What is the automorphism group of $\mathbb P^1 \times \mathbb P^1$?

Question

I know that the automorphism group of the projective space $\mathbb P^n$ is $PGL(n+1)$. What is it for $\mathbb P^1 \times \mathbb P^1$? Apart from individual automorphisms of the factors $\mathbb P^1$, there is the switch automorphism as well. I do not know how to incorporate this and write down the automorphism group. Please help.

Answer 1

The slick way (edit: not as slick as I thought!) to answer this is to use the Picard group of $\mathbf P^1 \times \mathbf P^1$, that is, the group of line bundles, or linear equivalence classes of divisors. (If there's an easier way, I'd be glad to hear it.)

In this case we have $\operatorname{Pic} \left( \mathbf P^1 \times \mathbf P^1 \right) \cong \mathbf Z \oplus \mathbf Z$, with generators given by the classes of the divisors $\pi_1^* (p)$ and $\pi_2^* (p)$, the pullback of a point from either of the two factors.

Now any automorphism of the variety must induce an automorphism of the Picard group, but moreover it must send effective divisors to effective divisors. It's not hard to check that any such automorphism must either preserve the generators above, or switch them.

First suppose $\phi$ is an automorphism that preserves the generators of the Picard group. Then it preserves the fibres of the maps $\pi_1$, $\pi_2$, and so it induces automorphisms of the two $\mathbf P^1$ factors. Moreover $\phi$ is determined by its action on those two factors, so the subgroup of such automorphisms is $PGL(2) \times PGL(2)$.

Finally, denote the switch automorphism $(p,q) \mapsto (q,p)$ by $s$. If $\phi$ is an automorphism that switches the generators of the Picard group, then $s \circ \phi$ is an automorphism that preserves them, so $s \circ \phi \in PGL(2) \times PGL(2)$. But $s$ is an involution, so $\phi = s \circ (s \circ \phi)$. So the automorphism group is a (nonabelian) $\mathbf Z_2$-extension of $PGL(2) \times PGL(2)$.

Supplementary remark: I suppose another argument would be to observe that $\mathbf P^1 \times \mathbf P^1$ is isomorphic to a smooth quadric in $\mathbf P^3$, and any automorphism of the quadric comes from an automorphism of the ambient space (because the quadric is embedded by a multiple of $-K$). One could then identify the subgroup of $PGL(4)$ preserving the quadratic form, and show it's isomorphic to the group I wrote down. Offhand, I don't know if this is easier or harder than the first method.