Proof. We consider the group $G=E^\times $ of invertible elements in E and its action by multiplication on $V=E$ . By semisimplicity, we may decompose E into a sum of non-trivial irreducible representations $E=\oplus _iV_i$ . Let $\pi _i\colon G\to {\mathrm {GL}}(V_i)$ denote the representation of G on $V_i$ . By [Reference He and de Saxcé24, Theorem 2.3], there is a neighborhood U of the identity in G and constants $\varepsilon = \varepsilon (E,\kappa ,\varepsilon _0)> 0$ and $s = s(E,\kappa ,\varepsilon ) \geq 1$ such that the following holds for all $\delta> 0$ sufficiently small. Let $A_0$ be a subset of U and $A_1$ a subset of $ {\mathrm {B}}_V(0,1)$ . Assume:

1. (i) for all $i = 1,\dotsc ,k$ , for all $\rho \geq \delta $ , $\mathcal {N}(\pi _i(A_0),\rho ) \geq \delta ^\varepsilon \rho ^{-\kappa }$ ;

2. (ii) for any linear subspace $W \subset V$ which is not G-invariant, there is $a \in A_0$ such that $d(a, {\mathrm {Stab}}_G(W)^\circ ) \geq \delta ^\varepsilon $ ;

3. (iii) for any proper G-invariant linear subspace $W \subset V$ , there is $a \in A_1$ such that $d(a,W) \geq \delta ^\varepsilon $ .

Then,

$$ \begin{align*} {\mathrm{B}}_V(0,\delta^{\varepsilon_0}) \subset \langle A_0,A_1\rangle_s + {\mathrm{B}}(0,\delta). \end{align*} $$

Here, $\langle A_0,A_1\rangle _s$ denotes the set of elements in V that can be obtained as sums of at most s products of at most s elements of $A_0$ and elements of $A_1 \cup (-A_1)$ . In the argument below, we apply this result with $\varepsilon $ replaced by $O(\varepsilon /\kappa )$ .

Our set A is not necessarily contained in the neighborhood U, but we may cover A by translates of U in E, and then, by the pigeonhole principle, there is $a \in A$ such that $A_0 = (A - a) \cap U$ satisfies

$$ \begin{align*} \mathcal{N}(A_0,\delta) \gg_U \delta^{O(\varepsilon)} \mathcal{N}(A,\delta). \end{align*} $$

This set $A_0$ satisfies $ {\mathrm {ANC}}(O(\varepsilon ),\kappa )$ at scale $\delta $ . This non-concentration condition applied to affine subspaces parallel to $\oplus _{j\neq i}V_j$ shows that the first condition above is verified. Moreover, if $W\subset E$ is not G-invariant, then the algebra generated by $ {\mathrm {Stab}}_G(W)$ is a proper subalgebra of E. In particular, it is included in a proper affine subspace of E, and by $ {\mathrm {ANC}}(O(\varepsilon ),\kappa )$ , there must exist a in $A_0$ such that $d(a, {\mathrm {Stab}}_G(W))\geq \delta ^{O(\varepsilon /\kappa )}$ ; so the second condition is also satisfied. To conclude, take $A_1=A_0$ , which satisfies the third condition with $\varepsilon $ replaced by $O(\varepsilon /\kappa )$ .

Proof of Theorem 2.4

For $K \geq 1$ , define

$$ \begin{align*} R_\delta(A,K) = \{ f \in E\otimes E^{\mathrm{op}} \,\vert\, \mathcal{N}(A + f A, \delta) \leq K \mathcal{N}(A,\delta) \}. \end{align*} $$

Let us show that $R_\delta (A,K)$ is almost stable under addition and multiplication. By Ruzsa’s covering lemma, if $f\in R_\delta (A,K)$ , there exists a set $X_f$ such that $\mathcal {N}(X_f,\delta )=O(K)$ and

$$ \begin{align*} fA \subset A-A + X_f. \end{align*} $$

Therefore, for $f_1,f_2$ in $\mathbb {R}_\delta (A,K)$ , one has

$$ \begin{align*} A+(f_1+f_2)A \subset A+f_1A+f_2A \subset 3A-2A + X_{f_1} + X_{f_2}. \end{align*} $$

With the Plünnecke–Ruzsa inequality, this yields $\mathcal {N}(A+(f_1+f_2)A,\delta )\leq K^{O(1)}\mathcal {N}(A,\delta )$ , that is, $f_1+f_2$ is in $R_\delta (A,K^{O(1)})$ . Similarly, $f_1f_2\in R_\delta (A,K^{O(1)})$ . By induction, this implies that for $s\in \mathbb {N}$ ,

$$ \begin{align*} \langle R_\delta(A,K)\rangle_s + {\mathrm{B}}_{E\otimes E^{\mathrm{op}}}(0,\delta) \subset R_\delta(A,K^{O_s(1)}). \end{align*} $$

Now assume for a contradiction that $B \cup \{1\} \subset R_\delta (A,\delta ^{-\varepsilon })$ . Since E is a semisimple algebra, $E \otimes E^{\mathrm {op}}$ is also one. Thus, by Proposition 2.5 applied to the set $B\subset E \otimes E^{\mathrm {op}}$ , for any $\varepsilon _0> 0$ , there is $s=s(E,\kappa ,\varepsilon _0) \geq 1$ such that

$$ \begin{align*} {\mathrm{B}}_{E\otimes E^{\mathrm{op}}}(0,\delta^{\varepsilon_0}) \subset \langle B\rangle_s + {\mathrm{B}}(0,\delta) \end{align*} $$

and therefore,

$$ \begin{align*} {\mathrm{B}}_{E\otimes E^{\mathrm{op}}}(0,\delta^{\varepsilon_0}) \subset \langle B\rangle_s + {\mathrm{B}}(0,\delta) \subset R_\delta(A,\delta^{-O_s(\varepsilon)}). \end{align*} $$

In particular, $\delta ^{\varepsilon _0} \in R_\delta (A,\delta ^{-O_s(\varepsilon )})$ . This certainly implies $\delta ^{-\varepsilon _0} \in R_\delta (A,\delta ^{-O_s(\varepsilon _0 + \varepsilon )})$ and then, using once more stability of $R_\delta (A,K)$ under product,

$$ \begin{align*} {\mathrm{B}}_{E\otimes E^{\mathrm{op}}}(0,1) \subset R_\delta(A,\delta^{-O_s(\varepsilon_0 + \varepsilon)}). \end{align*} $$

If $\varepsilon _0$ and $\varepsilon $ are chosen small enough, this contradicts Lemma 2.6 below.

Proof. The image of $E \otimes E^{\mathrm {op}}$ in $ {\mathrm {End}}(E)$ is equal to the image of $\bigoplus _{j = 1}^r E_j \otimes E_j^{\mathrm {op}}$ . Let $f_j$ , $j=1,\ldots ,r$ be a family of jointly independent random elements of $ {\mathrm {B}}_{E_j\otimes E_j^{\mathrm {op}}}(0,1)$ distributed according to the Lebesgue measure on $E_j \otimes E_j^{\mathrm {op}}$ , and set

$$ \begin{align*} f = f_1 + \cdots + f_r \end{align*} $$

regarded as a random element of $ {\mathrm {End}}(E)$ . In the following argument, probabilities and expectations are taken with respect to these random variables. For each j, since the algebra $E_j$ is simple, the action of $E_j \otimes E_j^{\mathrm {op}}$ on $E_j$ is irreducible. Hence, $E_j \otimes E_j^{\mathrm {op}}(y) = E_j$ for any non-zero $y \in E_j$ and consequently,

(2.1) $$ \begin{align} \text{ for all } \delta> 0, \text{ for all } x, y \in E_j,\quad \mathbb{P}[ \lVert f_j(y) - x\rVert \leq \delta] \ll \delta^{\dim E_j} \lVert y\rVert^{-\dim E_j}. \end{align} $$

Consider the map

$$ \begin{align*} \begin{array}{llcl} \varphi \colon & A \times A & \to & E\\ & (x,y) & \mapsto & x + fy \end{array} \end{align*} $$

The energy of the map $\varphi $ at scale $\delta> 0$ is defined as

$$ \begin{align*} \mathcal{E}_\delta(\varphi,A\times A) = \mathcal{N}( \{ (a,a',b,b') \in A \times A\times A\times A\ \,\vert\, \lVert\varphi(a,b) - \varphi(a',b')\rVert \leq \delta \}, \delta). \end{align*} $$

By the Cauchy–Schwarz inequality—see also [Reference He23, Lemma 12(i)],

$$ \begin{align*} \mathcal{N}(\varphi(A\times A),\delta) = \mathcal{N}(A + fA,\delta) \geq \frac{\mathcal{N}(A,\delta)^4}{\mathcal{E}_\delta(\varphi,A \times A)}. \end{align*} $$

Taking expectations and applying Jensen’s inequality, we find

(2.2) $$ \begin{align} \mathbb{E}[\mathcal{N}(A + fA,\delta) ] \geq \frac{\mathcal{N}(A,\delta)^4}{\mathbb{E}[\mathcal{E}_\delta(\varphi,A \times A) ]}, \end{align} $$

so it suffices to bound $\mathbb {E}[\mathcal {E}_\delta (\varphi ,A \times A) ]$ from above.

For that, let $\tilde {A}$ be a maximal $\delta $ -separated subset of A. By [Reference He23, Lemma 12(ii)],

$$ \begin{align*} \mathbb{E}[\mathcal{E}_\delta(\varphi,A \times A)] \leq \sum_{x,y,x',y' \in \tilde{A}} \mathbb{P}[ f(y'-y) \in {\mathrm{B}}(x - x',5\delta)]. \end{align*} $$

Let $\rho = \delta ^{{\kappa }/({\dim E+ \kappa })}$ . We split the sum into two parts according to whether

$$ \begin{align*} \text{for all } j = 1,\dotsc, r,\quad \lVert\pi_j(y' - y)\rVert \geq \rho. \end{align*} $$

If this is the case, then equation (2.1) implies

$$ \begin{align*} \mathbb{P}[ f(y'-y) \in {\mathrm{B}}(x - x',5\delta)] \ll \delta^{\dim E} \rho^{-\dim E}. \end{align*} $$

Otherwise, there is $j \in \{1,\dotsc ,r\}$ such that $\pi _j(y') \in {\mathrm {B}}(\pi _j(y),\rho )$ . For fixed y, the number of such $y'$ in $\tilde {A}$ is

$$ \begin{align*} \# (\tilde{A} \cap \pi_j^{-1}({\mathrm{B}}(\pi_j(y),\rho))) \ll \mathcal{N}( A \cap \pi_j^{-1}({\mathrm{B}}(\pi_j(y),\rho)),\delta) \leq \rho^\kappa \mathcal{N}(A,\delta). \end{align*} $$

Moreover, for fixed $y,y'$ , and x, we have

$$ \begin{align*} \sum_{x' \in \tilde{A}} \mathbb{P}[ f(y'-y) \in {\mathrm{B}}(x - x',5\delta)] \ll 1 \end{align*} $$

because the balls $ {\mathrm {B}}(x',5\delta )$ have overlap multiplicity at most $O(1)$ . Putting these considerations together, we obtain

$$ \begin{align*} \mathbb{E}[\mathcal{E}_\delta(\varphi,A \times A)] &\ll \delta^{\dim E} \rho^{-\dim E} \mathcal{N}(A,\delta)^4 + \rho^\kappa \mathcal{N}(A,\delta)^3 \\ &\leq ( \delta^{\kappa} \rho^{-\dim E} + \rho^\kappa ) \mathcal{N}(A,\delta)^3\\ &\ll \delta^{{\kappa^2}/({\dim E+ \kappa})} \mathcal{N}(A,\delta)^3. \end{align*} $$

Combined with equation (2.2), this finishes the proof of the lemma.

Proof. For the first item, let $\tilde {A}$ be a maximal $\delta $ -separated subset of A. The normalized counting measure on $\tilde {A}$ satisfies the desired property. The second item is slightly more subtle. Since the normalized restriction of $\eta $ to A satisfies $ {\mathrm {ANC}}(2\varepsilon ,\kappa )$ , we may assume without loss of generality that $A = {\mathrm {supp}}\ \eta $ . Let $i_{\min }$ be the largest integer such that $2^{i_{\min }} \leq \delta ^{2\varepsilon \dim V}$ . For every integer $i \geq i_{\min }$ , set

$$ \begin{align*} A_{i,0} = \bigg\{ a \in A \mid 2^{i-1} < \frac{\eta({\mathrm{B}}(a,2\delta))}{\lvert{\mathrm{B}}(0,\delta)\rvert} \leq 2^i \bigg\} \end{align*} $$

and then

$$ \begin{align*} A_{-,0} = A \setminus \bigcup_{i \geq i_{\min}} A_{i,0}. \end{align*} $$

Next, for every $i \geq i_{\min }$ , set $A_i = A_{i,0}^{(\delta )}$ and also $A_- = A_{-,0}^{(\delta )}$ . By this construction,

(2.3)

and

(2.4)

Note that $A_i$ is empty whenever $i \geq -({\log \lvert {\mathrm {B}}(0,\delta )\rvert }/{\log 2}) + 1$ . Thus, integrating equation (2.3) and recalling $ {\mathrm {supp}}\ \eta \subset B(0,\delta ^{-\varepsilon })$ from $ {\mathrm {ANC}}(\varepsilon ,\kappa )$ for $\eta $ , we obtain some $i \geq i_{\min }$ such that $ 2^i \lvert A_i\rvert \geq \delta ^\varepsilon. $ Fix this i and set $A'=A_{i,0}$ . If W is a proper affine subspace in V and $\rho \geq \delta $ , we can bound using equation (2.4) and $ {\mathrm {ANC}}(\varepsilon ,\kappa )$ for $\eta $ ,

and using the above lower bound on $2^i\lvert A_i\rvert $ , we get

$$ \begin{align*} \mathcal{N}(A'\cap W^{(\rho)},\delta) & \ll \delta^{-\dim V} \lvert A_i\rvert \delta^{-2\varepsilon}\rho^\kappa\\ & \ll \delta^{-2\varepsilon} \rho^\kappa \mathcal{N}(A_i,\delta). \end{align*} $$

This shows that $A'$ satisfies $ {\mathrm {ANC}}(3\varepsilon ,\kappa )$ at scale $\delta $ .

Proof. We use the following rough comparison notation: for positive quantities f and g, we write $f \lesssim g$ for $f \leq \delta ^{-O(\varepsilon )} g$ , and $f \sim g$ for $f \lesssim g$ and $g \lesssim f$ .

We have

$$ \begin{align*} \lVert\mu*\eta_\delta \boxminus \mu*\eta_\delta\rVert_2 \gtrsim \lVert\mu*\eta \boxminus \mu*\eta\rVert_{2,\delta} \gtrsim \lVert\eta_\delta\rVert_2. \end{align*} $$

As in the proof of Lemma 2.9, we can approximate $\eta _\delta $ using dyadic level sets: there are $\delta $ -discretized sets (a $\delta $ -discretized set is a union of balls of radius $\delta $ ) $(A_i)_{i\geq 0}$ in $ {\mathrm {B}}_V(0,\delta ^{-\varepsilon })$ such that $A_i$ is empty for $i \gg \log (1/\delta)$ and

(2.5)

By the pigeonhole principle, there are $i,j\geq 0$ such that

In the last inequality, we used and $\lvert \det g\rvert \sim 1$ for all $g \in {\mathrm {supp}}\ \mu $ . By the right-hand inequality in equation (2.5), we have

$$ \begin{align*} 2^i\lvert A_i\rvert \lesssim 1 \quad \text{and} \quad 2^j\lvert A_j\rvert^{{1}/{2}} \lesssim \lVert\eta_\delta\rVert_2, \end{align*} $$

and similarly

$$ \begin{align*} 2^j\lvert A_j\rvert \lesssim 1 \quad \text{and} \quad 2^i\lvert A_i\rvert^{{1}/{2}} \lesssim \lVert\eta_\delta\rVert_2. \end{align*} $$

By Young’s inequality and the estimate on $\det g$ , we have for all $g,g' \in {\mathrm {supp}}\ \mu $ ,

Thus, by the pigeonhole principle again, there exists $g_0 \in {\mathrm {supp}}\ \mu $ and a set $B_0 \subset {\mathrm {supp}}\ \mu $ such that $\mu (B_0)\gtrsim 1$ and for all $g \in B_0$ ,

By the above estimates, this implies

By the Balog–Szemerédi–Gowers lemma [Reference Tao37, Theorem 6.10], for each $g \in B_0$ , there are $\delta $ -discretized subsets $A_g \subset A_i$ and $A^{\prime }_g \subset A_j$ such that

$$ \begin{align*} \lvert A_g\rvert \sim \lvert A_i\rvert,\quad \lvert A^{\prime}_g\rvert \sim \lvert A_j\rvert,\quad \text{ and}\quad \mathcal{N}(g_0A_g - gA^{\prime}_g,\delta)\lesssim \mathcal{N}(g_0A_g,\delta)^{{1}/{2}} \mathcal{N}(gA^{\prime}_g,\delta)^{{1}/{2}}. \end{align*} $$

Set $X=A_i\times A_j$ and $X_g=A_g\times A^{\prime }_g\subset X$ , and write

This shows that there exists $g_1$ and $B_1\subset B_0$ such that $\mu (B_1)\gtrsim 1$ for all g in $B_1$ , $\lvert X_{g_1}\cap X_{g}\rvert \gtrsim \lvert X\rvert $ . Equivalently,

$$ \begin{align*} \text{ for all } g \in B_1,\quad \lvert A_{g_1} \cap A_g\rvert \sim \lvert A_{g_1}\rvert \sim \lvert A_g\rvert \text{ and } \lvert A^{\prime}_{g_1} \cap A^{\prime}_g\rvert \sim \lvert A^{\prime}_{g_1}\rvert \sim \lvert A^{\prime}_g\rvert. \end{align*} $$

For subsets $A,A' \in V$ , write $A \approx A'$ if $\mathcal {N}(A-A',\delta ) \lesssim \mathcal {N}(A,\delta )^{1/2} \mathcal {N}(A',\delta )^{1/2}$ . Ruzsa’s triangle inequality [Reference Tao and Vu38, Lemma 2.6] is still valid for covering numbers at scale $\delta $ , so if $A\approx A'$ and $A'\approx A"$ , then $A\approx A"$ , and moreover, if $A\approx A'$ for some sets A and $A'$ , then $A\approx A$ and $A'\approx A'$ .

The above shows that for every $g\in B_0$ , $g_0A_g\approx gA_g'$ . This implies $g_0A_g\approx g_0A_g$ , and since g is $\delta ^{-\varepsilon }$ -Lipschitz, $A_g\approx A_g$ . Therefore, for $g_1\in B_0$ and $g\in B_1$ as above, we find $A_{g_1}\approx A_{g_1}\cap A_g\approx A_g$ . Similarly, $A^{\prime }_{g_1}\approx A^{\prime }_g$ . Finally,

$$ \begin{align*} g_0A_{g_1} \approx g_0A_g \approx gA^{\prime}_g \approx gA^{\prime}_{g_1} \approx g g_1^{-1}g_0A_{g_1}, \end{align*} $$

showing that $A = g_0 A_{g_1}$ has all the desired properties.

For the last assertion, note that $\eta (A_i)\gtrsim 1$ . By the proof of Lemma 2.9(ii), $A_i$ satisfies $ {\mathrm {ANC}}(O(\varepsilon ),\kappa )$ , and hence so do $A_{g_1}$ and A. Note also that

$$ \begin{align*} \mathcal{N}(A,\delta)\sim \mathcal{N}(A_{g_1},\delta)\sim \mathcal{N}(A_i,\delta) \sim \delta^{-\dim V}\lvert A_i\rvert \sim \delta^{-\dim V}\lVert\eta\rVert_{2,\delta}^2.\\[-39pt] \end{align*} $$

Proof of Proposition 2.8

Let $\mu $ be the image measure of $\eta \otimes \eta $ in $ {\mathrm {GL}}(E)$ , so that $\mu * \eta \boxminus \mu * \eta = \eta * \eta * \eta \boxminus \eta * \eta * \eta $ . We argue by contradiction as follows. Assuming

$$ \begin{align*} \lVert\mu*\eta \boxminus \mu*\eta\rVert_{2,\delta}> \delta^{\varepsilon} \lVert\eta\rVert_{2,\delta}, \end{align*} $$

we shall construct sets $A,B$ satisfying all assumptions of Theorem 2.4, with $\kappa $ and $\varepsilon $ replaced by $\kappa /2$ and $O(\varepsilon )$ but violating its conclusion. By Lemma 2.10, there is a subset $A \subset E$ satisfying $ {\mathrm {ANC}}(O(\varepsilon ),\kappa )$ at scale $\delta $ and an element $g_1 \in {\mathrm {supp}}\ \mu $ such that

$$ \begin{align*} \delta^{-\kappa + O(\varepsilon)}\leq \mathcal{N}(A,\delta) \leq \delta^{- \dim E + \kappa - O(\varepsilon)} \end{align*} $$

and

$$ \begin{align*} \mu( \{ g \in {\mathrm{GL}}(V) \,\vert\, \mathcal{N}(A - gg_1^{-1}A,\delta) \leq \delta^{-O(\varepsilon)}\mathcal{N}(A,\delta) \} ) \geq \delta^{O(\varepsilon)}. \end{align*} $$

By definition of $\mu $ , we may write $g_1=a_1\otimes b_1$ , and the above inequality becomes

(2.6) $$ \begin{align} (\eta * a_1^{-1})\otimes (b_1^{-1} *\eta)( \{ (a,b) \in E \times E \,\vert\, \mathcal{N}(A - aAb,\delta) \leq \delta^{-O(\varepsilon)}\mathcal{N}(A,\delta) \} ) \geq \delta^{O(\varepsilon)}. \end{align} $$

Since $a_1,b_1 \not \in S_E(\delta ^\varepsilon )$ , the measures $\eta * a_1^{-1}$ and $b_1^{-1} *\eta $ satisfy $ {\mathrm {ANC}}(O(\varepsilon ),\kappa )$ at scale $\delta $ . Moreover, Lemma 2.11 below shows that $(\eta * a_1^{-1})\ \dot \otimes\ (b_1^{-1} *\eta )$ satisfies $ {\mathrm {ANC}}(O(\varepsilon ),{\kappa }/{2})$ . By Lemma 2.9(ii), there exists a subset $B\subset {\mathrm {supp}}\ ((\eta * a_1^{-1})\ \dot \otimes\ (b_1^{-1} *\eta ))$ satisfying $ {\mathrm {ANC}}(O(\varepsilon ),{\kappa }/{2})$ . Equation (2.6) shows that $\mathcal {N}(A+fA,\delta )\leq \delta ^{-O(\varepsilon )}\mathcal {N}(A,\delta )$ for all f in B. Since $\eta $ is supported on $E\setminus S_E(\delta ^\varepsilon )$ , one also has $\mathcal {N}(fA,\delta )\geq \delta ^{O(\varepsilon )}\mathcal {N}(A,\delta )$ , and so by Plünnecke’s inequality, one also has $\mathcal {N}(A+A,\delta )\leq \delta ^{-O(\varepsilon )}\mathcal {N}(A,\delta )$ and, in turn,

$$ \begin{align*} \mathcal{N}(A+A,\delta) + \mathcal{N}(A+fA,\delta) \leq \delta^{-O(\varepsilon)}\mathcal{N}(A,\delta). \end{align*} $$

Thus, A and B violate the conclusion of Theorem 2.4 with parameters $\kappa /2$ and $O(\varepsilon )$ . This yields the desired contradiction, provided $\varepsilon $ is chosen small enough.

Proof. Let $v_1$ and $v_2$ be independent random variables taking values respectively in $V_1$ and $V_2$ , and distributed according to $\eta _1$ and $\eta _2$ . To establish $ {\mathrm {ANC}}(2\varepsilon ,{\kappa }/{2})$ for $\eta _1\ \dot \otimes\ \eta _2$ , it is enough to show that for any linear form $\varphi \in (V_1 \otimes V_2)^*$ with $\lVert \varphi \rVert = 1$ , any $t \in \mathbb {R}$ , and any $\rho \geq \delta $ , we have

(2.7) $$ \begin{align} \mathbb{P}[\lvert\varphi(v_1 \otimes v_2) - t\rvert < \rho] \ll \delta^{-\varepsilon}\rho^{\kappa/2}. \end{align} $$

Note that $(V_1 \otimes V_2)^* = V_1^* \otimes V_2^*$ . Hence, letting $(\psi _1,\dotsc ,\psi _d)$ be a orthonormal basis of $V_1^*$ , we can write $\varphi \in (V_1 \otimes V_2)^*$ as

$$ \begin{align*}\varphi = \psi_1 \otimes \varphi_1 + \cdots + \psi_d \otimes \varphi_d,\end{align*} $$

where $\varphi _1, \dotsc , \varphi _d \in V_2^*$ are uniquely determined. Moreover,

(2.8) $$ \begin{align} 1 = \lVert\varphi\rVert^2 = \lVert\varphi_1\rVert^2 + \cdots + \lVert\varphi_d\rVert^2. \end{align} $$

On the one hand, when $v_2$ is fixed, the map

$$ \begin{align*} v_1 \mapsto \varphi(v_1 \otimes v_2) = \sum_{i=1}^d \psi_i(v_1) \varphi_i(v_2) \end{align*} $$

is the linear form $ \sum _{i=1}^d \varphi _i(v_2) \psi _i \in V_1^*$ , which has norm $\sum _{i=1}^d \lvert \varphi _i(v_2)\rvert ^2$ . Thus, by independence of $v_1$ and $v_2$ and property $ {\mathrm {ANC}}(\varepsilon ,\kappa )$ for $\eta _1$ , we can estimate the conditional probability

(2.9) $$ \begin{align} \mathbb{P}\bigg[ \lvert\varphi(v_1 \otimes v_2) - t\rvert < \rho \,\bigg\vert\, \sum_{i=1}^d \lvert\varphi_i(v_2)\rvert^2 \geq \rho^{1/2}\bigg] \leq \delta^{-\varepsilon}\rho^{\kappa/2}. \end{align} $$

On the other hand, by property $ {\mathrm {ANC}}(\varepsilon ,\kappa )$ for $\eta _2$ , for each $i = 1,\dotsc ,d$ ,

$$ \begin{align*} \mathbb{P}[\lvert\varphi_i(v_2)\rvert \leq \rho^{1/2}\lVert\varphi_i\rVert] \leq \delta^{-\varepsilon} \rho^{\kappa/2}. \end{align*} $$

Hence, recalling equation (2.8),

(2.10) $$ \begin{align} \mathbb{P}\bigg[\sum_{i=1}^d \lvert\varphi_i(v_2)\rvert^2 < \rho\bigg] \leq d\delta^{-\varepsilon} \rho^{\kappa/2}. \end{align} $$

Inequalities (2.9) and (2.10) together imply equation (2.7) and finish the proof of the lemma.

Proof. By [Reference He, Lakrec and Lindenstrauss28, Lemma B.6], if $\eta $ , $\eta '$ , $\eta "$ are probability measures on E, then the Fourier transform of $\eta * (\eta ' \boxminus \eta ') * \eta "$ takes non-negative real values and moreover,

$$ \begin{align*} \text{ for all } \xi \in E^*,\quad \lvert(\eta * \eta' * \eta")^{\wedge}(\xi)\rvert^{2} \leq (\eta * (\eta' \boxminus \eta') * \eta")^{\wedge}(\xi). \end{align*} $$

By a simple scaling argument, we see that the same holds when $\eta $ , $\eta '$ , $\eta "$ are finite measures with total mass $\eta (E), \eta '(E), \eta "(E) \leq 1$ . Using this inequality m times with measure $\eta '=\eta ^{*3}$ , so that $\mu =\eta '\boxminus \eta '$ , we get

$$ \begin{align*} \widehat{\mu^{*m}}(\xi) \geq \lvert\widehat{\mu^{*(m-1)}*\eta^{*3}}(\xi)\rvert^2 \geq \cdots \geq \lvert\widehat{\eta^{*3m}}(\xi)\rvert^{2^m}.\\[-37pt] \end{align*} $$

Proof of Theorem 2.2

First case: $\eta _1=\cdots =\eta _s=\eta $ .

For a measure $\eta $ satisfying $ {\mathrm {NC}}(\varepsilon ,\kappa ,\tau )$ at scale $\delta $ , we write $ {\mathrm {ess}}(\eta )$ to denote the essential part of $\eta $ , defined as a measure on E satisfying:

1. (i) $ {\mathrm {ess}}(\eta ) \leq \eta $ and $ {\mathrm {ess}}(\eta )(E) \geq \eta (E) - 3 \delta ^\tau $ ;

2. (ii) $ {\mathrm {ess}}(\eta )$ is supported on $ {\mathrm {B}}(0,\delta ^{-\varepsilon }) \setminus S_E(\delta ^\varepsilon )$ ;

3. (iii) $ {\mathrm {ess}}(\eta )$ satisfies $ {\mathrm {ANC}}(\varepsilon ,\kappa )$ at scale $\delta $ .

The second and third conditions from $ {\mathrm {NC}}_0$ are invariant under translation, so that if $\mu $ is a measure satisfying $ {\mathrm {NC}}_0(\varepsilon ,\kappa ,\tau )$ and $\nu $ any measure supported on $ {\mathrm {B}}_E(0,\delta ^{-\varepsilon })$ , then $\mu \boxplus \nu $ always satisfies $ {\mathrm {NC}}_0(2\varepsilon ,\kappa ,\tau )$ . Therefore, if $\eta $ and $\eta '$ satisfy $ {\mathrm {NC}}(\varepsilon ,\kappa ,\tau )$ at scale $\delta $ , then $\eta \boxplus \eta '$ satisfy $ {\mathrm {NC}}(O(\varepsilon ),\kappa ,{\tau }/{2})$ at scale $\delta $ , with essential part $ {\mathrm {ess}}(\eta \boxplus \eta ')= {\mathrm {ess}}(\eta )\boxplus {\mathrm {ess}}(\eta ')$ . Similarly, $\eta \boxminus \eta '$ and $\eta * \eta '$ satisfy $ {\mathrm {NC}}(O(\varepsilon ),\kappa ,{\tau }/{2})$ . We may therefore define inductively $\eta _0 = {\mathrm {ess}}(\eta )$ , and for $k \geq 0$ ,

$$ \begin{align*} \eta_{k+1} = {\mathrm{ess}}( \eta_k^{*3} \boxminus \eta_k^{*3} ), \end{align*} $$

to get, for each $k \geq 0$ :

1. (i) $\eta _k(E) \geq 1 - \delta ^{{\tau }/{O_k(1)}}$ ;

2. (ii) $\eta _k$ is supported on $ {\mathrm {B}}(0,\delta ^{-O_k(\varepsilon )}) \setminus S_E(\delta ^{O_k(\varepsilon )})$ ;

3. (iii) $\eta _k$ satisfies $ {\mathrm {ANC}}(O_k(\varepsilon ),\kappa )$ at scale $\delta $ .

Note that $ {\mathrm {ANC}}(O_k(\varepsilon ),\kappa )$ implies

$$ \begin{align*} \lVert\eta_k\rVert_{2,\delta}^2 \leq \delta^{- \dim E + \kappa - O_k(\varepsilon)}. \end{align*} $$

Set $\kappa '={\kappa }/{3}$ . By Proposition 2.8 applied with $\kappa '$ instead of $\kappa $ , there exists $\varepsilon _1 = \varepsilon _1(E,\kappa ')$ such that, provided $\varepsilon>0$ is small enough, we have for each $0 \leq k \leq \lceil( {\dim E})/{\varepsilon _1} \rceil $ , either $\lVert \eta _k\rVert ^2_{2,\delta } \leq \delta ^{-\kappa '}$ or

$$ \begin{align*}\lVert\eta_{k+1}\rVert_{2,\delta}^2 \leq \delta^{\varepsilon_1} \lVert\eta_k\rVert_{2,\delta}^2.\end{align*} $$

Hence, there exists $s \leq \lceil( {\dim E})/{\varepsilon _1} \rceil $ such that

$$ \begin{align*} \lVert\eta_s\rVert^2_{2,\delta} \leq \delta^{-\kappa'}. \end{align*} $$

By Lemma 2.13 applied with $\kappa '$ instead of $\kappa $ , for $\xi \in E^*$ with $\delta ^{-1 + \varepsilon } \leq \lVert \xi \rVert \leq \delta ^{-1 - \varepsilon }$ ,

$$ \begin{align*} \lvert\widehat{\eta_s^{*2}}(\xi)\rvert \leq \delta^{{\kappa'}/{O(1)} -O(\varepsilon)} \leq \delta^{{\kappa'}/{O(1)}} \leq \delta^{\tau}, \end{align*} $$

provided $\varepsilon $ is chosen sufficiently small. Now, a first application of Lemma 2.12 to $\mu =\eta _{s-1}^{*3}\boxminus \eta _{s-1}^{*3}$ with $m=2$ yields

$$ \begin{align*} \widehat{\eta_s^{*2}}(\xi)+\delta^{{\tau}/{O_s(1)}} \geq \widehat{\mu^{*2}}(\xi) \geq \lvert\widehat{\eta_{s-1}^{*2\cdot 3}}(\xi)\rvert^{2^2}. \end{align*} $$

A second application of the same lemma to $\mu _1=\eta _{s-2}^{*3}\boxminus \eta _{s-2}^{*3}$ with $m=2\cdot 3$ gives

$$ \begin{align*} \widehat{\eta_{s-1}^{*2\cdot 3}}(\xi) + \delta^{{\tau}/{O_s(1)}} \geq \widehat{\mu_1^{*2\cdot 3}}(\xi) \geq \lvert\widehat{\eta_{s-2}^{*2\cdot 3^2}}(\xi)\rvert^{2^{2\cdot 3}} \end{align*} $$

and repeating this process s times, we finally obtain

$$ \begin{align*} \lvert\widehat{\eta_0^{*2 \cdot 3^s}}(\xi)\rvert^{O_s(1)} \leq \lvert\widehat{\eta_s^{*2}}(\xi)\rvert + \delta^{{\tau}/{O_s(1)}} \leq \delta^\tau + \delta^{{\tau}/{O_s(1)}} \leq \delta^{{\tau}/{O_s(1)}}. \end{align*} $$

This allows to conclude

$$ \begin{align*} \lvert\widehat{\eta^{*2 \cdot 3^s}}(\xi)\rvert \leq \lvert\widehat{\eta_0^{*2 \cdot 3^s}}(\xi)\rvert + O_s(\delta^\tau) \leq \delta^{\varepsilon \tau}, \end{align*} $$

provided $\varepsilon $ is sufficiently small. This proves the theorem, with parameter $s(E,\kappa )=2\cdot 3^s$ .

General case. To deduce the general case from the previous one, we follow [Reference He, Lakrec and Lindenstrauss28, Proof of Theorem B.3]. In short, one applies the previous case to the measures

$$ \begin{align*} \eta_\unicode{x3bb} = \unicode{x3bb}_1(\eta_1\boxminus\eta_1) + \cdots + \unicode{x3bb}_s(\eta_s\boxminus\eta_s), \end{align*} $$

where $\unicode{x3bb} =(\unicode{x3bb} _1,\ldots ,\unicode{x3bb} _s)$ in $\mathbb {R}_+^s$ is such that $\unicode{x3bb} _1+\cdots +\unicode{x3bb} _s\leq 1$ . The Fourier decay for $\eta _1*\cdots *\eta _s$ can be deduced from that of $\eta _\unicode{x3bb} *\cdots *\eta _\unicode{x3bb} $ for every $\unicode{x3bb} $ using the fact that Fourier coefficients of $\eta _\unicode{x3bb} *\cdots *\eta _\unicode{x3bb} $ can be written as polynomials in $\unicode{x3bb} $ whose coefficients are essentially Fourier coefficients of $\eta _1*\cdots *\eta _s$ . The reader is referred to [Reference He, Lakrec and Lindenstrauss28] for details.

Proof. Let $\xi \in E^*$ with $\delta ^{-1+\varepsilon }\leq \lVert \xi \rVert \leq \delta ^{-1-\varepsilon }$ . Since all the measures $\eta _i^{\boxplus D}\boxminus \eta _i^{\boxplus D}$ , $i=1,\ldots ,s$ satisfy $ {\mathrm {NC}}(\varepsilon ,\kappa ,\tau )$ at scale $\delta $ , Theorem 2.2 shows that

$$ \begin{align*} \lvert((\eta_1^{\boxplus D}\boxminus\eta_1^{\boxplus D}) * \dotsm * (\eta_s^{\boxplus D}\boxminus\eta_s^{\boxplus D}))^{\wedge}(\xi)\rvert \leq \delta^{\varepsilon \tau}. \end{align*} $$

Applying [Reference He, Lakrec and Lindenstrauss28, Lemma B.6] repeatedly s times, we see that

$$ \begin{align*} & \lvert((\eta_1^{\boxplus D}\boxminus\eta_1^{\boxplus D}) * \dotsm * (\eta_s^{\boxplus D}\boxminus\eta_s^{\boxplus D}))^{\wedge}(\xi)\rvert\\ & \quad \geq \lvert((\eta_1^{\boxplus D}\boxminus\eta_1^{\boxplus D}) * \dotsm * (\eta_{s-1}^{\boxplus D}\boxminus\eta_{s-1}^{\boxplus D})*\eta_s)^{\wedge}(\xi)\rvert^{2D}\\ & \quad \geq \cdots\\ & \quad \geq \lvert(\eta_1 * \dotsm * \eta_s)^{\wedge}(\xi)\rvert^{(2D)^s} \end{align*} $$

so that

$$ \begin{align*} \lvert(\eta_1 * \dotsm * \eta_s)^{\wedge}(\xi)\rvert \leq \delta^{{\varepsilon \tau}/{(2D)^s}}.\\[-37pt] \end{align*} $$