TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Compatible complex structures exist on symplectic vector spaces

Statement

Every finite-dimensional symplectic vector space $(V, ω)$ admits an $ω$ -compatible complex structure. More precisely, every chosen inner product $k$ on $V$ canonically determines one.

Facts & Assumptions

Given: A finite-dimensional symplectic vector space $(V, ω)$ and an inner product $k$ on $V$ .

[F1]

A non-negative self-adjoint endomorphism of a finite-dimensional inner product space has a unique non-negative square root. A non-negative operator has a unique non-negative square root.

[F2]

Compatibility means that $J^{2} = - I$ and $g_{J} (u, v) = ω (u, J v)$ is an inner product. Compatible complex structure on a symplectic vector space.

Proof

technique · direct

1.1

Nondegeneracy of $ω$ uniquely defines an invertible $A \in End (V)$ by $k (u, v) = ω (u, A v)$ . Skew-symmetry of $ω$ gives $k (u, A^{- 1} v) = - k (A^{- 1} u, v)$ , so $A^{*} = - A$ . Hence $- A^{2} = A^{*} A$ is positive definite.

givenalgebra

2.1

Let $P = (- A^{2})^{1 / 2}$ be the positive square root from [F1]. Since $A$ commutes with $- A^{2}$ , it preserves each eigenspace of $- A^{2}$ ; on that eigenspace $P$ is multiplication by the positive square root of the eigenvalue, so $P$ commutes with $A$ . Define $J = A P^{- 1}$ . Then $J^{2} = A^{2} P^{- 2} = - I$ .

F1step 1.1algebra

3.1

Since $P$ is positive definite and commutes with $A$ , $ω (u, J v) = ω (u, A P^{- 1} v) = k (u, P^{- 1} v) = k (P^{- 1 / 2} u, P^{- 1 / 2} v) .$ This is symmetric and positive definite, so [F2] makes $J$ compatible. For $V = 0$ the same formulas give the unique endomorphism, and all conditions are vacuous.

F1F2step 1.1step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)