PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Compact-group symplectic actions admit an invariant compatible almost-complex structure

Statement

Assume the Axiom of Choice and ${A C}_{ω}$ . Let a compact Lie group $G$ act symplectically on a symplectic manifold $(M, ω)$ . Then $M$ carries a $G$ -invariant almost-complex structure $J$ compatible with $ω$ : that is, $J^{2} = - id$ , $ω (J u, J v) = ω (u, v)$ for all tangent vectors, and $(u, v) \mapsto ω (u, J v)$ is a Riemannian metric on $M$ which is also $G$ -invariant.

Facts & Assumptions

Given: the Axiom of Choice, ${A C}_{ω}$ , a compact Lie group $G$ acting symplectically on $(M, ω)$ .

[A1]

The Axiom of Choice is The Axiom of Choice and ${A C}_{ω}$ is countable choice.

[A2]

AC is used to obtain the normalized Haar measure and the background Riemannian metric, and ${A C}_{ω}$ is inherited from the fundamental-field interface of the action; no other choice is made.

[F1]

$G$ has a unique regular Borel probability measure $μ$ invariant under left and right translations and inversion, and $\int_{G} f (h x) d μ (x) = \int_{G} f (x) d μ (x)$ for integrable $f$ . Normalized Haar measure on a compact Lie group, Haar integration is translation and conjugation invariant.

[F2]

Every smooth manifold admits a Riemannian metric. Assuming countable choice, every smooth manifold admits a Riemannian metric.

[F3]

A smooth self-adjoint positive-definite bundle endomorphism has a unique smooth self-adjoint positive-definite square root. Positive-definite bundle endomorphisms have smooth positive square roots.

[F4]

The action is symplectic: $a_{g}^{*} ω = ω$ for all $g$ , where $a_{g} (p) = g \cdot p$ . Symplectic and Hamiltonian Lie-group actions.

Proof

technique · direct

1.1A2F1F2

Choose a background Riemannian metric $h_{0}$ on $M$ by [F2] and put $h_{p} (u, v) : = \int_{G} (a_{g}^{*} h_{0})_{p} (u, v) d μ (g) .$ The integrand is smooth in $(g, p)$ and the integral is a finite-dimensional parameter integral, so $h$ is a smooth symmetric bilinear form; it is positive definite because the average of positive numbers is positive, and nondegenerate accordingly.

2.1step 1.1F1

The metric $h$ is $G$ -invariant: for $k \in G$ , invariance of Haar under left translation gives $h_{k \cdot p} (d (a_{k}) u, d (a_{k}) v) = \int_{G} (a_{g k}^{*} h_{0})_{p} (u, v) d μ (g) = \int_{G} (a_{g}^{*} h_{0})_{p} (u, v) d μ (h^{- 1} g) = h_{p} (u, v) .$

2.2step 1.1

Define a bundle endomorphism $A$ by $ω_{p} (u, v) = h_{p} (A_{p} u, v)$ ; it exists and is unique because $h_{p}$ is nondegenerate. It is invertible because $ω_{p}$ is nondegenerate, and it is skew-adjoint for $h$ : expanding $ω_{p} (u, v) + ω_{p} (v, u) = 0$ gives $h_{p} ((A_{p} + A_{p}^{*}) u, v) = 0$ for all $u, v$ , hence $A^{*} = - A$ . Therefore $- A^{2} = A^{*} A$ is $h$ -positive-definite, and it commutes with $A$ .

3.1step 2.2F3

By step 2.2 the endomorphism $- A^{2} = A^{*} A$ is self-adjoint and positive definite, so [F3] gives its unique smooth self-adjoint positive-definite square root; set $J : = A (- A^{2})^{- 1 / 2} .$ Since $A$ commutes with $- A^{2}$ and with its functional calculus, $J^{2} = A^{2} (- A^{2})^{- 1} = - id$ .

4.1step 3.1

Compatibility: from $J^{2} = - id$ and $A^{*} = - A$ one computes $ω (J u, J v) = ω (u, v)$ and that $(u, v) \mapsto ω (u, J v)$ is symmetric; positivity follows from $ω (u, J u) = h (A u, J u) = h ((- A^{2})^{1 / 2} u, u) > 0$ for $u \neq 0$ , so $g_{ω} (u, v) : = ω (u, J v)$ is a Riemannian metric.

5.1step 3.1step 4.1F4

Invariance: both $h$ and $ω$ are $G$ -invariant, so $A$ is $G$ -equivariant: $h (A d (a_{g}) u, d (a_{g}) v) = ω (d (a_{g}) u, d (a_{g}) v) = ω (u, v) = h (A u, v) = h (d (a_{g}) A u, d (a_{g}) v)$ for all $v$ , whence $A d (a_{g}) = d (a_{g}) A$ by nondegeneracy of $h$ . Hence $- A^{2}$ is $G$ -equivariant, its unique positive square root is $G$ -equivariant by uniqueness, and $J = A (- A^{2})^{- 1 / 2}$ is $G$ -equivariant. In particular $J$ and $g_{ω}$ are $G$ -invariant.

6.1step 4.1step 5.1A1∎

Steps 1.1--2.1 produce an invariant Riemannian metric, steps 2.2--3.1 produce a smooth almost-complex structure $J$ , step 4.1 verifies compatibility with $ω$ , and step 5.1 verifies $G$ -invariance; this proves the claim.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

30 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)
Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)