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Constant flat summands of positive real monads

Statement

Let

O (- 1) \otimes U \overset{α}{\to} O \otimes S \overset{β}{\to} O (1) \otimes U^{'}

be a locally split linear monad on $P^{3} (C)$ with locally free middle cohomology $F$ . Suppose a nondegenerate Hermitian form $H$ on $S$ , linear in the first variable, satisfies $β (z) = α (j z)^{†} H$ for $j (λ_{1}, λ_{2}, μ_{1}, μ_{2}) = (- {\overset{ˉ}{λ}}_{2}, {\overset{ˉ}{λ}}_{1}, - {\overset{ˉ}{μ}}_{2}, {\overset{ˉ}{μ}}_{1})$ . Assume $H^{0} (P^{3}, O) = C$ , $H^{q} (P^{3}, O (- 1)) = 0$ for every $q \geq 0$ , and $H^{0} (P^{1}, O (- 1)) = 0$ . Define $L = ⋂_{i} \ker β_{i} \subseteq S$ and $M = \sum_{i} im α_{i} \subseteq S$ , where $α (z) = \sum_{i} z_{i} α_{i}$ and $β (z) = \sum_{i} z_{i} β_{i}$ . If the induced Hermitian form on $H^{0} (F ∣_{ℓ})$ is positive definite for every $σ$ -invariant real line $ℓ$ , with line form the constant value $x \mapsto b_{x} (s (x), t (σ x))$ from the preceding first-order real-structure definition, then $H^{0} (F) = L = M^{⊥_{H}}$ , $H ∣_{L}$ is positive definite, and $S = L \oplus M$ is an $H$ -orthogonal decomposition. The monad and its pairing split as the constant summand $L \otimes O$ plus the reduced monad $U \otimes O (- 1) \to M \otimes O \to U^{'} \otimes O (1)$ ; the reduced cohomology has no global sections and remains positive on every real line. The conclusion includes $U = 0$ , when $L = S$ and $M = 0$ .

Statement references

Analytic negative-twist vanishings for the positive real monad

Unique chain lifting and sigma-Hermitian monad duality

Barth–Hulek, Section 3, Proposition 4

Verdier, Exposé VII, §§6.10–6.12

Facts & Assumptions

Given: The stated monad, the nondegenerate Hermitian form $H$ , the real-line pairing, and the analytic scalar vanishings.

[A1]

AC is the choice base inherited from the sheaf-cohomology supplier. The Axiom of Choice

[F1]

The analytic cohomology groups of $O$ and $O (- 1)$ have the values stated in the hypothesis, and $H^{0} (P^{1}, O (- 1)) = 0$ . Analytic negative-twist vanishings for the positive real monad

[F2]

A linear monad morphism with its specified map on middle cohomology has the unique chain lift from the preceding lemma; its sigma-dual identities are $β (z) = α (j z)^{†} H$ and $β (j z)^{†} = - H α (z)$ . Unique chain lifting and sigma-Hermitian monad duality

[F3]

The pointwise $σ$ -Hermitian pairing and its constant line-section form are defined in the first-order real-structure definition. Restriction of a locally split exact sequence to a real line remains exact; on the constant orthogonal complement of its Euler block, the section form is $H ∣_{E_{z}}$ . First-order families of positive real linear monads

[F4]

A short exact sequence of sheaves gives a natural long exact sequence in the same sheaf cohomology groups. Long exact sequence of sheaf cohomology

Proof

Given: The hypotheses and the definitions of $L$ and $M$ in the Statement.

1.1A1F1F4algebra

Let $K = \ker β$ . The locally split sequence $0 \to U \otimes O (- 1) \to K \to F \to 0$ and the vanishings [F1], together with [F4], give an isomorphism $H^{0} (K) \to H^{0} (F)$ . Since $K \subseteq S \otimes O$ and $H^{0} (O) = C$ , every global section of $K$ is a constant vector $s \in S$ . Its image in $F$ is a global section exactly when $β (z) s = 0$ for every $z$ , equivalently $β_{i} s = 0$ for every coefficient. Hence $H^{0} (F) = L$ .

1.2F2algebra

If $l \in L$ , then $β (z) l = 0$ for all $z$ . By [F2], $β (z) = α (j z)^{†} H$ , so $H (l, α (j z) u) = 0$ for every $u$ and $z$ ; since $j$ is bijective, $l ⊥_{H} M$ . Conversely, if $s ⊥_{H} M$ , then $α (j z)^{†} H s = 0$ for every $z$ , hence $β (z) s = 0$ and $s \in L$ . Thus $L = M^{⊥_{H}}$ , and nondegeneracy of $H$ gives $M = L^{⊥_{H}}$ .

2.1F1F3step 1.2algebra

For a real line $ℓ$ , a constant vector $l \in L$ defines a section of $F ∣_{ℓ}$ . If that section is zero, exactness of $0 \to U \otimes O_{ℓ} (- 1) \to K ∣_{ℓ} \to F ∣_{ℓ} \to 0$ and $H^{0} (ℓ, O_{ℓ} (- 1)) = 0$ force $l = 0$ . Hence $L \to H^{0} (F ∣_{ℓ})$ is injective. Positivity of the given section pairing implies $H (l, l) > 0$ for every nonzero $l \in L$ . Therefore $L$ is nondegenerate and $S = L \oplus M$ orthogonally.

3.1F3step 1.1step 2.1algebra∎

Every coefficient image of $α$ lies in $M$ by definition, and every coefficient of $β$ kills $L$ by definition. Thus the complex is the direct sum of $0 \to L \otimes O \to 0$ and the displayed reduced monad on $M$ . The induced cohomology splitting is $F = (L \otimes O) \oplus F_{M}$ and is orthogonal for $H$ . On each real line, the original section form is positive definite and the $L$ summand is a positive orthogonal subspace by step 2.1; its orthogonal complement, the section space of $F_{M} ∣_{ℓ}$ , is therefore positive definite. Since $H^{0} (F) = L$ by step 1.1, the direct sum decomposition implies $H^{0} (F_{M}) = 0$ . If $U = 0$ , then $α = β = 0$ , so $M = 0$ and $L = S$ , which gives the stated boundary case.

Constant flat summands of positive real monads

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources