Let $V$ be a rational Hodge structure of degree $k$. Precisely, $V$ is a finite dimensional $\mathbb{Q}$-vector space whose complexification admits a decomposition $V_\mathbb{C} = \oplus_{p+q=k} V^{p,q} $ such that $ \overline{V^{p,q}} = V^{q,p} $.
A polarization on the rational Hodge structure $V$ is a $\mathbb{Q} $-bilinear form $q: V\times V \to \mathbb{Q}$ with $\mathbb{C} $-linear extension $q_\mathbb{C} $, which is symmetric if $k$ is even, alternating if $k$ is odd, such that the Hodge decomposition $V_\mathbb{C} = \oplus_{p+q=k} V^{p,q}$ is orthogonal with respect to $q_\mathbb{C}$ and the associated Hermitian form $(\alpha,\beta) \mapsto q_\mathbb{C}(\sqrt{-1}^{p-q}\alpha, \bar{\beta}) $ is positive definite on $V^{p,q}$.
It's well known that for any compact Kähler manifold $X$ of dimension $n$, and rational Kähler class $[\omega] \in H^2(X,\mathbb{Q}) \cap H^{1,1}(X)$ (then $X$ will be a projective manifold by the Kodaira embedding theorem), the Hodge-Riemann bilinear form $q(\alpha,\beta)=(-1)^{\frac{k(k-1)}{2}}\int_X \omega^{n-k} \wedge \alpha \wedge \beta $ defines a polarization on the degree $k$ rational Hodge structure $H^k(X,Q)_{\rm prim}$, where $$H^k(X,\mathbb{Q})_{\rm prim} = {\rm Ker}\Big(\cup [\omega]^{n-k+1} : H^k(X,\mathbb{Q}) \to H^{2n-k+2}(X,\mathbb{Q}) \Big).$$
Now for any projective manifold $X$, the Lefschetz decomposition theorem $H^k(X,\mathbb{Q}) = \oplus_{2r\le k}[\omega]^r \cup H^{k-2r}(X,\mathbb{Q})_{\rm prim}$ decomposite $H^k(X,\mathbb{Q})$ into a direct sum of polarizable sub Hodge structures. Here I simply say that $H^k(X, \mathbb{Q})$ is split-polarizable.
Conversely, for a compact Kähler manifold $X$ with all the $H^k(X,\mathbb{Q})$ split-polarizable, is $X$ projective?
The only example I know is the complex torus $T$. The isogeny class of $T$ is determined by its first degree rational Hodge structure $H^1(T,\mathbb{Q})$, and $T$ is projective if and only if $H^1(T, \mathbb{Q})$ is polarizable.
I think the answer is NO in the general case. But I can't give an example of polarization without the help of a rational Kähler class.
Question: Is there a compact Kähler non-projective manifold $X$ with some $H^k(X,\mathbb{Q})$ split-polarizable?
Thanks a lot for your answers!
