I am reading these notes notes by Popa. I did some reading of this post but I didn't exactly answer my question.
It is claimed in example 4.5 that analytic classes sit inside $H^{p,p}(X)$ for various values of $p$. I do not know how to prove such a fact / nor do I see why it needs to be true.
Recall that an analytic class essentially is a cohomology class that comes from an algebraic subvariety of $X$ where $X$ is say a smooth complex manifold. If $V$ is a subvariety of codimension $p$, then $[V]\in H_{2n-2p}(X,\mathbb{C})$ is clear to me. By Poincare duality $[V]\in H^{2p}(X,\mathbb{C})$. Then I can apply the Hodge decomposition to get $[V]\in\bigoplus_{i+j=2p} H^{i,j}(X)$. However, I see no reason to have $[V]$ sitting precisely in $H^{p,p}(X)$.
What am I missing here / can someone provide a geometric picture to keep in mind?
Let $i,j$ be integers such that $i+j=2n-2p$ and $i>n-p=\dim_{\mathbb{C}}\,V$. Let $\omega \in H^{i,j}(X)$. Then $[V] \wedge \omega \in H^{2n}$ is (up to a constant) $\int_{V}{\omega_{|V}}$. But locally, by definition, $\omega$ is a wedge product of $i$ local holomorphic coordinates and $j$-antiholomorphic ones, with $i> \dim_{\mathbb{C}}$. So said wedge product must vanish when restricted to $V$, thus $[V] \wedge \omega=0$.
In other words, cup-product with $[V]$ annihilates the $H^{i,j}(X)$ with $i+j=2n-2p$ and $i>n-p$. For a similar reason, cup-product with $[V]$ annihiates the $H^{i,j}(X)$ with $i+j=2n-2p$ and $j > n-p$.
Thus the projection of $[V]$ to any $H^{u,v}(X)$ with $u+v=2p$ and $(u,v) \neq (p,p)$ is zero, so $[V]\in H^{p,p}(X)$.