Which line bundles on a product of projective spaces are ample?

Question

We know that $\text{Pic}(\mathbb{P}^r \times \mathbb{P}^s) \cong \mathbb{Z}^2$ via the pullbacks of the hyperplane bundles from the factors, namely $\mathcal{O}_{\mathbb{P}^r \times \mathbb{P}^s}(1,0)$ and $\mathcal{O}_{\mathbb{P}^r \times \mathbb{P}^s}(0,1)$.

Along this identification, which elements $(a,b) \in \mathbb{Z}^2$ are ample? Is it

1. all $(a,b)$ with $a, b \geq 0$ and at least one positive? or
2. only those "diagonal" such which in addition satisfy $a = b$? or
3. something else?

Which $(a,b)$ are globally generated? Is it all with $a,b \geq 0$?

Thanks!

Answer 1

$\mathcal O_{\mathbb P^r \times \mathbb P^s}(a,b)$ is ample if and only if $a > 0$ and $b > 0$.

Moreover, $a > 0$, $b > 0$ is also the condition for $\mathcal O_{\mathbb P^r \times \mathbb P^s}(a,b)$ to be very ample, meaning that the global sections of $\mathcal O_{\mathbb P^r \times \mathbb P^s}(a,b)$ define a closed embedding of $\mathbb P^r \times \mathbb P^s$ into projective space. Since a line bundle is ample if and only if some positive power of it is very ample, it suffices for the purpose of your question to check very ampleness.

Being globally generated is a weaker condition than very ampleness; it means that the global sections of $\mathcal O_{\mathbb P^r \times \mathbb P^s}(a,b)$ define a morphism from $\mathbb P^r \times \mathbb P^s$ into projective space, though this morphism doesn't necessarily need to be a closed embedding because it may fail to separate points or separate tangent vectors. We will see that $a \geq 0$ and $b \geq 0$ is the condition for $\mathcal O_{\mathbb P^r \times \mathbb P^s}(a,b)$ to be globally generated.

For example, a basis for the global sections of $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(1,1)$ is $\{x_0y_0, x_0y_1, x_1y_0, x_1y_1 \}$, where $[x_0:x_1]$ and $[y_0:y_1]$ are the homogeneous coordinates on the two $\mathbb P^1$'s. We can use these global sections to define the Segre embedding $\mathbb P^1 \times \mathbb P^1 \to\mathbb P^3$, $$( [x_0:x_1],[y_0:y_1]) \mapsto [x_0y_0: x_0y_1:x_1y_0:x_1y_1]. $$

A basis for the global sections of $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(1,2)$ is $\{ x_0y_0^2, x_0y_0y_1, x_0y_1^2, x_1y_0^2, x_1y_0y_1, x_1y_1^2 \}$. These define the following embedding of $\mathbb P^1 \times \mathbb P^1$ into $\mathbb P^5$: $$([x_0:x_1],[y_0:y_1]) \mapsto [x_0y_0^2: x_0y_0y_1: x_0y_1^2: x_1y_0^2: x_1y_0y_1: x_1y_1^2],$$ This can be understood as embedding the second $\mathbb P^1$ into $\mathbb P^2$ via the second Veronese embedding, then embedding $\mathbb P^1 \times \mathbb P^2$ into $\mathbb P^5$ via the Segre embedding.

A basis for the global sections of $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(1,0)$ are $x_0$ and $x_1$. They do define a morphism to $\mathbb P^1$, $$([x_0:x_1],[y_0:y_1]) \mapsto [x_0: x_1],$$ but this morphism is not injective, so is not a closed embedding, hence $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(1,0)$ is globally generated, but not very ample.

The line bundle $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(0,0)$ is even simpler. Its vector space of global sections is one dimensional, spanned by $1$. This defines a rather trivial morphism to $\mathbb P^0$, $$([x_0:x_1],[y_0:y_1]) \mapsto [1].$$ The conclusion is that $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(0,0)$, too, is globally generated but not very ample.

Finally, $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(2,-1)$ has no global sections at all, so its global sections do not define any morphism to projective space, hence $\mathcal O_{\mathbb P^1 \times \mathbb P^1}(2,-1)$ is neither globally generated nor very ample.

I hope you can see how these examples generalise.