Detecting same-slot inner products by smoothing and Fourier energy

This chapter reconstructs the same-slot branch of Lemma 6.1 on page 40 of H, using the one-parameter return argument of Lemma 4.3 on pages 16–18. The source is page 40 of the fixed October 5 version of An $L^3$ bound for the trilinear Hilbert transform.

We retain explicitly the arbitrarily large horizontal shift caused by the interval centre, instead of assuming an unquantified closure under translation. The analytic conclusion is again a bound on the frequency increment of every small return, to which the rational fast-splitting argument then applies. This proof does not use the two observation coordinates or the Bessel bound for side functions from the distinct-slot argument.

Back to the guide to H

1. The original columns and the required contradiction

Fix a slot $i\in\{0,1,2,3\}$ with $c_i\in\{-1,3,-3,1\}$ . The two bounded physical columns are

\phi_v(y)=e(c_i\lambda_vy^2+\beta_vy) \sum_{n_v\in\mathbb Z^{d_v}}\chi_v(\theta_vy-n_v) e\!\left(c_i(\theta_vy)^TC_v(\theta_vy-n_v)\right),

and $\phi_w$ of the same form. Assume $\|\phi_v\|_\infty,\|\phi_w\|_\infty\le1$ . The complexity parameter $H\ge\mathrm e^2$ from Definition 4.1 of the source bounds

d_v,d_w\le\log H,\quad \operatorname{ht}(C_v),\operatorname{ht}(C_w)\le H,

The amplitudes are supported in their respective boxes $[-H,H]^d$ , and the required first derivatives are at most $H^{C_0}$ . All $\theta,\lambda,\beta$ are arbitrary real parameters.

Fix a length $r>0$ and centre $x_I$ for an interval $I$ . Suppose that, for some $0<\tau_2\le1$ ,

\left|\frac1r\int_I\phi_v(y)\overline{\phi_w(y)}\,\mathrm{d}y\right|>\tau_2. \tag{Large inner product}

We will construct a splitting of the form (2) for the curvature difference data. If the graph-edge threshold $H_{\mathrm{ed}}$ is chosen in advance above the uniform construction bound, the large inner product forces the two tags to have distance at most $1$ .

2. Normalize without bounding the interval centre

Put $y=x_I+rz$ with $-1/2\le z\le1/2$ . Define

V=r(\theta_v,\theta_w),\quad a=x_I(\theta_v,\theta_w),\quad C=\operatorname{diag}(C_v,-C_w),\quad \Lambda=r^2(\lambda_v-\lambda_w).

Also write

\widetilde C=c_iC,\quad \widetilde\Lambda=c_i\Lambda,\quad \chi(q_v,q_w)=\chi_v(q_v)\overline{\chi_w(q_w)}.

The product is exactly

\Psi(z)=\phi_v(x_I+rz)\overline{\phi_w(x_I+rz)} =e(\widetilde\Lambda z^2+bz+c) \sum_{n\in\mathbb Z^{d_v+d_w}}\chi(a+Vz-n) e\!\left((a+Vz)^T\widetilde C(a+Vz-n)\right). \tag{Model with centre shift}

The real parameters $b,c,a$ may be arbitrarily large. The amplitude itself has not been translated or enlarged, so its support and first derivative retain the original quasipolynomial bounds. Also, $|\Psi|\le1$ .

Let the integer $P$ be the product of all reduced denominators of $C_v,C_w$ . Then $P\widetilde C$ and $PC$ are integer matrices, and

\log P\le(d_v^2+d_w^2)\log H\le C(1+\log H)^3. \tag{Clearing denominators}

3. Smooth the interval indicator with a quantitative error

Take a fixed nonnegative smooth transition function and smooth in boundary strips of length $O(\eta)$ , where $\eta=\tau_2/100$ . This gives $\vartheta_\eta\in C_c^\infty((-1/2,1/2))$ satisfying

0\le\vartheta_\eta\le1,\quad \|\mathbf1_{[-1/2,1/2]}-\vartheta_\eta\|_1\le4\eta,\quad \|\vartheta_\eta'\|_\infty\le C/\eta.

If full Gevrey bounds are needed, use the Gevrey transition function fixed in the source. Only the first derivative bound above is used here.

Define one fixed function

F(z)=\vartheta_\eta(z)\Psi(z).

The bound $|\Psi|\le1$ and the large inner product give

|\widehat F(0)|>\tau_2-4\eta>3\tau_2/4, \quad \widehat F(\xi)=\int_{\mathbb R}F(z)e(-\xi z)\,\mathrm{d}z. \tag{Initial Fourier value}

Also, $F$ is supported in a fixed interval of length $1$ , with $\|F\|_\infty\le1$ and $\|F\|_2\le1$ . The inequality

|e(-\xi z)-e(-\xi'z)|\le2\pi|\xi-\xi'|\,|z|

implies

|\widehat F(\xi)-\widehat F(\xi')| \le2\pi|\xi-\xi'|\int|zF(z)|\,\mathrm{d}z \le\pi|\xi-\xi'|. \tag{Fourier Lipschitz}

This Lipschitz estimate comes from the first moment on the fixed support, not from $F'$ . Arbitrarily large oscillation frequencies do not enlarge its constant.

4. The return formula with a large centre shift

Take

h\in\mathbb R,\quad m\in P\mathbb Z^d,\quad \delta=Vh-m,\quad d=d_v+d_w.

Write $\widetilde A=\widetilde C-\widetilde C^T$ and define

\widetilde\Omega(h,m)= (2\widetilde\Lambda+V^T\widetilde CV)h-V^T\widetilde Am.

For an integer summation index, put $u=a+Vz$ and $q=u-n$ . Translate $z$ to $z+h$ and the summation index to $n+m$ . The horizontal phase difference is exactly

u^T\widetilde C\delta+m^T\widetilde Cu +\delta^T\widetilde Cq +(m+\delta)^T\widetilde C\delta-m^T\widetilde Cn.

The last term is an integer. The first two terms equal

u^T(\widetilde CVh-\widetilde Am) =a^T(\widetilde CVh-\widetilde Am) +zV^T(\widetilde CVh-\widetilde Am).

The linear increment of the ordinary quadratic phase combines with the last $z$ term to give exactly $\widetilde\Omega z$ . The constant phase can be taken to be

\gamma(h,m)=\widetilde\Lambda h^2+bh +a^T(\widetilde CVh-\widetilde Am) +(m+\delta)^T\widetilde C\delta. \tag{Constant phase}

Both $a^T(\cdots)$ and $m^T\widetilde C\delta$ can be large, but they remain inside $e(\gamma)$ .

The only variable phase error is $\delta^T\widetilde Cq$ , whose size on the amplitude support is bounded by $D\|\delta\|_2$ . Translating the amplitude gives an error of the same order. At most $(2H+5)^d$ integer indices contribute, and $d\le2\log H$ . Together with the derivative bound for $\vartheta_\eta$ , this allows a choice

D\le\exp\!\left(C(1+\log H+\log(1/\tau_2))^C\right),\quad D\ge10,

such that whenever $|h|+\|\delta\|_2\le1/3$ ,

F(z+h)=e(\gamma(h,m))e(\widetilde\Omega(h,m)z)F(z)+E_{h,m}(z),

\|E_{h,m}\|_1+\|E_{h,m}\|_\infty \le D(|h|+\|\delta\|_2). \tag{Return of the fixed function}

The error is supported in the union of the original smoothed interval and a small translate, a set of absolutely bounded length. The uniform bound therefore gives the stated $L^1$ bound as well.

5. Choose the number of multiples before the return threshold

Take

\rho=\frac{\tau_2}{100D},\quad M=\left\lceil100D^2\tau_2^{-2}\rho^{-1}\right\rceil,\quad \varepsilon=\frac{\tau_2}{100DM}. \tag{Parameter order}

These quantities are fixed before choosing the return to be tested. If

|h|+\|Vh-m\|_2\le\varepsilon,

then the first $M$ multiples of the return satisfy the conditions of Section 4, because $M\varepsilon\le1/100$ . Moreover,

\widetilde\Omega(\ell h,\ell m)=\ell\widetilde\Omega(h,m).

Taking Fourier transforms in the return identity for the fixed function gives

e(\xi\ell h)\widehat F(\xi) =e(\gamma_\ell)\widehat F(\xi-\ell\widetilde\Omega) +\widehat E_{\ell h,\ell m}(\xi).

Set $\xi=0$ . The initial Fourier value and the $L^1$ error bound give, for every $1\le\ell\le M$ ,

|\widehat F(-\ell\widetilde\Omega)| \ge|\widehat F(0)|-D\ell\varepsilon > \tau_2/2. \tag{Several large values}

All these large values belong to the same $\widehat F$ . No additional identification of test functions belonging to different multiples is needed.

6. Disjoint frequency intervals contradict the total energy bound

Suppose $|\widetilde\Omega|\ge1$ . The centres $-\ell\widetilde\Omega$ are pairwise separated by at least $1$ . Since $\rho<1/100$ , the intervals of radius $\rho$ , of which there are $M$ , are disjoint.

The Fourier Lipschitz bound and $D\ge10$ imply, throughout these intervals,

|\widehat F(\xi)|>\tau_2/2-D\rho>\tau_2/3.

Hence

\|\widehat F\|_2^2 \ge M(2\rho)\frac{\tau_2^2}{9} \ge\frac{200}{9}D^2>D^2.

But Plancherel's identity with the present Fourier normalization gives

\|\widehat F\|_2^2=\|F\|_2^2\le1\le D^2,

This is a contradiction. Therefore every return below the same threshold satisfies

|\widetilde\Omega(h,m)|<1. \tag{Same-slot return bound}

There is no additional constant depending on the interval centre, scale, or linear modulation.

7. Remove the slot coefficient from the curvature data

Apply the rational fast-splitting argument to

(\widetilde C,\widetilde\Lambda,V,P,\varepsilon,B_\Omega=1).

It gives a rational space $U$ , an orthogonal splitting $V=V_f+V_s$ , and

\widetilde A|_{U\times U}=0,\quad \|V_s\|_2\le H',

\left|\widetilde\Lambda+\frac12V^T\widetilde C_{\mathrm{sy}}V+ V_f^T\widetilde C_{\mathrm{sk}}V_s\right|\le H'.

Since $c_i\ne0$ , the first condition is equivalent to $A|_{U\times U}=0$ . The effective curvature is exactly $c_i\Lambda_*$ , where

\Lambda_*=\Lambda+\frac12V^TC_{\mathrm{sy}}V+ V_f^TC_{\mathrm{sk}}V_s.

Because $|c_i|\ge1$ , the same $H'$ bounds $|\Lambda_*|$ . The projections and their heights are unchanged. Multiplication by the fixed integer $c_i$ costs at most a fixed power in the input complexity.

Combining the denominator-clearing, smoothing, and return-threshold bounds yields

H'\le\exp\!\left(C(1+\log H+\log(1/\tau_2))^C\right). \tag{Uniform edge bound}

Thus, with $H_{\mathrm{ed}}\ge H'$ chosen in advance from the uniform complexity of all allowed pure columns, a large inner product gives an edge of the graph. Taking the contrapositive,

d_s(v,w)>1\quad\Longrightarrow\quad \left|\frac1{|I|}\int_I\phi_v\overline{\phi_w}\right|\le\tau_2

holds at every depth $s$ on each whole cell $I$ .

8. Scope and the connection to endpoint Bessel bounds

The conclusion concerns physical inner products. Orthogonality of private coordinates from different groups must follow separately from the label data and endpoint separation of the masks. The physical estimate is not an unconditional estimate in the enhanced space.

With the classical input and detailed deduction of the rational fast-splitting chapter, this completes the same-slot branch of Lemma 6.1. Together with the two-slot return argument and the same fast-splitting result, it gives the two detection conclusions of Lemma 6.1. Exceptional sets for colouring, templates with the same tag, and the full catalog construction remain separate arguments.

The zero-dimensional horizontal case $d=0$ is also covered. The integer sum has one term, all horizontal matrix terms are empty, and the return formula contains only the ordinary quadratic phase. Smoothing on a finite interval and the Fourier energy contradiction require no change. The fast-splitting chapter treats this boundary case separately.