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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 $$Aut(Q)\cong(Aut(\mathbb{P^{1}_{1}})\times Aut(\mathbb{P^{1}_{2}}))\rtimes S_2 ?$$

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    $\begingroup$ Any automorphism must preserve the intersection product, so it either preserves or exchanges the two rulings. Using this you should be able to prove what you want. $\endgroup$ – quim Mar 11 '14 at 14:28
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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.

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Here is another way to see this. The automorphism group of any quadric hypersuface $$Q(x) = 0 \subset \mathbb{P}^n,$$ is exactly the projective orthogonal group $\textrm{PO}(Q)$ of $Q$.

The key point now is that we have the coincidence that $\textrm{SL}(2) \times \textrm{SL}(2)$ is a double cover of $\textrm{SO}(4)$. On taking suitable projective quotients and carefully keeping track of the $2$-torsion, one arrives at the isomorphism $$\textrm{PO}(4) \cong \textrm{PGL}(2) \times \textrm{PGL}(2) \rtimes S_2,$$ and the result is proved on noticing that $\textrm{Aut}(\mathbb{P}^1) \cong \textrm{PGL}(2).$

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You can see the question in this other way. Take an automorphism $\phi:Q\rightarrow Q$ and compose it with one of the two projections $\pi_i\circ \phi$. Then $$\pi_i\circ\phi:Q\rightarrow\mathbb{P}^1$$ is a fibration with connected fibers. Therefore $\pi_i\circ\phi$ necessarily factorizes through $\pi_{j_{i}}$ with $j_i\in\{1,2\}$. Associating to $\phi$ the permutation $\{i\mapsto j_i\}$ we get a surjactive morphism of groups $$f:Aut(Q)\rightarrow S_2.$$ On the other hand if $\pi_i\circ\phi$ factorized through $\pi_i$ this means than $\phi$ comes from an automorphism in $Aut(\mathbb{P}^1)\times Aut(\mathbb{P}^1)$. Therefore we have an exact sequence $$0\mapsto Aut(\mathbb{P}^1)\times Aut(\mathbb{P}^1)\rightarrow Aut(Q)\rightarrow S_2\mapsto 0.$$ Now, it is enough to obeserve that $f$ admits a sections to conclude that $Aut(Q)\cong (Aut(\mathbb{P}^1)\times Aut(\mathbb{P}^1))\rtimes S_2$.

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  • $\begingroup$ Why a fibration with connected fibres necessarily factorizes through $\pi_{j_i}$ ? $\endgroup$ – Jérémy Blanc Mar 20 '14 at 14:57
  • $\begingroup$ $\pi_i\circ\phi$ is a fibration over $\mathbb{P}^1$ with fiber isomorphic to $\mathbb{P}^1$. There are just two such fibrations on $Q$, namely the two projections. $\endgroup$ – F_L Mar 20 '14 at 16:16
  • $\begingroup$ Yes but why? I think that it is precisely the main point of the question. Once you have this, everything follows directly. $\endgroup$ – Jérémy Blanc Mar 20 '14 at 16:24
  • $\begingroup$ The fiber of the fibration $\pi_i\circ\phi$ is a smooth rational curve because $\phi$ is an automorphism. On the other hand the fiber is the divisor inducing $\pi_i\circ\phi$. Let us write the fiber as $F = aL+bR$. We must have $F^2= ab = 0$. Therefore either $a = 0$ or $b = 0$. The genus of $F$ is $g = ab-a-b+1 = 0$. Then $a =0$ implies $b = 1$ and $b =0$ implies $a = 1$. In the first case we get $\pi_1$ and in the second $\pi_2$. $\endgroup$ – F_L Mar 20 '14 at 17:51

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