Source
Harvey, Lipshutz & Williams (2024), LIT-257, §4 (eqs. 22–26), Proposition 3, App. A.1 (eqs. 27–53) and Fig. 2, read in NOTE-834. The Procrustes–Bures equivalence, P(X̃,Ỹ) = d_B(X̃X̃ᵀ, ỸỸᵀ), is Harvey, Larsen & Williams, LIT-250, Thm 1 (THEORY-004).
What was actually shown
Take the optimal readouts of eq. 1, w* = G(X)⁻¹Xᵀz/M, with G(X) = aC_X + bI, and targets with E[zzᵀ] = I. Then the average decoding distance is E‖Xw* − Yv*‖² = ‖K_X − K_Y‖²_F (Cor. 1). Normalize X̃ = XG(X)^{−1/2}/√M, so that X̃X̃ᵀ = K_X, and let P be the Procrustes distance, zero-padded to equal widths. With α = Tr K_X, β = Tr K_Y and R_Δ = ‖K_X − K_Y‖²_*/‖K_X − K_Y‖²_F:
(α + β) − √((α + β)² − R_Δ E‖Xw* − Yv*‖²) ≤ P(X̃,Ỹ)² ≤ √(R_Δ E‖Xw* − Yv*‖²) (eq. 25),
equivalently P⁴/R_Δ ≤ E‖Xw* − Yv*‖² ≤ (2(α + β)P² − P⁴)/R_Δ (eq. 26).
- The lower bound on P² comes from a Fuchs–van de Graaf-type argument. ‖αuuᵀ − βvvᵀ‖* = √((α + β)² − 4αβ⟨u,v⟩²) (Lemma 1). Choosing representatives in the orthogonal orbits so that ⟨vec X̃, vec Ỹ⟩ is the fidelity, partial-trace monotonicity gives ‖K_X − K_Y‖* ≤ √((α + β)² − 4F²).
- The upper bound on P² is d_B² ≤ ‖K_X^{1/2} − K_Y^{1/2}‖²_F ≤ ‖K_X − K_Y‖_* (Powers–Størmer).
- Lemma 2. With equal traces, R_Δ ≥ 2. A shorter proof: a trace-zero difference has nuclear norm 2S, S its positive eigenvalue sum, and squared Frobenius norm at most 2S² (NOTE-834).
Small P forces small decoding distance. Since R_Δ ≥ 1, E‖Xw* − Yv*‖² ≤ 2(α + β)P².
The converse fails without control of R_Δ. Take trace-one kernels K_X = Π₁/r and K_Y = Π₂/r, with Π₁, Π₂ orthogonal rank-r projections. The fidelity is 0, so P² = 2, the maximum under α = β = 1, while E‖Xw* − Yv*‖² = 2/r. That is the reader’s example (NOTE-834), and it is the case R_Δ = 2r. The paper shows the same region with random positive semidefinite pairs (Fig. 2D) and states the asymmetry in words (§4).
What could have failed: the two distances could have been equivalent with dimension-free constants. The participation ratio in eq. 25 is where they are not.
What this does not say
- Not which distance is right. The paper calls Procrustes “a more strict notion of geometric dissimilarity” and does not argue that either tracks function better.
- Not anything beyond K_z = I. Both bounds use Cor. 1’s identity, so they hold for targets uncorrelated across conditions, with unit variance.
- Not anything about nonlinear readouts, classifiers or the population limit. Linear regression readouts at finite M only.
- Not tight in general. Saturation is stated for rank-one kernels (lower bound) and for commuting kernels with spectra in {0,1} (upper bound). Elsewhere the slack is not characterized.
- Not about trained networks. Fig. 2 uses random ensembles of 50 × 50 PSD matrices.
Connections
- THEORY-004 supplies the premise: rotation-aligned distance equals Bures distance between kernels. Here the Bures distance is re-expressed through decoding.
- THEORY-008 holds the identity E‖Xw* − Yv*‖² = ‖K_X − K_Y‖²_F, which this account bounds against Procrustes.
- THEORY-233 is the parallel envelope between CKA and normalized Bures similarity (LIT-250), with the kernels’ ranks as the constant. Here the constant is the participation ratio of the difference. Both bounds come from the Fuchs–van de Graaf route, and both say the Euclidean and Bures geometries of kernels agree only up to a dimension-like factor.
Note of 2026-10-19: eqs. 25–26, derived directly
With the notation above, write Δ = K_X − K_Y and D = E‖Xw* − Yv*‖². All matrices are real.
- Cor. 1. Xw* = XG(X)⁻¹Xᵀz/M = K_X z, so D = E‖Δz‖² = Tr(Δ E[zzᵀ] Δ) = ‖Δ‖²_F when E[zzᵀ] = I.
- Procrustes is Bures (THEORY-004). P² = α + β − 2F, with F the fidelity ‖K_X^{1/2}K_Y^{1/2}‖_*, which is also the maximum over orthogonal Q of ⟨X̃, ỸQ⟩.
- Upper bound on P². F ≥ Tr(K_X^{1/2}K_Y^{1/2}), so P² ≤ ‖K_X^{1/2} − K_Y^{1/2}‖²_F. For Powers–Størmer, let S = K_X^{1/2}, T = K_Y^{1/2}, E = S − T with eigenpairs (dᵢ, eᵢ), and J = sign(E). Then K_X − K_Y = ½(E(S + T) + (S + T)E), and J commutes with E, so Tr(JΔ) = Σᵢ |dᵢ|⟨eᵢ, (S + T)eᵢ⟩ ≥ Σᵢ |dᵢ||⟨eᵢ, (S − T)eᵢ⟩| = ‖E‖²_F, because S and T are positive semidefinite. Since ‖J‖op ≤ 1, P² ≤ ‖E‖²_F ≤ ‖Δ‖* = √(R_Δ D).
- Lower bound on P². Replace Ỹ by ỸQ for the optimal Q, which leaves K_Y unchanged. Let u = vec X̃/√α and v = vec Ỹ/√β in ℝ^M ⊗ ℝ^N, so ⟨u, v⟩ = c = F/√(αβ). The partial trace over ℝ^N sends αuuᵀ to K_X and βvvᵀ to K_Y. It is a trace-norm contraction: ‖Tr_N H‖* = sup over ‖W‖op ≤ 1 of Tr((W ⊗ I)H) ≤ ‖H‖*. On span{u, v}, with v = cu + sw, the matrix αuuᵀ − βvvᵀ has trace α − β and determinant −αβs² ≤ 0. Its nuclear norm is therefore |λ₁ − λ₂| = √((α − β)² + 4αβs²) = √((α + β)² − 4F²), which is Lemma 1. So R_Δ D = ‖Δ‖²* ≤ (α + β)² − (α + β − P²)². Since α + β − P² = 2F ≥ 0, this is the left side of eq. 25.
- Eq. 26 rearranges eq. 25. Since ‖·‖_* ≥ ‖·‖_F, R_Δ ≥ 1 and D ≤ 2(α + β)P².
Uhlmann’s theorem enters only through the maximisation in THEORY-004. The other steps are elementary and hold for real matrices as stated.