Proof of Theorem 2.1.

(assuming Propositions 5.1 and 5.2) First observe that, with $D = C_3 r^2$, the conditions required in Propositions 5.1 and 5.2 will be satisfied if $N> e^{C r^4 \log r}$ for a sufficiently large C. Also, in proving Theorem 2.1, we may clearly assume that r is sufficiently large.

First note that by Proposition 5.2 and equation (5.3), we have

$$\begin{align*}\mathbb{P}_{\theta, \mathbf{e}} \big( \operatorname{Red}_{\theta, \mathbf{e}} \ \mbox{has no} \ N^{1/r}\mbox{-term progression} \big) \geqslant 1 - O(N^{-1}). \end{align*}$$

This and Proposition 5.1 imply that there is some choice of $\theta , \mathbf {e}$ (in fact, a random choice works with very high probability) for which simultaneously $\operatorname {Blue}_{\theta , \mathbf {e}}$ has no 3-term progression and $\operatorname {Red}_{\theta , \mathbf {e}}$ has no $N^{1/r}$-term progression. This completes the proof of Theorem 2.1.

Proof. We have the parallelogram law

$$\begin{align*}2 \Vert y \Vert_2^2 = \Vert x + 2y \Vert_2^2 + \Vert x\Vert_2^2 - 2 \Vert x+y \Vert_2^2.\end{align*}$$

Applying this with $x = (1 + \sigma (e))u$, $y = (1 + \sigma (e)) v$ gives

$$\begin{align*}\Vert (1 + \sigma(e)) v \Vert_2^2 = \Vert y \Vert_2^2 \leqslant \rho^2 - (\rho - N^{-4/D})^2 < \frac{1}{10} N^{-4/D}.\end{align*}$$

The result now follows from equation (4.4).

Proof. We can address (1) and (2) separately and then take the intersection of the corresponding sets of $\theta $.

(1) Suppose that $\xi _1,\dots , \xi _{4r}$ are linearly independent over $\mathbf {Q}$. Then as x varies uniformly over $\mathbf {T}^D$, $(\xi _1\cdot x, \dots , \xi _{4r} \cdot x)$ is equidistributed (the proof is the same as in Proposition 4.1). It follows that for fixed n,

$$\begin{align*}\mu \{ \theta : \Vert n \xi_i \cdot \theta \Vert_{\mathbf{T}} \leqslant D^{C_2 D} X^{-1} \; \ \mbox{for}\ i = 1,\dots, 4r \} = (2D^{C_2 D})^{4r} N^{-4}.\end{align*}$$

By the union bound, the measure of $\theta $ for which there exists $n \leqslant N$ such that $\Vert n \xi _i \cdot \theta \Vert \leqslant D^{C_2 D} N^{-1/r}$ for $i = 1,\dots , 4r$ is $\leqslant (2D^{C_2 D})^{4r} N^{-3}$, which is comfortably less than $N^{-2}$ with our assumptions on $r, D$ and N.

If item (1) fails, then there must be some choice of $\xi _1,\dots , \xi _{4r} \in \mathbf {Z}^D$, $|\xi _i| \leqslant D^{C_2}$ for which the preceding statement holds. The number of choices for this tuple is at most $\big ((3D^{C_2})^{D} \big )^{4r} < N$. The result follows by another application of the union bound.

(2) The proof is easier to read if we set $c := \frac {1}{10}$ throughout (also a similar claim with any $c> 0$ would suffice for our purposes). It is enough to show that

(7.1)$$ \begin{align} \mu_{\mathbf{T}^D} \{ \alpha : \# \{ n \leqslant X : \Vert \alpha n \Vert_{\mathbf{T}^D} \leqslant X^{-1/D}\} \geqslant X^{1-c} \} \leqslant X^{-c^2 D/2}. \end{align} $$

Then one may take a union bound over all $\alpha = \theta d$, $d = 1,\dots , N$, noting that $N X^{-c^2 D} < N^{-1}$ with our assumptions on D and r (recall that $X = N^{1/r}$). To prove equation (7.1), we employ an inductive approach based on the following claim.

Claim. Suppose that $\alpha ' \in \mathbf {T}^{D'}$ and

(7.2)$$ \begin{align} \# \{ n \leqslant X : \Vert \alpha' n \Vert_{\mathbf{T}^{D'}} \leqslant X^{-(1 - c)/D'} \} \geqslant X^{1 -c}. \end{align} $$

Then there is some $\xi \in \mathbf {Z}^{D'}$, $0 < |\xi | \leqslant 2X^{3c/D'}$, such that $\Vert \xi \cdot \alpha ' \Vert _{\mathbf {T}} \leqslant 2X^{3c - 1}$.

Proof of claim. Inside the proof of the claim, we drop the dashes for clarity (if we did not include them in the statement, it would make the subsequent deduction of Proposition 7.1 (2) confusing). By Lemma 2.5, there is a smooth cutoff $\chi : \mathbf {T}^D \rightarrow [0,\infty )$ satisfying

1. 1. $\chi (x) \geqslant 1$ for $\Vert x \Vert _{\mathbf {T}^D} \leqslant X^{-(1-c)/D}$;

2. 2. $\int \chi \leqslant 5^D X^{c-1}$;

3. 3. $\hat {\chi }(\xi ) = 0$ for $|\xi | \geqslant X^{(1-c)/D}$.

Then if equation (7.2) holds (remember, we have dropped the dashes),

$$ \begin{align*} X^{1-c} \leqslant \sum_{n \leqslant X} \chi(\alpha n) & = \sum_{\xi} \hat{\chi}(\xi) \sum_{n \leqslant X} e(\xi \cdot \alpha n) \\[6pt] & \leqslant 5^DX^{c-1} \sum_{|\xi| \leqslant X^{(1-c)/D}}| \sum_{n \leqslant X} e(\xi \cdot \alpha n)| \\[6pt] & < \frac{1}{2}X^{2c-1} \sum_{|\xi| \leqslant X^{(1-c)/D}}\min(X, \Vert \xi \cdot \alpha \Vert_{\mathbf{T}}^{-1}).\end{align*} $$

(The factor $\frac {1}{2}$ is a minor technical convenience for later.) Therefore,

$$\begin{align*}2X^{2 - 3c} < \sum_{|\xi| \leqslant X^{(1-c)/D}} \min(X, \Vert \xi \cdot \alpha \Vert_{\mathbf{T}}^{-1}).\end{align*}$$

The contribution from those $\xi $ with $\Vert \xi \cdot \alpha \Vert _{\mathbf {T}} \geqslant X^{3c - 1}$ is bounded above by $(3 X^{(1 - c)/D})^D X^{1 - 3c} < X^{2 - 3c}$. Writing $\Xi \subset [-X^{1/D}, X^{1/D}]^D$ for the set of $\xi $, $|\xi | \leqslant X^{1/D}$, with $\Vert \xi \cdot \alpha \Vert _{\mathbf {T}} \leqslant X^{3c - 1}$, it follows that $X^{1 - 3c} < |\Xi |$. By the pigeonhole principle, dividing $[-X^{1/D}, X^{1/D}]^D$ into $X^{1 - 3c}$ boxes of side length $2X^{3c/D}$, we see that there are distinct $\xi _1, \xi _2 \in \Xi $ with $| \xi _1 - \xi _2| \leqslant 2X^{3c/D}$. Taking $\xi := \xi _1 - \xi _2$ completes the proof of the claim.

Let us resume the proof of Proposition 7.1 (2). Set $D_0 := \lceil (1 - c)D\rceil $. We prove, by induction on $j = 0,1,\dots , D - D_0$, the following statement: The measure of all $\alpha ^{(j)} \in \mathbf {T}^{D_0 + j}$ such that

(7.3)$$ \begin{align} \# \{ n \leqslant X : \Vert \alpha^{(j)} n \Vert_{\mathbf{T}^{D_0 + j}} \leqslant X^{-1/D} \} \geqslant X^{1-c} \end{align} $$

is at most $X^{- cj}$. This result is trivial for $j = 0$, and the case $j = D - D_0$ gives the result we are trying to prove: that is, equation (7.1), which implies Proposition 7.1 (2) as explained above.

To deduce the case j from $j-1$, apply the claim with $D' = D_0 + j$. Since $D' \geqslant (1 - c)D$, the condition in equation (7.3) implies that

$$\begin{align*}\# \{ n \leqslant X : \Vert \alpha^{(j)} n \Vert_{\mathbf{T}^{D'}} \leqslant X^{-(1-c)/D'} \} \geqslant X^{1 - c},\end{align*}$$

and so by the claim there is some $\xi \in \mathbf {Z}^{D'}$, $0 < |\xi | \leqslant X^{3c/D_0}$ such that $\Vert \xi \cdot \alpha ^{(j)} \Vert _{\mathbf {T}} \leqslant 2X^{3c - 1}$. Suppose that the last nonzero coordinate of $\xi $ is $\xi _{D'} = \xi _{D_0 + j}$ (the other possibilities can be treated similarly with very minor notational changes). Form $\alpha ^{(j-1)} \in \mathbf {T}^{D_0 + j - 1}$ by restricting $\alpha ^{(j)}$ to the first $D_0 + j - 1$ coordinates (i.e., by dropping the last coordinate $\alpha ^{(j)}_{D_0 + j}$). Then, since $\Vert \alpha ^{(j-1)} n \Vert _{\mathbf {T}^{D_0 + j -1}} \leqslant \Vert \alpha ^{(j)} n \Vert _{\mathbf {T}^{D_0 + j}}$, the hypothesis in equation (7.3) is satisfied by $\alpha ^{(j-1)}$. By the inductive hypothesis, the measure of possible $\alpha ^{(j-1)}$ is at most $X^{-c(j-1)}$. For each such $\alpha ^{(j-1)}$, and for each fixed $\xi $, the final coordinate satisfies $\Vert \gamma + \xi _{D_0 + j} \alpha ^{(j)}_{D_0 + j} \Vert _{\mathbf {T}} = \Vert \xi \cdot \alpha ^{(j)}\Vert _{\mathbf {T}} \leqslant 2X^{3c - 1}$, where $\gamma $ depends only on $\alpha ^{(j-1)}$ and the first $D_0 + j - 1$ coordinates of $\xi $. As $\alpha ^{(j)}_{D_0 + j}$ varies uniformly over $\mathbf {T}$, so does $\xi _{D_0 + j} \alpha ^{(j)}_{D_0 + j}$, so the probability of this event is $\leqslant 4X^{3c - 1}$.

Summing over all possible choices of $\xi $ (of which there are at most $(3 X^{3c/D_0})^{D}$), it follows that the total measure of $\alpha ^{(j)}$ satisfying equation (7.3) is bounded above by

$$\begin{align*}4X^{3c - 1} \cdot (3 X^{3c/D_0})^{D} \cdot X^{-c(j-1)} < X^{-cj}.\end{align*}$$

(The key calculation for this last step is that $3c - 1 + \frac {3c}{1 - c} < - c$, which is certainly true for $c = \frac {1}{10}$, and we used the assumption that $N> D^{D^2}$ to comfortably absorb the $3^D$ term into a tiny power of $X = N^{1/r}$.)

This completes the inductive step, and hence equation (7.3) is true for all j. As previously remarked, the case $j = D - D_0 \geqslant cD/2$ gives Proposition 7.1 (2).

Question 7.2. Define $\Theta ' \subset \mathbf {T}^{r^2}$ to be the set of all $\theta $ for which $\# \{ n \leqslant N^{1/r} : \Vert \theta d n \Vert _{\mathbf {T}^{r^2}} \leqslant N^{-1/r}\} \leqslant N^{1/10r}$ for all $d \leqslant N^{1 - 1/r}$. Is $\mu _{\mathbf {T}^{r^2}}(\Theta ') \geqslant \frac {1}{2}$, for N large enough in terms of r?

Proof of Proposition 5.3.

Set $\alpha := \theta d$. Set $n_1 := n_0 + \lfloor \frac {X}{4} \rfloor d$ so that $P_{\operatorname {init}} \supset \{ n_1 + nd : |n| \leqslant X/5\}$. Set

(8.1)$$ \begin{align} \Lambda := \{ \xi \in \mathbf{Z}^D : |\xi| < \rho^{-3}, \Vert \xi \cdot \alpha \Vert_{\mathbf{T}} \leqslant \rho^{-2D} X^{-1}\}. \end{align} $$

By the definition of $\Theta $ (specifically, item (1) of Proposition 7.1, and assuming that $C_2 \geqslant 12$), we have $\dim _{\mathbf {Q}}\Lambda < 4r$. By Proposition 4.1, there is some j such that

(8.2)$$ \begin{align} \Vert \xi \cdot (x_j - \theta n_1) \Vert_{\mathbf{T}} \leqslant 10^{-2} \end{align} $$

for all $\xi \in \Lambda $. We claim that $\theta P_{\operatorname {init}}$ intersects the ball $x_j + \pi (B_{\rho /10}(0))$. To this end, take a function $\chi : \mathbf {T}^D \rightarrow \mathbf {R}$ with the following properties:

1. 1. $\chi (x) \leqslant 0$ outside of $\pi (B_{\rho /10}(0))$;

2. 2. $\hat {\chi }$ is real and nonnegative;

3. 3. $\hat {\chi }(\xi )$ is supported on $|\xi | \leqslant \rho ^{-3}$;

4. 4. $\int \chi = 1$;

5. 5. $\int |\chi | \leqslant 3$.

Such a function is constructed in Lemma 2.6. Take also a function $w : \mathbf {Z} \rightarrow [0,\infty )$ satisfying

1. (a) w is supported on $[-X/5, X/5]$;

2. (b) $\hat {w} : \mathbf {T} \rightarrow \mathbf {C}$ is real and nonnegative;

3. (c) $\sum _{n \in \mathbf {Z}} w(n) \geqslant X$;

4. (d) $|\hat {w}(\beta )| \leqslant 2^5 X^{-1} \Vert \beta \Vert _{\mathbf {T}}^{-2}$ for all $\beta \in \mathbf {T}$.

For this, one can take a Fejér kernel: see Lemma 2.2 for details.

Then it is enough to show that

(8.3)$$ \begin{align} \sum_{n \in \mathbf{Z}} w(n)\chi(\theta (n_1 + nd) - x_j)> 0. \end{align} $$

Indeed, if this holds, then there must be some $n \in \operatorname {Supp}(w)$ (and hence $|n| \leqslant X/5$) such that $\chi (\theta (n_1 + nd) - x_j)> 0$, which means $\theta (n_1 + nd) - x_j \in \pi (B_{\rho /10}(0))$. Thus $n' := n_1 + nd$ lies in $P_{\operatorname {init}}$ and $\theta n' \in x_j + \pi (B_{\rho /10}(0))$, as required.

It remains to establish equation (8.3). By Fourier inversion on $\chi $ (recalling that $\alpha = d \theta $), the LHS of equation (8.3) is

(8.4)$$ \begin{align} \nonumber \sum_{\xi \in \mathbf{Z}^D} \hat{\chi}(\xi) e(\xi \cdot (n_1 \theta - & x_j)) \sum_{n \in \mathbf{Z}} w(n) e(\xi \cdot \alpha n) \\ & = \sum_{\xi \in \mathbf{Z}^D} \hat{\chi}(\xi) e(\xi \cdot (n_1 \theta - x_j))\hat{w}(-\xi \cdot \alpha) \nonumber \\ & = \sum_{\xi \in \mathbf{Z}^D} \hat{\chi}(\xi) \cos (2\pi\xi \cdot (n_1 \theta - x_j))\hat{w}(-\xi \cdot \alpha) , \end{align} $$

where the last line follows by taking real parts, noting that the LHS of equation (8.3) is real, as are $\hat {\chi }$, $\hat {w}$. Note that by (3), the sum here is supported where $|\xi | \leqslant \rho ^{-3}$. We now divide into the contributions from the following three different classes of $\xi $: (i) $\xi = 0$; (ii) $\xi \in \Lambda $ and (iii) $|\xi | \leqslant \rho ^{-3}$ but $\xi \notin \Lambda $.

(i) The contribution from $\xi = 0$ is $\hat {\chi }(0) \hat {w}(0) = (\int \chi ) (\sum _{n \in \mathbf {Z}} w(n)) \geqslant X$, using the properties of $\chi $ and w listed above.

(ii) If $\xi \in \Lambda $, then by equation (8.2), we have $\cos (2\pi \xi \cdot (n_1 \theta - x_j))> 0$. Since $\hat {\chi }$, $\hat {w}$ are both real and nonnegative, the contribution of these terms to equation (8.4) is nonnegative.

(iii) By the triangle inequality, the contribution to equation (8.4) from these $\xi $ is bounded above by

(8.5)$$ \begin{align} \sum_{\xi \notin \Lambda, |\xi| \leqslant \rho^{-3}} |\hat{\chi}(\xi)| |\hat{w}(-\xi \cdot \alpha) | \leqslant \big( \sum_{|\xi| \leqslant \rho^{-3}} |\hat{\chi}(\xi)|\big) \sup_{\Vert \beta \Vert_{\mathbf{T}}> \rho^{-2D} X^{-1}} |\hat{w}(\beta)|, \end{align} $$

where the second step follows by the definition in equation (8.1) of $\Lambda $. Now from (5) above, we have $|\hat {\chi }(\xi )| \leqslant \int |\chi | \leqslant 3$ for all $\xi $, so

(8.6)$$ \begin{align} \sum_{|\xi| \leqslant \rho^{-3}} |\hat{\chi}(\xi)| \leqslant 3 (3\rho^{-3})^D < 2^{-6}\rho^{-4D} \end{align} $$

(here, of course, we have used the fact that D is sufficiently large). By property (4) of w, we have

(8.7)$$ \begin{align} \sup_{\Vert \beta \Vert_{\mathbf{T}}> \rho^{-2D} X^{-1}} |\hat{w}(\beta)| \leqslant 2^5 \rho^{4D} X.\end{align} $$

Combining equations (8.5), (8.6) and (8.7), we see that the total contribution from $\xi $ in (iii) is at most $X/2$ in magnitude.

Summing the estimates we have obtained under (i), (ii) and (iii), it follows from equation (8.4) that the LHS of equation (8.3) is at least $X + 0 - X/2 = X/2$ and so is indeed positive, which is what we aimed to prove.

Proof of Proposition 5.4.

(assuming Propositions 9.1, 9.2 and 9.3) We start by invoking Proposition 9.1. The contribution to $\prod _{i = 1}^s L_i$ from those i such that $L_i \leqslant X^{1/160 D}$ is at most $X^{1/160}$. Removing these i (which may involve reducing s), we may assume that $X^{1/160 D} \leqslant L_i \leqslant X^{C_1/D}$ for all i. Dividing into $O(\log C_1) < C_1/160$ ranges, we may further assume that that there is some L,

(9.2)$$ \begin{align} X^{1/2^8 D} \leqslant L \leqslant X^{C_1/D}, \end{align} $$

such that $L_i \in [L, L^{1 + 1/48}]$ for all i, at the expense of weakening Proposition 9.1 (1) to

(9.3)$$ \begin{align} \prod_{i = 1}^s L_i \geqslant X^{1/C_1}. \end{align} $$

The upper bound in equation (9.2) then implies that

(9.4)$$ \begin{align} s \geqslant D/C_1^2. \end{align} $$

Now recall that our task is to show that, with high probability in $\mathbf {e}$, for all $y \in \pi (B_{\rho /10}(0))$, the orbit $y + \theta \dot {P}$ contains a point in $\pi (A_{\mathbf {e}})$, where

$$\begin{align*}A_{\mathbf{e}} := \{ x \in \mathbf{R}^D : \rho - N^{-4/D} < \Vert (1 + \sigma(\mathbf{e})) x \Vert_2 \leqslant \rho \}.\end{align*}$$

Let $z := \pi ^{-1}(y) \in \mathbf {R}^D$, thus

(9.5)$$ \begin{align} \Vert z \Vert_2 \leqslant \rho/10.\end{align} $$

Recall that $v_i := \pi ^{-1}(\theta d n_i)$. Thus, by Proposition 9.1, any element of the form $\pi ( \sum _i \ell _i v_i )$, $\ell _i < L_i$, lies in $\theta \dot {P}$ (since $\pi ( \sum _i \ell _i v_i ) = (\sum _i \ell _i n_i) d \theta $). Thus it suffices to show that with probability $1 - O(N^{-3})$ in the random choice of $\mathbf {e}$, any set of the form

$$\begin{align*}\{ z + \sum_i \ell_i v_i : \ell_i \in \mathbf{N}, 0 \leqslant \ell_i < L_i \}, \quad \Vert z \Vert_2 < \rho/10, \end{align*}$$

has nontrivial intersection with $A_{\mathbf {e}}$; that is to say, there is some choice of the $\ell _i$ such that

(9.6)$$ \begin{align} \rho - N^{-4/D} < \Vert (I + \sigma(\mathbf{e})) (z + \sum_i \ell_i v_i ) \Vert_2 < \rho.\end{align} $$

In other words, it is enough that

(9.7)$$ \begin{align} 1 - N^{-4/D} \leqslant q_{a(\mathbf{e}), b(\mathbf{e}), c(\mathbf{e})}(\frac{\ell_1}{L_1}, \dots, \frac{\ell_{s}}{L_{s}}) \leqslant 1, \end{align} $$

where the quadratic form is given by

(9.8)$$ \begin{align} q_{a(\mathbf{e}), b(\mathbf{e}), c(\mathbf{e})}(\frac{\ell_1}{L_1}, \dots, \frac{\ell_{s}}{L_{s}}) := \rho^{-2} \Vert (I + \sigma(\mathbf{e}))(z + \sum_{i = 1}^{s} \ell_i v_i ) \Vert_2^2. \end{align} $$

One computes

$$\begin{align*}c(\mathbf{e}) := \rho^{-2} \Vert (I + \sigma(\mathbf{e})) z \Vert_2^2,\qquad b_i(\mathbf{e}) := 2\rho^{-2} \langle (I + \sigma(\mathbf{e})) z, (I + \sigma(\mathbf{e}))v_i\rangle L_i,\end{align*}$$

$$\begin{align*}a_{ij}(\mathbf{e}) := 2\rho^{-2} \langle (I + \sigma(\mathbf{e})) v_i, (I + \sigma(\mathbf{e})) v_j \rangle L_i L_j,\end{align*}$$

$$\begin{align*}a_{ii}(\mathbf{e}) := \rho^{-2} \Vert (I + \sigma(\mathbf{e})) v_i \Vert_2^2 L_i^2. \end{align*}$$

Set $Q := D^{10}$. We claim that the conditions $a_{ii}(\mathbf {e}) \geqslant 32$, $|a_{ij}(\mathbf {e})|$, $|b_i(\mathbf {e})| \leqslant Q$, $|c(\mathbf {e})| \leqslant \frac {1}{4}$ and $b_i(\mathbf {e})^2 - 4a_{ii}(\mathbf {e})c(\mathbf {e}) < 0$ of Proposition 9.3 are automatically satisfied.

To prove these, recall equation (4.4) that

(9.9)$$ \begin{align} \frac{1}{2} \leqslant \Vert I + \sigma(\mathbf{e}) \Vert \leqslant 2, \end{align} $$

and observe the bounds

(9.10)$$ \begin{align} \frac{1}{L_i} = \Vert \theta d n_i \Vert_{\mathbf{T}^D} = \Vert v_i \Vert_{\infty} \leqslant \Vert v_i \Vert_2 \leqslant D^{1/2}\Vert v_i \Vert_{\infty} = \frac{D^{1/2}}{L_i}. \end{align} $$

Using equations (9.9) and (9.10) and recalling that $\rho = D^{-4}$ with D large, we have the following. First,

$$\begin{align*}a_{ii}(\mathbf{e}) \geqslant \frac{1}{4} \rho^{-2} \Vert v_i \Vert_2^2 L_i^2> 32.\end{align*}$$

Second, by Cauchy-Schwarz,

$$\begin{align*}|a_{ij}(\mathbf{e})| \leqslant 2\rho^{-2}\Vert I + \sigma(\mathbf{e}) \Vert^2 \Vert v_i \Vert_2 \Vert v_j \Vert_2 L_i L_j \leqslant \frac{8D}{\rho^2} < Q.\end{align*}$$

Third,

$$\begin{align*}|b_i(\mathbf{e})| \leqslant 2 \rho^{-2} \Vert I + \sigma(\mathbf{e}) \Vert^2 \Vert z \Vert_2 \Vert v_i \Vert_2 L_i < Q.\end{align*}$$

Fourth, since $\Vert z \Vert _2 \leqslant \rho /10$, and by equation (9.9), we have

$$\begin{align*}c(\mathbf{e}) \leqslant \rho^{-2}\Vert I + \sigma(\mathbf{e}) \Vert^2 \Vert z \Vert^2_2 < \frac{1}{4}.\end{align*}$$

Finally, the discriminant condition $b_i(\mathbf {e})^2 - 4a_{ii}(\mathbf {e})c(\mathbf {e}) < 0$ is automatic from the positive-definiteness of the quadratic form (and is also immediate using Cauchy-Schwarz from the formulae above).

With all the relevant conditions having been verified, we are in a position to apply Proposition 9.3 with $Q = D^{10}$, with L as selected above (satisfying equation (9.2)), $s \geqslant D/C_1^2$ by equation (9.4) and the lengths $L_1,\dots , L_s \in [L, L^{1+1/48}]$ as given above and with $B := 2^{15}C_1^2 r$. We must first check that the application is valid by confirming that $L \geqslant (QB)^{C_2 B}$ and $s \geqslant C_1 B^2$. Using the assumption that $N> D^{D^2}$ and recalling that $D = C_3 r^2$, one may check using equations (9.2) and (9.4) that this is indeed the case, if $C_3$ is big enough.

Thus all of the conditions are satisfied. Now observe that with the choice of B we have made (and in view of the lower bound in equation (9.2), and recalling that $X = N^{1/r}$), the fact that $B \geqslant 2^{10} r$ then implies that

(9.11)$$ \begin{align} L^{-B} \leqslant N^{-4/D}, \end{align} $$

the importance of the right-hand side here being that this is the width of our ellipsoidal annuli.

Proposition 9.3 therefore tells us that indeed $q_{a(\mathbf {e}), b(\mathbf {e}), c(\mathbf {e})}(\frac {\ell _1}{L_1}, \dots , \frac {\ell _{s}}{L_{s}})$ takes values in $[1 -N^{-4/D}, 1]$ (that is, equation (9.7), and hence equation (9.6) hold) provided that $a(\mathbf {e}) := (a_{ij}(\mathbf {e}))_{1 \leqslant i \leqslant j \leqslant s} \in \Sigma $, where $\Sigma $ is the event appearing in Proposition 9.3. To complete the proof of Proposition 5.4, it therefore suffices to show that

(9.12)$$ \begin{align} \mathbb{P}_{\mathbf{e}} (a(\mathbf{e}) \in \neg \Sigma) \ll N^{-3}. \end{align} $$

Now, using the fact that $X = N^{1/r}$ together with equations (9.2) and (9.4), we see that if a is chosen randomly as in Proposition 9.3 (that is, uniformly from $[-Q,Q]^{s(s+1)/2}$ with diagonal terms $\geqslant 32$), then, from the conclusion of Proposition 9.3, we have

(9.13)$$ \begin{align} \mathbb{P}_a(a \in \neg \Sigma) \ll L^{- s B/16} \leqslant N^{-\frac{2^{-8}}{D} \cdot \frac{1}{16} \cdot \frac{D}{C_1^2} \cdot B \cdot \frac{1}{r}} = N^{-8}. \end{align} $$

Now, as $\mathbf {e}$ varies uniformly, $a(\mathbf {e})$ may not be close to a uniformly random element of $[-Q, Q]^{s(s+1)/2}$, so equations (9.12) and (9.13) are not trivially comparable. To link the two statements we invoke Proposition 9.2, taking the $w_i$ to be the normalised $v_i$s, as in the conclusion of Proposition 9.1. Observe that $a(\mathbf {e}) = \psi (f(\mathbf {e}))$, where f is the map in Proposition 9.2 and $\psi : \mathbf {R}^{s(s+1)/2} \rightarrow \mathbf {R}^{s(s+1)/2}$ is the diagonal linear map

$$\begin{align*}(\psi(\mathbf{x}))_{ij} = \rho^{-2} \Vert v_i \Vert_2 \Vert v_j \Vert_2 L_i L_j x_{ij}. \end{align*}$$

Recall that $\Vert v_i \Vert _{\infty } = L_i^{-1}$, so the determinant of $\psi $ is at least $1$ (in fact, much bigger).

Therefore, we have, by Proposition 9.2 and Lemma 9.1 (3),

$$ \begin{align*} \mathbb{P}_{\mathbf{e}}(a(\mathbf{e}) \in \neg\Sigma) & = \mathbb{P}_{\mathbf{e}} ( f(\mathbf{e}) \in \psi^{-1}(\neg \Sigma)) \\ & \leqslant D^{4D^2} \operatorname{vol}(w_1,\dots, w_s)^{-D - 1} \mu (\psi^{-1}(\neg\Sigma)) \\ & \leqslant D^{4D^2}\cdot D^{C_1D(D+1)} \cdot \mu(\psi^{-1}(\neg \Sigma)) \\ & \leqslant D^{4D^2}\cdot D^{C_1D(D+1)} \cdot (2Q)^{s(s+1)/2} \cdot \mathbb{P}_a(a \in \neg \Sigma). \end{align*} $$

In the last step, we used the fact that the determinant of $\psi ^{-1}$ is at most $1$. Recalling equation (9.13), the fact that $Q = D^{10}$ and $s \leqslant D$, we see that this is $< N^{-3}$ provided that $N \geqslant D^{C_2 D^2}$, if $C_2$ is chosen sufficiently large. This concludes the proof of Proposition 5.4.

Proof. If $w_1,\dots , w_{m-1}$ are linearly dependent then this is clear, so suppose they are not. Let the foot of the perpendicular from $w_m$ to $\operatorname {Span}_{\mathbf {R}}(w_1,\dots , w_{m-1})$ be $x_1 w_1 + \dots + x_{m-1} w_{m-1}$. Then the fact that $v = w_m - x_1 w_1 - \dots - x_{m-1} w_{m-1}$ is orthogonal to $w_1,\dots , w_{m-1}$ gives us $m-1$ linear relations

(10.1)$$ \begin{align} \langle w_1, w_i\rangle x_1 + \cdots + \langle w_{m-1}, w_i\rangle x_{m-1} = \langle w_m, w_i\rangle,\end{align} $$

$i = 1,\dots , m-1$. Writing $y := \Vert v \Vert _2^2$ for the length of the perpendicular, Pythagoras’s theorem gives

(10.2)$$ \begin{align} \langle w_1, w_m\rangle x_1 + \dots + \langle w_{m-1}, w_m \rangle x_{m-1} + y = \langle w_m, w_m\rangle. \end{align} $$

Combining equations (10.1) and (10.2) into an $m \times m$ system of equations in the variables $x_1,\dots , x_{m-1}, y$, it follows from Cramer’s rule that $y = \frac {G(w_1,\dots , w_m)}{G(w_1,\dots , w_{m-1})}$, which is equivalent to the stated result.

Proof. Choose $w_{m+1},\dots , w_n$ to be orthogonal to one another and to $w_1,\dots , w_m$.

Proof. We perform the following algorithm for as long as possible. Start, at stage 1, with $w_1$. At the ith stage, we will have (after reordering) $w_1,\dots , w_i$. If

$$\begin{align*}\operatorname{dist}(w_j, \operatorname{Span}(w_1,\dots, w_i)) \leqslant \delta^{1/k}\end{align*}$$

for all $j \in \{1,\dots ,m\}$, then stop. If this happens for some $i \leqslant k$, then we have (2). Otherwise, we may reorder $w_{i+1},\dots , w_m$ so that $\operatorname {dist}(w_{i+1},\operatorname {Span}(w_1,\dots , w_i))> \delta ^{1/k}$. By Lemma 10.1 (and a simple induction), we have $\operatorname {vol}(w_1,\dots , w_{i+1})> \delta ^{i/k}$. If this continues as far as $i = k$, then we have (1).

Proof. We may assume, by adding elements if necessary, that $\dim V = \lfloor (1 - \varepsilon ) D\rfloor $. Set $m := \lceil \varepsilon D\rceil $, $\delta := D^{-C_1}$ and $S := \mathcal {N}_{D^{-C_1}}(V) \cap B_{1/10}(0)$. Let $(v_i)_{i = 1}^D$ be an orthonormal basis for $\mathbf {R}^D$ with $V = \operatorname {Span}_{\mathbf {R}}(v_{m+1},\dots , v_D)$. Let $\psi : \mathbf {R} \rightarrow \mathbf {R}$ be a smooth cutoff satisfying the following conditions:

1. 1. $\psi \geqslant 0$ everywhere, and $\psi (x) \geqslant 1$ for $|x| \leqslant 1$;

2. 2. $\hat {\psi }(y) = 0$ for $|y| \geqslant 1$;

3. 3. $\int \psi \leqslant 5$.

For the proof that such a function exists; see Lemma 2.4. Now define $\chi : \mathbf {R}^D \rightarrow [0,\infty )$ by

(11.2)$$ \begin{align} \chi(x) = \prod_{i = 1}^{D} \psi(\delta^{-1_{i \leqslant m}}\langle x, v_i\rangle), \end{align} $$

where the notation means that the scale factor $\delta ^{-1}$ is included only for $i \leqslant m$ (that is, in the directions orthogonal to V) and is otherwise set equal to 1. If $x \in \mathcal {N}_{\delta }(V)$ and $i \leqslant m$, then

$$\begin{align*}|\langle x, v_i \rangle| \leqslant \big(\sum_{i = 1}^m |\langle x, v_i \rangle|^2\big)^{1/2} = \operatorname{dist}(x, V) \leqslant \delta,\end{align*}$$

and if $x \in B_{1/10}(0)$, then $|\langle x, v_i \rangle | < 1$ for all i. It follows from these observations and property (1) of $\psi $ that $\chi (x) \geqslant 1$ for $x \in S$, and therefore

(11.3)$$ \begin{align} \# \big( \theta d[X] \cap \pi(S) \big) \leqslant \sum_{n \leqslant X} \sum_{\lambda \in \mathbf{Z}^D} \chi(\theta d n + \lambda). \end{align} $$

The Poisson summation formula tells us that for any $t \in \mathbf {T}^D$, we have

$$\begin{align*}\sum_{\lambda \in \mathbf{Z}^D} \chi(\lambda + t) = \sum_{\xi \in \mathbf{Z}^D} \hat{\chi}(\xi) e( \xi \cdot t).\end{align*}$$

Substituting into equation (11.3) therefore implies that

(11.4)$$ \begin{align} \# \big( \theta d [X] \cap \pi(S) \big) \leqslant \sum_{\xi \in \mathbf{Z}^D} \hat{\chi}(\xi) \sum_{n \leqslant X} e( \xi \cdot \theta d n). \end{align} $$

To proceed further, we need to understand the Fourier transform $\hat {\chi }$, particularly at points of $\mathbf {Z}^D$. First, it follows from property (3) of $\psi $ above that

(11.5)$$ \begin{align} \Vert \hat{\chi} \Vert_{\infty} \leqslant \int \chi = \delta^m (\int \psi)^D \leqslant 5^D \delta^{m}. \end{align} $$

Next, expanding in the orthonormal basis $(v_j)_{j = 1}^D$ and then changing variables, we have

$$ \begin{align*} \hat{\chi}(\gamma) & = \int_{\mathbf{R}^D} \prod_{j = 1}^D \psi (\delta^{-1_{j \leqslant m}} \langle x , v_j\rangle) e(-\langle x, v_j\rangle \langle \gamma, v_j \rangle) dx \\ & = \int_{\mathbf{R}^D} \prod_{j = 1}^D \psi(\delta^{-1_{j \leqslant m}} t_j) e(-t_j \langle \gamma ,v_j \rangle) dt \\ & = \delta^{m} \prod_{j = 1}^D \hat{\psi} (\delta^{1_{j \leqslant m}} \langle \gamma , v_j\rangle). \end{align*} $$

Therefore, from property (2) of $\psi $, we see that

(11.6)$$ \begin{align} \operatorname{Supp}(\hat{\chi}) \subset B := \{ \gamma \in \mathbf{R}^D : | \langle \gamma , v_j \rangle | \leqslant \delta^{-1_{j \leqslant m}} \; \ \mbox{for all } \ j \}. \end{align} $$

First note that this implies (rather crudely) that if $\hat {\chi }(\xi ) \neq 0$ for some $\xi \in \mathbf {Z}^D$, then

(11.7)$$ \begin{align} |\xi|^2 < \Vert \xi \Vert_2^2 = \sum_{i} |\langle \xi, v_i \rangle|^2 \leqslant \frac{D}{\delta^2} < (\frac{D}{\delta})^2. \end{align} $$

Second, to analyse equation (11.4), we need a bound on $\#(\operatorname {Supp} (\hat {\chi }) \cap \mathbf {Z}^D)$. To get such a bound, note that if $\gamma \in \operatorname {Supp}(\chi )$ and if $u \in [0,1]^D$, then by equation (11.6) (and since D is big),

$$\begin{align*}|\langle \gamma + u, v_j \rangle| \leqslant \delta^{-1_{j \leqslant m}} + |\langle u , v_j \rangle | \leqslant \frac{D}{10} \delta^{-1_{j \leqslant m}} \end{align*}$$

for all j. Therefore, the disjoint union of cubes $(\operatorname {Supp}(\hat {\chi }) \cap \mathbf {Z}^D) + [0,1)^D$ is contained in the cuboid

$$\begin{align*}\{ x \in \mathbf{R}^D : | \langle x , v_j \rangle | \leqslant \frac{D}{10} \delta^{-1_{j \leqslant m}} \; \mbox{for all} \ j\}, \end{align*}$$

which has volume $(D/5)^D \delta ^{-m}$. It follows that $\# (\operatorname {Supp}(\chi ) \cap \mathbf {Z}^D) \leqslant (D/5)^D \delta ^{-m}$, so by equation (11.5),

(11.8)$$ \begin{align} \sum_{\xi \in \mathbf{Z}^D} |\hat{\chi}( \xi)| \leqslant D^D. \end{align} $$

Let us return to the main task of estimating the right-hand side of equation (11.4). Summing the geometric series on the right of equation (11.4), we have

(11.9)$$ \begin{align} \# \big( \theta d [X] \cap \pi(S) \big) \leqslant \sum_{\xi \in \mathbf{Z}^D} |\hat{\chi}(\xi)| \min(X, \Vert \xi \cdot \theta d \Vert_{\mathbf{T}}^{-1}). \end{align} $$

Now, by equation (11.8), the contribution from $\xi $ with $\Vert \xi \cdot \theta d \Vert _{\mathbf {T}}> D^D \delta ^{-m} X^{-1}$ is at most $\delta ^{m}X \leqslant D^{-\varepsilon C_1D}X$. It therefore follows from equations (11.5), (11.7) and (11.9) that

(11.10)$$ \begin{align} \# \big( \theta d [X] \cap \pi(S) \big) \leqslant 5^D \delta^{m} X \# \Omega+ D^{-\varepsilon C_1D}X,\end{align} $$

where

(11.11)$$ \begin{align} \Omega := \{ \xi \in \mathbf{Z}^D: |\xi| < D/\delta, \Vert \xi \cdot \theta d \Vert_{\mathbf{T}} \leqslant D^D \delta^{-m} X^{-1}\} .\end{align} $$

Since $\theta $ is diophantine, it follows from Proposition 7.1 (1) (the definition of diophantine) that $\dim \Omega < 4r$, assuming that $C_2 \geqslant C_1 + 1$.

It follows from Lemma A.3 that

(11.12)$$ \begin{align} \# \Omega \leqslant 20^D (4r)^{D/2}(D/\delta)^{4r}. \end{align} $$

To conclude the proof, we combine equations (11.10) and (11.12) and bound the resulting terms crudely. Since $D = C_3 r^2$ and r is large, crude bounds for the terms in equation (11.12) show that

$$\begin{align*}\# \big( \theta d [X] \cap \pi(S)\big) \leqslant D^D \delta^{m-4r} X + D^{-\varepsilon C_1D}X.\end{align*}$$

Using the assumption that $m \geqslant 8r$, the first term is bounded above by $D^D\delta ^{m/2}X \leqslant D^{1-C_1\varepsilon D/2}X$. The proposition follows.

Proof. In $\mathbf {R} \times \mathbf {R}^D$, consider the lattice

$$\begin{align*}\Lambda = \mathbf{Z} (\frac{1}{X}, \alpha_1,\dots, \alpha_D) \oplus (\{0\} \times \mathbf{Z}^D)\end{align*}$$

and the centrally symmetric convex body

$$\begin{align*}K := [-D^{-D}, D^{-D}] \times [-\frac{1}{2}, \frac{1}{2}]^D.\end{align*}$$

We have $\operatorname {vol} (K) = 2D^{-D}$ and $\det (\Lambda ) = \frac {1}{X}$, so Minkowski’s second theorem tells us that the successive minima $\lambda _1,\dots , \lambda _{D+1}$ for K with respect to $\Lambda $ satisfy

(12.1)$$ \begin{align} \lambda_1 \cdots \lambda_{D+1} \leqslant \frac{2^{D+1} \det(\Lambda)}{\operatorname{vol}(K)} = (2D)^D \frac{1}{X}. \end{align} $$

Consider a directional basis $\mathbf {b}_1,\dots , \mathbf {b}_{D+1}$ for $\Lambda $ with respect to K. Thus $\mathbf {b}_i \in \Lambda $, the $\mathbf {b}_i$ are linearly independent and $\mathbf {b}_i \in \lambda _i K$ (but the $\mathbf {b}_i$ need not be an integral basis for $\Lambda $). Write

$$\begin{align*}\mathbf{b}_i = (\frac{n_i}{X}, n_i \alpha - m_{i}),\end{align*}$$

where $n_i \in \mathbf {Z}$ and $m_i \in \mathbf {Z}^D$. Replacing $\mathbf {b}_i$ by $-\mathbf {b}_i$ if necessary, we may assume that $n_i \geqslant 0$. We take these $n_i$s to be the ones in Lemma 12.1. Set

$$\begin{align*}L^{\prime}_i := \frac{1}{(D+1) \lambda_i}.\end{align*}$$

We now verify statements (1), (2) and (3) in the lemma. Item (3) is a straightforward consequence of the definition of the $L^{\prime }_i$ and equation (12.1):

$$\begin{align*}\prod_{i = 1}^{D+1} L^{\prime}_i = (D+1)^{-D - 1} \big(\prod_{i=1}^{D+1} \lambda_i \big)^{-1} \geqslant (D+1)^{-D-1} (2D)^{-D} X> D^{-3D} X.\end{align*}$$

Item (2) follows from the fact that $\mathbf {b}_i \in \lambda _i K$; looking at the last D coordinates, one sees that this means that $\Vert n_i \alpha \Vert _{\mathbf {T}^D} \leqslant \lambda _i/2 < 1/L^{\prime }_i$. Finally we turn to (1). We have

(12.2)$$ \begin{align} \ell_1 \mathbf{b}_1 + \dots + \ell_{D+1} \mathbf{b}_{D+1} = \big(\frac{1}{X} \sum_{i=1}^{D+1} \ell_i n_i , \sum_{i=1}^{D+1} \ell_i (n_i \alpha - m_i)\big). \end{align} $$

Since $\mathbf {b}_i \in \lambda _i K$, comparing first coordinates we see that if $0 \leqslant \ell _i < L^{\prime }_i$, then

$$\begin{align*}0 \leqslant \frac{1}{X} \sum_{i=1}^{D+1} \ell_i n_i \leqslant D^{-D} \sum_{i=1}^{D+1} \ell_i \lambda_i \leqslant D^{-D}.\end{align*}$$

This is one part of statement (1). For the statement about distinctness, suppose that $\sum _{i=1}^{D+1} \ell _i n_i = \sum _{i=1}^{D+1} \ell ^{\prime }_i n_i$ with $0 \leqslant \ell _i, \ell ^{\prime }_i < L^{\prime }_i$. Then $\sum _{i=1}^{D+1} (\ell _i - \ell ^{\prime }_i) \mathbf {b}_i \in \{0\} \times \mathbf {Z}^D$ (by equation (12.2) and its analogue for the $\ell ^{\prime }_i$). However,

$$\begin{align*}\Vert \sum_{i=1}^{D+1} (\ell_i - \ell^{\prime}_i) \mathbf{b}_i \Vert_{\infty} \leqslant \sum_{i=1}^{D+1} L^{\prime}_i \lambda_i \cdot \frac{1}{2} < 1.\end{align*}$$

It follows that $\sum _{i=1}^{D+1} (\ell _i - \ell ^{\prime }_i) \mathbf {b}_i = 0$ and hence, since the $\mathbf {b}_i$ are linearly independent, that $\ell _i = \ell ^{\prime }_i$ for all i.

Proof. Suppose that this is not the case for some index set I. Then certainly (by equation (12.3)) $\prod _{i \notin I} L^{\prime }_i < X^{1/80}$, so by Lemma 12.1 (3), we have

$$\begin{align*}\prod_{i \in I} L^{\prime}_i \geqslant D^{-3D} X^{1 - 1/80}> X^{1 - 1/40}.\end{align*}$$

Now look at all sums $\sum _{i \in I} \ell _i n_i$ with $0 \leqslant \ell _i \leqslant \frac {1}{2D} X^{-1/D} L^{\prime }_i$. By Lemma 12.1 (1) and (2), these sums are all distinct, and for each one

$$\begin{align*}\Vert \theta d \sum_{i \in I} \ell_i n_i \Vert_{\mathbf{T}^D} \leqslant \sum_{i \in I} \ell_i \Vert \theta d n_i \Vert_{\mathbf{T}^D} \leqslant X^{-1/D}.\end{align*}$$

However, Proposition 7.1 (2) tells us that

$$\begin{align*}\# \{ n \leqslant X : \Vert \theta d n \Vert_{\mathbf{T}^D} \leqslant X^{-1/D} \} \leqslant X^{9/10}.\end{align*}$$

It follows that

$$\begin{align*}\big( \frac{X^{-1/D}}{2D} \big)^{2^{-7} D} X^{1 - 1/40} \leqslant \prod_{i \in I}(\frac{X^{-1/D}}{2D} L^{\prime}_i) \leqslant X^{9/10},\end{align*}$$

which is a contradiction.

Proof of Proposition 9.1.

The idea is to take the $n_i$ output by Lemma 12.1 but discard indices i, which cause items (2), (3) and (4) of Proposition 9.1 to be violated, whilst ensuring that the lower bound (1) is maintained. When we say that (2), (3) or (4) holds for indices in a set I, this has the obvious meaning (namely, for (2) we mean that $\sum _{i \in I} n_i L_i \leqslant X/2$, and for (3) that $\operatorname {vol} ((w_i)_{i \in I}) \geqslant D^{-C_1 D}$). Note that properties (2), (3) and (4) are all hereditary: that is to say if they hold for indices in I, then they also hold for indices in $I'$, for any subset $I' \subset I$. For (2) and (4), this is obvious; for (3), it follows from Corollary 10.2.

This, together with Lemma 12.2, allows us to treat (2), (3) and (4) of Proposition 9.1 essentially separately. We will show, for each $n \in \{2,3,4\}$, that there is a set $I_{(n)} \subset [D+1]$ of indices,

(12.4)$$ \begin{align} |I^c_{(2)} |,|I^c_{(3)}| \leqslant 2^{-9} D, \; \; |I^c_{(4)}| \leqslant 2^{-8} D, \end{align} $$

such that item $(n)$ of Proposition 9.1 holds on $I_{(n)}$. If we set $I := I_{(2)} \cap I_{(3)} \cap I_{(4)}$, then properties (2), (3) and (4) all hold on I, and by Lemma 12.2, we have the lower bound $\prod _{i \in I} L_i \geqslant X^{1/80}$. Relabelling so that $I = \{1,\dots , s\}$, Proposition 9.1 then follows.

The arguments for properties (2) and (3) share some common features, which we introduce now. Consider the set

(12.5)$$ \begin{align} B = \{ \sum_{i=1}^{D+1} \ell_i n_i : 0 \leqslant \ell_i \leqslant \frac{L^{\prime}_i}{20D^2}\}. \end{align} $$

All the elements in this presentation of B are distinct by Lemma 12.1 (1), so (rather crudely)

(12.6)$$ \begin{align} |B| \geqslant D^{-6D} X \end{align} $$

by Lemma 12.1 (3) and the fact that D is large. Recall that $v_i = \pi ^{-1}(\theta d n_i)$ (that is, the unique smallest lift under the projection map $\pi : \mathbf {R}^D \rightarrow \mathbf {T}^D$, so in particular $\Vert v_i \Vert _{\infty } = \Vert \theta d n_i \Vert _{\mathbf {T}^D}$) and $w_i = v_i/\Vert v_i \Vert _2$ is the associated unit vector. Therefore, since $\pi $ is a homomorphism,

$$\begin{align*}\theta d ( \sum_{i=1}^{D+1} \ell_i n_i) = \pi (x),\end{align*}$$

where

(12.7)$$ \begin{align} x = \sum_{i=1}^{D+1} \ell_i v_i = \sum_{i=1}^{D+1} \ell_i \Vert v_i \Vert_2 w_i. \end{align} $$

By equation (12.3) (and the fact that $\ell _i < L^{\prime }_i/20D^2 \leqslant L_i/20D^2$), we have

(12.8)$$ \begin{align} |\ell_i| \Vert v_i \Vert_2 \leqslant \frac{L_i}{20D^2} \cdot D^{1/2} \Vert v_i \Vert_{\infty} = \frac{L_i}{20D^2} \cdot D^{1/2} \Vert \alpha n_i \Vert_{\mathbf{T}^D} = \frac{1}{20 D^{3/2}}, \end{align} $$

and so in particular

(12.9)$$ \begin{align} \Vert x \Vert_2 \leqslant (D+1) \frac{1}{20 D^{3/2}} < \frac{1}{10}. \end{align} $$

Now we look at properties (2), (3) and (4) of Proposition 9.1 separately.

Property (2). Set $I_{(2)} := \{ i : L_i \leqslant D^{C_1} L^{\prime }_i.\}$, then property (2) holds on $I_{(2)}$, since

$$\begin{align*}\sum_{i \in I_{(2)}} n_i L_i \leqslant D^{C_1} \sum_{i \in I_{(2)}} n_i L^{\prime}_i \leqslant D^{C_1-D} X < \frac{X}{2},\end{align*}$$

by Lemma 12.1 (1). It remains to prove equation (12.4): that is to say, that $|I_{(2)}^c| \leqslant 2^{-9} D$. Suppose not, and take $V := \operatorname {Span}_{\mathbf {R}}((w_i)_{i \in I_{(2)}})$. Suppose that equation (12.4) fails; then $\dim V \leqslant (1 - 2^{-9})D$. Consider an element $b = \sum _i \ell _i n_i \in B$, with B defined in equation (12.5). As explained in equation (12.9) above, we have $\theta d b = \pi (x)$, with $x \in B_{1/10}(0)$. From equation (12.7), since the $w_i$ are unit vectors, we see that

(12.10)$$ \begin{align} \operatorname{dist}(x, V) \leqslant \sum_{i \notin I_{(2)}} |\ell_i |\Vert v_i \Vert_2. \end{align} $$

Now observe that if $i \notin I_{(2)}$, then

$$ \begin{align*} |\ell_i| \Vert v_i \Vert_2 \leqslant \frac{L^{\prime}_i \Vert v_i \Vert_2}{20D^2} \leqslant \frac{L^{\prime}_i \Vert v_i \Vert_{\infty} }{20 D^{3/2}} & = \frac{L^{\prime}_i \Vert \theta d n_i \Vert_{\mathbf{T}^D}}{20 D^{3/2}} \\[6pt] & = \frac{L^{\prime}_i}{20 D^{3/2} L_i} < \frac{1}{2}D^{-1-C_1},\end{align*} $$

and so from equation (12.10), we have $\operatorname {dist}(x, V) < D^{-C_1}$. We have shown that $\theta d B \subset \pi ( N_{D^{-C_1}}(V) \cap B_{1/10}(0))$, so by equation (12.6) and the fact that $B \subset [X]$, we have

(12.11)$$ \begin{align} \# \{ \theta d [X] \cap \pi( N_{D^{-C_1}}(V) \cap B_{1/10}(0))\} \geqslant D^{-6D} X. \end{align} $$

On the other hand, Lemma 11.1 (with $\varepsilon = 2^{-9}$, since $\dim V \leqslant (1 - 2^{-9}) D$) tells us that

(12.12)$$ \begin{align} \# \{ \theta d [X] \cap \pi(\mathcal{N}_{D^{-C_1}}(V) \cap B_{1/10}(0))\} \leqslant 2D^{( 1-2^{-10} C_1)D} X. \end{align} $$

If $C_1$ is big enough, equations (12.11) and (12.12) contradict one another, so we were wrong to assume that $|I_{(2)}^c|> 2^{-9} D$.

Property (3). Suppose that there does not exist a set $I_{(3)} \subset [D+1]$, $|I_{(3)}| \geqslant D +1 - 2^{-9} D$, with $\operatorname {vol} ((w_i)_{i \in I_{(3)}}) \geqslant D^{-C_1 D}$. Then applying Lemma 10.4, we see that there is a subspace $V \leqslant \mathbf {R}^D$, $\dim V < D(1 - 2^{-9})$, such that

(12.13)$$ \begin{align} \operatorname{dist} (w_i, V) \leqslant D^{-C_1} \end{align} $$

for all i. We now proceed much as before. Consider an element $b = \sum _i \ell _i n_i \in B$, with B defined in equation (12.5). As explained above, we have $\theta d b = \pi (x)$, with $x \in B_{1/10}(0)$. Now, using equations (12.7), (12.8) and (12.13), we see that

(12.14)$$ \begin{align} \operatorname{dist} (x, V ) \leqslant \sum_i |\ell_i| \Vert v_i \Vert_2 \operatorname{dist}(w_i, V) \leqslant D^{-C_1}. \end{align} $$

We have shown that $\theta d B \subset \pi ( N_{D^{-C_1}}(V) \cap B_{1/10}(0))$, so once again we conclude from equation (12.6) and the fact that $B \subset [X]$ that

$$\begin{align*}\# \{ \theta d [X] \cap \pi( N_{D^{-C_1}}(V) \cap B_{1/10}(0))\} \geqslant D^{-6D} X.\end{align*}$$

Once again, this contradicts Lemma 11.1 (if $C_1$ is large enough). Thus we were wrong to assert that $I_{(3)}$ does not exist.

Property (4). Set $J := \{ i \in I_{(2)} : L_i \geqslant X^{C_1/D} \}$. Since $J \subset I_{(2)}$, we have

$$\begin{align*}\prod_{i \in J} L_i \leqslant D^{C_1 (D+1)} \prod_i L^{\prime}_i < D^{C_1 (D+1)} X < X^2, \end{align*}$$

by Lemma 12.1 (1) and the assumption on N. If $C_1 \geqslant 2^{11}$, then it follows that $|J| \leqslant 2D/C_1 < 2^{-10} D$, so if we set $I_{(4)} := I_{(2)} \setminus J$, then the required bound equation (12.4) follows.

Finally – a very minor point – we note that property (3) implies that $w_1,\dots , w_s$ are linearly independent, so $s \leqslant D$. This completes the proof of Proposition 9.1.

Proof. , we first handle the case $s = D$ (which we will prove with the slightly stronger constant $D^{3D^2}$) and then deduce the case $s < D$ from it. Suppose, for the moment, that $s = D$.

We write $f(x) = f_2(f_1(x))$ as a composition of two maps

$$\begin{align*}f_1 : \mathbf{R}^{D(D+1)/2} \rightarrow \mathbf{R}^{D(D+1)/2}: \quad f_1(x) = \sigma^{-1} \big((I + \sigma(x))^T (I + \sigma(x)) \big) \end{align*}$$

(note here that the transpose is superfluous, since $\sigma (\mathbf {x})$ is symmetric) and

$$\begin{align*}f_2 : \mathbf{R}^{D(D+1)/2} \rightarrow \mathbf{R}^{D(D+1)/2}: \quad f_2(x) = \sigma^{-1} \big( W^T \sigma(x) W), \end{align*}$$

where $W_{ij} = (w_j)_i$ (that is, the ith coordinate of $w_j$ in the standard basis). Checking that f is indeed the composition of these two maps amounts to checking that $(W^T A^T A W)_{ij} = \langle A w_i, A w_j \rangle $ for any matrix A, which is an easy exercise.

We claim that $f_1$ is injective on the domain $[-\frac {1}{D^4}, \frac {1}{D^4}]^{D(D+1)/2}$. If we have $f_1(x_1) = f_1(x_2)$, then $(I + \sigma (x_1))^2 = (I + \sigma (x_2))^2$. Suppose that $I + \sigma (x_i) = U_i^{-1} \Delta _i U_i$ for $i = 1,2$, with the $U_i$ orthogonal and $\Delta _i$ diagonal with entries nonincreasing down the diagonal. If v is a unit vector and x lies in the domain, then $\Vert \sigma (x) v \Vert _2 \leqslant D \Vert x \Vert _{\infty } \leqslant D^{-3}$, so all eigenvalues of $I + \sigma (x_i)$ (that is, entries of $\Delta _i$) are extremely close to $1$ and in particular positive. Therefore, $\Delta _1^2, \Delta _2^2$ are also diagonal with entries nonincreasing down the diagonal. Moreover $U_1^{-1} \Delta ^2_1 U_1 = U_2^{-1} \Delta ^2_2 U_2$, so $\Delta ^2_1$ and $\Delta ^2_2$ are similar matrices and therefore the same. It follows that $\Delta _1 = \Delta _2$, and also $U_1 U_2^{-1}$ commutes with $\Delta _1^2$ ($= \Delta _2^2$). However, a diagonal matrix $\Delta $ with positive entries and its square $\Delta ^2$ have the same centraliser (block matrices based on equal diagonal entries), and hence $U_1 U_2^{-1}$ commutes with $\Delta _1$ ($= \Delta _2$), which means that $I + \sigma (x_1) = U_1^{-1} \Delta _1 U_1 = U_2^{-1} \Delta _1 U_2 = U_2^{-1} \Delta _2 U_2 = I + \sigma (x_2)$ and so $x_1 = x_2$. The claim follows.

For the two maps $f_1, f_2$, we additionally claim that

1. 1. $f_1$ is differentiable on $[-\frac {1}{D^4} , \frac {1}{D^4} ]^{D(D+1)/2}$, with Jacobian bounded below by $1$;

2. 2. $f_2$ is linear, with $|\det f_2| = |\det W|^{D+1}$.

The proposition then follows by change of variables, in fact with the more precise constant $(2D^4)^{D(D+1)/2}$ for the D-dependence, noting that $\operatorname {vol}(w_1,\dots , w_D) = |\det (W^T W)|^{1/2} = |\det W|$.

Proof of (1). This can be established by direct coordinatewise calculation. Indeed, it is easy to check that

$$\begin{align*}(f_1(\mathbf{x}))_{ij} = 1_{i = j} + 2 x_{ij} + q_{ij}(\mathbf{x}),\end{align*}$$

where $q_{ij}(\mathbf {x})$ is a quadratic form with at most D terms, each with coefficient bounded by $1$. Thus

(13.2)$$ \begin{align} \big| \frac{\partial (f_1)_{ij}}{\partial x_{uv}}(\mathbf{e}) - 2 \cdot 1_{(i,j) = (u,v)} \big| \leqslant 2D \Vert \mathbf{e} \Vert_{\infty}. \end{align} $$

To bound the Jacobian of $f_1$ below, we need a lower bound for determinants of perturbations of the identity. Using Fredholm’s identity $\det (I + E) = \exp (\sum _{k = 1}^{\infty } \frac {(-1)^{k-1}}{k} \mbox {tr}(E^k))$, one can check that if E is an n-by-n matrix with entries bounded in absolute value by $\varepsilon < \frac {1}{2n}$, then $\det (I+E) \geqslant e^{-2 n \varepsilon }$. (In fact, the stronger bound $1 - n \varepsilon $ for $\varepsilon \leqslant \frac {1}{n}$ was shown by Ostrowksi [Reference Ostrowski17, Eq (5.5)] in 1938; for our purposes, fairly crude bounds would suffice.)

It follows from this, equation (13.2) and the fact that $\vert \mathbf {e} \Vert _{\infty } \leqslant D^{-4}$ that if D is large, the Jacobian of $f_1$ is indeed greater than $1$.

Proof of (2). If A is a $D \times D$ matrix, define $f_2^A : \mathbf {R}^{D(D+1)/2} \rightarrow \mathbf {R}^{D(D+1)/2}$ by $f_2^A(x) = \sigma ^{-1}(A^T \sigma (x) A)$. It is clear that $f_2^A$ is linear in x for fixed A and also that $f_2^{AA'} = f_2^{A'} \circ f_2^A$ for all $A, A'$. Thus, to prove that $|\det (f_2^A)| = |\det A|^{D+1}$ for all A (and in particular for $A = W$), it suffices, by the existence of singular value decomposition, to prove this in the cases (i) A is diagonal and (ii) A is orthogonal.

When A is diagonal with diagonal entries $\lambda _1,\dots , \lambda _D$, then one sees by inspection that $(f_2^A(x))_{ij} = \lambda _i \lambda _j x_{ij}$, which renders the result clear in this case. When A is orthogonal, $f_2^A$ preserves a nondegenerate quadratic form, namely $\psi (x) := \Vert \sigma (x) \Vert _{\operatorname {HS}}$, where $\Vert \cdot \Vert _{\operatorname {HS}}$ is the Hilbert-Schmidt norm. Therefore, in this case $|\det f_2^A| = 1$.

This completes the proof in the case $D = s$. Now suppose that $s < D$. In this case, we first complete $w_1,\dots , w_{s}$ to a set $w_1,\dots , w_D$ of unit vectors with

(13.3)$$ \begin{align} \operatorname{vol}(w_1,\dots, w_D) \geqslant \operatorname{vol}(w_1,\dots, w_{s}). \end{align} $$

For the proof that this is possible, see Corollary 10.3.

Let $\tilde f : \mathbf {R}^{D(D+1)/2} \rightarrow \mathbf {R}^{D(D+1)/2}$ be

$$\begin{align*}\tilde f(x) = \sigma^{-1} \big( (\langle (I + \sigma(x)) w_i, (I + \sigma(x))w_j\rangle)_{1 \leqslant i, j \leqslant D} \big);\end{align*}$$

then $f = \pi \circ \tilde f$, where $\pi : \mathbf {R}^{D(D+1)/2} \rightarrow \mathbf {R}^{s(s+1)/2}$ is the natural projection: that is,

$$\begin{align*}\pi((a_{ij})_{1 \leqslant i \leqslant j \leqslant D}) = (a_{ij})_{1 \leqslant i \leqslant j \leqslant s}.\end{align*}$$

The case $s =D$ just established shows that for any measurable $\tilde U$,

(13.4)$$ \begin{align} \mathbb{P}_{\mathbf{e}} (\tilde f(\mathbf{e}) \in \tilde U) \leqslant D^{3D^2} \operatorname{vol}(w_1,\dots, w_D)^{-D-1} \tilde{\mu}(\tilde U), \end{align} $$

where $\tilde \mu $ is the Lebesgue measure on $\mathbf {R}^{D(D+1)/2}$.

Now suppose that $U \subset \mathbf {R}^{s(s+1)/2}$. Then, since (by equation (4.4)) $\Vert \tilde f (\mathbf {e}) \Vert _{\infty } \leqslant 8$ for all $\mathbf {e} \in [-\frac {1}{D^4} , \frac {1}{D^4} ]^{D(D+1)/2}$, we see using equation (13.4) that

$$ \begin{align*} \mathbb{P}_{\mathbf{e}} (f(\mathbf{e}) \in U) & = \mathbb{P}_{\mathbf{e}} (\tilde f(\mathbf{e}) \in \pi^{-1}(U)) \\ & = \mathbb{P}_{\mathbf{e}}(\tilde f(\mathbf{e}) \in \pi^{-1}(U) \cap [-8,8]^{D(D+1)/2}) \\ & \leqslant D^{3D^2} \operatorname{vol}(w_1,\dots, w_D)^{-D-1} \tilde \mu(\pi^{-1}(U) \cap [-8,8]^{D(D+1)/2}) \\ & \leqslant D^{4D^2} \operatorname{vol}(w_1,\dots, w_D)^{-D-1} \mu(U). \end{align*} $$

To complete the proof, apply equation (13.3).

Proof. The idea is to compare W with a random (symmetric) matrix Z from the Gaussian Orthogonal Ensemble (GOE) and then compute using the joint density function for eigenvalues there, which is explicit. The density function of $\mbox {GOE}(n)$ is $f(Z) = c_n e^{-\frac {1}{4} \operatorname {tr} (Z^2)}$, where $c_n = 2^{-n/2} (2\pi )^{-n(n+1)/4}$ (see [Reference Anderson, Guionnet and Zeitouni2, equation 2.5.1]). If W has entries in $[-\frac {1}{n},\frac {1}{n}]$, then $\operatorname {tr} W^2 \leqslant 1$, so $f(W) \geqslant c_n e^{-1/4}$. Therefore, we have the comparison estimate

(14.1)$$ \begin{align} \mathbb{P}_W (|\det W| \leqslant \delta) \leqslant (\frac{n}{2})^{n(n+1)/2} c_n^{-1} e^{1/4} \mathbb{P}_Z (|\det Z| \leqslant \delta).\end{align} $$

It remains to estimate the right-hand side. For this, we use the well-known formula (see [Reference Anderson, Guionnet and Zeitouni2, Theorem 2.5.2]) for the joint eigenvalue density of $\mbox {GOE}(n)$, together with the fact that the determinant is the product of the eigenvalues. This tells us that $\mathbb {P}_Z(|\det Z| \leqslant \delta )$ is a normalising constant times

$$\begin{align*}I := \int_{\substack{ \lambda_1 \geqslant \lambda_2 \geqslant \dots \geqslant \lambda_n \\ |\lambda_1 \cdots \lambda_n| \leqslant \delta}} \prod_{i < j} |\lambda_i - \lambda_j| e^{-\frac{1}{4} \sum_{i = 1}^n \lambda_i^2} d\lambda_1 \dots d\lambda_n .\end{align*}$$