Are there smooth proper varieties $X$ and $Y$ over $\mathbb{F}_p$ of the same dimension such that $X(\mathbb{F}_p)=Y(\mathbb{F}_p)$ but Frobenius has different traces on $H^i(X, \mathbb{Q}_l)$ and $H^i(Y, \mathbb{Q}_l)$ for some $i$?
If I understand correctly we can take $X=C\times C$ and $Y=C\times C'$ where $C$ is a curve with no points and $C'$ is a curve with a point. Then $X(\mathbb{F}_p)=Y(\mathbb{F}_p)=0$ but the trace on $H^1(X, \mathbb{Q}_l)$ is$$p+1-C(\mathbb{F}_p)+p+1-C(\mathbb{F}_p)=2(p+1)$$while on $H^1(Y, \mathbb{Q}_l)$ it is$$p+1-C(\mathbb{F}_p)+p+1-C'(\mathbb{F}_p)<2(p+1)$$so we win.
An example where $X$ and $Y$ only have even-dimensional cohomology would be great. I guess the dimension has to be at least four because $H^0$ is predetermined and we have Poincaré duality. We can imitate the above example with K3 surfaces instead of curves but is there a K3 surface with no points?
