Proof. By [Reference Ramanathan14, Theorem 3.11] the ring of global sections $\bigoplus _{k \in \mathbb {N}} H^0(X, {\mathcal O}_X(kM))$ is generated in degree one, so M is very ample and defines an embedding

$$ \begin{align*}i : X \hookrightarrow \mathbb{P}^N \end{align*} $$

such that $M = X \cap H$ with $H \subset \mathbb {P}^N$ a hyperplane. By [Reference Ramanathan14, Theorem 3.11(i), Remark 3.12] the variety $X \subset \mathbb {P}^N$ is an intersection of quadrics, so we can apply [Reference Beauville and Mérindol1, corollaire 2, remarque (3)] to get the statement. For the convenience of the reader, let us sketch their remarkable argument: note first that the splitting map induces a map . Since $M \in |{\mathcal O}_X(1)|,$ the splitting of the normal sequence is thus given by a non-zero morphism . Consider the exact sequence

The section j induces a section such that $u(s)=0$ . Now consider the twisted Euler sequence restricted to M:

Since $\dim M \geq 2,$ we have by Kodaira’s theorem $H^0(M, {\mathcal O}_M(-1))=H^1(M, {\mathcal O}_M(-1))=0$ . Thus, the restriction map

is an isomorphism. Repeating the argument for the twisted Euler sequence, we obtain an isomorphism

Thus, up to linear coordinate change, the section s is the restriction of an Euler vector field $l(x) \frac {\partial }{\partial X_0}$ with l a linear form.

Step 1. We claim that if $x \in M$ , then the projective tangent space $\mathbb {P}(T_{X,x}) \subset \mathbb {P}^N$ passes through the point $p:= (1:0: \dots :0)$ .

Let $V \subset \mathbb {P}^N$ be the vanishing set of the Euler vector field, that is, the union of the hyperplane $l=0$ and a point. Note that $M \not \subset V$ since otherwise $s=0$ . In order to show the claim, we can assume without loss of generality that $x \in (M \setminus V)$ , the general case will just follow by passing to the limit.

We can choose homogeneous polynomials $F_1, \ldots , F_k$ vanishing on X such that ${\mathbb {P}(T_{X,x}) = \cap _{j=1}^k \mathbb {P}(T_{V(F_j), x}),}$ where $\mathbb {P}(T_{V(F_j), x}) \subset \mathbb {P}^N$ is the hyperplane defined by the equation $\sum _{i=0}^N \frac {\partial F_j}{\partial X_i}(x) X_i=0$ . It is clearly sufficient to show that

$$ \begin{align*}p \in \mathbb{P}(T_{V(F_j), x}). \end{align*} $$

Yet this is equivalent to showing that $\frac {\partial F_j}{\partial X_0}(x)=0$ . Since $x \not \in V$ , the vector field $l(x) \frac {\partial }{\partial X_0}$ does not vanish in a neighborhood of x. Yet the condition $u(s)=0$ is equivalent , so the derivation $\frac {\partial F_j}{\partial X_0}(x)$ vanishes.

Step 2. We have $p \not \in M$ . We argue by contradiction. By Step 1, we have $p \in \mathbb {P}(T_{X,x})$ for every $x \in M$ . Since $M = X \cap H$ is smooth, we have $\mathbb {P}(T_{M,x}) = \mathbb {P}(T_{X,x}) \cap H$ for every $x \in M$ . Thus, we have $p \in \mathbb {P}(T_{M,x})$ for every $x \in M$ . Take now $H_1, \ldots , H_{\dim M-1}$ general hyperplane sections passing through p. Then, $C:= M \cap H_1 \cap \cdots H_{\dim M-1}$ is a smooth curve such that $p \in \mathbb {P}(T_{C,x})$ for every $x \in C$ . By [Reference Hartshorne6, Chapter IV, Proposition 3.9] the curve C is a line and therefore $(X, {\mathcal O}_X(M)) \simeq (\mathbb {P}^n, {\mathcal O}_{\mathbb {P}^n}(1))$ , a contradiction to our assumption.

Step 3. Conclusion. Let $\sigma :\mathbb {P}^N \rightarrow \mathbb {P}^N$ be the involution fixing pointwise the point p and the hyperplane H. Let $Q \subset \mathbb {P}^N$ be a quadric that contains X, then $\sigma (Q)=Q$ by [Reference Beauville and Mérindol1, Corollary 1]. Since X is an intersection of quadrics, we obtain that $\sigma (X)=X$ and therefore the sought involution is $\sigma _X:=\sigma \vert _X$ . We are left to show that the fixed locus of $\sigma _X$ is M, that is, we have to show that $p \not \in X$ . Yet again by [Reference Beauville and Mérindol1, Corollary 1] a quadric Q with $X \subset Q$ and $p \in Q$ is singular in p. Thus, X would be an intersection of quadric cones with vertex the point p, so itself a cone. Yet X is smooth, so we get again $(X, {\mathcal O}_X(M)) \simeq (\mathbb {P}^n, {\mathcal O}_{\mathbb {P}^n}(1))$ , the final contradiction.

Proof. By [Reference Hwang and Mok7, Main Theorem] the manifold Y is isomorphic to $\mathbb P^n$ . The ramification divisor is a rational homogeneous space because it is normal split in X. Thus, the branch locus $B \simeq R \subseteq \mathbb P^n$ is homogeneous and therefore of degree at most two. Moreover, B is divisible by 2 in , so B is a smooth quadric in $\mathbb P^n$ . The covering $\tau $ is thus induced by the data , where H is the hyperplane class. The ramification formula $K_X=\tau ^*(-(n+1)H+H)$ implies that $-K_X$ is divisible by n. By the Kobayashi–Ochiai criterion, this shows that X is a quadric.

Proof. If $X \simeq \mathbb {P}^n$ , this is van den Ven’s theorem, so assume $X \not \simeq \mathbb {P}^n$ . By Proposition 3.2, there is an involution $\sigma \colon X\to X$ whose fixed locus is M. By Corollary 3.3, the variety X is a smooth quadric and M is a hyperplane section.

Proof. Without any assumption on the dimension, observe first that $K_M= (K_X+M)|_M$ is not nef since the normal split divisor $M \subset X$ is rational homogeneous [Reference Floris and Höring5]. Applying the cone theorem to the Fano manifold $X,$ we obtain that there exists an extremal ray $\Gamma $ such that $(K_X+M) \cdot \Gamma <0$ , denote the corresponding contraction by $f: X \rightarrow Y$ .

1st case. Assume that $\dim X-\dim Y=1$ . Then, the general fiber F is $\mathbb {P}^1$ , so $K_X \cdot F=-2$ and $M \cdot F>0$ implies that $M \cdot F=1$ . Thus, M is a rational section of f. Since M is homogeneous, the birational morphism $M \rightarrow Y$ is an isomorphism, hence, M is even a section. By [Reference Kollár10, Appendix, Lemma (1)] this implies that $\dim X \leq 2$ . Thus, we have $\dim X=2$ and M is a section of the ruling f.

2nd case. Assume that $\dim X-\dim Y>1$ . Note that this implies $\dim X>2$ since otherwise we would have $\rho (X)=1$ . Let F be a general fiber of F. Since $(F \cap M) \subset F$ is an ample divisor of dimension at least one, it is connected. Thus, f induces a fibration $f|_M:M \rightarrow Z$ . Since M is rational homogeneous, the fibration $f|_M$ has maximal rank. Thus, a theorem of Sommese [Reference Sommese15, Proposition V] implies that $\dim X \geq 2 \dim Y$ .

Proof. Since X is rational homogeneous, the fibration f is smooth. Thus, by [Reference Codogni, Fanelli, Svaldi and Tasin2, Theorem 2.2] the local system is a local system on Z with finite monodromy. Since Z is simply connected, it is a trivial local system. The statement follows.

Proof. Fix a $k \in \mathbb {N}$ such that $\sigma ^k = \mathrm {id}$ .

We start with (1). Let R be an extremal ray in , then by assumption $\sigma _*(R)=R$ . Choose $0 \neq v\in R$ , then v is an eigenvector with eigenvalue $\lambda \in \mathbb {R}$ . Since $\sigma ^k=id$ , we have $\sigma _*v=\pm v $ . Yet , so we have $\sigma _*v=v $ . The extremal rays of span $N_1(X)_{\mathbb R}$ , therefore, $\sigma _*$ is the identity.

As for (2), we first prove that $\tau ^*\colon N^1(Y)_{\mathbb R}\to N^1(X)_{\mathbb R}$ is an isomorphism. The map $\tau ^*$ is always injective. To show that it is surjective, let $[D]\in N^1(X)_{\mathbb R}$ . Since $\sigma _*$ and therefore $\sigma ^*$ is the identity, we have

$$ \begin{align*}[kD]=\left(\sum_0^{k-1} (\sigma^i)^*[D]\right)=\tau^*\tau_*[D], \end{align*} $$

proving that $D=\tau ^*\tau _*[1/k D]$ .

Moreover, $\tau ^*$ induces an isomorphism between the nef cones of X and Y. Passing to the dual cones, we get the claim.

Proof. The fibrations $f_i$ are $\mbox {Aut}^{\circ }(X)$ -equivariant by Blanchard’s lemma, so it is clear that f is also equivariant. For every point $(z_1, z_2) \in f(X)$ , we have $f^{-1} ((z_1, z_2))= f_1^{-1} (z_1) \cap f_2^{-1} (z_2)$ . Since the extremal rays are distinct, their fibers intersect in at most finitely many points. This implies that $f\colon X \to f(X)$ is finite. Since $\mbox {Aut}^{\circ }(X)$ acts transitively on $X,$ the equivariance of f implies that $f(X)$ is smooth and $f: X \rightarrow f(X)$ is étale. Since $f(X)$ is rationally connected and smooth, it is simply connected. Thus, the étale map $f: X \rightarrow f(X)$ is an isomorphism.

Proof. The Mori cone of a Fano manifold is generated by its extremal rays and $\sigma _*$ acts by permutation on them. If the permutation is trivial, then $\sigma _*$ is the identity by Lemma 3.7(1).

Thus, we can assume that there exists an extremal ray $\Gamma _1$ such that $\Gamma _2 := \sigma _* \Gamma _1$ is distinct from $\Gamma _1$ . Let $f_i\colon X\to Z_i$ for $i=1,2$ be the two $Aut^{\circ }(X)$ -equivariant fibrations induced by the two extremal rays of $NE(X)$ . By Lemma 3.9, the map

$$ \begin{align*}f:=f_1 \times f_2:X \rightarrow Z_1 \times Z_2 \end{align*} $$

is an isomorphism onto its image.

By the rigidity lemma, there is an isomorphism $\sigma _Z$ making the following diagram commute:

Consider now the involution

$$ \begin{align*}\begin{array}{rcl} i\,{:}\, Z_1\times Z_2&\to &Z_1\times Z_2\\ (z_1,z_2)&\mapsto&(\sigma_Z^{-1}z_2,\sigma_Z z_1). \end{array} \end{align*} $$

Let f be the embedding of Lemma 3.9. The image $f(M) \simeq M \subset Z_1 \times Z_2$ is contained in the fixed locus of i.

Moreover, i is conjugated by the isomorphism $(id,\sigma _Z^{-1})\colon Z_1\times Z_2\to Z_1\times Z_1$ to the involution

$$ \begin{align*}\begin{array}{l@{\,}rcl} \iota: & Z_1\times Z_1&\to &Z_1\times Z_1\\ & (z_1,z_2)&\mapsto&(z_2,z_1). \end{array} \end{align*} $$

The fixed locus of $\iota $ is the diagonal $\Delta \simeq Z_1$ , so we obtain that $\dim M \leq \dim Z_1 \leq \dim X-1$ . Since M is a divisor in $X,$ we have equality and $M \simeq Z_1$ . In particular, we have $\rho (M)=\rho (Z_1)=\rho (X)-1$ .

By the Lefschetz hyperplane theorem, the restriction map $N^1(X) \rightarrow N^1(M)$ is injective if $\dim X \geq 3$ . Thus, we obtain $\dim X \leq 2$ and therefore $X\cong \mathbb {P}^1\times \mathbb {P}^1$ (note that the case of a projective space is excluded by the condition that $\sigma _*$ is not the identity).

Proof. Since $n \geq 4,$ we know by Remark 3.8 that $\rho (X)=\rho (Y)$ .

Since M is rational homogeneous, it is clear that $-K_M=(-K_X-M)|_M$ is ample, but it is not so clear for $-K_X-M$ itself. Arguing by contradiction, we set $0 < \lambda \leq 1$ such that $-K_X - \lambda M$ is a nef, but non-ample divisor class. Note that in general $\lambda \in \mathbb {R}$ , but since the Mori cone of the Fano manifold X is rational polyhedral, we have $\lambda \in \mathbb {Q}$ . Moreover, by the cone theorem, there exists an extremal ray $\Gamma $ such that $(-K_X-\lambda M) \cdot \Gamma =0$ , and we denote by

$$ \begin{align*}f: X \rightarrow T \end{align*} $$

its contraction. Since X is rational homogeneous, the fibration is of fiber type and a locally trivial fibration with general fiber F with $\rho (F)=1$ (by Lemma 3.6).

Since $\sigma _*$ acts as the identity on $N_1(X)$ , the image of the extremal ray $\tau _* \Gamma $ is an extremal ray in $NE(Y)$ by Lemma 3.7(2), and we denote its contraction by

$$ \begin{align*}g: Y \rightarrow U. \end{align*} $$

By the rigidity lemma, we have an induced morphism $\bar \tau :T \rightarrow U$ such that

commutes. Since $\rho (X)=\rho (Y),$ we have

$$ \begin{align*}\rho(T)=\rho(X)-1=\rho(Y)-1=\rho(U). \end{align*} $$

Thus, the morphism $\bar \tau $ is finite, and we claim that it is actually an isomorphism. We argue by contradiction. Since $\tau $ is two-to-one, it is clear that $\bar \tau $ is then also two-to-one. In what follows, we denote by $G=g^{-1} (u)$ a general g-fiber.

1st case. Assume that the fibers of f and g have dimension one. In this case $G \simeq \mathbb {P}^1$ . Since $\bar \tau ^{-1} (u)$ consists of two points, the preimage $\tau ^{-1} (G)$ has two connected components $F_1 \cup F_2$ mapping isomorphically onto G. Yet $-K_X \cdot G=2=-K_Y \cdot F_i$ since they are general fibers of the fibrations. Moreover, by the ramification formula, $-K_X = \tau ^*(-K_Y) - M$ . Since $\tau |_{F_i}$ is an isomorphism, we obtain $M \cdot F_i=0$ , a contradiction to the ampleness of M.

2nd case. Assume that the fibers of f and g have dimension at least two. In this case, the restriction of g to $M \subset Y$ has connected fibers, since the intersection $M \cap G \subset G$ is an ample divisor. Since M is rational homogeneous (something that we do not know about Y), we obtain that $g|_M: M \rightarrow U$ is smooth and U is smooth and rational homogeneous by Blanchard’s lemma.

The morphism $\bar \tau \colon T\to U$ is a finite morphism of degree two between rational homogeneous, and in particular Fano, varieties. Therefore, it cannot be étale.

Since $\bar \tau $ is of degree two and not étale, there exists by purity of branch, a prime divisor $B \subset U$ such that $\bar \tau ^* B$ is not reduced. Thus, $f^* \bar \tau ^* B$ has a non-reduced irreducible component. Since g is an elementary Mori contraction, the preimage $g^{-1} B$ is irreducible. As a consequence of the Graber–Harris–Starr theorem, the irreducible divisor $g^* B$ is reduced. Yet $\tau ^* g^* B = f^* \bar \tau ^* B$ is not reduced, so $g^* B$ is an irreducible component of the branch locus of $\tau $ . Yet $\tau $ ramifies exactly along the ample divisor M, a contradiction.

This finishes the proof of the claim that $\bar \tau $ is an isomorphism.

Thus, for any point $u \in U$ , we have an induced degree two cover

$$ \begin{align*}\tau_F: F := f^{-1} (\bar \tau^{-1}(u)) \rightarrow G := g^{-1} (u) \end{align*} $$

that ramifies along $F \cap M$ . Choose $u \in U$ general so that G is smooth. The fiber F is rational homogeneous by Example 2.3. Since $\rho (F)=1,$ we can apply Corollary 3.3 to obtain that F is a quadric $Q^d$ and $G \simeq \mathbb {P}^d$ . Moreover, the divisor $(F \cap M) \subset F$ is a hyperplane section, that is, an element of $|{\mathcal O}_Q(1)|$ .

If $\dim F \geq 2,$ this gives an immediate contradiction: by assumption, $K_F+\lambda M_F$ is trivial. Yet for a quadric $-K_Q \simeq {\mathcal O}_Q(\dim Q)$ , so we obtain $\lambda = \dim Q>1$ , a contradiction.

Thus, we have $\dim F=1$ , in this case, the ramification formula yields

$$ \begin{align*}2 = -K_X \cdot F = \tau^* (-K_Y) \cdot F - M \cdot F = -K_Y \cdot 2G - M \cdot F = 4 - M \cdot F, \end{align*} $$

and therefore $M \cdot F=2$ . Since $X \rightarrow T$ is a $\mathbb {P}^1$ -bundle and and $M \rightarrow T$ is finite (since M is rational homogeneous), a result of Kollár [Reference Kollár10, Appendix, Lemma (2)] implies $\dim T \leq 2$ , the final contradiction.

Proof. Assume by contradiction that $\rho (X)\geq 2$ . Let $\Gamma $ be any extremal ray in , and let $f: X \rightarrow T$ be its contraction.

1st case. Assume first that $\dim X-\dim T=1$ . By Lemma 3.11, we have ${(-K_X-M) \cdot F>0}$ , which implies that

$$ \begin{align*}2 = -K_X \cdot F> M \cdot F \geq 1. \end{align*} $$

Thus, $M \cdot F=1$ , and M being rational homogeneous by example 2.1 and by [Reference Floris and Höring5], it is a section of f. Yet M is ample, so Kollár’s result [Reference Kollár10, Appendix, Lemma (1)] gives $\dim T \leq 1$ , a contradiction.

2nd case. Assume now that $\dim X-\dim T>1$ . Applying Proposition 3.5, we see that we even have $\dim X \geq 2 \dim T$ . Since $\rho (X) \geq 2,$ there are at least two elementary contractions $f_1$ and $f_2$ , and their general fibers $F_1$ and $F_2$ intersect in at most finitely many points. By the first case and the estimate above implies that

$$ \begin{align*}\dim F_1 = \dim F_2 = \frac{1}{2} \dim X. \end{align*} $$

In particular, the morphism $(f_1,f_2)\colon X\to T_1\times T_2$ is surjective and therefore an isomorphism by Lemma 3.9.

Since $\sigma _*$ is the identity, for $i=1,2,$ there are involutions $\sigma _i$ for $i=1,2$ and a commutative diagram

But this implies that the fixed locus of $\sigma $ is contained in $p_1^{-1}Fix(\sigma _1)\cap p_2^{-1}Fix(\sigma _2) $ , a contradiction because the fixed locus of $\sigma $ is a non-empty ample divisor.

Proof. Assume by contradiction that $\rho (X)\geq 2$ and let $f_i\colon X\to T_i$ for $i=1,2$ be two extremal contractions. Since the fibers of $f_1$ and $f_2$ intersect in finitely many points, there are, up to renumbering $f_1$ and $f_2$ , two possibilities: either $\dim T_1=1$ and $\dim T_2=2$ , or $\dim T_1=\dim T_2=2$ .

Since $\rho (X)\geq 2$ , by Proposition 3.2, there is an involution $\sigma $ whose fixed locus is M. We denote by Y the quotient of X by the action of $\sigma $ , and by $\tau \colon X\to Y$ the degree two morphism ramified along M. Moreover, by [Reference Floris and Höring5] the divisor M is rational homogeneous.

1st case: Assume that $\dim T_1=1$ and $\dim T_2=2$ . Since X has an elementary Mori contraction onto a curve, we have $\rho (X)=2$ . Thus, Lemma 3.9 implies that $X \simeq \mathbb {P}^1 \times \mathbb {P}^2$ . Since the fibers of $f_1$ and $f_2$ have different dimensions, the involution $\sigma _*$ acts as the identity and by Lemma 3.7, we have $\rho (Y)=\rho (X)=2$ .

The ample divisor M is an element of $|{\mathcal O}_{\mathbb {P}^1 \times \mathbb {P}^2}(a,b)|$ for some $a \geq 1, b \geq 1$ . Since M is rational homogeneous, the induced fibration

$$ \begin{align*}f_1: M \rightarrow \mathbb{P}^1 \end{align*} $$

must be a $\mathbb {P}^1$ -bundle structure. In particular, by the adjunction formula, we have $b \leq 2$ . Since $K_M^2=8$ for every Hirzebruch surface, a short intersection computation allows to exclude the case $b=2$ . Thus, we have $b=1$ and it is not difficult to see that a general element in ${\mathcal O}_X(a,1)$ with $a \in 2 \mathbb {N}$ is indeed isomorphic to $\mathbb {P}^1 \times \mathbb {P}^1$ . We claim that such an $M \subset X$ is never normal split and argue by contradiction. Recall that there is a degree two cover

$$ \begin{align*}\tau:\mathbb{P}^1 \times \mathbb{P}^2 \rightarrow Y \end{align*} $$

that ramifies along M. By the ramification formula,

$$ \begin{align*}\tau^* (-K_Y) \simeq -K_X + M \end{align*} $$

is ample, so Y is Fano. Moreover, we compute

$$ \begin{align*}(-K_Y)^3 = \frac{1}{2} (-K_X+M)^3 = 8 (2+a). \end{align*} $$

Since $(-K_Y)^3 \leq 64$ for any smooth Fano threefold [Reference Mori and Mukai13, Corollary 11], we see that $a=2$ or $a=4$ .

If $a=2,$ we have $(-K_Y)^3=32$ , so by [Reference Mori and Mukai13, Table 2] the threefold Y is a blow-up of $\mathbb {P}^3$ . Yet we know that the Mori contractions of Y are fiber spaces, a contradiction.

If $a=4,$ we have $(-K_Y)^3=48$ , so by [Reference Mori and Mukai13, Table 2] the threefold Y is the flag manifold $\mathbb {P}(T_{\mathbb {P}^2})$ . Yet for this threefold, both Mori contractions have one-dimensional fibers, a contradiction to $\dim T_1=1$ .

2nd case: Assume that $\dim T_1=\dim T_2=2$ . The surfaces $T_1$ and $T_2$ are rational homogeneous of the same Picard rank. Since M is a rational homogeneous surface, we have $\rho (M)\leq 2$ . We get by the Lefschetz theorem,

$$ \begin{align*}\rho(T_i)=\rho(X)-1 \leq \rho(M)-1\leq 1. \end{align*} $$

Thus, $T_1\cong T_2\cong \mathbb {P}^2$ . Now, the Mori–Mukai classification (or Lemma 3.9) implies that $X \simeq \mathbb {P}(T_{\mathbb {P}^2}) \subset \mathbb {P}^2 \times \mathbb {P}^2$ .

Since $M \subset X$ is ample, we have $M \in |{\mathcal O}_X(a,b)|$ with $a \geq 1, b \geq 1$ . Again, recall that there is a degree two morphism $\tau \colon X \rightarrow Y$ ramified along M. The threefold Y is Fano and

$$ \begin{align*}(-K_Y)^3 = \frac{1}{2} (-K_X+M)^3 = \frac{3}{2} (2+a)(2+b)(4+a+b). \end{align*} $$

Since $a \geq 1, b \geq 1,$ we get $(-K_Y)^3 \geq 81$ , a contradiction.