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

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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

1.1

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.

A2F1F2

2.1

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

step 1.1F1

2.2

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 $A$ .

step 1.1

3.1

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, $J^{2} = A^{2} (- A^{2})^{- 1} = - id$ .

step 2.2F3

4.1

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.

step 3.1

5.1

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.

step 3.1step 4.1F4

6.1

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.

step 4.1step 5.1A1∎

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Sources

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