Passing fast spaces along a finite path

This chapter gives a close reading of Lemma 6.2 in H, on page 41 of the fixed October 5 version of An $L^3$ bound for the trilinear Hilbert transform. The splitting condition used below is equation (2) on page 16.

We assume that every edge of the path already has the splitting in (2). We show how to combine these edge splittings into an endpoint splitting, keeping track of the height bounds. The argument in Lemma 6.1 that detects an edge from a correlation is a separate input.

Back to the guide to H

1. Put all the data at one scale

Write the path as $v_0,\ldots,v_L$. The $j$th tag has horizontal dimension $d_j$, a real vector $\theta_j$, a rational matrix $C_j^{\mathrm{tag}}$, and a real number $\lambda_j$. Fix a physical scale $r>0$ and define

There is no bound on the size of $V_j$ or $\lambda_j$. Write the symmetric and skew-symmetric parts of a matrix as

We always write a rational number $p/q$ in lowest terms with $q>0$; its height is $\max(|p|,q)$. The height of a matrix is the largest height of its entries. The projection onto the zero subspace is assigned height $1$.

For edge $e=1,\ldots,L$, take the coordinate restriction

The superscript $\mathrm{tag}$ always denotes a matrix belonging to a single tag. Without this superscript, $C_e$ denotes the difference matrix for an edge.

2. The precise assumptions on each edge

Edge $e$ supplies a rational subspace $U_e$ and orthogonal projections $P_{f,e},P_{s,e}=I-P_{f,e}$. Set

Assume that some $B\ge2$ bounds the heights of all tag matrices and edge projections, and that

Absorbing $L+1$ and $n+1$ into $B$ only enlarges the bound. In H we can take

We will obtain the same kind of splitting in the endpoint coordinates, with bound

where $C$ is an absolute constant. The exponent does not involve the arbitrarily large real parameters $\theta_j,\lambda_j$.

3. The cost of one rational elimination

We first check a calculation that will be used repeatedly. Let $T$ be a rational matrix with at most $2N$ rows and $N$ columns, whose entries have height at most $B\ge2$. The product $q$ of all denominators satisfies

Choose a maximal independent set of rows of $M$ to form a matrix $A$ with $s$ rows. If $s=0$, the kernel is the whole space and the conclusions below are immediate. If $s>0$, then

is a positive definite integer matrix, so $\det G\ge1$. Put $X=B^{2N^2+1}$. Its entries and minors satisfy

The adjugate formula therefore gives entrywise bounds for

The absolute values of their entry numerators, their denominators, and their operator norms are all at most

The right inverse here applies to the image of the independent row system $A$. This does not assert that a system with redundant rows has a right inverse on the whole target space.

We will also use the following version. If $W$ is a rational matrix with at most $2N$ rows and $N$ columns, and its entries have height at most $B$, then the orthogonal projection onto $\operatorname{im}W$ has a bound of the same form. The proof is identical: clear denominators, choose independent columns, and use $W_0(W_0^TW_0)^{-1}W_0^T$. The chosen columns need not be nearly orthogonal.

4. A joint fast space and a controlled slow vector

In the joint coordinate space, define

This is the common kernel of the rational equations

Stack these equations into a matrix $T$. Its number of rows is

Its entries have height at most $B$. The preceding section bounds the rational height of $P_U$ by $Q_n$.

Set

We do not need a bound for $\|V\|$. What matters is the residual of each equation at $V$. Indeed,

Using the denominator clearing and independent row matrix $A$ from the preceding section, we have $\|AV\|_2\le q\sqrt L\,B$, while

Consequently,

The last inequality absorbs $L\le B$. Define $\widetilde F_e=R_eF$ and $\widetilde S_e=R_eS$. Then $\widetilde F_e\in U_e$, and all the $\widetilde S_e$ satisfy the same bound.

5. Changing the fast vector without reintroducing large parameters

Fix an edge and temporarily suppress its index. For fixed $V$, write

If $F,\widetilde F\in U_e$, put $\Delta=\widetilde F-F\in U_e$. Skew-symmetry gives $x^TC_{\mathrm{sk}}x=0$, so

Now write $V=F+S_e$. Since $U_e$ is isotropic for $A_e=2C_{\mathrm{sk},e}$,

The second equality is the essential point: its right-hand side contains only the original slow part for the edge. Also,

so $\|\Delta\|_2$ is bounded. A matrix norm is bounded by the dimension times the largest absolute entry. Thus, after replacing the edge fast vector by the joint fast vector, the effective curvature of each edge is still at most $\exp(C(n+1)^3\log(2B))$.

6. Cancellation of the intermediate coordinate blocks

Write the joint vectors in tag blocks as $F=(F_0,\ldots,F_L)$ and $S=(S_0,\ldots,S_L)$. After the replacement, the effective curvature of edge $e$ is

Here $V_e,F_e,S_e$ are the blocks for tag $e$, rather than the edge vectors denoted by these symbols earlier. Upon summing, every intermediate block occurs once with each sign and cancels exactly. Denote the endpoint restriction by $R_\partial$ and set

We obtain

with absolute value at most $L$ times the bound for one edge, which can still be absorbed into the stated bound. Independently chosen edge fast vectors would not cancel in this way. The cancellation uses the single choice of each tag block made in Section 4.

7. The endpoint fast space is an image

Define

This is an image space, not the intersection obtained by setting the intermediate coordinates to zero. Given $u',v'\in U'$, choose respective lifts $u,v\in U$. For every edge,

Adding these identities cancels the intermediate blocks and leaves

Thus $U'$ is isotropic for the endpoint skew-symmetric form. The lifts need not be unique.

Rationality and height also require a calculation. The image of the matrix $R_\partial P_U$ is $U'$, and its entries have height at most

The endpoint dimension is at most $n$ when $L\ge1$. Applying the column-space version of Section 3 gives the following height bound for the endpoint orthogonal projection $P_f'$:

This is the cost of the two eliminations. Merely asserting that the image space has bounded height would leave this estimate unjustified.

8. Endpoint orthogonalization is controlled by the slow part

Set

Since $F_0'\in U'$, the best-approximation property of orthogonal projection gives

Apply the calculation from Section 5 once more, now to the endpoint form:

Replacing $V_\partial$ by $S_0'$ here uses isotropy and $F_0',\Delta'\in U'$. Hence the endpoint quantity

satisfies a bound of the same form. The sixth-power dependence on dimension in the projection height, together with the finitely many products and sums, fits within the twelfth-power bound in Section 2.

9. Degenerate cases and the parameters in H

If some $d_j=0$, the corresponding blocks are empty and the formulas still apply. If $n=0$, it is enough to sum the scalar curvature bounds on the edges. If $L=0$, the two endpoints are two copies of the same tag coordinates; take the diagonal space

Its orthogonal projection has blocks $I/2$, the vector $V_\partial=(V_0,V_0)$ lies entirely in the fast space, and the difference curvature vanishes. This handles separately the case in which the endpoint dimension can be $2n$.

In the application in H, the dimension of a single tag, $\log H_{\mathrm{ed}}$, and $\log H_{\mathrm{basic}}$ are bounded by fixed polynomials in $k$, and $L\le\operatorname{poly}(P,k)$. Hence $n\le\operatorname{poly}(P,k)$, giving

This is the conclusion of Lemma 6.2. Here $P$ is the scale parameter on page 39, not the integer used later to clear denominators in the return calculation.

10. The role of this argument in the proof

We have shown that rational fast splittings on the edges produce an endpoint splitting with an explicit finite height bound. The proof uses two rational eliminations, two replacements justified by isotropy, and exact cancellation in the joint coordinates.

A separate analytic argument must explain why a non-negligible correlation between two tags gives an edge satisfying (2) in the first place. The argument for distinct slots first bounds the frequency increments of every sufficiently small return. Polar bodies and successive minima are then needed to organize these returns into a fast space. That is another part of Lemma 6.1.