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

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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.

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Graphs of linear maps and Lagrangian relations

Statement

Let $A : (V, ω_{V}) \to (W, ω_{W})$ be linear. Its graph is Lagrangian in $V^{-} \oplus W$ , equipped with $- ω_{V} \oplus ω_{W}$ , if and only if $A$ is a symplectic isomorphism. In particular, a symplectic embedding into a strictly larger symplectic space has an isotropic, but not Lagrangian, graph.

Facts & Assumptions

Given: Symplectic vector spaces $(V, ω_{V})$ and $(W, ω_{W})$ and a linear map $A : V \to W$ .

[F1]

In a $2 N$ -dimensional symplectic space, an isotropic subspace is Lagrangian exactly when it has dimension $N$ . Equivalent characterizations of Lagrangian subspaces.

Proof

technique · direct

1.1

On graph vectors one has $(- ω_{V} \oplus ω_{W}) ((u, A u), (v, A v)) = - ω_{V} (u, v) + ω_{W} (A u, A v)$ . Thus the graph is isotropic exactly when $A^{*} ω_{W} = ω_{V}$ .

algebra

2.1

If the graph is Lagrangian, [F1] and $\dim graph A = \dim V$ give $2 \dim V = \dim V + \dim W$ , hence $\dim V = \dim W$ . Step 1.1 also says $A$ preserves the forms, which makes $A$ injective by nondegeneracy; equal dimensions make it an isomorphism.

F1step 1.1algebra

3.1

Conversely, if $A$ is a symplectic isomorphism, step 1.1 makes its graph isotropic and its dimension is half that of $V^{-} \oplus W$ , so [F1] makes it Lagrangian. If instead $A$ is a symplectic embedding with $\dim W > \dim V$ , step 1.1 still gives isotropy but the half-dimension equality fails. The zero spaces cause no exception.

F1step 1.1step 2.1∎

Depends on

Equivalent characterizations of Lagrangian subspaces

Used by

Nothing in the library uses this result yet.

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)