Let $Q\cong\mathbb{P^{1}\_{1}}\times\mathbb{P^{1}\_{2}}\subset\mathbb{P}^{3}$ be a smooth quadric surface. We have the following two actions on $Q$:
$$S\_2\times Q\rightarrow Q,\; (\sigma,(x,y))\mapsto\sigma(x,y)$$
where $S\_2$ is the symmetric group of order $2$, and
$$Aut(\mathbb{P^{1}\_{1}})\times Aut(\mathbb{P^{1}\_{2}})\times Q\rightarrow Q,\;((f,g),(x,y))\mapsto (f(x),g(y)).$$
These two actions do not commute. Furthermore $h^{0}(Q,T\_{Q}) = 6$, that is $dim(Aut(Q)) = 6$.


Is it true that $Aut(Q)$ is the semi-direct product
$$Aut(Q)\cong(Aut(\mathbb{P^{1}\_{1}})\times Aut(\mathbb{P^{1}\_{2}}))\rtimes S\_2 ?$$



Automorphisms of a smooth quadric surface $Q\subset\mathbb{P}^{3}$
Yes, it is true. This is easily seen by looking for example at the Picard group, generated by $C$ and $D$, the two fibres of the projections. Since $C^2=D^2=0$ and$C\cdot D=1$, the only curves of self intersection $0$ are multiple of $C$ or $D$, and the irreducible ones are equivalent to $C$ or $D$. Composing an automorphism by the exchange of coordinates you fix then $C$ and $D$ and thus act on each factor via an automorphism. This gives the result.


