The static quadratic-chaos input and the two-tail obstruction

The time-zero version of two-color covariance control is already a strong statement. The intrinsic estimate used here is Theorem 25.1, from Letwin’s July 2026 version-1 preprint Letwin, 2026. Its application to a localized posterior separates intrinsic quadratic control from the covariance-alignment problem created by unwhitening. The quantified two-tail configuration at the end of the chapter is the static obstruction that both branches of the fixed-cut approach must respect: it fixes the covariance weight of the near-Cheeger variant and marks the boundary of what slice-wise verification can establish for the all-cut variant.

Let $X\sim\nu$ be isotropic and log-concave, and put

Z=XX^T-I.

(25.1)

For a color

g=\frac{\one_E-p}{\sqrt{pq}}, \qquad p=\nu(E),\quad q=1-p,

(25.2)

one computes

\E[gZ]=\sqrt{pq}\left(\Sigma_E-\Sigma_F+(q-p)\delta\delta^T\right).

(25.3)

At exact balance, $p=q=1/2$ , this is $\frac12(\Sigma_E-\Sigma_F)$ .

Corollary 25.1 (Whitened two-color control and its exact alignment loss — Proved · agent review (gpt-6-astra, 2026-10-01) · cor:qcts-source · Correction)

Let $\nu$ be centered and log-concave with positive-definite covariance $A$ , let $E$ have mass $p\in(0,1)$ , put $s=p(1-p)$ , and let $K$ be the translation-invariant two-color contrast from Proposition 32.1. Then

s\norm{A^{-1/2}KA^{-1/2}}_{\HS}^2\le8, \qquad s\norm K_{\HS}^2\le8\lmax(A)^2.

(25.7)

If in addition $p\in[1/3,2/3]$ , then

s\norm G_{\HS}^2\le17\lmax(A)^2.

(25.8)

At exact balance, where $K=G$ , the constant in the last display is 8.

Proof 25.1

Whiten $Y=A^{-1/2}X$ . For any symmetric $N$ , Proposition 32.1 and Cauchy–Schwarz give

s\inner{A^{-1/2}KA^{-1/2}}{N}^2 \le \Var\bigl(Y^TNY\bigr) \le 8\norm N_{\HS}^2,

(25.9)

where the last step is Theorem 25.1. Taking the supremum over $\norm N_{\HS}\le1$ proves the first estimate in (25.7). Since $K=A^{1/2}(A^{-1/2}KA^{-1/2})A^{1/2}$ , the ideal property of the Hilbert–Schmidt norm gives the second estimate. Finally $G=K-(q-p)\delta\delta^T$ , so on the coarse balanced window, using $s\ge2/9$ , $\abs{q-p}\le1/3$ , and $r=s\abs{\delta}^2\le\lmax(A)$ from $B\preceq A$ ,

s\norm G_{\HS}^2 \le2s\norm K_{\HS}^2+2s(q-p)^2\abs{\delta}^4 \le16\lmax(A)^2+\frac{2(q-p)^2}{s}r^2 \le17\lmax(A)^2.

(25.10)

At balance the correction term vanishes.

Proof 25.2

Assume (25.5). For $\norm M_{\HS}=1$ ,

\inner{\E[gZ]}{M} =\E\left[g\left(X^TMX-\Tr M\right)\right] \le \sqrt{\Var(X^TMX)}\le\sqrt C .

(25.12)

Taking the supremum over $M$ and using (25.3) gives the two-color bound at balance; the slightly unbalanced case is handled by the explicit $(q-p)\delta\delta^T$ term and $B\preceq A$ .

Conversely, fix $M$ with $\norm M_{\HS}=1$ and set $Y=X^TMX-\Tr M$ . Since $\nu$ is isotropic, its convex support has nonempty interior and $\nu$ is absolutely continuous there. The polynomial $x\mapsto x^TMx$ is nonconstant, because $M\ne0$ , and each of its level sets has Lebesgue measure zero. Thus the law of $Y$ is atomless, and there is a median $m$ for which the deterministic cut $E=\{Y\ge m\}$ has mass exactly $1/2$ .

For this cut, $g=2\one_E-1=\operatorname{sgn}(Y-m)$ almost surely and $\E g=0$ , so

\E[gY]=\E[g(Y-m)]=\E\abs{Y-m}.

(25.13)

The balanced two-color estimate and (25.3) give $\norm{\E[gZ]}_{\HS}\le\sqrt C$ , hence $\E\abs{Y-m}=\inner{\E[gZ]}M\le\sqrt C$ . The Carbery–Wright reverse moment inequality for polynomials of degree at most two Carbery & Wright, 2001 now gives

\sqrt{\Var(Y)} \le\norm{Y-m}_{L^2} \lesssim\norm{Y-m}_{L^1} \lesssim\sqrt C.

(25.14)

Taking the supremum over symmetric $M$ with $\norm M_{\HS}=1$ proves $\Var(X^TMX)\lesssim C\norm M_{\HS}^2$ .

Why ordinary thin shell did not suffice¶

The thin-shell theorem controls the radial quadratic form:

\Var(\abs X^2)\le Cn,

(25.15)

which is (25.5) only for $M=I/\sqrt n$ . Applying thin shell to every projection $P$ gives

\Var(X^TPX)\le C\operatorname{rank}(P).

(25.16)

For a positive semidefinite matrix $M=\int_0^\infty P_s\dd s$ , where $P_s=\one_{\{M\ge s\}}$ , Minkowski yields

\norm{X^TMX-\Tr M}_{L^2} \le C\int_0^\infty\sqrt{\operatorname{rank}(P_s)}\dd s.

(25.17)

The right-hand side is the Lorentz $\ell_{2,1}$ norm of the eigenvalue sequence of $M$ , bounded by $C\sqrt{\log n}\norm M_{\HS}$ . Thus projection thin-shell estimates by themselves give only

\Var(X^TMX)\lesssim \log n\,\norm M_{\HS}^2,

(25.18)

a factor $\log n$ away from the dimension-free conclusion of Theorem 25.1.

Projection tests cannot remove the logarithm¶

The logarithm is not merely an artifact of the integration. There is an abstract positive operator $T$ on symmetric matrices such that

\inner{P}{TP}\le C\operatorname{rank}(P)

(25.19)

for every orthogonal projection $P$ , while $\norm T_{\op}\simeq\log n$ .

Let

H_n=\sum_{i=1}^n\frac1i, \qquad N=\diag\left(\frac1{\sqrt{H_n}},\frac1{\sqrt{2H_n}},\ldots,\frac1{\sqrt{nH_n}}\right),

(25.20)

so that $\norm N_{\HS}=1$ , and define

T(M)=H_n\inner{M}{N}N.

(25.21)

Ky Fan’s principle gives, for every rank- $r$ projection $P$ ,

\inner{P}{N}\le\sum_{i=1}^r\frac1{\sqrt{iH_n}}\le2\sqrt{\frac r{H_n}}.

(25.22)

Therefore $\inner{P}{TP}\le4r$ , whereas $\norm T_{\op}=H_n\simeq\log n$ . This is not a log-concave counterexample. It shows that projection tests alone cannot imply quadratic-chaos thin shell; Letwin’s proof escapes the obstruction by using moment measures and a matrix Stein kernel rather than only projection data.

Static lesson¶

A fixed-cut argument cannot rely only on radial information or on projection tests. Theorem 25.1 supplies the full intrinsic quadratic-chaos estimate, but Corollary 25.1 shows exactly what whitening loses: the Euclidean Riccati source may still be amplified by $\lmax(A_t)^2$ . The remaining task is therefore dynamic or geometric control of the alignment with $A_t$ , attached to the fixed cut $E$ .

The quantified two-tail obstruction and the weight calibration¶

The following sharpens the qualitative two-tail warning above by tracking the excess of the configuration, not only its covariance contrast. It is the static obstruction used in covariance-weighted form by the near-Cheeger variant (Chapter The fixed cut: the near-Cheeger variant) and the configuration whose dynamical occupation the all-cut variant must control (Chapter The fixed cut: the mass martingale and the Carleson estimate).

Proposition 25.2 (Quantified two-tail configuration — Proved · agent review (gpt-6-astra, 2026-10-04) · prop:two-tail · Correction)

Let $\Lambda\ge1$ , $\nu_\Lambda=N(0,\diag(\Lambda,1,\dots,1))$ on $\R^n$ , and

E_\Lambda=\bigl\{x:\abs{x_1}\ge a\sqrt\Lambda\bigr\}, \qquad a=\Phi^{-1}(3/4)\approx0.6745,

(25.23)

so that $\nu_\Lambda(E_\Lambda)=\tfrac12$ . Write $\varphi$ for the standard Gaussian density. Then:

(i) $\delta=0$ , hence $r=D=0$ and $K=G$ ;

(ii) $G=8a\varphi(a)\,\Lambda\,e_1e_1^T$ , hence $\calS_{\nu_\Lambda}(E_\Lambda)/s=16a^2\varphi(a)^2\Lambda^2\approx0.735\,\Lambda^2$ ;

(iii) $\nu_\Lambda^+(E_\Lambda)=2\varphi(a)\Lambda^{-1/2}$ and, exactly,

e_0(E_\Lambda)=\bigl(2\varphi(a)-\varphi(0)\bigr)\Lambda^{-1/2} \approx0.237\,\Lambda^{-1/2}\to0;

(25.24)

(iv) the relative excess is constant uniformly in $\Lambda$ and $n$ :

\frac{e_0(E_\Lambda)}{I_{\nu_\Lambda}(1/2)}\ =\ \frac{2\varphi(a)}{\varphi(0)}-1 \ \approx\ 0.593 .

(25.25)

Consequently, no inequality of the slice-wise form $\calS_\nu(E)/s\le C_0+C_1r+\beta D+C_2\,e(E)$ , with constants independent of the measure, can hold for all log-concave $\nu$ and all sets of mass $\tfrac12$ .

Proof 25.3

(i) is symmetry. (ii): conditioning on $\{\abs Z\ge a\}$ for $Z=X_1/\sqrt\Lambda\sim N(0,1)$ affects only the first coordinate, and the Gaussian integrations $\int_a^\infty z^2\varphi=a\varphi(a)+1-\Phi(a)$ , $1-\Phi(a)=\tfrac14$ , give $\E[Z^2\mid\abs Z\ge a]=1+4a\varphi(a)$ and $\E[Z^2\mid\abs Z<a]=1-4a\varphi(a)$ , hence $G_{11}=8a\varphi(a)\Lambda$ with all other entries zero; then $\calS/s=s\norm G_\HS^2=\tfrac14\cdot64a^2\varphi(a)^2\Lambda^2$ . (iii): the boundary consists of the two hyperplanes $\{x_1=\pm a\sqrt\Lambda\}$ , each of boundary density $\varphi(a)\Lambda^{-1/2}$ . To compute the excess, write $\nu_\Lambda=T_\#\gamma_n$ with $\norm T_\op=\sqrt\Lambda$ . Gaussian isoperimetry and whitening give $I_{\nu_\Lambda}(1/2)\ge\varphi(0)\Lambda^{-1/2}$ , while the long-axis halfspace $\{x_1\le0\}$ attains the reverse inequality. Thus $I_{\nu_\Lambda}(1/2)=\varphi(0)\Lambda^{-1/2}$ , proving (iii) and (iv). The final claim follows by letting $\Lambda\to\infty$ : the left side grows like $\Lambda^2$ , the right side is $C_0+C_2O(\Lambda^{-1/2})$ .

Remark 25.2 (Weight calibration and the rate

T^{1+\gamma}

)

Three consequences.

(a) No slice-wise proof. At $t=0$ the initial measure is isotropic and the configuration of Proposition 25.2 cannot occur; but anisotropic posteriors $\mu_t$ with $\lmax(A_t)=\Lambda\gg1$ are exactly the states the dynamics may visit, and there the slice-wise inequality fails. Any proof of the Stein-trace estimate must therefore show that such configurations carry negligible expected occupation — a covariance-occupation statement of the same nature as the Carleson problem itself. Thus the same stress configuration constrains both the excess-propagation and operator-to-trace proposals; no equivalence between them is asserted.

(b) The weight $(1+\norm A_\op)^{5/2}$ is calibrated, not chosen. Matching powers of $\Lambda$ ( $\calS/s\asymp\Lambda^2$ against $e\asymp\Lambda^{-1/2}$ ) identifies $(1+\norm{A}_\op)^{5/2}$ as the minimal pure power of $\lmax(A)$ multiplying absolute excess that is statically consistent with this slice inequality: indeed $e_0\Lambda^{5/2}=(2\varphi(a)-\varphi(0))\Lambda^2$ , and the ratio of the leading $\calS/s$ coefficient to this coefficient is approximately 3.11. The uniform value in (iv) of the relative excess is consistent evidence: two-tail sets are never relatively near-minimal in the Gaussian model, so a stability theory normalized at the Cheeger scale of the posterior is not contradicted by the example.

(c) What a $T^{1+\gamma}$ rate would require. By the Brascamp–Lieb cap (23.10) the weight is at most $(1+t^{-1})^{5/2}$ pathwise. If that cap were saturated deterministically, the pointwise condition $\E e_t\lesssim t^{5/2+\gamma}$ would be sufficient to produce a $T^{1+\gamma}$ integral; the formerly suggested $t^{3/2}$ rate would instead leave the nonintegrable factor $t^{-1}$ . A realistic proof would more likely need a joint rare-event bound coupling excess to large $\norm{A_t}_\op$ , rather than separate pointwise estimates. Thus the $T^{1+\gamma}$ term in (28.12) is a deliberately strong requirement, not a consequence of the static power matching in part (b).

What this chapter argues against¶

The two remarks below name the proof shapes the computations of this chapter argue against. Each records what a proof should not try to do, not a theorem; later chapters cite them as heuristic barriers, never as a step in a proof.

References¶

Letwin, B. (2026). The KLS Constant is O(\log1/4 n).
Carbery, A., & Wright, J. (2001). Distributional and Lq Norm Inequalities for Polynomials over Convex Bodies in \mathbbRn. Mathematical Research Letters, 8(3), 233–248. 10.4310/MRL.2001.v8.n3.a1