Relative flat vs flat direct image Let $Y$ be a Noetherian scheme.
Let $\mathscr{F}$ be a coherent sheaf on $\mathbb{P}^n \times Y$.
Denote $\pi: \mathbb{P}^n \times Y \rightarrow Y $ canonical projection.
We have two notions.


*

*$\pi_* \mathscr{F}$ is flat sheaf on $Y$

*$\mathscr{F}$ is flat over $Y$
Question What is relation between these two notion? Does one of them imply another?
I remind you that $\mathscr{F}$ is flat over $Y$ at point $x \in  \mathbb{P}^n \times Y$ if if the stalk $\mathscr{F}_x$ is a flat module $\mathcal{O}_y$, where $y = \pi(x)$. $\mathscr{F}$ is flat over $Y$ if it is flat at every point.
Comment. 
If you replace $\mathbb{P}^n \times Y$ by any scheme $X$ then there is obvious contrexample. Let $X$ be a point and $Y$ be $\mathbb{A}^1$. Any sheaf on $X$ is flat over $Y$ but direct image is not flat unless $\mathcal{F}$ is zero.
 A: One answer is that $\mathcal{F}$ is flat over $Y$ iff $\pi_*(\mathcal{F}(k))$ is flat over $Y$ for all sufficiently large twists $k$. 
Proof: if $\mathcal{F}$ is flat, a sufficiently large twist has vanishing higher cohomology at each stalk, since the twist needed to kill cohomology can be taken uniform w/r/t the Hilbert polynomial. Then by cohomology and base change (e.g. EGA III.2 corollary 7.9.9) $\pi_*\mathcal{F}(k)$ is locally free, which here is equivalent to flat.
In the other direction, if $\pi_*(\mathcal{F})$ is locally free actually already implies $\mathcal{F}$ being locally free. Indeed, we see the stalks of the former are the sections of the latter on projective fibers. So certainly any point will have a free open neighborhood.
In the other direction, if $\pi_*(\mathcal{F}(k))$ is locally free for sufficiently large twists, then since the stalks of $\mathcal{F}$ are the global sections of the sheaf restricted to the fibers, and for sufficiently large $k$ $\mathcal{F}(k)$ will be globally generated, the result follows
