Source
Ovsjanikov, Ben-Chen, Solomon, Butscher & Guibas (2012), LIT-993, Remarks 4.1–4.3, Theorem 5.1 and its proof (App. A), and §5.1, as read in NOTE-822.
What was actually shown
Equivalence. A bijection T: M → N induces T_F(f) = f∘T⁻¹, which is linear, and T is recovered from T_F because indicator functions go to indicator functions (Remarks 4.1–4.2). In bases, T_F is a matrix C (Remark 4.3).
Orthogonality (Theorem 5.1(1), checked). If T preserves volume, the integral defining C_ij and the one defining the matrix D of T⁻¹ have the same integrand after a change of variables, so C = Dᵀ. Since CD = I, C is orthogonal and T⁻¹ is represented by Cᵀ. The same holds when the discrete bases are orthonormal in the standard inner product.
Eigenspaces (Theorem 5.1(2) and §5.1). An isometry commutes with the Laplace–Beltrami operator, so C_ij ≠ 0 only between eigenfunctions of the same eigenvalue; C is block-diagonal, and diagonal if no eigenvalue repeats. The individual eigenvectors are not preserved (signs and bases within a repeated eigenvalue can change); the eigenspaces are.
What could have failed: the orthogonality argument needs only the volume condition; a volume-changing bijection gives a non-orthogonal C, which is why the paper uses an unweighted Laplacian where it helps compactness.
What this does not say
- Nothing about maps that are not bijections. The operator runs between two spaces, but every structural result assumes T is invertible. Inverting a truncated matrix for a non-bijective map is a numerical step without guarantee.
- Not that truncation is stable. That the span of the first n eigenfunctions survives a near-isometry is cited to Kato (1995), with a gap between the nth and (n + 1)st eigenvalues; it is not shown. When the truncated block of an exact or near isometry stays orthogonal is QUESTION-138.
- Not which matrices come from point maps. The paper calls that “a difficult open problem” (§5.5).
- Not that near-isometries are nearly diagonal in general. The 94% sparsity is measured on TOSCA cats, and the paper says C stops being diagonal quickly under non-isometric deformation.
Connections
- QUESTION-052 asks for a signature stable under a kind’s admissible transformations. For isometries, this gives one: the Laplacian eigenspaces, as spaces, not as bases, which is CLAIM-164’s point. It gives nothing for the directed case the question asks about, so it is not filed as an answer.
- THEORY-239 obtains singular functions of an operator between two spaces from a joint law. Functional maps are another operator between two spaces, whose structure comes from a bijection and an intrinsic Laplacian, not from a pair distribution.
- THEORY-009 says self-adjointness in spectral self-supervised learning comes from a symmetric pair distribution. Here the analogue of symmetry is volume preservation, which makes the inverse the adjoint.
Note of 2026-10-19: the three statements, re-derived
For a measurable bijection T: M → N between compact Riemannian manifolds, with T_F(f) = f∘T⁻¹:
- Equivalence. T_F is linear and sends 1_A to 1_(T(A)), so it fixes T(A) for every measurable A, and with it T up to null sets. If T is continuous, it fixes T pointwise.
- Orthogonality. If T preserves volume, change of variables gives ‖f∘T⁻¹‖²_N = ∫_N |f(T⁻¹y)|² dy = ∫_M |f|² dx, so T_F is unitary from L²(M) to L²(N). In orthonormal bases its matrix C satisfies CᵀC = I, and its inverse is the matrix D of T⁻¹, so C = Dᵀ. That is Theorem 5.1(1), which NOTE-822 checked.
- Eigenspaces. A Riemannian isometry pulls back the metric, and the Laplace–Beltrami operator is built from the metric alone, so Δ_N(f∘T⁻¹) = (Δ_M f)∘T⁻¹. So T_F maps the λ-eigenspace of M onto the λ-eigenspace of N, and C_ij = 0 whenever λ_i ≠ λ_j, since eigenspaces of distinct eigenvalues are orthogonal. The paper cites this as well known. It is the naturality of the Laplacian under isometries.
For an exact isometry the truncation question is easy when n ends at a spectral gap. C is block-diagonal by eigenspace, so its leading n × n block is a direct sum of orthogonal blocks. The open part is the near-isometric case, which is QUESTION-138.