Existence of a morphism between two toric varieties

Question

Does there exist a morphism between the blow-up of $\mathbb{P}^3$ in four general points and $\mathbb{P}^1\times\mathbb{P}^1$? If not why?

Answer 1

(Building on Pedro Montero's comment.) I don't know much about studying general such morphisms, but toric morphisms are easy to think about.

Four general points is $PGL(4)$-equivalent to the four $T$-invariant points. I picture blowing them up polytopally, rather than fanly, by cutting the four corners off a tetrahedron. Call that 8-sided polytope $P$.

If there were such a morphism, we could pull back an ample line bundle on $(\mathbb P^1)^2$ to a nef line bundle on $\widetilde{\mathbb P^3}$. Its corresponding moment polytope (polygon!) would be a degeneration of $P$.

If we try and flatten $P$, we push a hexagonal face (one of the original $\mathbb P^2$s, blown up at three points) into its opposite triangular face (the exceptional divisor lying over the point that wasn't on that $\mathbb P^2$), and the resulting squashed $P$ is a triangle, not a hexagon (as Pedro was effectively asserting).

I'm pretty certain that these four squashings are the only ways to degenerate $P$ to a polygon, each giving a map to $\mathbb P^2$ and none to $(\mathbb P^1)^2$. So there are no toric morphisms.

Answer 2

I don't think this is possible. Let $F_1$ and $F_2$ be two disjoint fibers of the projection $\mathbb{P}^1\times\mathbb{P}^1\rightarrow \mathbb{P}^1$. Suppose that there were a surjective morphism $$m\,:\,Bl_{p_1,p_2,p_3,p_4}\mathbb{P}^3\rightarrow \mathbb{P}^1\times\mathbb{P}^1.$$ Then $S_1:=m^{-1}(F_1)$ and $S_2:=m^{-1}(F_2)$ are two disjoint surfaces in $Bl_{p_1,p_2,p_3,p_4}\mathbb{P}^3$. Let $\overline{S_1}$ and $\overline{S_2}$ be their images in $\mathbb{P}^3$. Since two surfaces in $\mathbb{P}^3$ intersect in at least a curve, there is no way the blow-up at four points can make them disjoint. Contradiction.

Answer 3

This is just a small fanny addition to Allen Knutson's polytopal answer:

If you find a (toric) map to $(\mathbb{P}^1)^2$, you automatically get two possible (toric) maps from $\widetilde{\mathbb{P}^3}$ to $\mathbb{P}^1$. But a toric map $\widetilde{\mathbb{P}^3}\to\mathbb{P}^1$ corresponds to a hyperplane in $\mathbb{Q}^3$ separating all maximal cones of the fan of $\widetilde{\mathbb{P}^3}$ into two groups (positive and negative side).

Let $\Sigma\subseteq\mathbb{Q}^3$ be the fan of $\mathbb{P}^3$. Taking a non-trivial linear map $f:\mathbb{Q}^3\to\mathbb{Q}^1$ you will always find a two-dimensional cone $\sigma\in\Sigma$ such that $f(\sigma)=\mathbb{Q}^1$, effectively telling you that there is no toric map $\mathbb{P}^3\to\mathbb{P}^1$ (for visualization consider $\mathbb{P}^2\to\mathbb{P}^1$) . Blowing up in the torus invariant points leaves the two-dimensional cones of $\Sigma$ unchanged.