Proof. The curve $\Gamma _1$ is an intersection of six quadrics in $\mathbb {P}^4$ ; therefore, it is an intersection of five quadrics in $Q_1$ . Hence, if $E_1$ is the exceptional divisor of the blowup $\pi _1 \colon X_{{\bar {\mathsf {k}}}} \to Q_1$ and $H_1$ is the pullback of the hyperplane class of $Q_1$ , the linear system $|2H_1-E_1|$ on $X_{\bar {\mathsf {k}}}$ is four-dimensional and base-point free. Therefore, this linear system defines a morphism $\pi _2 \colon \operatorname {{\mathrm {Bl}}}_{\Gamma _1}(Q_1)\to \mathbb {P}^4$ ; moreover, standard intersection theory gives $(2H_1 - E_1)^3 = 2$ . Hence, the image of $\pi _2$ (which is not contained in a hyperplane by definition) is a quadric $Q_2 \subset \mathbb {P}^4$ and $\pi _2$ is birational. Since $-K_{X_{\bar {\mathsf {k}}}}$ is ample on the fibres of $\pi _2$ and $\unicode{x3c1} (X_{\bar {\mathsf {k}}}) = 2$ , we see that $\pi _2$ is an extremal Mori contraction. By [Reference Mori16], the quadric $Q_2$ is smooth and $\pi _2$ is the blowup of a curve, which must be a linearly normal smooth rational quartic curve. For the last statement, just note that $H_1 + (2H_1 - E_1) = 3H_1 - E_1$ is the anticanonical class of $X_{\bar {\mathsf {k}}}$ .

Proof. For type $\mathsf {X}_{(4,4)}$ , this follows from Lemma 2.2, and for the other types, the first statement follows from the Lefschetz hyperplane theorem and the second from adjunction and the description of Table 1.

Proof. Part (ii) is proved in Lemma 2.2. So, assume X is a variety of either of types $\mathsf {X}_{(2,2,2)}$ , $\mathsf {X}_{(3,3)}$ or $\mathsf {X}_{(1,1,1,1)}$ . Birationality of the projection $\pi _{\widehat {\imath }}$ is clear from the descriptions of Table 1; and it also follows that all fibres of $\pi _{\widehat {\imath }}$ are linear subspaces in $\mathbb {P}^n$ and $-K_{X_{\bar {\mathsf {k}}}}$ restricts to each of them as the hyperplane class by (2.1.2). Also, it is easy to see that the image of $\pi _{\widehat {\imath }}$ is smooth in all cases (for type $\mathsf {X}_{(2,2,2)}$ , if $W_{\widehat {\imath }} \subset \mathbb {P}^2 \times \mathbb {P}^2$ is singular, then its preimage in $\mathbb {P}^2 \times \mathbb {P}^2 \times \mathbb {P}^2$ is singular along a plane, hence, $X_{\bar {\mathsf {k}}}$ , which is the intersection of this preimage with two other divisors, must be singular; and for types $\mathsf {X}_{(3,3)}$ and $\mathsf {X}_{(1,1,1,1)}$ , the image is just $\mathbb {P}^3$ or $\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1$ , respectively).

By Lemma 2.3, the relative Picard number of $\pi _{\widehat {\imath }}$ is $1$ and $-K_{X_{\bar {\mathsf {k}}}}$ is ample, hence, $\pi _{\widehat {\imath }}$ is an extremal Mori contraction. Since both the source and target of $\pi _{\widehat {\imath }}$ are smooth, it follows from [Reference Mori16] that the morphism $\pi _{\widehat {\imath }}$ is either the blowup of a smooth curve or the blowup of a smooth point. In the latter case, the restriction of $-K_{X_{\bar {\mathsf {k}}}}$ to the nontrivial fibre $\mathbb {P}^2$ of $\pi _{\widehat {\imath }}$ would be isomorphic to ${\mathscr {O}}_{\mathbb {P}^2}(2)$ , contradicting to the above observation, hence, $\pi _{\widehat {\imath }}$ is the blowup of a smooth curve.

The remaining assertions are easy and left to the reader (see also [Reference Mori and Mukai15]).

Proof. The classes $H_i$ are pullbacks of ample classes on $\mathbb {P}^n$ , hence, semiample, and they generate $\operatorname {Pic}(X_{\bar {\mathsf {k}}})$ by Lemma 2.3. If $\Lambda _i$ is the class of a nontrivial fibre of $\pi _{\widehat {\imath }}$ , we have

$$ \begin{align*} H_j \cdot \Lambda_i= \delta_{ij}, \end{align*} $$

therefore, $H_j$ generate the rays of the nef cone. It follows that the Galois group permutes the $H_i$ , hence, its action on $\operatorname {Pic}(X_{\bar {\mathsf {k}}})$ factors through the permutation group. Transitivity of the subgroup ${\mathrm {G}}_X \subset \mathfrak {S}_r$ follows from the equality $\unicode{x3c1} (X) = 1$ .

Proof. If $\Pi \subset X_{\bar {\mathsf {k}}}$ is an H-plane, the restriction $(H_1 + \dots + H_r)\vert _{\Pi }$ is the line class. Since all the $H_i$ are nef, it follows that $H_j\vert _{\Pi } \sim 0$ for all $j \ne i$ and some i, hence, $\Pi $ is contracted to a point by the projection $\pi _{\widehat {\imath }}$ . It remains to note that the fibres of $\pi _{\widehat {\imath }}$ are at most one-dimensional by Lemma 2.4.

Proof. Since the classes $H_i$ are semiample, it follows from (2.1.2) that for each ${\bar {\mathsf {k}}}$ -line L on X, there is a unique i, such that $L \cdot H_i = 1$ and $L \cdot H_j = 0$ for $j \ne i$ (i.e. $[L] = \Lambda _i$ in the notation of Lemma 2.5). Thus, L is contracted by the projection $\pi _{\widehat {\imath }}$ , hence, it is equal to a fibre of the exceptional divisor of this projection. Taking into account the description of the projections $\pi _{\widehat {\imath }}$ from Lemma 2.4, we obtain the description of $\mathrm {F}_1(X_{\bar {\mathsf {k}}})$ .

Further, the description of the normal bundle of L follows from the exact sequence

$$ \begin{align*} 0 \longrightarrow {\mathscr{N}}_{L/E_{\widehat{\imath}}} \longrightarrow {\mathscr{N}}_{L/X_{\bar{\mathsf{k}}}} \longrightarrow {\mathscr{N}}_{E_{\widehat{\imath}}/X_{\bar{\mathsf{k}}}}\vert_L \longrightarrow 0, \end{align*} $$

because the first term is trivial and the last is ${\mathscr {O}}_L(-1)$ . Finally, factorisation of the Galois action on the set of connected components of $\mathrm {F}_1(X_{\bar {\mathsf {k}}})$ and its transitivity follow from Lemma 2.5.

Proof. By Lemma 2.7 for each ${\bar {\mathsf {k}}}$ -point x of X, there is at most one line from each of the connected components of the Hilbert scheme $\mathrm {F}_1(X_{\bar {\mathsf {k}}})$ passing through x. This proves that $\mathrm {F}_1(X_{\bar {\mathsf {k}}},x)$ is finite and reduced and gives the bound for its length.

Now, assume x is a point of X defined over ${\mathsf {k}}$ , and let L be a ${\bar {\mathsf {k}}}$ -line through x. Then for any $g \in {\mathrm {G}}({\bar {\mathsf {k}}}/{\mathsf {k}})$ , the line $g(L)$ also passes through x. Transitivity of the Galois action on the set of components of $\mathrm {F}_1(X_{\bar {\mathsf {k}}})$ then implies that there is a unique line of each type through x, hence, the length of $\mathrm {F}_1(X_{{\bar {\mathsf {k}}}},x)$ is r and the ${\mathrm {G}}({\bar {\mathsf {k}}}/{\mathsf {k}})$ -action on $\mathrm {F}_1(X_{\bar {\mathsf {k}}},x)$ factors through ${\mathrm {G}}_X$ and is transitive.

Proof. First, note that no conic on X is contracted by the projections $\pi _{\widehat {\imath }}$ , since by Lemma 2.4 any reduced connected curve contracted by $\pi _{\widehat {\imath }}$ is a line, and lines do not support nonreduced conics by (2.2.1) and [Reference Kuznetsov, Prokhorov and Shramov12, Remark 2.1.7]. Therefore, we deduce from (2.1.2), that for each ${\bar {\mathsf {k}}}$ -conic $C \subset X_{{\bar {\mathsf {k}}}}$ , there is a pair of indices $1 \le i_1 < i_2 \le r$ , such that

(2.3.1) $$ \begin{align} H_{i_1} \cdot C = H_{i_2} \cdot C = 1 \qquad\text{and}\qquad H_j \cdot C = 0\quad\text{for } j \not\in \{i_1,\,i_2\}. \end{align} $$

If $r \ge 3$ , that is, if X is of type $\mathsf {X}_{(2,2,2)}$ or $\mathsf {X}_{(1,1,1,1)}$ , such C is contracted by one of the projections

(2.3.2) $$ \begin{align} \pi_i \colon X_{\bar{\mathsf{k}}} \longrightarrow \mathbb{P}^2_{\bar{\mathsf{k}}} \qquad\text{or}\qquad \pi_{i_1,i_2} \colon X_{\bar{\mathsf{k}}} \longrightarrow \mathbb{P}^1_{\bar{\mathsf{k}}} \times \mathbb{P}^1_{\bar{\mathsf{k}}}, \end{align} $$

respectively. It is easy to see that the maps (2.3.2) are flat conic bundles, hence, C is a fibre of one of them, and, therefore, $\mathrm {F}_2(X_{\bar {\mathsf {k}}})$ is the disjoint union of $\mathbb {P}^2_{\bar {\mathsf {k}}}$ or of $\mathbb {P}^1_{\bar {\mathsf {k}}} \times \mathbb {P}^1_{\bar {\mathsf {k}}}$ , respectively.

Assume X is of type $\mathsf {X}_{(4,4)}$ . Applying Corollary A.2 twice, we obtain a morphism

$$ \begin{align*} \varphi = (\varphi_1, \varphi_2) \colon \mathrm{F}_2(X_{\bar{\mathsf{k}}}) \longrightarrow \Gamma_1 \times \Gamma_2 \end{align*} $$

that takes a smooth conic $C \subset X_{\bar {\mathsf {k}}}$ to the unique pair of lines $(L_2,L_1)$ of different types, such that $C \cap L_i \ne \varnothing $ . We will show that $\varphi $ is an isomorphism.

First, note that by (2.3.1), if $C \subset X$ is a conic, then $\pi _1(C) \subset Q_1$ and $\pi _2(C) \subset Q_2$ are lines, and by Lemma 2.4 (ii), they intersect the curves $\Gamma _1$ and $\Gamma _2$ , respectively. Thus, by Corollary A.2, for $x_1 \in \Gamma _1$ if $[C] \in \varphi _1^{-1}(x_1)$ , the line $\pi _1(C) \subset Q_1$ passes through $x_1$ . Since any line on $Q_1$ through $x_1$ lies in the embedded tangent space to $Q_1$ at $x_1$ , and the intersection of this tangent space with $Q_1$ is a two-dimensional quadratic cone with vertex at $x_1$ , it follows that

$$ \begin{align*} \varphi_1^{-1}(x_1) \cong \mathbb{P}^1 \end{align*} $$

for any $x_1 \in \Gamma _1$ . Since $\Gamma _2 \cong \mathbb {P}^1$ also, the morphism $\varphi $ is a morphism of $\mathbb {P}^1$ -bundles over $\Gamma _1$ and, to show that it is an isomorphism, it is enough to check that it is birational.

So, consider a general pair $(L_2,L_1)$ of lines on X of different types. It follows from (2.3.1) and Lemma 2.4 (ii) that $\bar {L}_1 := \pi _1(L_1)$ is a line on $Q_1$ bisecant to $\Gamma _1$ , $x_1 := \pi _1(L_2)$ is a point on $\Gamma _1$ and by Corollary A.2, the preimage $\varphi ^{-1}(L_2,L_1)$ is the Hilbert scheme of lines $L \subset Q_1$ passing through $x_1$ and intersecting $\bar {L}_1$ . By genericity, we may assume $x_1 \not \in \bar {L}_1$ (i.e. that the lines $L_1$ and $L_2$ do not intersect). Then any line L as above is contained in the intersection of the plane spanned by $\bar {L}_1$ and $x_1$ with $Q_1$ , which is equal to the union of the line $\bar {L}_1$ with a residual line. Therefore, L must be equal to the residual line, hence, the scheme $\varphi ^{-1}(L_2,L_1)$ consists of a single point, so $\varphi $ is birational, and, hence, it is an isomorphism.

Since the embedded tangent space to $Q_1$ at a general point $x \in Q_1$ intersects the quartic curve $\Gamma _1$ at four points, there are four lines on $Q_1$ through x intersecting $\Gamma _1$ , hence, the universal conic is dominant of degree $4$ over $X_{\bar {\mathsf {k}}}$ .

Finally, assume X is of type $\mathsf {X}_{(3,3)}$ . By (2.3.1) and Lemma 2.4 (iii), the image of C with respect to the blowup $\pi _i \colon X_{\bar {\mathsf {k}}} \to \mathbb {P}^3$ is a line intersecting the curve $\Gamma _{i} \subset \mathbb {P}^3$ at two points. This defines a morphism

$$ \begin{align*} \mathrm{F}_2(X_{\bar{\mathsf{k}}}) \longrightarrow \operatorname{{\mathrm{Sym}}}^2\Gamma_{i}, \end{align*} $$

and it is easy to see that it is an isomorphism. It is also easy to see that for a general point $x \in \mathbb {P}^3$ , there are seven lines passing through x and bisecant to $\Gamma _1$ ; therefore, the universal conic on $X_{\bar {\mathsf {k}}}$ is dominant of degree $7$ over $X_{\bar {\mathsf {k}}}$ .

Proof. Let $L_1$ and $L_2$ be the irreducible components (over ${\bar {\mathsf {k}}}$ ) of the conic $\Theta $ . The argument of Lemma 2.9 shows that $L_i$ are lines of two different types and

$$ \begin{align*} \mathrm{F}_2(X_{\bar{\mathsf{k}}},\Theta) = \mathrm{F}_2(X_{\bar{\mathsf{k}}},L_1) \cup \mathrm{F}_2(X_{\bar{\mathsf{k}}},L_2). \end{align*} $$

Recall that by Lemma 2.7, the curves $\Gamma _1$ and $\Gamma _2$ can be identified with the two connected components of $\mathrm {F}_1(X_{\bar {\mathsf {k}}})$ and the isomorphism $\mathrm {F}_2(X_{\bar {\mathsf {k}}}) \cong \Gamma _1 \times \Gamma _2$ of Lemma 2.9 is defined by taking a conic C to the unique pair of lines of different types intersecting C. This means that

$$ \begin{align*} \mathrm{F}_2(X_{\bar{\mathsf{k}}},\Theta) \cong (\Gamma_1 \times [L_1]) \cup ([L_2] \times \Gamma_2) \subset \Gamma_1 \times \Gamma_2; \end{align*} $$

thus, $\mathrm {F}_2(X_{\bar {\mathsf {k}}},\Theta )$ is the union of two rulings of the surface $\mathrm {F}_2(X)$ and we have the equality $\mathscr {C}_2(X_{\bar {\mathsf {k}}},\Theta ) = \mathscr {C}_2(X_{\bar {\mathsf {k}}},L_1) \cup \mathscr {C}_2(X_{\bar {\mathsf {k}}},L_2)$ .

Furthermore, it follows from the description of Lemma 2.9 that the natural morphism $\mathscr {C}_2(X_{\bar {\mathsf {k}}},L_2) \to X_{\bar {\mathsf {k}}}$ is birational onto the hyperplane section tangent to $Q_1$ at the point $\pi _1(L_2)$ ; it contains the line $\pi _1(L_1)$ with multiplicity $1$ and has multiplicity $2$ at the point $\pi _1(L_2)$ . Similarly, the morphism $\mathscr {C}_2(X_{\bar {\mathsf {k}}},L_1) \to X_{\bar {\mathsf {k}}}$ is birational onto the hyperplane section containing the line $\pi _2(L_2)$ with multiplicity $1$ and having multiplicity $2$ at the point $\pi _2(L_1)$ . Thus, the morphism $\mathscr {C}_2(X,\Theta ) \to X$ is birational onto a divisor of class $(H_1 - L_1 - 2L_2) + (H_2 - L_2 - 2L_1) = H - 3\,\Theta $ .

Proof. First, consider a threefold X of type $\mathsf {X}_{(2,2,2)}$ . If C is a rational normal cubic curve and $H_i \cdot C = 0$ for some i, then C is contracted by one of the conic bundles (2.3.2), hence, the curve C is supported on a fibre of (2.3.2). But the conormal bundle of any such fibre is trivial, hence, it cannot support a nonreduced curve of arithmetic genus $0$ and degree more than $2$ . This means that we have $H_i \cdot C = 1$ for each i, and we conclude from this, and Lemma 2.4, that the image of C under the map $\pi _{1,2} \colon X_{\bar {\mathsf {k}}} \to W_{1,2}$ is a rational curve of bidegree $(1,1)$ intersecting the curve $\Gamma _{1,2}$ and passing through x. The argument analogous to that of Corollary A.2 shows that there is a morphism

$$ \begin{align*} \varphi_{1,2} \colon \mathrm{F}_3(X,x) \longrightarrow \Gamma_{1,2} \end{align*} $$

that takes a twisted cubic curve $C \subset X$ to the unique point $x_{1,2} \in \Gamma _{1,2}$ , such that $C \cap \pi _{1,2}^{-1}(x_{1,2}) \ne \varnothing $ . This morphism is an isomorphism, because on $W_{1,2}$ , there is a unique curve of bidegree $(1,1)$ through a given pair of points (unless they lie on a fibre of either of the projections $W_{1,2} \to \mathbb {P}^2_{\bar {\mathsf {k}}}$ , in which case, x lies on a line in X). The same argument proves isomorphisms of $\mathrm {F}_3(X,x)$ with the curves $\Gamma _{1,3}$ and $\Gamma _{2,3}$ .

Next, consider a threefold of type $\mathsf {X}_{(4,4)}$ . If $H_i \cdot C = 0$ for some i, then C is contracted by $\pi _i$ , hence, is supported on a line. But the conormal bundle of a line is globally generated by (2.2.1), hence, a line cannot support a nonreduced curve of arithmetic genus $0$ and degree more than $1$ . This means that C has bidegree $(1,2)$ or $(2,1)$ . In the first case, the image of C under $\pi _1$ is a line on the quadric $Q_1$ passing through x; hence, the corresponding component of $\mathrm {F}_3(X,x)_{\bar {\mathsf {k}}}$ is isomorphic to $\mathbb {P}^1_{\bar {\mathsf {k}}}$ . It also follows that the corresponding component of $\mathscr {C}_3(X,x)$ is a Hirzebruch surface that maps birationally onto the hyperplane section of $Q_1$ tangent at x, that is, a divisor of class $H_1$ passing through x with multiplicity $2$ . The second component is described analogously. The total divisor class of the image $R_x$ of $\mathscr {C}_3(X,x) \to X$ is $H_1 + H_2 - 4x$ , that is, it is the anticanonical class passing through x with multiplicity $4$ .

Next, consider a threefold of type $\mathsf {X}_{(3,3)}$ . The same argument as above shows that C has bidegree $(1,2)$ or $(2,1)$ . In the first case, the image of C under $\pi _1$ is a line on $\mathbb {P}^3_{\bar {\mathsf {k}}}$ passing through x and intersecting the curve $\Gamma _1$ . Since for any point of $\Gamma _1$ there is a unique line through it and x, the corresponding component of $\mathrm {F}_3(X_{\bar {\mathsf {k}}},x)$ is isomorphic to $\Gamma _1$ . Analogously, the second component is isomorphic to $\Gamma _2$ .

Finally, consider a threefold of type $\mathsf {X}_{(1,1,1,1)}$ . Then, of course, $H_i \cdot C = 0$ for some i. The argument used for threefolds of type $\mathsf {X}_{(2,2,2)}$ shows this cannot hold for two distinct i. So, assume this holds for $i = 1$ . By Lemma 2.4 (iv), the image of C under the map $\pi _{1,2,3}$ is a curve of multidegree $(0,1,1)$ on $\mathbb {P}^1_{\bar {\mathsf {k}}} \times \mathbb {P}^1_{\bar {\mathsf {k}}} \times \mathbb {P}^1_{\bar {\mathsf {k}}}$ intersecting the curve $\Gamma _{1,2,3}$ and passing through x. In other words, it is a curve of bidegree $(1,1)$ on the surface $\mathbb {P}^1_{\bar {\mathsf {k}}} \times \mathbb {P}^1_{\bar {\mathsf {k}}}$ passing through x and either of the two points of intersection of $\Gamma _{1,2,3}$ with this surface (note that these points cannot collide because otherwise x would lie on a line in X). Therefore, there are two pencils of such curves. Using the same argument for other i, we see that altogether there are eight pencils of twisted cubic curves on X passing through x.

Proof. For each $u_i \in V_i$ , denote by $\bar {u}_i \in \bar {V}_i$ the image of $u_i$ under the linear projection from the fixed vector $v_i \in V_i$ . Then the rational map $\psi _0 \colon Y \dashrightarrow Y^+$ is given by the formula

$$ \begin{align*} (u_1,\dots,u_r) \longmapsto (\bar{u}_1,\dots,\bar{u}_r). \end{align*} $$

This map is regular on the open orbit $Y^{\circ } \subset Y$ (given by the conditions $\bar {u}_i \ne 0$ for all indices $1 \le i \le r$ ), and it extends regularly to the orbits $Y_i^{\circ } \subset Y$ of codimension $1$ (given by the condition $\bar {u}_i = 0$ for some $p + 1 \le i \le p+ q$ and $\bar {u}_j \ne 0$ for all $j \ne i$ ).

Now, consider the rational G-equivariant map

(3.1.11) $$ \begin{align} \hat\psi_0 \colon Y \dashrightarrow {\hat{Y}}^+, \qquad (u_1,\dots,u_r) \longmapsto \left((\bar{u}_1,\dots,\bar{u}_r), \sum_{|I| \ge r - 1} \bigotimes_{i \in I} \bar{u}_i\right). \end{align} $$

Here, we consider the summand $\otimes _{i \in I} \bar {u}_i$ as a point in the fibre of the line bundle ${\mathscr {O}}(-h_I)$ and their sum for $|I| \ge r - 1$ as a point in the fibre (of the projectivisation) of the vector bundle ${\mathscr {E}}$ . Obviously, the map $\hat \psi _0$ induces an isomorphism of the open orbit $Y^{\circ } \subset Y$ onto the open orbit $\mathbb {P}_{Y^+}({\mathscr {E}}) \setminus \mathbb {P}_{Y^+}(\bar {{\mathscr {E}}}_{\varnothing })$ in ${\hat {Y}}^+$ and contracts each orbit $Y_i^{\circ }$ of codimension $1$ to the section $s_i(Y^+) \subset {\hat {Y}}^+$ , $p + 1 \le i \le p + q$ .

Now, consider the composition $\hat \psi = \hat \psi _0 \circ \sigma \colon {\tilde {Y}} \dashrightarrow {\hat {Y}}^+$ . The restriction of $\hat \psi $ to the exceptional divisor E is given by

(3.1.12) $$ \begin{align} E &= \mathbb{P}(\bar{V}_1 \oplus \dots \oplus \bar{V}_r) \dashrightarrow {\hat{Y}}^+, \nonumber\\ (\bar{u}_1 + \dots + \bar{u}_r) &\longmapsto \left((\bar{u}_1,\dots,\bar{u}_r), \sum_{|I| = r - 1} \bigotimes_{i \in I} \bar{u}_i\right). \end{align} $$

It maps $E_{\varnothing }^{\circ } \subset {\tilde {Y}}$ isomorphically onto the G-orbit $({\tilde {Y}}^+_{\varnothing })^{\circ } = \mathbb {P}_{Y^+}(\bar {{\mathscr {E}}}) \setminus \left ( \bigcup _{i=1}^r \mathbb {P}_{Y^+}(\bar {{\mathscr {E}}}_i) \right )$ of codimension $1$ . By the above arguments, it also gives an isomorphism of open G-orbits and contracts the orbits ${\tilde {Y}}_i^{\circ } \cong Y_i^{\circ }$ , $p + 1 \le i \le p + q$ , to the sections $s_i(Y^+) \subset {\hat {Y}}^+$ . Therefore, $\hat \psi $ induces a birational isomorphism

$$ \begin{align*} \psi \colon {\tilde{Y}} \dashrightarrow \operatorname{{\mathrm{Bl}}}_{s_{p+1}(Y^+) \sqcup \dots \sqcup s_{p+q}(Y^+)}(\mathbb{P}_{Y^+}({\mathscr{E}})) = {\tilde{Y}}^+. \end{align*} $$

Finally, it is easy to see that the induced map ${\tilde {Y}}_i^{\circ } \to E_i^{\circ }$ is an isomorphism for all indices $p + 1 \le i \le p + q$ . This gives the commutative diagram (3.1.8) and proves that $\psi $ is small.

The first two lines in (3.1.9) follow easily from the formulas (3.1.11), (3.1.12) and (3.1.2). The last line follows from the equality of the canonical classes of ${\tilde {Y}}$ and ${\tilde {Y}}^+$ expressed in terms of $H_i$ and E on the one hand, and $h_i$ , h and $e_i$ on the other hand.

Finally, (3.1.10) follows from (3.1.9).

Proof. First, we prove that for any $y \in Y({\mathsf {k}})$ , the diagram (3.1.8) is defined over ${\mathsf {k}}$ . The divisor classes $H = \sum _{i=1}^r H_i$ and $H' := \sum _{i=1}^p H_i$ on $Y_{\bar {\mathsf {k}}}$ are Galois-invariant, and since we have $Y({\mathsf {k}}) \ne \varnothing $ by assumption, we conclude that they are defined over ${\mathsf {k}}$ . Also ${\tilde {Y}}$ and E are defined over ${\mathsf {k}}$ as y is a ${\mathsf {k}}$ -point. Therefore, the divisor classes

$$ \begin{align*} \sum_{i=1}^p h_i = H' - pE, \qquad h = H - (r - 1)E \qquad\text{and}\qquad -\sum_{i=p+1}^{p+q} e_i = H - H' - qE \end{align*} $$

(which, by Theorem 3.1, are equal to the strict transforms of the classes that are ample on $Y^+$ , relatively ample for ${\hat {Y}}^+ \to Y^+$ and for ${\tilde {Y}}^+ \to {\hat {Y}}^+$ , respectively) are defined over ${\mathsf {k}}$ , hence, the varieties $Y^+$ , ${\hat {Y}}^+$ and ${\tilde {Y}}^+$ , equal to the images of ${\tilde {Y}}$ under the maps given by their appropriate linear combinations, are defined over ${\mathsf {k}}$ , as well as the remaining arrows in the diagram.

Now to prove ${\mathsf {k}}$ -rationality of Y, we argue by induction in $\dim (Y) = \sum n_i$ . If the dimension is zero, there is nothing to prove. So, assume the dimension is positive and consider the diagram (3.1.8). By Theorem 3.1, the variety $Y^+$ is a ${\mathsf {k}}$ -form of the product $Y_{\bar {\mathsf {k}}}^+ = \mathbb {P}^{n_1-1}_{\bar {\mathsf {k}}} \times \mathbb {P}^{n_2-1}_{\bar {\mathsf {k}}} \times \dots \times \mathbb {P}^{n_r-1}_{\bar {\mathsf {k}}}$ . By the Nishimura lemma (see [Reference Nishimura18]), we have $Y^+({\mathsf {k}}) \ne \varnothing $ , hence, $Y^+$ is ${\mathsf {k}}$ -rational by the induction assumption. Furthermore, the morphism ${\hat {Y}}^+ \to Y^+$ is a ${\mathsf {k}}$ -form of a projective bundle and, by (3.1.10), the strict transform of the exceptional divisor E of ${\tilde {Y}}$ provides for it a relative hyperplane section. But E is defined over ${\mathsf {k}}$ , therefore, ${\tilde {Y}}^+$ is rational over $Y^+$ , hence, it is ${\mathsf {k}}$ -rational. It remains to note that the morphisms $\sigma $ , $\psi $ and $\tilde \sigma _+$ in (3.1.8) are birational, hence, Y is ${\mathsf {k}}$ -rational as well.

Proof. By Corollary 3.3 the diagram (3.1.8) is defined over ${\mathsf {k}}$ , so it is enough to check that the complete intersection ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ is equal to the strict transform of $\operatorname {{\mathrm {Bl}}}_y(X_{\bar {\mathsf {k}}})$ . First, note that the assumption that y is a smooth point of X implies that the strict transform $\operatorname {{\mathrm {Bl}}}_y(X_{\bar {\mathsf {k}}})$ of $X_{\bar {\mathsf {k}}}$ in ${\tilde {Y}}$ is the complete intersection of the divisors ${\tilde {D}}_{\alpha }$ . Furthermore, the first part of the assumptions (3.2.1) implies that the intersection of $\operatorname {{\mathrm {Bl}}}_y(X_{\bar {\mathsf {k}}})$ with the open G-orbit in ${\tilde {Y}}_{\bar {\mathsf {k}}}$ is dense in $\operatorname {{\mathrm {Bl}}}_y(X_{\bar {\mathsf {k}}})$ . Therefore, the strict transform of $\operatorname {{\mathrm {Bl}}}_y(X_{\bar {\mathsf {k}}})$ in ${\tilde {Y}}^+_{\bar {\mathsf {k}}}$ is contained in ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ . So, it remains to check that ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ is irreducible of dimension $\dim (X_{\bar {\mathsf {k}}})$ . This is definitely true for the intersection of ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ with the complement of the union of projective subbundles $\mathbb {P}_{Y^+_{\bar {\mathsf {k}}}}(\bar {{\mathscr {E}}}_i)$ , because the map $\psi $ defines an isomorphism of this complement with an open subset of ${\tilde {Y}}_{\bar {\mathsf {k}}}$ . On the other hand, the second part of the assumptions (3.2.1) gives a bound for the dimension of the intersections with these projective subbundles, which implies the irreducibility.

Proof. Let x be a ${\mathsf {k}}$ -point of X. By definition, X is a smooth divisor of bidegree $(1,1)$ in a ${\mathsf {k}}$ -form Y of $\mathbb {P}^2 \times \mathbb {P}^2$ . Since the birational isomorphism $\psi \colon \operatorname {{\mathrm {Bl}}}_x(Y) = {\tilde {Y}} \dashrightarrow {\tilde {Y}}^+ = \mathbb {P}_{Y^+}({\mathscr {E}})$ is small by Theorem 3.1, it follows that X is birational to a ${\mathsf {k}}$ -form of a divisor

$$ \begin{align*} {\tilde{X}}^+_{\bar{\mathsf{k}}} \subset \mathbb{P}_{\mathbb{P}^1 \times \mathbb{P}^1}({\mathscr{E}}) = \mathbb{P}_{\mathbb{P}^1 \times \mathbb{P}^1}({\mathscr{O}}(-1,-1) \oplus {\mathscr{O}}(-1,0) \oplus {\mathscr{O}}(0,-1)), \end{align*} $$

which by (3.1.10) has type $H_1 + H_2 - E = h$ . Any such divisor corresponds to a morphism

$$ \begin{align*} \xi \colon {\mathscr{O}}(-1,-1) \oplus {\mathscr{O}}(-1,0) \oplus {\mathscr{O}}(0,-1) \longrightarrow {\mathscr{O}}. \end{align*} $$

Furthermore, the divisor ${\tilde {X}}^+ \subset {\tilde {Y}}^+$ comes with a morphism $\sigma _+ \colon {\tilde {X}}^+ \to Y^+$ defined over ${\mathsf {k}}$ . By the Nishimura lemma, we have ${\tilde {X}}^+({\mathsf {k}}) \ne \varnothing $ , hence, $Y^+({\mathsf {k}}) \ne \varnothing $ , and since $Y^+$ is a ${\mathsf {k}}$ -form of $\mathbb {P}^1 \times \mathbb {P}^1$ , it is ${\mathsf {k}}$ -rational by Corollary 3.3. Finally, the general fibre of the morphism $\sigma _+ \colon {\tilde {X}}^+ \to Y^+$ is a $1$ -dimensional linear section of a form of a projective plane, hence, it is isomorphic to $\mathbb {P}^1$ , hence, ${\tilde {X}}^+$ is rational over $Y^+$ , hence, is ${\mathsf {k}}$ -rational, hence, so is X.

Proof. Let x be a ${\mathsf {k}}$ -point of X. By definition, $X_{\bar {\mathsf {k}}}$ is a complete intersection of three divisors in $Y_{\bar {\mathsf {k}}} = \mathbb {P}(V_1) \times \mathbb {P}(V_2) \times \mathbb {P}(V_3) \cong \mathbb {P}^2 \times \mathbb {P}^2 \times \mathbb {P}^2$ of multidegree $(1,1,0)$ , $(1,0,1)$ and $(0,1,1)$ , respectively. Denote by

$$ \begin{align*} F_{12} \in V_1^{\vee} \otimes V_2^{\vee}, \qquad F_{13} \in V_1^{\vee} \otimes V_3^{\vee}, \qquad F_{23} \in V_2^{\vee} \otimes V_3^{\vee}, \end{align*} $$

their equations. We apply Proposition 3.5; for this, we consider the intersection

$$ \begin{align*} &{\tilde{X}}^+_{\bar{\mathsf{k}}} \subset \mathbb{P}_{\mathbb{P}^1 \times \mathbb{P}^1 \times \mathbb{P}^1}({\mathscr{E}}) \\&\quad= \mathbb{P}_{\mathbb{P}^1 \times \mathbb{P}^1 \times \mathbb{P}^1}({\mathscr{O}}(-1,-1,-1) \oplus {\mathscr{O}}(-1,-1,0) \oplus {\mathscr{O}}(-1,0,-1) \oplus {\mathscr{O}}(0,-1,-1) ) \end{align*} $$

of the three strict transforms of the above divisors, which, by (3.1.10), have types

$$ \begin{align*} H_1 + H_2 - E = h - h_3, \qquad H_1 + H_3 - E = h - h_2, \qquad H_2 + H_3 - E = h - h_1, \end{align*} $$

hence, correspond to a morphism of vector bundles

$$ \begin{align*} \xi \colon {\mathscr{O}}(-1,-1,-1) \oplus {\mathscr{O}}(-1,-1,0) \oplus {\mathscr{O}}(-1,0,-1)& \oplus {\mathscr{O}}(0,-1,-1)\longrightarrow \\ &\!\!\!\!\!\!\longrightarrow {\mathscr{O}}(0,0,-1) \oplus {\mathscr{O}}(0,-1,0) \oplus {\mathscr{O}}(-1,0,0). \end{align*} $$

It is easy to see that $\xi $ is given by the matrix

(4.2.1) $$ \begin{align} \xi = \begin{pmatrix} \bar{F}_{12} & 0 & F_{12}(-,v_2) & F_{12}(v_1,-) \\ \bar{F}_{13} & F_{13}(-,v_3) & 0 & F_{13}(v_1,-) \\ \bar{F}_{23} & F_{23}(-,v_3) & F_{23}(v_2,-) & 0 \end{pmatrix}, \end{align} $$

where we write $x = (v_1,v_2,v_3)$ , choose splittings $V_i = {\bar {\mathsf {k}}} v_i \oplus \bar {V}_i$ , write $\bar {F}_{ij}$ for the restriction of $F_{ij}$ to $\bar {V}_i \otimes \bar {V}_j$ and consider $F_{ij}(v_i,-)$ and $F_{ij}(-,v_j)$ as linear functions on $\bar {V}_j$ and $\bar {V}_i$ by restriction.

Let us check the dimension conditions (3.2.1). Since $(Y_i)_{\bar {\mathsf {k}}}$ is a fibre of the projection

$$ \begin{align*} \pi_{i} \colon Y_{\bar{\mathsf{k}}} = \mathbb{P}^2 \times \mathbb{P}^2 \times \mathbb{P}^2 \longrightarrow \mathbb{P}^2, \end{align*} $$

it follows from Lemma 2.9 that $X_{\bar {\mathsf {k}}} \cap (Y_i)_{\bar {\mathsf {k}}}$ is a conic, hence, the first part of the dimension conditions is satisfied. To check the second part, we need to show that the restriction

$$ \begin{align*} \xi_{2,3} \colon {\mathscr{O}}(-1,0,-1) \oplus {\mathscr{O}}(0,-1,-1) \longrightarrow {\mathscr{O}}(0,0,-1) \oplus {\mathscr{O}}(0,-1,0) \oplus {\mathscr{O}}(-1,0,0) \end{align*} $$

of $\xi $ to the last two summands of ${\mathscr {E}}$ (given by the last two columns of (4.2.1)) cannot be everywhere degenerate and similarly for the restrictions $\xi _{1,3}$ and $\xi _{1,2}$ . Assuming that $\xi _{2,3}$ is everywhere degenerate, we conclude from (4.2.1) that

$$ \begin{align*} F_{12}(v_1,-) &= F_{13}(v_1,-) = 0 \quad\text{or}\\F_{12}(-,v_2) &= F_{23}(v_2,-) = 0 \quad\text{or}\\ F_{13}(v_1,-) &= F_{23}(v_2,-) = 0. \end{align*} $$

In any case, it would follow that at least two of the bilinear forms $F_{i,j}$ are degenerate, hence, at least two of the divisors $W_{i,j} \subset \mathbb {P}(V_i) \times \mathbb {P}(V_j)$ (defined by the equation $F_{i,j}$ ) are singular, which contradicts Lemma 2.4 (i).

Thus, the conditions (3.2.1) are satisfied and we conclude from Proposition 3.5 that X is ${\mathsf {k}}$ -birational to a ${\mathsf {k}}$ -form of the complete intersection ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ .

Finally, the subvariety ${\tilde {X}}^+ \subset {\tilde {Y}}^+$ comes with a morphism $\sigma _+ \colon {\tilde {X}}^+ \to Y^+$ defined over ${\mathsf {k}}$ . By the Nishimura lemma, we have ${\tilde {X}}^+({\mathsf {k}}) \ne \varnothing $ , hence, $Y^+({\mathsf {k}}) \ne \varnothing $ , and since $Y^+$ is a ${\mathsf {k}}$ -form of $\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1$ , it is ${\mathsf {k}}$ -rational by Corollary 3.3. Moreover, the general fibre of the morphism $\sigma _+ \colon {\tilde {X}}^+ \to Y^+$ is a zero-dimensional linear section of a form of a projective space, hence, this morphism is birational, hence, ${\tilde {X}}^+$ is ${\mathsf {k}}$ -rational, hence, so is X.

Proof. Let x be a ${\mathsf {k}}$ -point of X. By definition, the variety X is a smooth divisor of multidegree $(1,1,1,1)$ in a ${\mathsf {k}}$ -form Y of $\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1$ . The birational isomorphism $\psi \colon \operatorname {{\mathrm {Bl}}}_x(Y) = {\tilde {Y}} \dashrightarrow {\tilde {Y}}^+ = \operatorname {{\mathrm {Bl}}}_{s_1,s_2,s_3,s_4}(\mathbb {P}^4)$ is small by Theorem 3.1, so it follows that X is birational to a ${\mathsf {k}}$ -form of a divisor

$$ \begin{align*} {\tilde{X}}^+_{\bar{\mathsf{k}}} \subset \operatorname{{\mathrm{Bl}}}_{s_1,s_2,s_3,s_4}(\mathbb{P}^4) ,\end{align*} $$

which, by (3.1.10), has type $H_1 + H_2 + H_3 + H_4 - E = 3h - 2 \sum _{i=1}^4 e_i$ , in particular, ${\tilde {X}}^+$ is a cubic hypersurface. Moreover, the exceptional divisor $E \cap \operatorname {{\mathrm {Bl}}}_x(X) \subset {\tilde {Y}} = \operatorname {{\mathrm {Bl}}}_x(Y)$ of the blowup $\operatorname {{\mathrm {Bl}}}_x(X) \to X$ is an irreducible ${\mathsf {k}}$ -rational surface birational to a ${\mathsf {k}}$ -form of the complete intersection of ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ with the linear span $\mathbb {P}^3 \subset \mathbb {P}^4$ of the points $s_i$ .

Let us prove that ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ is not a cone. Indeed, ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ is smooth away from the linear span $\mathbb {P}^3$ of the $s_i$ , because the map $\psi $ from Theorem 3.1 is an isomorphism over its complement and $X_{\bar {\mathsf {k}}}$ is smooth, so if ${\tilde {X}}^+_{\bar {\mathsf {k}}}$ is a cone, its vertex belongs to the $\mathbb {P}^3$ . But then its intersection with the $\mathbb {P}^3$ (which has been shown to be an irreducible ${\mathsf {k}}$ -rational surface) is itself a cone and has a singular point at each of the $s_i$ . But it is easy to see that any such cone is reducible; this contradiction proves the claim. Now, we conclude that ${\tilde {X}}^+$ is ${\mathsf {k}}$ -unirational by [Reference Kollár9, Theorem 1.2].

Proof. The dimension is estimated in [Reference Kuznetsov and Prokhorov10, Lemma 5.4(i) and (iii)]. To prove that ${\mathscr {M}}$ has no fixed components, note that the linear system $|kE|$ is zero-dimensional for any $k \ge 0$ (since E is the exceptional divisor of a blowup), hence, the only possibility for a fixed component of ${\mathscr {M}}$ is provided by the divisor E with some multiplicity. So, assume

$$ \begin{align*} |H - (m+1)E|=(a - m - 1)E + |H - aE|, \end{align*} $$

where $a\ge m + 2$ and E is not a fixed component of the linear system $|H - aE|$ . Since the linear system $|H - mE|$ is base-point free and $|H - aE|$ has no fixed components, using (5.1.2) we obtain

$$ \begin{align*} 0 \le (H - aE)^2\cdot (H - mE)= 2g-2-a^2(m^2 - 3m + 4) + 4am(m-2). \end{align*} $$

When $m = 2$ , this gives $a^2 \le 14$ , hence, $a \le 3$ , and when $m = 1$ , this gives $(a+1)^2 \le 15$ , hence, $a \le 2$ . In both cases, this contradicts the assumption $a \ge m + 2$ .

Proof. We use notation (5.1.1). Recall that the anticanonical class $H = -K_X$ is very ample and the image of the anticanonical embedding $X \subset \mathbb {P}^{g + 1}$ is an intersection of quadrics (see Theorem 2.1). The anticanonical morphism $\phi \colon {\tilde {X}} \to \mathbb {P}^{g-m-1}$ cannot contract a divisor D, because by [Reference Kuznetsov and Prokhorov10, Lemmas 5.5 and 5.7], this divisor is then a fixed component of ${\mathscr {M}}$ , but by Lemma 5.2, this linear system has no fixed components. Therefore, the required link exists and is defined over ${\mathsf {k}}$ by [Reference Kuznetsov and Prokhorov10, Lemmas 5.5 and 5.7].

Proof. Since $\phi $ is crepant and $\psi $ is a flop, the morphism $\phi _+$ is crepant as well. Moreover, the anticanonical linear system $|-K_{\tilde {X}}|$ is base-point free by [Reference Kuznetsov and Prokhorov10, Lemma 5.7], hence, $|-K_{{\tilde {X}}^+}|$ is base-point free as well.

On the other hand, we have $(H - mE) \cdot \Upsilon _i = -K_{{\tilde {X}}} \cdot \Upsilon _i = 0$ , and since $H \cdot \Upsilon _i> 0$ , we conclude that $E \cdot \Upsilon _i> 0$ . Therefore, for $M \in {\mathscr {M}}$ we have

$$ \begin{align*} M \cdot \Upsilon_i = (H - (m+1)E)\cdot \Upsilon_i = -E \cdot \Upsilon_i < 0. \end{align*} $$

If $M^+ \in {\mathscr {M}}^+$ is the strict transform of M, this implies that $M^+ \cdot \Upsilon _i^+> 0$ by one of the definitions of a flop (see, for example, [Reference Kollár and Mori7, Definition 6.10]). Now if $M^+$ is not nef, it is negative on the extremal ray $\mathrm {R}$ corresponding to the contraction $\sigma _+ \colon {\tilde {X}}^+ \to X^+$ . Since the canonical class is also negative on $\mathrm {R}$ , the contraction $\sigma _+$ cannot be small (see [Reference Benveniste2, Theorem 0] or [Reference Mori and Prokhorov17, Corollary 6.3.4]), hence, curves in $\mathrm {R}$ sweep a subvariety of ${\tilde {X}}^+$ of dimension $\ge 2$ , hence, the base locus of ${\mathscr {M}}^+$ is at least two-dimensional, which contradicts Lemma 5.2. This proves that $M^+$ is nef.

Now, we have $-K_{\tilde {X}} \cdot {\tilde {C}} = (H - mE) \cdot {\tilde {C}} = 1$ by (5.2.1) and (5.2.2), hence, $-K_{{\tilde {X}}^+} \cdot {\tilde {C}}^+ = 1$ . On the other hand, since a general divisor H meets C away from the indeterminacy locus of the map $X\dashrightarrow \tilde X^+$ , we have $H^+\cdot {\tilde {C}}^+ \ge H\cdot C=m+1$ and so $E^+\cdot {\tilde {C}}^+ \ge 1$ . Thus,

$$ \begin{align*} M^+ \cdot {\tilde{C}}^+ = (-K_{{\tilde{X}}^+} - E^+) \cdot {\tilde{C}}^+ = 1 - E^+ \cdot {\tilde{C}}^+ \le 0. \end{align*} $$

Since $M^{+}$ is nef, $M^+ \cdot {\tilde {C}}^+= 0$ .

Combining the above computations, we conclude that the nef cone of ${\tilde {X}}^+$ is generated by $-K_{{\tilde {X}}^+}$ and $M^+$ and the Mori cone is generated by $\Upsilon _i^+$ and ${\tilde {C}}^+$ . The rest of the lemma follows from the Mori contraction theorem [Reference Mori16, Theorems 3.1 and 3.2].

Proof of Theorem 5.1

Let X be a Fano threefold of type $\mathsf {X}_{(4,4)}$ . By Proposition 5.3, there exists a Sarkisov link (1.2.1), and it remains to describe the contraction $\sigma _+$ . Since $\sigma _+$ is an extremal contraction, we have $\unicode{x3c1} (X^+) = \unicode{x3c1} ({\tilde {X}}^+) - 1 = \unicode{x3c1} ({\tilde {X}}) - 1 = \unicode{x3c1} (X)$ , hence, $\unicode{x3c1} (X^+) = 1$ . Similarly, we have $\unicode{x3c1} (X^+_{\bar {\mathsf {k}}}) \le \unicode{x3c1} ({\tilde {X}}^+_{\bar {\mathsf {k}}}) - 1 = \unicode{x3c1} ({\tilde {X}}_{\bar {\mathsf {k}}}) - 1 = \unicode{x3c1} (X_{\bar {\mathsf {k}}}) = 2$ , hence

(5.2.3) $$ \begin{align} \unicode{x3c1}(X^+_{\bar{\mathsf{k}}}) \le 2. \end{align} $$

On the other hand, in the case $\mathrm {F}_1(X,x) \ne \varnothing $ , the varieties ${\tilde {X}}$ and ${\tilde {X}}^+$ are not smooth and, arguing as in the proof of [Reference Kuznetsov and Prokhorov10, Theorem 5.9], we obtain

(5.2.4) $$ \begin{align} \operatorname{{\mathrm{rk}}} \operatorname{{\mathrm{Cl}}}({\tilde{X}}^+) = 2, \qquad \operatorname{{\mathrm{rk}}} \operatorname{{\mathrm{Cl}}}({\tilde{X}}^+_{\bar{\mathsf{k}}}) = 5 - m. \end{align} $$

Since $\phi $ and $\phi _+$ are crepant morphisms, the projection formula implies that any triple intersection product of divisor classes on ${\tilde {X}}^+$ which includes $K_{{\tilde {X}}^+}$ is equal to the analogous triple product on ${\tilde {X}}$ , so using (5.1.2), we compute (recall that $g = {\mathrm {g}}(X) = 15$ )

(5.2.5) $$ \begin{align} (-K_{\tilde X^+})^3 &= 2(g - m - 3) {} = 24 - 2m,\nonumber\\ (-K_{\tilde X^+})^2\cdot E^+ &= 4,\\ (-K_{\tilde X^+})\cdot (E^+)^2 &= -2.\nonumber \end{align} $$

On the other hand, by Lemma 5.4 and primitivity of $H^+ - (m + 1)E^+$ , we have

(5.2.6) $$ \begin{align} H^+ - (m + 1)E^+ = \sigma_+^*A^+, \end{align} $$

where $A^+$ is the ample generator of the Picard group of $X^+$ . We have

(5.2.7) $$ \begin{align} \begin{aligned}(\sigma_+^*A^+)^2 \cdot(-K_{{\tilde{X}}^+}) &= (H^+ - (m + 1)E^+)^2\cdot (-K_{X^+})\\&= 2(g - m - 8) = 14 - 2m> 0, \end{aligned}\end{align} $$

therefore, $\sigma _+$ is not a del Pezzo fibration. Similarly, if $\sigma _+$ is a conic bundle, it follows that

$$ \begin{align*} (A^+)^2 = 7 - m, \end{align*} $$

hence, $X^+$ is a smooth quintic or sextic del Pezzo surface, which of course contradicts the inequality (5.2.3). Therefore, the morphism $\sigma _+$ is birational.

By Lemma 5.4, the morphism $\sigma _+$ contracts the strict transform $R^+$ of the divisor swept by curves C, that is, the strict transform of the divisor $R_x \subset X$ if $\mathrm {F}_1(X,x) = \varnothing $ , or of the divisor $R_{\Theta (x)}$ otherwise. In both cases, Lemmas 2.11 and 2.12 show that $R^+$ has over ${\bar {\mathsf {k}}}$ two irreducible components swapped by the Galois group. Therefore, it follows from (5.2.4) that

$$ \begin{align*} \operatorname{{\mathrm{rk}}} \operatorname{{\mathrm{Cl}}}(X^+) = 1, \qquad \operatorname{{\mathrm{rk}}} \operatorname{{\mathrm{Cl}}}(X^+_{\bar{\mathsf{k}}}) = 3 - m, \end{align*} $$

and $\sigma _+$ is the blowup of two ${\bar {\mathsf {k}}}$ -curves or two ${\bar {\mathsf {k}}}$ -points. Furthermore, by Lemmas 2.11 and 2.12, we have

$$ \begin{align*} R^+ \sim H^+ - (m + 2)E^+. \end{align*} $$

Denoting by $i_+$ the index of $X^+$ and by $a_+$ the discrepancy of the exceptional divisor $R^+$ of $\sigma _+$ , and computing the anticanonical class of ${\tilde {X}}^+$ in two ways, we obtain the equality

$$ \begin{align*} H^+ - mE^+ = i_+(H^+ - (m+1)E^+) - a_+(H^+ - (m+2)E^+). \end{align*} $$

Solving this equation, we obtain $i_+ = 2$ and $a_+ = 1$ . Thus, $X^+$ is a Fano threefold of index $2$ and $\sigma _+$ is either the blowup of a ${\mathsf {k}}$ -irreducible curve $B^+$ with two ${\bar {\mathsf {k}}}$ -components or of two rational double ${\bar {\mathsf {k}}}$ -points on $X^+$ swapped by the Galois action. Moreover, using the equality from (5.2.7), we obtain

$$ \begin{align*} 14 - 2m= (\sigma_+^*A^+)^2\cdot (-K_{{\tilde{X}}^+}) = (\sigma_+^*A^+)^2 \cdot (2\sigma_+^*A^+ - R^+) = 2(\sigma_+^*A^+)^3, \end{align*} $$

hence, $X^+$ is a quintic or sextic del Pezzo threefold, respectively. Finally, if $X^+$ is singular, its class group $\operatorname {{\mathrm {Cl}}}(X^+)$ has rank greater than $1$ (see [Reference Prokhorov20, Theorem 1.7]), which contradicts to the equality $\operatorname {{\mathrm {rk}}}\operatorname {{\mathrm {Cl}}}(X^+) = 1$ obtained above. Thus, $X^+$ is smooth and $\sigma _+$ is the blowup of a curve $B^+$ , such that $B^+_{{\bar {\mathsf {k}}}}$ has two irreducible ${\bar {\mathsf {k}}}$ -components swapped by the Galois group.

To finally compute the degree of $B^+$ , recall that $R^+$ is the exceptional divisor of $\sigma _+$ . Note that on the one hand, equalities (5.2.5) and (5.2.6) imply that

$$ \begin{align*} (\sigma_+^*A^+) \cdot (-K_{{\tilde{X}}^+})^2 = 2g - 2m - 10 = 20 - 2m, \end{align*} $$

and on the other hand, this expression is equal to

$$ \begin{align*} (\sigma_+^*A^+) \cdot (2\sigma_+^*A^+ - R^+)^2 = 4(\sigma_+^*A^+)^3 + (\sigma_+^*A^+) \cdot (R^+)^2 = 4(7 - m)- \deg(B^+) \end{align*} $$

(where the degree is computed with respect to $A^+$ ). Thus,

$$ \begin{align*} \deg(B^+) = 8 - 2m, \end{align*} $$

hence, $B^+$ is a quartic or sextic curve with two connected ${\bar {\mathsf {k}}}$ -components (swapped by the Galois action), that is, a union of two conics or two cubic curves.

Proof. Let $x \in X({\mathsf {k}})$ be a ${\mathsf {k}}$ -point. First, assume that $\mathrm {F}_1(X,x) = \varnothing $ . Then, by Theorem 5.1 (i), the variety X is birational to a smooth quintic del Pezzo threefold $X^+$ . But $X^+$ is ${\mathsf {k}}$ -rational by [Reference Kuznetsov and Prokhorov10, Theorem 3.3], hence, so is X.

Now, assume that $\mathrm {F}_1(X,x) \ne \varnothing $ . Then, by Theorem 5.1 (ii), the variety X is birational to a smooth Fano threefold $X^+$ of type $\mathsf {X}_{(2,2)}$ . Moreover, by the Nishimura lemma, we have $X^+({\mathsf {k}}) \ne \varnothing $ . Therefore, $X^+$ is ${\mathsf {k}}$ -rational by Proposition 4.1, hence, so is X.

Proof. The discriminant divisor in $\mathbb {P}(V_1^{\vee } \otimes V_2^{\vee })$ , that is, the divisor parameterising degenerate bilinear forms, is a quartic hypersurface, hence, $\Gamma $ is a quartic curve or the entire plane. To prove that $\Gamma $ is a smooth curve, we can work over ${\mathsf {k}}'$ , and it is enough to show that the tangent space to $\Gamma $ at any point is one-dimensional. Assume to the contrary, that the tangent space at a point $[a] \in \mathbb {P}(A)$ is two-dimensional; then

1. (i) either the bilinear form $a(-,-) \in V_1^{\vee } \otimes V_2^{\vee }$ has corank at least $2$ ,

2. (ii) or a has corank $1$ , and if the vectors $v_1 \in V_1$ and $v_2 \in V_2$ generate its left and right kernels, respectively, the point $([v_1],[v_2]) \in \mathbb {P}(V_1) \times \mathbb {P}(V_2)$ belongs to $X_{{\mathsf {k}}'}$ .

In case (i), if $K_1 \subset V_1$ and $K_2 \subset V_2$ are the left and right kernels of a (they have dimension $\ge 2$ by assumption), then the form a vanishes on $\mathbb {P}(K_1) \times \mathbb {P}(K_2)$ , hence, the intersection $\big (\mathbb {P}(K_1) \times \mathbb {P}(K_2)\big ) \cap X_{{\mathsf {k}}'}$ is a codimension-2 linear section of $\mathbb {P}(K_1) \times \mathbb {P}(K_2)$ , hence, it is nonempty. Therefore, in case (i), similarly to the case (ii), there is a point $([v_1],[v_2]) \in X_{{\mathsf {k}}'}$ , such that $v_1$ and $v_2$ belong to the left and right kernels of some a. Then, the hyperplane section of $\mathbb {P}(V_1) \times \mathbb {P}(V_2)$ by the hyperplane corresponding to a is singular at $([v_1],[v_2])$ , hence, $X_{{\mathsf {k}}'}$ is also singular at this point.

Proof. The fibre of the projection $\pi _1$ over a point $[v_1] \in \mathbb {P}(V_1)$ is the intersection of the projectivisations of the orthogonals of $v_1$ with respect to all bilinear forms $a \in A$ . Therefore, it has positive dimension if and only if $v_1$ belongs to the left kernel of one of the forms. Furthermore, if $v_1$ belongs to the left kernel of two distinct forms in A, the fibre of $\pi _1$ over $[v_1]$ contains a plane, which contradicts Corollary 2.6. This means that the morphism

$$ \begin{align*} \gamma_i \colon \Gamma_{{\mathsf{k}}'} \longrightarrow \mathbb{P}(V_i), \qquad a \longmapsto \operatorname{{\mathrm{Ker}}}_i(a), \end{align*} $$

where $\operatorname {{\mathrm {Ker}}}_1$ and $\operatorname {{\mathrm {Ker}}}_2$ denote the left and right kernels of the bilinear form a, respectively, is an isomorphism $\Gamma _{{\mathsf {k}}'} \to \Gamma _i$ .