Generalizing the square theorem

Question

Let $X$ and $Y$ be connected quasi-projective varieties over $\mathbf{C}$. Let $\mathcal{L}$ be an algebraic vector bundle over $X\times Y$. Let $p_2:X\times Y\rightarrow Y$ be the projection.

($\star$) Assume that for all $y_0\in Y$ one has that $\mathcal{L}|_{X\times \{y_0\}}$ is trivial.

In this case, the first (naive) intuition is that $\mathcal{L}\simeq p_2^*\mathcal{F}$ for some algebraic vector bundle $\mathcal{F}$ over $Y$.

Q1: Give an example where $(\star)$ is satified but where $\mathcal{L}$ is not the pullback of any algebraic vector bundle $\mathcal{F}$ over $Y$.

Q2: Under what (interesting) additional assumptions on $X$, $Y$ and $\mathcal{L}$ is it possible to conclude that $\mathcal{L}$ is the pullback of a vector bundle over $Y$?

For example, if $X$ is complete then the square theorem says that $\mathcal{L}$ is the pullback of a vector bundle over $Y$.

Answer 1

There are many examples. Take $X := \mathbb A^1$ and $Y := \mathbb A^2 \smallsetminus \{(0,0)\}$; since all vector bundles on $X$ and $Y$ are trivial, it is sufficient to give an example of a vector bundle on $U := X \times Y$ that is not trivial.

Consider the embedding $j \colon U \subseteq \mathbb A^3$. The are reflexive sheaves $F$ on $\mathbb A^3$ that are locally free on $\mathbb A^3 \smallsetminus \{(0,0, 0)\}$, but not at the origin. For example, you can take the sheaf associated with a second syzygy of the $\mathbb C[x,y,z]$-module $\mathbb C[x,y,z]/(x,y,z)$; this is not projective, because the projective dimension of $\mathbb C[x,y,z]/(x,y,z)$ over $\mathbb C[x,y,z]$ is 3, but it is reflexive, because every second syzygy is. The restriction $G$ of $F$ to $U$ is a non-trivial locally free sheaf on $U$. In fact, since $F$ is reflexive we have $F = j_*G$, so if $G$ were trivial $F$ would also be a free sheaf, which is not the case.