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Can one can omit the integrality condition in the Cattani-Deligne-Kaplan theorem?

That is, in Corollary 1.2 in the Cattani-Deligne-Kaplan paper, is the condition "$u\in V_s$ is integral" in "$u\in V_s$ is integral of type (0,0)" necessary?

For the convenience of the reader, let me cite the relevant results from the paper.

Let S be a nonsingular complex algebraic variety and V a variation of Hodge structures of weight 0 on S with polarization form Q. If C is the Weil operator: multiplication by $i^{p−q}$ on $V^{p,q}$, the hermitian form $h(u, v) = Q(Cu, v)$ is positive definite and makes the Hodge decomposition orthogonal. For $u$ a real element of type $(0, 0)$, $h(u, u) = Q(u, u)$. Fix an integer K and let $S^{ (K)}$ be the space of pairs $(s, u)$ with $s ∈ S$, $u ∈ V_s$ integral of type $(0, 0)$, and $Q(u, u) ≤ K$. It projects to S and arguments as above show that, locally on S, $S^{ (K)}$ is a finite disjoint sum of closed analytic subspaces. Our main result is:

Theorem 1.1. $S^{ (K)}$ is an algebraic variety, finite over S.

Corollary 1.2. Fix $s ∈ S$ and $u ∈ V_s$ integral of type $(0, 0)$. The germ of analytic subvariety of S where u remains of type $(0, 0)$, is algebraic.

Generally, I am interested in projections to $S$ of analytic subvarieties of $V$ defined in terms of Hodge decomposition, $H^{p,q}$'s etc, and I want to know when such projections are algebraic and/or analytic. I am happy to assume Hodge conjecture.

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The integrality condition is necessary to get finiteness, without it there may be infinitely many components. The point is that without integrality the condition $Q(u,u) \leq K$ becomes superfluous since that can always be arranged by scaling $u$ by an element of $\mathbb{Q}$. –  ulrich Jun 18 '12 at 3:24
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