Non-vanishing of cup product in cohomology Let $X$ be a smooth projective variety over $\mathbb{C}$, and let $\alpha \in H^{2k}(X)$ be an algebraic cohomology class. Let us fix an $l$ such that $H^l(X) \neq 0$ and $H^{l+2k}(X) \neq 0$.
The general question (perhaps a bit vague) is: does anybody know a sufficient condition on $X$  such that the intersection map
$
 \alpha \cup - : H^l(X) \to H^{l+2k}(X)
$
is NOT the zero map? 
In particular, I would be interested in one of the following two cases.
1) Both $h^l$ and $h^{l+2k}$ can be arbitrarily large (so that being the zero map is very unlikely, so to speak).
2) The number $2l+2k$ equals $2 \dim_{\mathbb{C}}(X)$, thus $H^l\cong H^{l+2k}$ by Poincarè duality.
Would it help to assume that $\alpha$ is the Euler class (top Chern class) of a vector bundle on $X$?
ADDED: I am interested in the case when $l$ is odd.
 A: Since you ask only for a sufficient condition, let $l=2n$ be even and suppose that $\alpha$
is effective i.e.
 a positive linear combination of fundamental classes of subvarieties $V_i$.
Let $[H]\in H^2(X)$ be the fundamental class of a hyperplane section. Then $\beta=\[H]^n\in H^l(X)$
is a class such that $\alpha\cup \beta\not=0$. To see this, note that after cupping with additional copies of $[H]$, we obtain the (positive weighted) sum of degrees of $V_i$.
Addendum (added in response to the edited question): The case where $l$  is odd is actually more interesting. Let me give an example
to show that the desired result can fail without additional hypotheses. 
Choose a smooth projective variety $X'$ with $\dim X'=n>1$ and  $H^l(X')\not=0$ with $l$ odd. Now blow up a smooth codimension $k+1>1$ subvariety $V$ to get $X$. Then $H^l(X')\subset H^l(X)$, so it's also nonzero. 
Let $E$ be the exceptional divisor. This is a $\mathbb{P}^{k}$ bundle over $V$.
Let $\alpha=[\mathbb{P}^k]$ (one of the fibres). This is a nonzero class, but $\alpha\cup:H^l(X)\to H^{l+2k}(X)$ is zero, because it factors through restriction to $\mathbb{P}^{k}$. Note that $\dim H^l$ and $\dim H^{l+2k}$ can be arbitrarily large.
A: Also, have a look at the Hard Lefschetz Theorem, if you have not seen it yet.
