A birational map of \(\PP^2\) into itself is called a plane Cremona transformation. One example, called a quadratic transformation, is the rational map \(\varphi: \PP^2 \dashrightarrow \PP^2\) given by \[ (a_0, a_1, a_2) \mapsto (a_1 a_2,\ a_0 a_2,\ a_0 a_1) \] when no two of \(a_0, a_1, a_2\) are \(0\).
Show that \(\varphi\) is birational, and is its own inverse.
Find open sets \(U, V \subseteq \PP^2\) such that \(\varphi: U \to V\) is an isomorphism.
Find the open sets where \(\varphi\) and \(\varphi^{-1}\) are defined, and describe the corresponding morphisms. See also (V, 4.2.3).
Solution
Write homogeneous coordinates as \([a_0:a_1:a_2]\) and put \[ \varphi([a_0:a_1:a_2]) = [a_1a_2:a_0a_2:a_0a_1] \] whenever the three displayed coordinates are not all zero.
1. On the dense open torus\[ T=D_+(a_0a_1a_2) \]we have\[ \varphi^2=\operatorname{id}_T. \]
If \(a_0a_1a_2\ne0\), then \[ \begin{aligned} \varphi^2([a_0:a_1:a_2]) &= [a_0^2a_1a_2:a_0a_1^2a_2:a_0a_1a_2^2]\\ &=[a_0:a_1:a_2], \end{aligned} \] after cancelling the common nonzero scalar \(a_0a_1a_2\). The same hypothesis shows that \(\varphi(T)\subseteq T\). Thus \(\varphi|_T\) is its own set-theoretic inverse.
2. The restriction\[ \varphi|_T:T\xrightarrow{\sim}T \]is an isomorphism; hence \(\varphi\) is birational and is its own inverse as a rational map.
The three quadratic forms \[ a_1a_2,\qquad a_0a_2,\qquad a_0a_1 \] have no common zero on \(T\), so they define a morphism \(T\to\PP^2\). Its image lies in \(T\), and step 1 shows that the same morphism is its inverse there. Therefore it is an isomorphism \(T\cong T\).
The open set \(T\) is nonempty and hence dense in the irreducible variety \(\PP^2\). An isomorphism on dense open subsets defines a birational map, and the identity \(\varphi^2=\operatorname{id}\) on \(T\) says that its inverse rational map is represented by the same quadratic transformation. This proves (a), and also answers (b) with \[ \boxed{U=V=T}. \]
3. The quadratic formula defines a morphism precisely on\[ D = \PP^2\setminus \{[1:0:0],[0:1:0],[0:0:1]\}. \]
The three defining quadrics vanish simultaneously exactly when at least two of \(a_0,a_1,a_2\) vanish. In projective space this happens at precisely the three coordinate vertices. Away from them, homogeneous forms of the same degree with no common zero define a morphism \[ \varphi_D:D\to\PP^2, \qquad [a_0:a_1:a_2] \longmapsto [a_1a_2:a_0a_2:a_0a_1]. \]
4. The rational map \(\varphi\) cannot be extended across any of the three omitted coordinate vertices.
By symmetry it suffices to consider \[ P_0=[1:0:0]. \] Suppose a morphism representing \(\varphi\) extended to an open neighborhood \(W\) of \(P_0\). After replacing \(W\) by \(W\cap D_+(a_0)\), assume \(W\subseteq D_+(a_0)\).
Let \[ L_1=Z(a_1), \qquad L_2=Z(a_2). \] Both lines pass through \(P_0\). On \(L_1\setminus\{P_0\}\) we have \(a_2\ne0\), so \[ \varphi([a_0:0:a_2])=[0:1:0]. \] Thus the restriction of the hypothetical extension to the irreducible open curve \(W\cap L_1\) is equal to the constant point \([0:1:0]\) on the dense open subset obtained by deleting \(P_0\). Because a point of \(\PP^2\) is closed, its inverse image is closed; hence density forces the value at \(P_0\) also to be \([0:1:0]\).
On the other hand, on \(L_2\setminus\{P_0\}\), \[ \varphi([a_0:a_1:0])=[0:0:1]. \] The same argument forces the value at \(P_0\) to be \([0:0:1]\). These values are distinct, a contradiction.
Permuting the coordinates gives the same contradiction at \([0:1:0]\) and \([0:0:1]\). Therefore none of the three vertices belongs to the domain of definition of the rational map.
5. Both \(\varphi\) and \(\varphi^{-1}\) have maximal domain \(D\), and on that domain both are represented by the same quadratic formula.
Steps <1>3–<1>4 show that \(D\) is exactly the maximal domain of \(\varphi\) in the sense of P-AGH42RATMAPDOM. By step 2 the inverse rational map satisfies \[ \varphi^{-1}=\varphi. \] It therefore has the same maximal domain and the same representing morphism \[ [a_0:a_1:a_2] \longmapsto [a_1a_2:a_0a_2:a_0a_1]. \] This proves (c).
6. Q.E.D.
Steps <1>1–<1>2 prove (a) and (b), while steps <1>3–<1>5 prove (c).
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P-AGH42RATMAPDOMLargest open set on which a rational map is a morphismuses