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Lemma A.12 - Gaussian head misalignment

Proved
FeatureDistortion.GaussianHeadMisalignment

by Minghui · Sep 26, 2026 · Mathlib c5ea003 (Lean v4.30.0)

linear-algebramachine-learningprobability

Notation: n=#{training examples}n = \#\{\text{training examples}\}n=#{training examples}, ddd is the input dimension, kkk the feature dimension, X:Rd→RnX:\mathbb R^d\to\mathbb R^nX:Rd→Rn the data map, YYY the labels, B:Rd→RkB:\mathbb R^d\to\mathbb R^kB:Rd→Rk the features, and v∈Rkv\in\mathbb R^kv∈Rk the head. Adjoint means Euclidean transpose. The loss is L^(v,B)=∥XB⊤v−Y∥2\widehat L(v,B)=\|XB^\top v-Y\|^2L(v,B)=∥XB⊤v−Y∥2, with no normalization. The probability model, when present, is explicitly specified below; deterministic flow statements involve no random data assumption.

For every natural number kkk, every u∈Rku\in\mathbb R^ku∈Rk, and every real number σ\sigmaσ, if u≠0u\neq0u=0 and σ>0\sigma>0σ>0, then for almost every v0v_0v0​ under the pushforward of standard Gaussian measure on Euclidean Rk\mathbb R^kRk by z↦σzz\mapsto\sigma zz↦σz, ∣⟨v0,u⟩2−⟨u,u⟩2∣>0\left|\langle v_0,u\rangle^2-\langle u,u\rangle^2\right|>0​⟨v0​,u⟩2−⟨u,u⟩2​>0. Equivalently, both equalities ⟨v0,u⟩=∥u∥2\langle v_0,u\rangle=\|u\|^2⟨v0​,u⟩=∥u∥2 and ⟨v0,u⟩=−∥u∥2\langle v_0,u\rangle=-\|u\|^2⟨v0​,u⟩=−∥u∥2 fail outside a set of measure zero. For k=0k=0k=0 the hypothesis u≠0u\neq0u=0 is impossible, so the implication is vacuous. For u=0u=0u=0 or σ≤0\sigma\leq0σ≤0, no almost-everywhere conclusion is required.

Formalization note: Qualitative source-derived consequence of Lemma A.12; no numerical anti-concentration constant is claimed. Source: Kumar, Raghunathan, Jones, Ma, and Liang, Fine-Tuning can Distort Pretrained Features and Underperform Out-of-Distribution, ICLR 2022, https://arxiv.org/pdf/2202.10054v1. Appendix A.3.2, PDF pp. 34--35, Lemma A.12, equations (A.123), (A.126)--(A.128). Source-backed parent: Section 3.4, PDF p. 10, Proposition 3.7, equations (3.10)--(3.11); Appendix A.7, PDF pp. 45--47.

Preamble
import Definitions.Def_FeatureDistortion_Model
open MeasureTheory Filter
open scoped Topology
Formal statement
namespace FeatureDistortion
theorem GaussianHeadMisalignment :
  ∀ (k : ℕ) (u : Vec k) (σ : ℝ), u ≠ 0 → 0 < σ →
    ∀ᵐ v₀ ∂gaussianHead k σ, 0 < alignmentError v₀ u := by sorry
end FeatureDistortion
Source
Kumar, Raghunathan, Jones, Ma, and Liang, Fine-Tuning can Distort Pretrained Features and Underperform Out-of-Distribution, ICLR 2022, https://arxiv.org/pdf/2202.10054v1. Appendix A.3.2, PDF pp. 34--35, Lemma A.12, equations (A.123), (A.126)--(A.128). Source-backed parent: Section 3.4, PDF p. 10, Proposition 3.7, equations (3.10)--(3.11); Appendix A.7, PDF pp. 45--47.
Read-back

What the Lean code literally says, in plain math · gpt-6

For every natural number kkk, every u∈Rku\in\mathbb R^ku∈Rk, and every real number σ\sigmaσ, if u≠0u\neq0u=0 and σ>0\sigma>0σ>0, then for almost every v0v_0v0​ under the pushforward of standard Gaussian measure on Euclidean Rk\mathbb R^kRk by z↦σzz\mapsto\sigma zz↦σz, ∣⟨v0,u⟩2−⟨u,u⟩2∣>0\left|\langle v_0,u\rangle^2-\langle u,u\rangle^2\right|>0​⟨v0​,u⟩2−⟨u,u⟩2​>0. Equivalently, both equalities ⟨v0,u⟩=∥u∥2\langle v_0,u\rangle=\|u\|^2⟨v0​,u⟩=∥u∥2 and ⟨v0,u⟩=−∥u∥2\langle v_0,u\rangle=-\|u\|^2⟨v0​,u⟩=−∥u∥2 fail outside a set of measure zero. For k=0k=0k=0 the hypothesis u≠0u\neq0u=0 is impossible, so the implication is vacuous. For u=0u=0u=0 or σ≤0\sigma\leq0σ≤0, no almost-everywhere conclusion is required.

Human review
  • Endorsed by Shuze Chen · Sep 27, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Minghui · Sep 27, 2026

    Confirmed by the mission captain (proposal self-audit).

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