2: assume that $p^{\ast }V^{\ast }$ is not a torsion point. Then an irreducible component of $S(Z)$ containing this point has dimension ${\geqslant}1$ .

Let us recall that for a line bundle $N\in \operatorname{Pic}^{0}(Z)$ we have an isomorphism of complex vector spaces

$$\begin{eqnarray}H^{0}(Z,\unicode[STIX]{x1D6FA}_{Z}^{1}~\otimes ~N)\simeq \overline{H^{1}(Z,N^{-1})},\end{eqnarray}$$

so

$$\begin{eqnarray}S(Z)=\{N\in \operatorname{Pic}^{0}(Z)\mid \dim H^{1}(Z,N^{-1})\geqslant 1\}.\end{eqnarray}$$

Beauville proves in [Reference BeauvilleBea92] that the positive-dimensional components of this set all come from maps to orbicurves. More precisely, [Reference BeauvilleBea92, Corollaire 2.3] says that there exists a fibration $f:Z\rightarrow C$ over a smooth projective curve of genus $g\geqslant 1$ such that $p^{\ast }V^{\ast }$ is trivial on every smooth fiber of $f$ .

The proof of the following lemma is modelled on the proof of [Reference BeauvilleBea92, Proposition 1.10].

Lemma 5.2. Let $Z$ be a smooth complex projective variety and let $C$ be a smooth complex projective curve. Let $f:Z\rightarrow C$ be a fibration with reduced fibers. Let $L\in \operatorname{Pic}^{\unicode[STIX]{x1D70F}}Z$ be a non-torsion line bundle, which is trivial on some smooth fiber of $f$ . If $0\neq \unicode[STIX]{x1D6FC}\in H^{0}(\unicode[STIX]{x1D6FA}_{Z}~\otimes ~L^{-1})$ then there exists a line bundle $L_{C}$ on $C$ and $\unicode[STIX]{x1D6FD}\in H^{0}(\unicode[STIX]{x1D6FA}_{C}~\otimes ~L_{C}^{-1})=\operatorname{Hom}(L_{C},\unicode[STIX]{x1D6FA}_{C})$ such that $L=f^{\ast }L_{C}$ and the composition

$$\begin{eqnarray}L=f^{\ast }L_{C}\longrightarrow ^{f^{\ast }\unicode[STIX]{x1D6FD}}f^{\ast }\unicode[STIX]{x1D6FA}_{C}\longrightarrow \unicode[STIX]{x1D6FA}_{Z}\end{eqnarray}$$

corresponds to $\unicode[STIX]{x1D6FC}$ .

Proof. First let us recall that existence of $L_{C}$ such that $L=f^{\ast }L_{C}$ follows from [Reference BeauvilleBea92, Proposition 1.2]. We have a short exact sequence

$$\begin{eqnarray}0\rightarrow H^{1}(C,L_{C})\rightarrow H^{1}(Z,L)\rightarrow H^{0}(C,R^{1}f_{\ast }L)\rightarrow 0\end{eqnarray}$$

coming from the Leray spectral sequence. Using the isomorphisms $H^{1}(Z,L)\simeq \overline{H^{0}(Z,\unicode[STIX]{x1D6FA}_{Z}^{1}~\otimes ~L^{-1})}$ and $H^{1}(C,L_{C})\simeq \overline{H^{0}(C,\unicode[STIX]{x1D6FA}_{C}^{1}~\otimes ~L_{C}^{-1})}$ , it is sufficient to show that $H^{0}(C,R^{1}f_{\ast }L)=0$ . But $H^{0}(C,R^{1}f_{\ast }L)=H^{0}(C,L_{C}~\otimes ~R^{1}f_{\ast }{\mathcal{O}}_{Z})$ and $R^{1}f_{\ast }{\mathcal{O}}_{Z}$ is dual to $f_{\ast }\unicode[STIX]{x1D714}_{Z/C}$ , so we need to check that $\operatorname{Hom}(f_{\ast }\unicode[STIX]{x1D714}_{Z/C},L_{C})=0$ . Let us recall that by [Reference FujitaFuj78, Theorem] (see also [Reference Catanese and DettweilerCD17, Theorem 17]), $f_{\ast }\unicode[STIX]{x1D714}_{Z/C}$ is a direct sum of an ample vector bundle and a direct sum of stable vector bundles of degree zero. Therefore $\operatorname{Hom}(f_{\ast }\unicode[STIX]{x1D714}_{Z/C},L_{C})\neq 0$ if and only if $L_{C}$ is isomorphic to one of the direct summands of $f_{\ast }\unicode[STIX]{x1D714}_{Z/C}$ . But by Deligne’s result, the rank $1$ summands of $f_{\ast }\unicode[STIX]{x1D714}_{Z/C}$ are torsion (see [Reference Catanese and DettweilerCD17, Corollary 21]) and $L_{C}$ is not torsion, so we get a contradiction.◻

Taking a semi-stable reduction of $f$ we can assume that all the fibers of $f$ are reduced. Then, by the above lemma, there exists a line bundle $L_{C}\in \operatorname{Pic}^{0}(C)$ such that $p^{\ast }V=f^{\ast }L_{C}$ and the section $\unicode[STIX]{x1D6FC}$ comes from a section of $H^{0}(C,\unicode[STIX]{x1D6FA}_{C}^{1}~\otimes ~L_{C}^{\ast })$ . This shows that $p^{\ast }V^{\ast }\rightarrow f^{\ast }\unicode[STIX]{x1D6FA}_{C}$ is an isomorphism at the generic point of $Z$ . Since $p^{\ast }V\rightarrow p^{\ast }M$ is also an isomorphism at the generic point of $Z$ , the sheaf $p^{\ast }M$ is contained in $f^{\ast }\unicode[STIX]{x1D6FA}_{C}\subset \unicode[STIX]{x1D6FA}_{Z}$ (since the fibers of $f$ are reduced, $f^{\ast }\unicode[STIX]{x1D6FA}_{C}$ is saturated in $\unicode[STIX]{x1D6FA}_{Z}$ ).

As before this implies that the pullback of our variation of Hodge structure to a fiber of $f$ splits into a direct sum of three rank $1$ variations of Hodge structure. Therefore Corollary 3.5 implies that $\unicode[STIX]{x1D70C}$ projectively factors through an orbicurve. Alternatively, the same arguments as before show that $((p^{\prime })^{\ast }M^{\prime })^{\otimes b}$ is a line bundle associated to a certain sum of rational multiples of fibers of $f$ (but not necessarily positive). Then by passing to a certain cover $Z^{\prime \prime }\rightarrow Z^{\prime }$ we conclude that the pullback of our original variation of Hodge structure to $Z^{\prime \prime }$ is isomorphic to the pullback of a variation of Hodge structure on a certain curve $C^{\prime }$ . So $\unicode[STIX]{x1D70C}$ virtually projectively factors through an orbicurve and we can again conclude by Corollary 3.5. This completes the proof of Theorem 4.1.◻

Proof. We claim that for all $i=1,\ldots ,n-1$ the map

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{i}:L_{i}\oplus L_{i+1}\rightarrow \unicode[STIX]{x1D6FA}_{X}\end{eqnarray}$$

induced from $\unicode[STIX]{x1D703}$ has rank $1$ . Indeed, in the local coordinates $\unicode[STIX]{x1D6FC}$ is given by two $1$ -forms $\unicode[STIX]{x1D714}_{1}$ and $\unicode[STIX]{x1D714}_{2}$ and the integrability of $\unicode[STIX]{x1D703}$ implies that $\unicode[STIX]{x1D714}_{1}\wedge \unicode[STIX]{x1D714}_{2}=0$ , so they are proportional and $\unicode[STIX]{x1D6FC}_{i}$ has rank $1$ . It follows that the map

$$\begin{eqnarray}\unicode[STIX]{x1D6FC}:\bigoplus _{i=1}^{n}L_{i}\rightarrow \unicode[STIX]{x1D6FA}_{X}\end{eqnarray}$$

also has rank $1$ . Let $M$ be the saturation of the image of $\unicode[STIX]{x1D6FC}$ in $\unicode[STIX]{x1D6FA}_{X}$ (i.e., the largest rank $1$ subsheaf of $\unicode[STIX]{x1D6FA}_{X}$ containing the image of $\unicode[STIX]{x1D6FC}$ ). Then $M$ is a line bundle such that $\unicode[STIX]{x1D6FA}_{X}/M$ is torsion-free.

By construction we have inclusions $L_{i}\subset M$ defining effective divisors $B_{i}$ . Let us set $e^{p,q}=c_{1}(E^{p,q})$ , $l_{i}=c_{1}(L_{i})$ and $b_{i}=c_{1}(B_{i})$ . Then we can write

$$\begin{eqnarray}m=c_{1}(M)=l_{1}+b_{1}=\cdots =l_{n}+b_{n}.\end{eqnarray}$$

By assumption we have

$$\begin{eqnarray}e^{n,0}+e^{n-1,1}+\cdots +e^{0,n}=0\end{eqnarray}$$

in the rational cohomology $H^{2}(X,\mathbb{Q})$ , so the classes $e^{p,q}$ can be written in terms of $l_{i}$ in the following way:

$$\begin{eqnarray}e^{n-i,i}=\frac{1}{n+1}(-l_{1}-2l_{2}-\cdots -il_{i}+(n-i)l_{i+1}+(n-i-1)l_{i+2}+\cdots +l_{n})\end{eqnarray}$$

for $i=0,\ldots ,n$ .

Slope $A$ -stability of $(E,\unicode[STIX]{x1D703})$ implies that

$$\begin{eqnarray}(e^{0,n}+e^{1,n-1}+\cdots +e^{i,n-i}).\,A<0\end{eqnarray}$$

for $i=0,\ldots ,n-1$ . This is equivalent to

$$\begin{eqnarray}(e^{n,0}+e^{n-1,1}+\cdots +e^{i,n-i}).\,A>0\end{eqnarray}$$

for $i=1,\ldots ,n$ . Rewriting these inequalities in terms of $L_{i}$ we get

$$\begin{eqnarray}N_{i}.A>0\end{eqnarray}$$

for $i=1,\ldots ,n$ . Since $L_{i}\leqslant M$ we have

$$\begin{eqnarray}\mathop{\sum }_{i=1}^{n}x_{i}N_{i}\leqslant \biggl(\mathop{\sum }_{i=1}^{n}\frac{i(n-i+1)(n+1)}{2}x_{i}\biggr)M\end{eqnarray}$$

for all non-negative rational numbers $x_{1},\ldots ,x_{n}$ . But by Lemma 4.4, $M|_{Y}$ is not big and hence the $\mathbb{Q}$ -divisor $\sum _{i=1}^{n}x_{i}N_{i}$ restricted to $Y$ is also not big. Therefore we have

$$\begin{eqnarray}\biggl(\mathop{\sum }_{i=1}^{n}x_{i}N_{i}\biggr)^{2}\leqslant 0\end{eqnarray}$$

for all non-negative rational numbers $x_{1},\ldots ,x_{n}$ (so also for all non-negative real numbers $x_{1},\ldots ,x_{n}$ ).◻

Proof. This is discussed in [Reference Corlette and SimpsonCS08, § 10] and [Reference SimpsonSim92, Theorem 5]. The basic idea goes back at least to Deligne’s ‘Travaux de Shimura’ [Reference DeligneDel71].

The discussion in the third and following paragraphs of the proof of Theorem 5 in [Reference SimpsonSim92] does not use the rigidity hypothesis of that theorem, but only the statement that for every embedding $\unicode[STIX]{x1D70E}:K\rightarrow \mathbb{C}$ the induced local system $V_{\unicode[STIX]{x1D70E}}$ is a polarizable VHS. One should assume there that $K$ is a Galois extension of $\mathbb{Q}$ (we can make that hypothesis), and the discussion gives the conclusion that the subfield $L\subset K$ generated by the traces of monodromy elements, has a uniquely defined complex conjugation operation.

We recall the reason. The field of traces that was denoted by $L$ in [Reference SimpsonSim92] will be denoted by $L^{\text{tr}}$ here, and the extension that we are looking for, denoted $L^{\prime }$ in [Reference SimpsonSim92], will be denoted by $L$ here.

Since $V_{\unicode[STIX]{x1D70E}}$ is polarizable, its complex conjugate is isomorphic to its dual. As we assume $V$ is absolutely irreducible, each $V_{\unicode[STIX]{x1D70E}}$ is irreducible so this isomorphism is unique up to a scalar.

Let $c:\mathbb{C}\rightarrow \mathbb{C}$ denote complex conjugation. It means that $V_{c\circ \unicode[STIX]{x1D70E}}\cong (V_{\unicode[STIX]{x1D70E}})^{\ast }$ . Suppose $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D70B}_{1}(X,x)$ , and let $\unicode[STIX]{x1D70C}_{K}^{\ast }$ denote the dual representation. Recall that $\unicode[STIX]{x1D70C}_{K}^{\ast }(\unicode[STIX]{x1D6FE})=(\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D6FE})^{-1})^{t}$ , so $\text{Tr}(\unicode[STIX]{x1D70C}_{K}^{\ast }(\unicode[STIX]{x1D6FE}))=\text{Tr}(\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D6FE}^{-1}))$ . From the isomorphism $V_{c\circ \unicode[STIX]{x1D70E}}\cong (V_{\unicode[STIX]{x1D70E}})^{\ast }$ we therefore get

$$\begin{eqnarray}(c\circ \unicode[STIX]{x1D70E})(\text{Tr}(\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D6FE})))=\unicode[STIX]{x1D70E}(\text{Tr}(\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D6FE}^{-1}))).\end{eqnarray}$$

As we are assuming that $K$ is a Galois extension of $\mathbb{Q}$ , the image of $c\circ \unicode[STIX]{x1D70E}$ is equal to the image of $\unicode[STIX]{x1D70E}$ and it makes sense to write $\unicode[STIX]{x1D70E}^{-1}\circ c\circ \unicode[STIX]{x1D70E}:K\rightarrow K$ (this notational shortcut was used without explanation in [Reference SimpsonSim92] and required the hypothesis that $K$ be Galois). We obtain

$$\begin{eqnarray}\unicode[STIX]{x1D70E}^{-1}\circ c\circ \unicode[STIX]{x1D70E}(\text{Tr}(\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D6FE})))=\text{Tr}(\unicode[STIX]{x1D70C}_{K}(\unicode[STIX]{x1D6FE}^{-1})).\end{eqnarray}$$

It follows that on the subfield $L^{\text{tr}}\subset K$ generated by the traces of monodromy elements, the map

$$\begin{eqnarray}c_{L^{\text{tr}}}:=\unicode[STIX]{x1D70E}^{-1}\circ c\circ \unicode[STIX]{x1D70E}:L^{\text{tr}}\rightarrow L^{\text{tr}}\end{eqnarray}$$

is well-defined and independent of $\unicode[STIX]{x1D70E}$ . In other words, $L^{\text{tr}}$ has a well-defined complex conjugation operation.

Therefore, $L^{\text{tr}}$ is either a totally real field, if $c_{L^{\text{tr}}}$ is the identity, or a totally imaginary quadratic extension of a totally real field $F:=(L^{\text{tr}})^{c_{L^{\text{tr}}}}$ otherwise.

Larsen’s lemma [Reference SimpsonSim92, Lemma 4.8] says that we can extend $L^{\text{tr}}$ to $L$ (as said before, that was denoted $L^{\prime }$ there), a totally imaginary quadratic extension of a totally real field, such that $V$ may be defined over $L$ . The proof uses a consideration of Brauer groups, and we refer to the discussion in [Reference SimpsonSim92, Lemma 4.8].

We have

$$\begin{eqnarray}V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C}=\bigoplus _{\unicode[STIX]{x1D70E}:L\rightarrow \mathbb{C}}V_{\unicode[STIX]{x1D70E}}.\end{eqnarray}$$

If $\unicode[STIX]{x1D70E}:L\rightarrow \mathbb{C}$ then it extends to $\unicode[STIX]{x1D70E}_{K^{\prime }}:K^{\prime }\rightarrow \mathbb{C}$ which then restricts to an embedding $\unicode[STIX]{x1D70E}_{K}:K\rightarrow \mathbb{C}$ , and $V_{\unicode[STIX]{x1D70E}}\cong V_{\unicode[STIX]{x1D70E}_{K}}$ . Therefore, our hypothesis that the $V_{\unicode[STIX]{x1D70E}_{K}}$ are VHS of weight $k$ implies that the $V_{\unicode[STIX]{x1D70E}}$ are VHS of weight $k$ .

Our hypothesis says that for any $\unicode[STIX]{x1D70E}$ , the Hodge types of $V_{\unicode[STIX]{x1D70E}}$ can be chosen within $\{(p,q):\,p+q=k\text{ and }p,q\geqslant 0\}$ . There could be a choice to be made in assigning the Hodge types, for instance if $V_{\unicode[STIX]{x1D70E}}$ is unitary then it has only a single Hodge type and that could be put at any $(p,q)$ in the desired set.

As noted in the discussion of [Reference Corlette and SimpsonCS08, § 10.2] we may choose the Hodge types in such a way that there is a $\mathbb{Q}$ -polarization. We may do the analogue of [Reference Corlette and SimpsonCS08, Lemma 10.3] for the general case as follows. Let $c_{L}:L\rightarrow L$ and $c:\mathbb{C}\rightarrow \mathbb{C}$ denote the complex conjugation automorphisms. For $\unicode[STIX]{x1D70E}:L\rightarrow \mathbb{C}$ put $\overline{\unicode[STIX]{x1D70E}}:=c\circ \unicode[STIX]{x1D70E}=\unicode[STIX]{x1D70E}\circ c_{L}$ . Choose a collection of embeddings $\unicode[STIX]{x1D70E}_{1},\ldots ,\unicode[STIX]{x1D70E}_{d}:L\rightarrow \mathbb{C}$ such that the $\overline{\unicode[STIX]{x1D70E}_{i}}$ are distinct from the $\unicode[STIX]{x1D70E}_{j}$ and the collection

$$\begin{eqnarray}\{\unicode[STIX]{x1D70E}_{1},\ldots ,\unicode[STIX]{x1D70E}_{d},\overline{\unicode[STIX]{x1D70E}}_{1},\ldots ,\overline{\unicode[STIX]{x1D70E}}_{d}\}\end{eqnarray}$$

is equal to the full collection of embeddings. For each $i$ choose a structure of VHS for $V_{\unicode[STIX]{x1D70E}_{i}}$ , and then put

$$\begin{eqnarray}V_{\overline{\unicode[STIX]{x1D70E}}_{i}}^{p,q}:=\overline{V_{\unicode[STIX]{x1D70E}_{i}}^{q,p}}.\end{eqnarray}$$

This determines the structures of VHS for $V_{\overline{\unicode[STIX]{x1D70E}}_{i}}$ in such a way that the property of [Reference Corlette and SimpsonCS08, Lemma 10.3] holds.

As in [Reference Corlette and SimpsonCS08, Proposition 10.2], there is a form

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}:V_{L}\times V_{L}\rightarrow L\end{eqnarray}$$

which is $\mathbb{Q}$ -bilinear, and $L$ -linear in the first variable and $L$ - $c_{L}$ -antilinear in the second variable that is to say $\unicode[STIX]{x1D6F7}(u,\unicode[STIX]{x1D706}v)=c_{L}(\unicode[STIX]{x1D706})\unicode[STIX]{x1D6F7}(u,v)$ . Also we may assume that it is $c_{L}$ - $(-1)^{k}$ -symmetric: $\unicode[STIX]{x1D6F7}(v,u)=(-1)^{k}c_{L}\unicode[STIX]{x1D6F7}(u,v)$ by multiplying by a purely imaginary element if necessary to get this sign right. The form $\unicode[STIX]{x1D6F7}$ is unique up to multiplying by an element of $F^{\ast }$ .

Put $\unicode[STIX]{x1D6F7}_{\mathbb{Q}}:=\text{Tr}_{L/\mathbb{Q}}\circ \unicode[STIX]{x1D6F7}$ , giving a $\mathbb{Q}$ -bilinear form

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{\mathbb{Q}}:V_{L/\mathbb{Q}}\times V_{L/\mathbb{Q}}\rightarrow \mathbb{Q}\end{eqnarray}$$

that is $(-1)^{k}$ -symmetric. This will be our polarization form, after having first adjusted $\unicode[STIX]{x1D6F7}$ by multiplying by an appropriate element of $F^{\ast }$ to be chosen below. Let

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}_{\mathbb{C}}:(V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C})\times (V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C})\rightarrow \mathbb{C}\end{eqnarray}$$

be the induced $\mathbb{C}$ -bilinear form. We would like to say that $\unicode[STIX]{x1D6F7}_{\mathbb{C}}$ pairs $V_{\unicode[STIX]{x1D70E}}$ with $V_{\overline{\unicode[STIX]{x1D70E}}}$ .

To see this, let $v\mapsto z.v$ be the action of $z\in L$ upon $v\in V_{L/\mathbb{Q}}$ . This extends to an action of $L$ on $V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C}$ by $\mathbb{C}$ -linear automorphisms. Then $V_{\unicode[STIX]{x1D70E}}\subset V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C}$ is the subset of vectors such that

$$\begin{eqnarray}z.v=\unicode[STIX]{x1D70E}(z)v\end{eqnarray}$$

for $z\in L$ . We have

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}(z.u,v)=\unicode[STIX]{x1D6F7}(u,(c_{L}(z)).v).\end{eqnarray}$$

Thus, if $u\in V_{\unicode[STIX]{x1D70E}}$ and $v\in V_{\unicode[STIX]{x1D70F}}$ then

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}_{\mathbb{C}}(z.u,v) & = & \displaystyle \unicode[STIX]{x1D6F7}_{\mathbb{C}}(\unicode[STIX]{x1D70E}(z)u,v)=\unicode[STIX]{x1D70E}(z)\unicode[STIX]{x1D6F7}_{\mathbb{C}}(u,v)=\unicode[STIX]{x1D6F7}_{\mathbb{C}}(z,(c_{L}(z)).v)\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6F7}_{\mathbb{C}}(u,\unicode[STIX]{x1D70F}(c_{L}(z))v)=\unicode[STIX]{x1D70F}(c_{L}(z))\unicode[STIX]{x1D6F7}_{\mathbb{C}}(u,v)=\overline{\unicode[STIX]{x1D70F}}(z)\unicode[STIX]{x1D6F7}_{\mathbb{C}}(u,v).\nonumber\end{eqnarray}$$

Therefore, if $\unicode[STIX]{x1D70E}(z)\neq \overline{\unicode[STIX]{x1D70F}}(z)$ then this expression must be zero. For $\unicode[STIX]{x1D70E}\neq \overline{\unicode[STIX]{x1D70F}}$ there is a $z\in L$ with $\unicode[STIX]{x1D70E}(z)\neq \overline{\unicode[STIX]{x1D70F}}(z)$ and we obtain $\unicode[STIX]{x1D6F7}_{\mathbb{C}}(u,v)=0$ . We conclude that $\unicode[STIX]{x1D6F7}_{\mathbb{C}}$ pairs $V_{\unicode[STIX]{x1D70E}}$ with $V_{\overline{\unicode[STIX]{x1D70E}}}$ .

The complex conjugation on $V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C}$ also sends $V_{\unicode[STIX]{x1D70E}}$ to $V_{\overline{\unicode[STIX]{x1D70E}}}$ , indeed if $z.v=\unicode[STIX]{x1D70E}(z)v$ then $z.\overline{v}=\overline{z.v}=\overline{\unicode[STIX]{x1D70E}(z)v}=\overline{\unicode[STIX]{x1D70E}}(z)v$ .

The Hodge types on $V_{\mathbb{C}}:=V_{L/\mathbb{Q}}~\otimes _{\mathbb{Q}}~\mathbb{C}$ are defined by

$$\begin{eqnarray}V_{\mathbb{C}}^{p,q}:=\bigoplus _{\unicode[STIX]{x1D70E}}V_{\unicode[STIX]{x1D70E}}^{p,q}.\end{eqnarray}$$

With our choice of Hodge types satisfying the property of [Reference Corlette and SimpsonCS08, Lemma 10.3], the complex conjugation operation satisfies $\overline{V_{\mathbb{C}}^{p,q}}=V_{\mathbb{C}}^{q,p}$ .

We get the hermitian form $(u,v)\mapsto \unicode[STIX]{x1D6F7}_{\mathbb{C}}(u,\overline{v})$ on $V_{\unicode[STIX]{x1D70E}}$ induced by $\unicode[STIX]{x1D6F7}_{\mathbb{Q}}$ . As the hermitian forms on the irreducible local system $V_{\unicode[STIX]{x1D70E}}$ are unique up to multiplication by a real scalar, we deduce that this hermitian form is either a polarization, or minus a polarization of the complex VHS $V_{\unicode[STIX]{x1D70E}}$ . By uniqueness, the hermitian form has a pure Hodge type, and this type has to be $(k,k)$ otherwise the Hodge types of $V_{\unicode[STIX]{x1D70E}}$ and $V_{\overline{\unicode[STIX]{x1D70E}}}$ could not be symmetric under $(p,q)\mapsto (q,p)$ .