The Campana–Peternell conjecture in dimension six

1. Separate the known Picard-number range

Projectivity gives $\rho(X)\ge1$. If $\rho(X)\ge2$, then $\rho(X)>6-5$, so Kanemitsu's classification applies. The manuscript uses this result directly without reclassifying contractions and fibres on this branch. It remains to consider $\rho(X)=1$.

[Kan16, Theorem 0.2, p. 1]

2. Reduce the pseudoindex using the minimal family

Choose a family $V$ of curves of least anticanonical degree $\iota$, with universal family $\pi:U\to V$ and evaluation $e:U\to X$. Nefness makes every curve free, and minimality of the degree prevents splitting. The known minimal-family theorem gives smooth projective $U,V$, a smooth $\mathbb P^1$-fibration $\pi$, and a smooth surjective evaluation of relative dimension $\iota-2$ with connected fibres. Proposition 2.1 in the manuscript further proves, for every $\pi$-fibre $C$,

Differentiating the map-space construction makes $H^0(T_U|_C)\to H^0(e^*T_X|_C)$ surjective. The two spaces have the same dimension, so the kernel bundle has vanishing $H^0$ and $H^1$. The splitting holds along every fibre, a fact needed later for descent of a bundle.

Proposition 2.1 · pp. 5–6 [Kan17, Proposition 1.6, p. 4]

The differential $\mathcal O(2)\to\varphi^*T_X$ gives $\iota\ge2$. If $\iota=2$, evaluation is an isomorphism and the resulting fibration $X\to V$ produces a nonzero nef divisor of degree zero on its curve fibres, contradicting $\rho(X)=1$. If $\iota=3$, the finite tangent map gives a one-dimensional VMRT; the Mok–Hwang classification only allows dimensions two, three and five, excluding six. If $\iota\ge6$, every rational curve $B$ satisfies $-K_X\cdot B\ge6$, and Dedieu–Höring gives $X\simeq\mathbb P^6$ or $Q^6$.

Proposition 2.2 · p. 6 [Mok08s, Theorem 15, print p. 199 / PDF p. 50] [DH17, Theorem 1.3, p. 1]

3. Join the two remaining branches

Proposition 3.1 handles $\iota=4$, and Proposition 6.2 handles $\iota=5$. They give rational homogeneity and $X\simeq\operatorname{Gr}(2,5)$, respectively. Once the pseudoindex-four argument leaves a $\mathbb P^2$-fibration as evaluation, the external Proposition 4.1 identifies $X$ as $\mathbb P^3$ or $\operatorname{LG}(3,6)$; dimension six selects the latter. This refinement adds no hypothesis to the main theorem. Every Picard-number and pseudoindex branch is now connected to the conclusion.

Proposition 3.1 · pp. 8–10 Proposition 6.2 · pp. 19–20 [Kan17, Proposition 4.1; Remark 4.2, p. 12] §7, proof of Theorem 1.1 · p. 20

1. Fix the factorization and relative degrees

Kanemitsu's results give either rational homogeneity or a factorization $e=g\circ f:U\to W\to X$, where $f,g,\pi$ are smooth $\mathbb P^1$-fibrations. The latter case comes with a smooth elementary Mori contraction $q:W\to Y$, with $\dim W=7$. Put $H=c_1(T_X)/4$, $h=g^*H$, $c=c_1(T_g)$, and for a $\pi$-fibre $C$ set $a=c_1(T_f)\cdot C$, $b=c\cdot f(C)$. The splitting $T_e|_C=\mathcal O(-1)^{\oplus2}$ gives $a+b=-2$ and $a\le-1$. A positive multiple of $c-bh$ is the pullback of an ample class from $Y$, while $g_*c=2$, $g_*c^2=0$ give $b\le0$. To exclude $b=0$, choose a curve $D\subset X$ for which the surface $g^{-1}(D)$ has two-dimensional image under $q$, contradicting its zero intersection number. Thus $a=b=-1$.

[Kan17, Theorem 3.3, p. 9] [Kan17, Proposition 4.1; Remark 4.2, p. 12] Lemma 2.3 · p. 7 Proposition 3.1 · pp. 8–10

2. Narrow the possible fibres

Relative degree $-1$ reduces the fibre candidates of $q$ to $\mathbb P^2,Q^3,K(G_2)$. In the $\mathbb P^2$ case, the two Leray–Hirsch decompositions and Poincaré duality on $Y$ give $b_6(X)=b_4(X)-1$, contradicting Hard Lefschetz. In the remaining cases, $z=(h+c)/2$ descends from $Y$; since $\dim Y\le4$, we have $z^5=0$.

The cohomological involution for $g$ sends $z$ to $h-z$ and negates top-degree integrals. The even-index relations give $m_1=m_2=5m_4$ and $m_3=3m_4$. Thus $m_4>0$ and $\dim Y=4$: the $K(G_2)$ branch disappears and $q$ is a $Q^3$-fibration.

[Kan17, Remark 2.3, p. 5] Proposition 3.1 · pp. 8–10

3. Contradict integrality using the orthogonal bundle

Fischer–Grauert gives local triviality. Using $\operatorname{Aut}(Q^3)=\operatorname{PGO}_5\simeq\operatorname{SO}_5$, the manuscript realizes $W$ as a relative quadric for an orthogonal rank-five bundle with trivial determinant, algebraizes it by GAGA, and obtains $q_*(-K_q)^4=0$ by adjunction. On the other hand, write $-K_q=3h-\lambda z$. Degrees on a $g$-fibre imply $\lambda\in\mathbb Z$. The projection formula gives

Since $m_3=3m_4$, this forces $\lambda=9/4$, a contradiction. The factorization is excluded, leaving the known homogeneous case. The manuscript also cites Watanabe's treatment of pseudoindex four; the argument described here is the manuscript's own intersection-theoretic alternative.

[Poc25, Theorem 1.1 (Fischer–Grauert), p. 1] Proposition 3.1 · pp. 8–10

1. Extract linear relations from the two maps

For $\delta=\pi^*\pi_*$, the involution for $\pi$ satisfies $\bar\alpha=\alpha-c\delta\alpha$ and $\delta(\alpha\gamma)=(\delta\alpha)\gamma+\bar\alpha\delta\gamma$. The actual line bundle $e^*(-K_X)\otimes(T_\pi^*)^{\otimes2}$ has degree one along the fibres, so its tensor product with $E$ is trivial on every $\pi$-fibre. Descending that bundle shows that $r,s,t$ also come from $V$. Weight-nine products descending to $V$ therefore vanish, as do base classes of weight at least seven pulled back from $X$.

Using simple connectedness of $X$ and Deligne's fixed-part results, Lemma 4.3 establishes triviality of $R^je_*\Omega^p_{U/X}$. Relative GRR applied to $\mathcal O_U$ and $E^*$ then gives

Expanding these relations in a common weight yields $Av=0$. The final family consists of vanishing integrals; it need not consist of zero cohomology classes.

Lemma 2.3 · p. 7 Lemmas 4.1–4.2 · pp. 10–11 Lemma 4.3; (4.4) · p. 12 [Del71, Theorem 4.1.1(ii), print p. 40 / PDF p. 37] [Del71, Corollary 4.1.2, print p. 42 / PDF p. 39] [BS58, §7, Riemann–Roch theorem, print p. 113 / PDF p. 18]

2. The projection formula supplies the product structure

For invariant monomials $G_j$ of weights three and four, integration along $e$ lands in $H^0(X,\mathbb Q)$ and $H^2(X,\mathbb Q)=\mathbb Qh_X$. Each image is therefore determined by a scalar: $e_*G_j=D\eta_jh_X^{w_j-3}$, where $D=L^3>0$. Normalize the intersection numbers of the weight-six base monomials $M_i$ as $\xi_i$. The projection formula identifies the selected coordinates, divided by the single positive number $D\int_Xh_X^6$, with $\xi_i\eta_j$. This product structure, in addition to the linear equations, is what supplies a nonlinear constraint.

§§4.4–4.5; (4.11)–(4.19) · pp. 13–15

3. Combine modular ranks with a rational certificate

Multiply $N(\xi_i\eta_j)=0$ by $\eta_a\eta_b$, identifying repeated triples to form a linear solution space $K_3$. The calculation over $\mathbb F_{10007}$ gives kernel dimension six in 6820 coordinates, implying the rational upper bound $\dim K_3\le6$. Six rational columns satisfying every double-slice equation are then exhibited; a minor equal to $-16/5\ne0$ proves that they form a basis. The functional

vanishes on that basis. Returning to product coordinates and using $\xi_0=\eta_0=1$ gives $(u-4)(3u-8)=0$. The argument does not infer a rational identity merely from a finite-field computation. Appendix A supplies an executable certificate for the finite calculation.

§§5.1–5.3 · pp. 16–18 Appendix A · pp. 22–25

1. The tangent image spans the whole space

Suppose the affine cone over $\mathcal C_x=\tau_x(F)$ spans a proper subspace $W_x\subset T_xX$. Then $4\le\dim W_x\le5$. If the tangent bivectors $\alpha\wedge v$ do not span $\bigwedge^2W_x$, a nonzero alternating form annihilates them, and the four-dimensional tangent spaces to the cone are isotropic. The rank bound leaves only $\dim W_x=5$ and a form of rank two. But projection of the cone to the two-dimensional symplectic quotient has an irreducible conical image of dimension at most one, hence a line, contradicting the spanning property. Thus the bivectors span the whole exterior square.

One Hwang–Mok proposition now gives integrability of the spanned distribution; the other gives nonintegrability under $b_2(X)=1$ and linear degeneracy of the VMRT. This contradiction proves nondegeneracy of $\mathcal C_x\subset\mathbb P^5$. Smoothness of the VMRT has not yet been used.

Lemma 6.1 · p. 19 [HM99, Propositions 1.2.1–1.2.2, print pp. 356–357 / PDF pp. 7–8]

2. Sectional genus determines both degrees

The intersection $B$ of two general members of the ample globally generated system $|L|$ is a smooth connected curve: use Bertini and connectedness from Kodaira vanishing. Adjunction and the degree formula for a finite morphism give

The value $u=4$ would give negative genus. For $u=8/3$, the equation becomes $g(B)=1-L^3/3$, forcing $L^3=3$, $\deg(\tau_x)=1$, and $\deg(\mathcal C_x)=3$.

Proposition 6.2 · pp. 19–20 [EGHP06, Theorem 0.1, p. 2]

3. Use normality before recognizing the Segre image

The minimal-degree classification makes $\mathcal C_x$ a three-dimensional rational normal scroll of degree three, still possibly a cone. The manuscript realizes its coordinate ring as the invariant ring for a $\mathbb G_m$-action with weights $(-a_1,-a_2,-a_3,1,1)$ and proves normality. Only then does the finite birational map $\tau_x$ become an isomorphism, transferring smoothness of $F$ to the image. Excluding cones gives $a_i>0$, and $\sum a_i=3$ implies $(a_1,a_2,a_3)=(1,1,1)$. Thus $\mathcal C_x=\mathbb P^1\times\mathbb P^2\hookrightarrow\mathbb P^5$.

This is the projectivization of rank-one maps in $T_A\operatorname{Gr}(2,5)=\operatorname{Hom}(A,\mathbb C^5/A)$. The model $\operatorname{Gr}(2,5)$ is an irreducible compact Hermitian symmetric space of rank two. Picard number one and projective equivalence of the VMRT at a general point verify the hypotheses of Mok's Main Theorem, giving $X\simeq\operatorname{Gr}(2,5)$.

[EGHP06, Theorem 0.1, p. 2] Proposition 6.2 · pp. 19–20 [Mok08r, Main Theorem, §2.1, p. 10]

1. Classify fibres on a maximal-irregularity cover

Finite étale covers preserve the Kähler condition and analytic nefness. Surjectivity of their Albanese maps bounds irregularity by $\dim X$, so the integer maximum $q$ is attained. Demailly–Peternell–Schneider gives a smooth Albanese surjection on that cover with connected Fano fibres. Their tangent bundles are nef and their dimension is $d$. Use Theorem 1.1 for $d=6$, the known classification with $\rho(F)\ge1>d-5$ for $1\le d\le5$, and a point for $d=0$. If $X$ itself has dimension zero, connectedness makes it a point from the outset.

[DPS94, Proposition 3.9, p. 25] [DPS94, Theorem 3.14, p. 31] [Kan16, Theorem 0.2, p. 1] §7.2 · pp. 21–22

2. Pass through local rigidity to split the cover

The vanishing $H^1(F,T_F)=0$ gives local deformation rigidity, making the fibre type locally constant and therefore constant on connected $A$. Fischer–Grauert makes $\alpha$ holomorphically locally trivial. For $d=0$, $\alpha$ is itself an isomorphism. Since $\widetilde A=\mathbb C^q$ is Stein and contractible, Claudon–Höring–Kollár gives $\widetilde{X'}\simeq\widetilde F\times\mathbb C^q$. Rational homogeneous $F$ is simply connected, and a connected finite étale cover has the same ordinary universal cover, yielding the asserted product for $\widetilde X$.

[BB96, §3.7, print pp. 17–18 / PDF pp. 18–19] [Poc25, Theorem 1.1 (Fischer–Grauert), p. 1] [CHK13, Theorem 4.4, p. 11] §7.2 · pp. 21–22

3. Quasi-projectivity, holomorphic convexity and Remmert reduction

Projectivity of $F$ makes $F\times\mathbb A^q$ quasi-projective. Projection to $\mathbb C^q$ is proper with connected compact fibres, and every holomorphic function is constant on each fibre. Its direct image of the structure sheaf is consequently $\mathcal O_{\mathbb C^q}$, so this map to a Stein base is the Remmert reduction. The argument includes $q=0$, with a point as base. Finally, $0\le q\le\dim X$ makes $d\le6$ automatic when $\dim X\le6$.

§7.2 · pp. 21–22

1. Make the positive Einstein constant equal to one

The main theorem in dimension six and Kanemitsu's result in dimensions one through five make $X$ rational homogeneous. Apply the Alekseevskii–Perelomov construction as recalled in Demailly §1 to obtain a Kähler form with $\rho(\sigma)=\lambda\sigma$, $\lambda>0$. Constant rescaling leaves the Ricci form unchanged, so $\omega=\lambda\sigma$ satisfies $\rho(\omega)=\omega$. In the stated convention the Ricci form represents $2\pi c_1(-K_X)$, which also proves the cohomological conclusion.

[Kan16, Theorem 0.2, p. 1] [Dem18, §1 (Alekseevskii–Perelomov construction), p. 1] Corollary 7.1 · p. 21