Song–Zhang v1: polynomial coefficients control the spectral gap

Part of the first version of Song–Zhang, Chapter Song–Zhang, first version: polynomial estimates and curvature; the reading order is on the full proofs page.

Overview. This reconstructs Section 5 of the version-pinned Song & Zhang, 2026. The argument starts from a first eigenfunction, iterates a normalized inverse square root followed by differentiation, and bounds the mass removed by centering. A combinatorial recovery inequality converts polynomial testing into a bound on this loss. A convolution estimate absorbs all normalization defects, uniformly in the number of generations and the terminal polynomial degree. The only quantitative input beyond the profile in the statement is the universal coefficient estimate Theorem 7.1.

Analytic setup and normalized families¶

The analytic preparation Lemma 7.1, proved in Lemma 120.3, establishes the following facts for the present regular measure. The operator $H=-\Delta+\nabla W\cdot\nabla$ is the self-adjoint operator associated with the closed gradient form, its kernel is the constants, and its first positive eigenvalue $\lambda=C_P(\nu)^{-1}$ is attained. Its form domain is the weighted Sobolev space. On its operator domain,

In particular all second weak derivatives exist in $L^2$ and commute. Inverse powers act on centered functions. For centered form-domain $h$, $H^{-1/2}h\in\operatorname{Dom}(H)$, $\|H H^{-1/2}h\|_2^2=\|H^{1/2}h\|_2^2$, and $\|\nabla H^{-1/2}h\|_2^2=\|h\|_2^2$. Polynomial tests belong to the form domain. These facts apply componentwise to finite families, with summed squared norms.

Put $P_+h=h-\mathbb Eh$, $D=P_+\nabla H^{-1/2}$, and $Lh=\mathbb E[Xh]$. Both $D$ and $L$ are contractions: for $L$ use $\mathbb E(z\cdot X)^2\le|z|^2$ and duality. Testing the form against $x_i$ gives $\mathbb E\nabla H^{-1/2}h=LH^{1/2}h$.

For a nonzero centered form-domain family $u$, define

The scalar $\beta$ belongs to the entire family, not individual components. Cauchy--Schwarz gives $v^2\le b\|H^{1/2}u\|_2^2$, whence $\lambda\le\beta\le\|H^{1/2}u\|_2^2/v$ and $\chi\ge0$. Furthermore

satisfies the exact identity

This uses spectral calculus and the form identity; it does not commute a spatial derivative with an inverse spectral power.

Start with a centered unit first eigenfunction $F_0$. Set $u^j=DF_j$ and normalize $u^j$ as above to obtain $F_{j+1}$, normalizer $\beta_{j+1}$, and defect $\chi_j$. Write

The gradient before centering has norm squared $v_j$ and mean norm squared $p_j$. Bochner applied to $H^{-1/2}F_j$ bounds its derivative energy by $e_j-av_j$. Normalization removes exactly $\chi_j$. Thus

In particular $v_j\le1$, $e_j\le\lambda$, and $p_j\le\min(v_j,e_j)\le\lambda$. At the first step, even if $u^0=0$, $\lambda(1-p_0)\le\|H^{1/2}u^0\|_2^2\le\lambda-a$, so $a\le\lambda p_0\le\lambda^2$. The same inequalities hold with zero successors after a vanishing family, but normalizers will only be used on the nonvanishing stopped prefix constructed below. Each $u^j$ is in the form domain by Bochner; hence every operation on that prefix has the required domain.

Recovering a partially symmetric tensor¶

We will also use, for an $l$-slot array $Y$ and its adjacent transpositions $s_i$,

Indeed insertion sort writes each permutation with each generator at most $l$ times. For a product of isometries, telescope $I-U_1\cdots U_m$ as $\sum_{i=1}^mU_1\cdots U_{i-1}(I-U_i)$; all differences are thereby applied to the original array. Averaging the resulting bounds proves (2).

Polynomial tests, adjacent swaps, and the dyadic loss¶

Let $\mathcal A_k$ be the tensor-valued Appell polynomial, so $P_k[T]=\langle\mathcal A_k,T\rangle$, and put $Q_kh=\mathbb E[\mathcal A_kh]$. Duality gives $\|Q_k\|\le\sqrt{K_k}$. If $\mathsf P_{k+1}$ symmetrizes the $k$ polynomial indices and the newest derivative index, then for $j\ge1$

To check this, pair against an arbitrary symmetric $(k+1)$-tensor, use $\nabla P_{k+1}=(k+1)P_k$ with one free index, and test the form against $H^{-1/2}F_j$. Centering contributes zero because $\mathbb E\mathcal A_k=0$.

Suppose the normalizers in question satisfy $\beta_j\le b\lambda$, $b\ge1$. The normalization defect gives

The first term is symmetric in its newest two indices: $u^j$ is a centered gradient and taking another weak derivative gives a Hessian. Thus the newest swap has defect at most $2\sqrt{\chi_j/\lambda}$. Each subsequent map $D\sqrt{\beta}H^{-1/2}$ has norm at most $\sqrt b$ and commutes with permutations of old indices. Its scalar is kept fixed at the value for the original entire family, even when the map is applied to a difference. Numbering derivative slots newest first, the swap of slots $a,a+1$ in $u^J$ was created at $u^{J-a+1}$ and underwent exactly $a-1$ subsequent maps. Hence

No old defect has been differentiated.

Fix an integer $L\ge3$ and put

For $j\ge m_d$ and dyadic $d\ge2$, we claim

Here and below dyadic indices start at one. To prove (5), set $T_k=Q_ku^{j-k}$ at successive dyadic degrees. There are $k$ symmetric polynomial slots and $j-k+1\ge Lk$ derivative slots for $k<d$. Symmetrize the newest $Lk$ derivative slots to obtain $S_k$. The block lemma with $(s,l,q)=(k,Lk,2k)$ has recovery constant

Every denominator factor exceeds $(L-2)k$, which proves this bound. Repeated application of (3), whose inner projections are absorbed by the outer symmetrization, yields

Equations (2), (4) and $\|Q_k\|\le\sqrt{K_k}$ give

Thus $\|T_k\|\le B^k(b\lambda)^{k/2}k!/(2k)!\,\|T_{2k}\|+2B^kE_k$. The initial factor is $\sqrt{p_j}=\sqrt{\beta_j}\|T_1\|$. The preceding dyadic degrees sum to $k-1$, and the factorials telescope, so the coefficient before stage $k$ is at most

At the terminal stage $\|T_d\|\le\sqrt{K_d}$ because $\|u^{j-d}\|_2\le1$. Its contribution is $(A\lambda)^{d/2}c_d/B$. Multiplying the stage coefficient by $2B^kE_k$ gives exactly $t_k$ above. Every defect index is nonnegative: $j\ge(L+1)k-1$. This proves (5), including the boundary degree $d=2$.

Uniform absorption and stopping¶

Fences respected. The proposed node has no bounded_by edges. The argument proves a curvature-dependent general-test estimate, not CMH, a sharp third-moment constant, or a universal-time occupation estimate. In particular it does not invert any sufficient-condition implication in the brief. Its constants require $R\ge2^{40}\varepsilon^{-2}$ uniformly in the degree; retaining this threshold is essential when the theorem is iterated with depth-dependent $\varepsilon$. All polynomial assumptions are hypotheses of the displayed implication; the universal small-degree estimate is an actual dependency, not an assumed open antecedent. Independent review of this proof and that dependency is required.