LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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.

Finite symplectic generator factorization

Statement

Let $r \geq 1$ , $J_{0} = (\begin{array}{cc} 0 & I_{r} \\ - I_{r} & 0 \end{array})$ , and $Sp (2 r, R) = {S \in M_{2 r} (R) : S^{T} J_{0} S = J_{0}} .$ Every $S \in Sp (2 r, R)$ is a finite product of matrices of the following three types and their inverses: $U_{X} = (\begin{array}{cc} I & X \\ 0 & I \end{array}), L_{Y} = (\begin{array}{cc} I & 0 \\ Y & I \end{array}), D_{A} = (\begin{array}{cc} A & 0 \\ 0 & A^{- T} \end{array}),$ where $X, Y$ are real symmetric and $A$ is real invertible. Partial Fourier rotations, which act by $(q_{j}, p_{j}) \mapsto (- p_{j}, q_{j})$ on any selected set of coordinates and fix the others, are also products of these matrices. In addition, for every $S$ there is a product $G$ of these generators such that the upper-left block of $S G$ is invertible.

Facts & Assumptions

Given: $r \geq 1$ and a real matrix $S$ satisfying the displayed symplectic identity. Blocks are written in grouped coordinate order $z = (q, p)$ with $q, p \in R^{r}$ .

[A1]

The input matrix obeys $S^{T} J_{0} S = J_{0}$ for $J_{0} = (\begin{array}{cc} 0 & I_{r} \\ - I_{r} & 0 \end{array})$ , and $r \geq 1$ .

[F1]

Block multiplication gives, for $S = (\begin{array}{cc} A & B \\ C & D \end{array})$ , the equivalent relation $S J_{0} S^{T} = J_{0}$ and in particular $A B^{T} = B A^{T}$ ; the original relation $S^{T} J_{0} S = J_{0}$ gives $A^{T} C = C^{T} A$ and $A^{T} D - C^{T} B = I$ . These relations also imply that the first $r$ rows of $S$ have rank $r$ .

Proof

1.1givenA1F1algebra

Each listed matrix is symplectic by direct multiplication with $J_{0}$ : $U_{X}^{T} J_{0} U_{X} = J_{0}$ iff $X = X^{T}$ , $L_{Y}^{T} J_{0} L_{Y} = J_{0}$ iff $Y = Y^{T}$ , and $D_{A}^{T} J_{0} D_{A} = J_{0}$ . A quarter-turn on a subset is itself a block diagonal product of two-dimensional matrices $(\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$ and identities, so is symplectic; on one selected coordinate it equals the product $U_{- 1} L_{1} U_{- 1}$ in that coordinate. Thus all proposed generators are symplectic.

1.2givenF1algebra

First suppose the upper-left block $A$ is invertible. The block equations in [F1] imply $A^{- 1} B = (A^{- 1} B)^{T}$ and $C A^{- 1} = (C A^{- 1})^{T}$ by multiplying $A B^{T} = B A^{T}$ and $A^{T} C = C^{T} A$ by the relevant inverses. They also give $D = A^{- T} + C A^{- 1} B$ from $A^{T} D - C^{T} B = I$ . Therefore direct block multiplication yields $S = L_{C A^{- 1}} D_{A} U_{A^{- 1} B},$ which is the required factorization in this case.

1.3givenF1algebra

For arbitrary $A$ , let $k = rank A$ . Choose invertible real matrices $L, R$ with $L A R = diag (I_{k}, 0)$ , using row and column reduction, and replace $S$ by $S^{'} = D_{L} S D_{R}$ . This replacement is symplectic. Write its upper-right block as $B^{'}$ and partition rows and columns after the first $k$ . The identity $A^{'} B^{' T} = B^{'} A^{' T}$ in [F1], with $A^{'} = diag (I_{k}, 0)$ , forces the lower-left rectangular block of $B^{'}$ to vanish. Since the first $r$ rows of $S^{'}$ are independent, the bottom $r - k$ rows of $B^{'}$ restricted to its last $r - k$ columns form an invertible matrix $B_{22}$ . Right-multiply $S^{'}$ by the inverse partial Fourier rotation $F$ on those last coordinates, $(q_{j}, p_{j}) \mapsto (p_{j}, - q_{j})$ . It is a product of the inverse listed generators. The resulting upper-left block is block upper triangular with diagonal blocks $I_{k}$ and $- B_{22}$ , hence is invertible. This argument also covers $k = 0$ and $k = r$ (with the empty block omitted).

2.1step 1.1step 1.2step 1.3algebra∎

The matrix $S^{'} F$ has invertible upper-left block, so step 1.2 factors it into the stated generators. Since $S = D_{L}^{- 1} (S^{'} F) F^{- 1} D_{R}^{- 1}$ and the inverses of $D_{L}, D_{R}, F$ are of the same listed types, the same is true of $S$ . This proves the factorization for every symplectic matrix.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Arvind, B. Dutta, N. Mukunda and R. Simon, The Real Symplectic Groups in Quantum Mechanics and Optics, §§II and IV