TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Equivalent characterizations of Lagrangian subspaces

Statement

Let $(V, ω)$ be a real symplectic vector space of dimension $2 n$ . For a subspace $L \leq V$ , the following are equivalent:

$L = L^{ω}$ ;
$L$ is isotropic and $\dim L = n$ ;
$L$ is coisotropic and $\dim L = n$ ;
$L$ is maximal among isotropic subspaces.

Thus each condition characterizes the Lagrangian subspaces.

Facts & Assumptions

Given: A $2 n$ -dimensional real symplectic vector space $(V, ω)$ and $L \leq V$ .

[F1]

For every $W \leq V$ , $\dim W + \dim W^{ω} = 2 n$ . Symplectic double-orthogonal and dimension identities.

[F2]

Isotropic, coisotropic, and Lagrangian mean respectively $W \subseteq W^{ω}$ , $W^{ω} \subseteq W$ , and $W = W^{ω}$ . Isotropic, coisotropic, symplectic, and Lagrangian subspaces.

Proof

technique · direct

1.1

If $L = L^{ω}$ , [F1] gives $2 \dim L = 2 n$ ; [F2] then makes $L$ both isotropic and coisotropic. Thus condition 1 implies conditions 2 and 3.

F1F2

1.2

If condition 2 holds, then $L \subseteq L^{ω}$ and [F1] gives $\dim L^{ω} = n = \dim L$ , hence equality. If condition 3 holds, the reverse inclusion and the same dimension calculation likewise give equality. Thus conditions 2 and 3 each imply condition 1.

F1F2

1.3

A self-orthogonal $L$ is maximal isotropic: if an isotropic $K$ contains $L$ , then $K \subseteq K^{ω} \subseteq L^{ω} = L$ , hence $K = L$ .

F1F2

2.1

Conversely, suppose $L$ is maximal isotropic. If $L \neq L^{ω}$ , choose $v \in L^{ω} ∖ L$ ; alternation and $v \in L^{ω}$ make $L + R v$ a strictly larger isotropic subspace, which maximality forbids. Hence $L = L^{ω}$ . This argument also covers $n = 0$ , when $L = 0$ .

F2choosealgebra∎

Depends on

Used by

Product and opposite symplectic manifolds Example
The zero section and cotangent fibres as Lagrangians Example
A graph of a one-form is Lagrangian exactly when the form is closed Proposition
Graphs of linear maps and Lagrangian relations Proposition
Lagrangian submanifolds have half dimension Proposition
Regular common level sets are Lagrangian submanifolds Proposition

Dependency tree · two levels

3 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)