Proof. (1): For $\zeta _K(s)$ , $L(s,V)$ , $L(s,\boldsymbol {c})$ , $L(s,\boldsymbol {c},\bigwedge ^2)$ , this follows directly from [Reference Kahn35, proof of Theorem 5.58], since the corresponding $\ell $ -adic sheaves $\mathbb {Q}_\ell $ , $H^4_\ell (V)/H^4_\ell (\mathbb {P}^5)$ , $H^3_\ell (V_{\boldsymbol {c}})$ , $\bigwedge ^2{H^3_\ell (V_{\boldsymbol {c}})}$ are ‘weakly polarisable’ (by Poincaré duality on V and $V_{\boldsymbol {c}}$ , which gives a symmetric pairing on $H^4_\ell (V)$ and a skew-symmetric pairing on $H^3_\ell (V_{\boldsymbol {c}})$ , and thus a symmetric pairing on the tensor squares thereof). By Proposition 3.2 and the formula $\zeta _K(s) = 1/(1-q^{1-s})$ , the result then follows for $L(s,\boldsymbol {c},2)$ by (3.2).

(2): This is explained in [Reference Browning and Vishe12, § 3.4] using Swan conductors.

(3): This follows from [Reference Kahn35, Theorem 5.46].

(4): This follows from (1)–(3) and the Hadamard three circle theorem as in [Reference Browning and Vishe12, proof of Lemma 8.4].

(5): Let $s=\frac 12+\varepsilon +i\tau $ and integrate $\widehat R^s/L(s) \ll _{\varepsilon } \lvert \boldsymbol {c} \rvert ^\varepsilon \widehat R^{1/2+\varepsilon }$ over $\tau \in \mathbb {R}/\mathsf {T}\mathbb {Z}$ .

Proof. Let $M = H^3_\ell (V_{\boldsymbol {c}})$ . Poincaré duality gives a perfect skew-symmetric pairing

$$ \begin{align*}\psi\colon M \wedge M \to \mathbb{Q}_\ell(-3),\end{align*} $$

where $\mathbb {Q}_\ell (-3)$ denotes the Tate motive of weight $6$ . Since $\psi $ is surjective, its dual

$$ \begin{align*}\psi^\vee\colon \mathbb{Q}_\ell(3) \to M^\vee \wedge M^\vee\end{align*} $$

is injective. But $\psi $ is perfect, so that it induces an isomorphism

$$ \begin{align*}\psi'\colon M \to \operatorname{\mathrm{Hom}}(M, \mathbb{Q}_\ell(-3)) = M^\vee(-3).\end{align*} $$

After twisting $\psi ^\vee $ by $\mathbb {Q}_\ell (-6)$ , we thus obtain an injection $\mathbb {Q}_\ell (-3) \to M \wedge M$ . Thus $(M\wedge M)(3)$ (which is pure of weight $0$ ) has a nonzero space of $\operatorname {\mathrm {Gal}}(K^{\text {sep}}/K)$ -invariants. So by [Reference Lyons42, correction to Theorem 2.1] (which builds on [Reference Lafforgue40,Reference Lyons41]), it follows that the L-function $L(s,(M\wedge M)(3)) = L(s,\boldsymbol {c},\textstyle {\bigwedge ^2})$ has a pole at $s=1$ .

Proof. This follows directly from [Reference Kisin37, case (2) of Theorem 5.1].

Proof. We first show that $l(\varpi ,\boldsymbol {b})$ is locally constant. Let $\boldsymbol {b}, \boldsymbol {c}\in \mathcal {O}_\varpi ^6$ with $F^\ast (\boldsymbol {b})\ne 0$ and $\boldsymbol {c}\equiv \boldsymbol {b}\bmod {\varpi ^{1+l(\varpi ,\boldsymbol {b})}}$ . Then $F^\ast (\boldsymbol {c})\ne 0$ and $L_\varpi (s, \boldsymbol {c}) = L_\varpi (s, \boldsymbol {b})$ , and thus $l(\varpi ,\boldsymbol {c})\leqslant l(\varpi ,\boldsymbol {b})$ (since $\boldsymbol {a}\equiv \boldsymbol {c}\bmod {\varpi ^{1+l(\varpi ,\boldsymbol {b})}} \Rightarrow \boldsymbol {a}\equiv \boldsymbol {b}\bmod {\varpi ^{1+l(\varpi ,\boldsymbol {b})}}$ ). But then $\boldsymbol {b}\equiv \boldsymbol {c}\bmod {\varpi ^{1+l(\varpi ,\boldsymbol {c})}}$ , so $l(\varpi ,\boldsymbol {b})\leqslant l(\varpi ,\boldsymbol {c})$ , whence $l(\varpi ,\boldsymbol {b}) = l(\varpi ,\boldsymbol {c})$ . So if $\boldsymbol {b}$ lies in (3.3) for some $l\geqslant 0$ , then (3.3) is indeed closed under translation by $\varpi ^{l+1}\mathcal {O}_\varpi ^6$ .

We now turn to measures. Let $A\in \mathbb {Z}_{\geqslant 0}$ . The set $S_A = \{\boldsymbol {c}\in \mathcal {O}_\varpi ^6: v_\varpi (F^\ast (\boldsymbol {c}))\leqslant A\}$ is compact. Yet the function $\boldsymbol {b}\mapsto l(\varpi ,\boldsymbol {b})$ on $S_A$ is continuous. So (3.3) contains $S_A$ for all $l\gg _A 1$ , say. But by Corollary 2.8, the measure of $S_A$ tends to $1$ as $A\to \infty $ .

Proof. This result is directly analogous to [Reference Wang57, Proposition 6.1], and is purely local. For the reader’s convenience, we highlight the main points. Existence and multiplicativity of the limit follow from Lemma 3.4, the bound $\lvert \mu _{\boldsymbol {c}}(r) \rvert \ll _\varepsilon \lvert r \rvert ^\varepsilon $ (which follows from Proposition 3.1(3)) and the Chinese remainder theorem. The bound $\lvert \mu _{\boldsymbol {c}}(r) \rvert \ll _\varepsilon \lvert r \rvert ^\varepsilon $ implies (3.4). The estimates in (3.5) and (3.6) are subtler; they relate to the rank and homogeneity type [Reference Sarnak, Shin and Templier46, § 1] of our geometric family of L-functions $L(s,\boldsymbol {c})$ .

Recall the definition of $E^\natural _{\boldsymbol {c}}(k)$ from (2.15). The following hold uniformly over primes $\varpi \in \mathcal {O}^+$ (where $k_\varpi = \mathcal {O}/\varpi \mathcal {O}$ and $k_{\varpi ,2}$ is the unique quadratic field extension of $k_\varpi $ ):

(3.7) $$ \begin{align} \mathbb{E}_{\boldsymbol{c}\in k_\varpi^6}[E^\natural_{\boldsymbol{c}}(k_\varpi) \boldsymbol{1}_{\varpi\nmid F^\ast(\boldsymbol{c})}] &= \lambda_V(\varpi) \lvert \varpi \rvert^{-1/2} + O(\lvert \varpi \rvert^{-1}), \end{align} $$

(3.8) $$ \begin{align} \mathbb{E}_{\boldsymbol{c}\in k_\varpi^6}[E^\natural_{\boldsymbol{c}}(k_{\varpi,2}) \boldsymbol{1}_{\varpi\nmid F^\ast(\boldsymbol{c})}] &= 1 + O(\lvert \varpi \rvert^{-1}),\end{align} $$

(3.9) $$ \begin{align} \mathbb{E}_{\boldsymbol{c}\in k_\varpi^6}[E^\natural_{\boldsymbol{c}}(k_\varpi)^2 \boldsymbol{1}_{\varpi\nmid F^\ast(\boldsymbol{c})}] &= 1 + O(\lvert \varpi \rvert^{-1}). \end{align} $$

This is proven for prime fields in [Reference Wang53, Corollary 1.7]; the proof there carries over directly to arbitrary finite fields of characteristic $>3$ .

Finally, observe that if $\varpi \nmid F^\ast (\boldsymbol {c})$ , then $L_\varpi (s,\boldsymbol {c})^{-1} = \prod _{1\leqslant j\leqslant 10} (1 - \alpha _{\boldsymbol {c},j}(\varpi ) \lvert \varpi \rvert ^{-s})$ , so

$$ \begin{align*} \mu_{\boldsymbol{c}}(\varpi) &= -\sum_{1\leqslant j\leqslant 10} \alpha_{\boldsymbol{c},j}(\varpi) = E^\natural_{\boldsymbol{c}}(k_\varpi), \\ \mu_{\boldsymbol{c}}(\varpi^2) &= \sum_{1\leqslant i<j\leqslant 10} \alpha_{\boldsymbol{c},i}(\varpi) \alpha_{\boldsymbol{c},j}(\varpi) = \tfrac12(E^\natural_{\boldsymbol{c}}(k_\varpi)^2+E^\natural_{\boldsymbol{c}}(k_{\varpi,2})), \end{align*} $$

because $E^\natural _{\boldsymbol {c}}(k_\varpi ) = -\sum _{1\leqslant j\leqslant 10} \alpha _{\boldsymbol {c},j}(\varpi )$ and $E^\natural _{\boldsymbol {c}}(k_{\varpi ,2}) = -\sum _{1\leqslant j\leqslant 10} \alpha _{\boldsymbol {c},j}(\varpi )^2$ by the Grothendieck–Lefschetz trace formula. Thus (3.7)–(3.9) imply (3.5) and (3.6).

Proof assuming (R2)

By (3.1) and GRH, the left-hand side of (3.11) is independent of $\sigma _0>\frac 12$ . If $\widehat Z\ll 1$ then (3.11) is trivial, so suppose $Z\geqslant 2$ . We find, after shifting contours to $\Re (s) = \beta +\sigma (Z)\in (\tfrac 12, 2]$ and expanding squares using self-duality of $\Phi ^{\boldsymbol {c},1}$ , the left-hand side of (3.11) equals

$$ \begin{align*} \Sigma_0 := \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z} \mathbb{E}_{\Re(s_1),\Re(s_2)=\beta+\sigma(Z)}[ \Phi^{\boldsymbol{c},1}(s_1) \Phi^{\boldsymbol{c},1}(s_2) \cdot \widehat R^{s_1+s_2}]. \end{align*} $$

After switching the order of $\boldsymbol {c}$ and $\boldsymbol {s}$ in $\Sigma_0 $ , and plugging in (3.10) for each $\boldsymbol {s}$ , we get

(3.12) $$ \begin{align} \Sigma_0 = \Sigma_2 + O(\widehat Z^{6-6\beta} \Sigma_1^2), \end{align} $$

where $\Sigma_1 := \mathbb {E}_{\Re (s)=\beta +\sigma (Z)}[\lvert \widehat R^s \rvert ]=\widehat R^{\beta +\sigma (Z)}$ and

$$ \begin{align*} \Sigma_2 := \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z} \mathbb{E}_{\Re(s_1),\Re(s_2)=\beta+\sigma(Z)}[ \zeta_K(s_1+s_2) A_{F,2}(s_1,s_2) \cdot \widehat R^{s_1+s_2}]. \end{align*} $$

Since $R\leqslant 3Z$ and $\widehat {Z}^{1/Z} = q \ll 1$ , we certainly have $\Sigma_1 \ll \widehat Z^{3\beta } \widehat R^{1/2}$ .

Let $\delta =\frac {1}{20}$ ; then $\frac 12-\delta \geqslant \frac 13+\delta $ . In $\Sigma_2 $ we shift $\Re (s_1)$ from $\beta +\sigma (Z)$ to $\frac 12$ , then shift $\Re (s_2)$ from $\beta +\sigma (Z)$ to $\frac 12-\delta $ . This yields $\Sigma_2 = \Sigma_3 +O(\Sigma_4 ),$ where

$$ \begin{align*} \Sigma_3 := \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z} \mathbb{E}_{\Re(s_1)=\frac12}[ \tfrac{1}{\mathsf{T}} \tfrac{2\pi}{\log{q}} A_{F,2}(s_1, 1-s_1) \cdot \widehat R] \end{align*} $$

comes from the residue $(\log {q})^{-1}$ of $\zeta _K(s_1+s_2) = (1-q^{1-s_1-s_2})^{-1}$ at $s_2=1-s_1$ , and where we bound the contribution from $\Re (s_1) = \frac 12$ and $\Re (s_2) = \frac 12-\delta $ by

$$ \begin{align*} \Sigma_4 := \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z} \widehat R^{1-\delta} \ll \widehat Z^6 \widehat R^{1-\delta}, \end{align*} $$

on using the bounds $A_{F,2}(s_1,s_2)\ll 1$ and $\zeta _K(s_1+s_2)\ll _\delta 1$ . Next we note that

$$ \begin{align*} \begin{aligned} \Sigma_3 &\ll \widehat Z^6 \widehat R, \end{aligned} \end{align*} $$

since $A_{F,2}(s_1, 1-s_1) \ll 1$ in $\Sigma_3 $ . But

$$ \begin{align*}\Sigma_0 \leqslant O(\widehat Z^{6-6\beta} \Sigma_1^2) + \Sigma_3 + O(\Sigma_4), \end{align*} $$

by (3.12). Plugging in our estimates for $\Sigma_1 , \Sigma_3 $ and $\Sigma_4 $ , we finally get (3.11).

Proof assuming Conjecture 3.7

Since $R\in \mathbb {Z}$ , we may write

$$ \begin{align*}\boldsymbol{1}_{\lvert r \rvert=\widehat R} = \mathbb{E}_{\Re(s)=\sigma_0}[(\widehat R/\lvert r \rvert)^s]\end{align*} $$

for each $r\in \mathcal {O}^+$ . Now apply (3.11).

Proof. If $\widehat R\leqslant 1$ then (4.2) is trivial, so suppose $\widehat R>1$ . By Hölder’s inequality, we may assume A is an even integer. Let $\Sigma (Z,R)$ denote the left-hand side of (4.2) and put

$$ \begin{align*}\widetilde{\Sigma}(Z,R) := \Sigma(Z,R)/\widehat Z^6.\end{align*} $$

We claim that for $Z\geqslant AR$ we have

(4.3) $$ \begin{align} \widetilde{\Sigma}(Z,R) = \widetilde{\Sigma}(AR,R). \end{align} $$

To prove this, we write $\lvert \sum _{\lvert r \rvert =\widehat R} a_{\boldsymbol {c},2}(r) \rvert ^A = \sum _{\lvert r_1 \rvert =\dots =\lvert r_A \rvert =\widehat R} a_{\boldsymbol {c},2}(r_1)\cdots a_{\boldsymbol {c},2}(r_A)$ , on noting that $a_{\boldsymbol {c},2}(r)\in \mathbb {R}$ . If $r_1\cdots r_A = M$ , then (by smooth proper base change) the function $\boldsymbol {c}\mapsto a_{\boldsymbol {c},2}(r_1)\cdots a_{\boldsymbol {c},2}(r_A)$ is constant on each fibre of the map

$$ \begin{align*}\{\boldsymbol{c}\in \mathcal{S}_1: \lvert \boldsymbol{c} \rvert\leqslant \widehat Z,\; (M,F^\ast(\boldsymbol{c})) = 1\} \to \{\boldsymbol{c}\in (\mathcal{O}/M\mathcal{O})^6: \gcd(M,F^\ast(\boldsymbol{c})) = 1\}. \end{align*} $$

Each fibre has cardinality $(\widehat Z/\lvert M \rvert )^6$ , since $1<\lvert M \rvert \leqslant \widehat R^A\leqslant \widehat Z$ . The claim follows.

By GRH (Proposition 3.1(5)), we have $\Sigma (Z,R) \ll _{A,\varepsilon } \widehat Z^{6+\varepsilon } \widehat R^{A/4+\varepsilon }$ for all $\widehat Z>0$ . (The exponent is $A/4$ rather than $A/2$ , because the Euler product of $\Phi ^{\boldsymbol {c},2}(s)$ , as given in Definition 4.1, is supported on square moduli due to the appearance of $2s$ .) Thus $ \widetilde {\Sigma }(Z,R) \ll _{A,\varepsilon } \widehat R^{A/4+\varepsilon } $ for $Z\leqslant AR$ , and so for all Z, by (4.3).

Proof. If $\varpi \mid F^\ast (\boldsymbol {c})$ , then $\Phi ^{\text {G}}_\varpi (\boldsymbol {c},s) = 1$ . If $\varpi \nmid F^\ast (\boldsymbol {c})$ , then (2.17) implies

$$ \begin{align*}\Phi^{\text{G}}_\varpi(\boldsymbol{c},s) = 1 + S^\natural_\varpi(\boldsymbol{c}) \lvert \varpi \rvert^{-s} = 1 - \lambda_{\boldsymbol{c}}(\varpi) \lvert \varpi \rvert^{-s} - \lambda_V(\varpi) \lvert \varpi \rvert^{-1/2-s},\end{align*} $$

because $S^\natural _\varpi (\boldsymbol {c}) = E^\natural _{\boldsymbol {c}}(\mathcal {O}/\varpi \mathcal {O})-\lvert \varpi \rvert ^{-1/2}E^\natural _F(\mathcal {O}/\varpi \mathcal {O})$ by (2.16) and

$$ \begin{align*}\lambda_{\boldsymbol{c}}(\varpi) = \sum_{1\leqslant j\leqslant 10} \alpha_{\boldsymbol{c},j}(\varpi) = -E^\natural_{\boldsymbol{c}}(\mathcal{O}/\varpi\mathcal{O}), \quad \lambda_V(\varpi) = \sum_{1\leqslant j\leqslant 22} \alpha_{V,j}(\varpi) = E^\natural_F(\mathcal{O}/\varpi\mathcal{O}),\end{align*} $$

by the Grothendieck–Lefschetz trace formula. In either case, $a_{\boldsymbol {c},3}(\varpi ^l)\ll _\varepsilon \lvert \varpi \rvert ^{l\varepsilon }$ follows from the Ramanujan bound $\lvert \alpha \rvert \leqslant 1$ in Proposition 3.1(3).

Now suppose $\varpi \nmid F^\ast (\boldsymbol {c})$ . Multiplying $\Phi ^{\text {G}}_\varpi (\boldsymbol {c},s)$ by

$$ \begin{align*} \begin{aligned} L_\varpi(s,\boldsymbol{c}) &= 1 + \lambda_{\boldsymbol{c}}(\varpi) \lvert \varpi \rvert^{-s} + \lambda_{\boldsymbol{c}}(\varpi^2) \lvert \varpi \rvert^{-2s} + O(\lvert \varpi \rvert^{-3s}), \\ L_\varpi(\tfrac12+s,V) &= 1 + \lambda_V(\varpi) \lvert \varpi \rvert^{-1/2-s} + O(\lvert \varpi \rvert^{-1-2s}), \\ \zeta_K(2s) L_\varpi(2s,\boldsymbol{c},2) = L_\varpi(2s,\boldsymbol{c},\textstyle{\bigwedge^2}) &= 1 + \lambda_{\boldsymbol{c},\bigwedge^2}(\varpi) \lvert \varpi \rvert^{-2s} + O(\lvert \varpi \rvert^{-4s}), \end{aligned} \end{align*} $$

and discarding all terms of the form $O(\lvert \varpi \rvert ^{-s} \times \lvert \varpi \rvert ^{-2s})$ or $O(\lvert \varpi \rvert ^{-s} \times \lvert \varpi \rvert ^{-1/2-s})$ , we get

$$ \begin{align*}\Phi^{\boldsymbol{c},3}_\varpi(s) = 1 + O(\lvert \varpi \rvert^{-1/2-2s}) + O(\lvert \varpi \rvert^{-3s}),\end{align*} $$

since $\lambda _{\boldsymbol {c}}(\varpi )^2 = \lambda _{\boldsymbol {c}}(\varpi ^2) + \lambda _{\boldsymbol {c},\bigwedge ^2}(\varpi )$ . This completes the proof of the proposition.

Proof. By multiplicativity (of $\lvert a_{\boldsymbol {c},3} \rvert $ ) and positivity, we have

(4.4) $$ \begin{align} \sum_{r\in \mathcal{O}^+:\, \lvert r \rvert\leqslant \widehat R} \lvert a_{\boldsymbol{c},3}(r) \rvert \leqslant \sum_{d\in \mathcal{R}_{\boldsymbol{c}}^{\text{B}}:\, \lvert d \rvert\leqslant \widehat R} \lvert a_{\boldsymbol{c},3}(d) \rvert \sum_{e\in \mathcal{R}_{\boldsymbol{c}}^{\text{G}}:\, \lvert e \rvert\leqslant \widehat R} \lvert a_{\boldsymbol{c},3}(e) \rvert. \end{align} $$

But by Proposition 4.3, we have $a_{\boldsymbol {c},3}(d)\ll _\varepsilon \lvert d \rvert ^\varepsilon $ and

$$ \begin{align*} \sum_{e\in \mathcal{R}_{\boldsymbol{c}}^{\text{G}}:\, \lvert e \rvert\leqslant \widehat R} \frac{\lvert a_{\boldsymbol{c},3}(e) \rvert}{\widehat R^{1/3+\varepsilon}} \leqslant \sum_{e\in \mathcal{R}_{\boldsymbol{c}}^{\text{G}}} \frac{\lvert a_{\boldsymbol{c},3}(e) \rvert}{\lvert e \rvert^{1/3+\varepsilon}} \leqslant \prod_\varpi \left(1 + \frac{O(\lvert \varpi \rvert^{-1/2})}{\lvert \varpi \rvert^{2/3+2\varepsilon}} + \frac{O_\varepsilon(\lvert \varpi \rvert^\varepsilon)}{\lvert \varpi \rvert^{1+3\varepsilon}}\right) \ll_\varepsilon 1, \end{align*} $$

So by (4.4), the quantity to be estimated is

$$ \begin{align*} \ll_{A,\varepsilon} \widehat R^{(1/3+2\varepsilon) A} \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z} \#{ \{d\in \mathcal{R}_{\boldsymbol{c}}^{\text{B}}: \lvert d \rvert\leqslant \widehat R\} }^A. \end{align*} $$

Now the claimed bound follows from Lemma 2.3.

Proof assuming Conjecture 3.8

Let $\boldsymbol {c}\in \mathcal {S}_1$ and $R_1,R_2,R_3\in \mathbb {Z}_{\geqslant 0}$ . Writing

$$ \begin{align*}\Sigma^{\boldsymbol{c},j}(R_j) := \sum_{r_j\in \mathcal{O}^+:\, \lvert r_j \rvert=\widehat R_j} a_{\boldsymbol{c},j}(r_j)\end{align*} $$

for brevity, we have

$$ \begin{align*}\sum_{r\in \mathcal{R}_{\boldsymbol{c}}^{\text{G}}:\, \lvert r \rvert=\widehat R} S^\natural_r(\boldsymbol{c}) = \sum_{R_1+R_2+R_3=R}\, \prod_{1\leqslant j\leqslant 3} \Sigma^{\boldsymbol{c},j}(R_j), \end{align*} $$

since $\Phi ^{\text {G}} = \Phi ^{\boldsymbol {c},1} \Phi ^{\boldsymbol {c},2} \Phi ^{\boldsymbol {c},3}$ . Let $\beta := 2-\varepsilon $ . Let

$$ \begin{align*}W_1(R_1) := \widehat R_1^\eta, \quad W_2(R_1) := \widehat R_1^{-(\beta-1)\eta},\end{align*} $$

for some $\eta>0$ to be chosen. Let

$$ \begin{align*}\mathcal{I}_1 := \sum_{R_1+R_2+R_3=R} W_1(R_1).\end{align*} $$

Using Hölder’s inequality in the form $1 = \frac {\beta -1}{\beta } + \frac {1}{\beta }$ over the set $\{R_1+R_2+R_3=R\}$ (and then exponentiating by $\beta $ ), we obtain

(4.6) $$ \begin{align} \Big\lvert{ \sum_{r\in \mathcal{R}_{\boldsymbol{c}}^{\text{G}}:\, \lvert r \rvert=\widehat R} S^\natural_r(\boldsymbol{c}) }\Big\rvert^\beta \leqslant \mathcal{I}_1^{\beta-1} \sum_{R_1+R_2+R_3=R} W_2(R_1) \prod_{1\leqslant j\leqslant 3} \lvert \Sigma^{\boldsymbol{c},j}(R_j) \rvert^\beta, \end{align} $$

since $\beta \geqslant 1$ and $W_1^{\beta -1} W_2 = 1$ . Here

$$ \begin{align*}\mathcal{I}_1 \leqslant \sum_{R_2,R_3\geqslant 0} \frac{\widehat R^\eta}{(\widehat R_2\widehat R_3)^\eta} \ll_\eta \widehat R^\eta.\end{align*} $$

Upon summing (4.6) over $\lvert \boldsymbol {c} \rvert \leqslant \widehat Z$ , we thus find that the left-hand side of (4.5) is

(4.7) $$ \begin{align} \ll_\eta \sum_{R_1+R_2+R_3=R} (\widehat R/\widehat R_1)^{(\beta-1)\eta} \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z}\, \prod_{1\leqslant j\leqslant 3} \lvert \Sigma^{\boldsymbol{c},j}(R_j) \rvert^\beta. \end{align} $$

Let $(\gamma _1, \gamma _2, \gamma _3) := (2, 4\beta /\varepsilon , 4\beta /\varepsilon )$ . Then $\sum _{1\leqslant j\leqslant 3} \beta /\gamma _j = 1$ , so by Hölder over $\boldsymbol {c}$ , we obtain

(4.8) $$ \begin{align} \sum_{\boldsymbol{c}\in \mathcal{S}_1:\, \lvert \boldsymbol{c} \rvert\leqslant \widehat Z}\, \prod_{1\leqslant j\leqslant 3} \lvert \Sigma^{\boldsymbol{c},j}(R_j) \rvert^\beta \leqslant \prod_{1\leqslant j\leqslant 3} \mathscr{M}_j^{\beta/\gamma_j}, \end{align} $$

where $\mathscr {M}_j = \sum _{\boldsymbol {c}\in \mathcal {S}_1:\, \lvert \boldsymbol {c} \rvert \leqslant \widehat Z} \lvert \Sigma ^{\boldsymbol {c},j}(R_j) \rvert ^{\gamma _j}$ . By Conjecture 3.8 with $R_1$ in place of R, we have

$$ \begin{align*}\mathscr{M}_1 \ll \widehat Z^6 \widehat R_1, \end{align*} $$

since $R_1\leqslant R\leqslant 3Z$ by assumption. Also, by (4.2) and Corollary 4.4, we have

$$ \begin{align*}\mathscr{M}_2 \ll_{\gamma_2} \widehat Z^6 \widehat R_2^{11\gamma_2/40} \quad \text{ and }\quad \mathscr{M}_3 \ll_{\gamma_3} \widehat Z^6 \widehat R_3^{11\gamma_3/30}. \end{align*} $$

Therefore, if $R_1+R_2+R_3=R$ , then the right-hand side of (4.8) is

$$ \begin{align*} \ll_\beta \widehat Z^{\sum_j 6\beta/\gamma_j} \widehat R_1^{\beta/2} \widehat R_2^{11\beta/40} \widehat R_3^{11\beta/30} = \widehat Z^6 (\widehat R^{1/2} \widehat R_2^{-9/40} \widehat R_3^{-2/15})^\beta. \end{align*} $$

Plugging (4.8) into (4.7) now bounds the left-hand side of (4.5) by

(4.9) $$ \begin{align} \ll_{\eta,\beta} \sum_{R_1+R_2+R_3=R} \frac{(\widehat R/\widehat R_1)^{(\beta-1)\eta} \widehat Z^6 \widehat R^{\beta/2}}{(\widehat R_2^{9/40} \widehat R_3^{2/15})^\beta}. \end{align} $$

Let $\eta = \min (9/40, 2/15)$ . Then $R_1+R_2+R_3=R \Rightarrow (\widehat R/\widehat R_1)^\eta \leqslant \widehat R_2^{9/40} \widehat R_3^{2/15}$ . Since $\beta -1\geqslant 0$ , it follows that the quantity (4.9) is

$$ \begin{align*} \ll_\beta \sum_{R_1+R_2+R_3=R} \frac{\widehat Z^6 \widehat R^{\beta/2}}{\widehat R_2^{9/40} \widehat R_3^{2/15}} \leqslant \sum_{R_2,R_3\geqslant 0} \frac{\widehat Z^6 \widehat R^{\beta/2}}{\widehat R_2^{9/40} \widehat R_3^{2/15}} \ll \widehat Z^6 \widehat R^{\beta/2}. \end{align*} $$

This implies the desired inequality (4.5).

Hypothesis 5.3. Let $H(\boldsymbol {x})=(\frac {\partial ^{2}F(\boldsymbol {x})}{\partial x_i\partial x_j})$ be the Hessian associated to F.

1. (i) If $\boldsymbol {x}\in \operatorname {\mathrm {supp}}(w)$ , then $\det H(\boldsymbol {x})\neq 0$ .

2. (ii) There exist $\boldsymbol {x_0}\in K_\infty ^n$ with $\lvert \boldsymbol {x}_0 \rvert \leqslant 1$ , and an integer $L\geqslant 0$ , such that

   $$\begin{align*}w(\boldsymbol{x})=\begin{cases} 1 &\text{if }|\boldsymbol{x}-\boldsymbol{x}_0|<\widehat{L}^{-1},\\ 0&\text{else.} \end{cases} \end{align*}$$

Proof. In this proof all implied constants are allowed to depend on F. Upon making the change of variables $\boldsymbol {y}=t^L(\boldsymbol {x}-\boldsymbol {x}_0)$ , it follows that

$$\begin{align*}J_{F,w}(\gamma,\boldsymbol{w})=\widehat{L}^{-n}\psi(\boldsymbol{w}\cdot\boldsymbol{x_0})J_{G}(\gamma, t^{-L}\boldsymbol{w}), \end{align*}$$

where $G(\boldsymbol {y})=F(t^{-L}\boldsymbol {y}+\boldsymbol {x}_0)$ . It is clear that $H_G\ll H_F\ll 1$ . At this point we can use Lemma 5.1 to conclude that $J_{F,w}(\gamma , \boldsymbol {w})=0$ unless $|\boldsymbol {w}|\ll \text {max}\{1, |\gamma |\}$ . Furthermore, by Lemma 5.2

$$ \begin{align*} |J_{F,w}(\gamma, \boldsymbol{w})| &\leqslant \widehat{L}^{-n}\text{meas}\{\boldsymbol{y}\in \mathbb{T}^n\colon |\gamma \nabla G(\boldsymbol{y})+t^{-L}\boldsymbol{w}| \leqslant H_G\ \text{max}\{1,|\gamma|^{1/2}\} \}\\ & \leqslant \text{meas}\{\boldsymbol{x}\in\operatorname{\mathrm{supp}}(w)\colon |\gamma \nabla F(\boldsymbol{x})+\boldsymbol{w}|\ll \text{max}\{1, |\gamma|^{1/2}\}\}, \end{align*} $$

since $\nabla G(\boldsymbol {y}) = t^{-L}\nabla F(\boldsymbol {x})$ . Let us denote the set whose measure we want to estimate by $\Omega $ . Since $\boldsymbol {0}\not \in \operatorname {\mathrm {supp}}(w)$ , we must have $|\nabla F(\boldsymbol {x})|\asymp 1$ for all $\boldsymbol {x}\in \operatorname {\mathrm {supp}}(w)$ . If $|\boldsymbol {w}|\ll 1$ , then $\Omega $ is empty unless $|\gamma |\ll 1$ , so that we can use the trivial estimate $|J_{F,w}(\gamma ,\boldsymbol {w})|\leqslant 1$ to conclude that $J^\Gamma _{F,w}(\boldsymbol {w})\ll 1$ if $|\boldsymbol {w}|\ll 1 $ . In particular, we may assume $|\boldsymbol {w}|\gg 1$ for the rest of the proof. In this case the contribution from $|\gamma |\ll 1$ vanishes, since then $\Omega $ is empty, while for $|\gamma |\gg 1$ the defining inequality for $\Omega $ becomes

$$\begin{align*}|\gamma \nabla F(\boldsymbol{x})+\boldsymbol{w}|\ll |\gamma|^{1/2}. \end{align*}$$

This can only hold if $|\gamma |\asymp |\boldsymbol {w}|$ . From this we deduce the statement that $J_{F,w}(\gamma ,\boldsymbol {w})=0$ unless $|\gamma |\asymp |\boldsymbol {w}|$ .

It remains to give an upper bound. This will be similar to [Reference Browning and Vishe12, proof of Lemma 7.3]. Let us now assume that $\boldsymbol {x}, \boldsymbol {x}+\boldsymbol {x}'$ are both in $\Omega $ . Note that $\nabla F(\boldsymbol {x}+\boldsymbol {x}')-\nabla F (\boldsymbol {x})= H(\boldsymbol {x}+\boldsymbol {x}'/2)\boldsymbol {x}'$ , so that we must then have

(5.2) $$ \begin{align} |H(\boldsymbol{x}+\boldsymbol{x}'/2)\boldsymbol{x}'|\ll |\gamma|^{-1/2}. \end{align} $$

As $\det H(\boldsymbol {y})\neq 0$ for all $\boldsymbol {y}\in \operatorname {\mathrm {supp}}(w)$ , compactness of $\operatorname {\mathrm {supp}}(w)$ implies that we have $|\det H(\boldsymbol {y})|\gg 1$ for all $\boldsymbol {y}\in \operatorname {\mathrm {supp}}(w)$ . In addition, a straightforward computation shows that $\boldsymbol {x}+\boldsymbol {x}'/2\in \operatorname {\mathrm {supp}}(w)$ , so that we can multiply (5.2) from the left by the inverse of $H(\boldsymbol {x}+\boldsymbol {x}'/2)$ , whose entries have absolute value $O(1)$ , to conclude that $|\boldsymbol {x}'|\ll |\gamma |^{-1/2}\ll |\boldsymbol {w}|^{-1/2}$ . Finally, from this we obtain

$$ \begin{align*} J^\Gamma_{F,w}( \boldsymbol{w}) \ll \int_{|\gamma|\asymp |\boldsymbol{w}|}\text{meas}(\Omega)\text{d}\gamma \ll |\boldsymbol{w}|^{1-n/2} \end{align*} $$

when $|\boldsymbol {w}|\gg 1$ , as desired.

Proof. Let us write $\lambda =\lambda _1$ and show that for $\boldsymbol {w}$ fixed, the value of $J^\Gamma _{F,w}(\lambda \boldsymbol {w})$ only depends on the image of $\lambda $ in the quotient group $K_\infty ^\times / (1+t^{-L}\mathbb {T})$ . We have

$$ \begin{align*} J^\Gamma_{F,w}(\lambda\boldsymbol{w}) &= \int_{|\gamma|<\widehat{\Gamma}}\int_{K_\infty^n}w(\boldsymbol{x})\psi(\gamma F(\boldsymbol{x})+\lambda \boldsymbol{w}\cdot\boldsymbol{x})\text{d} \boldsymbol{x}\text{d}\gamma\\ & = |\lambda|^n \int_{K_\infty^n}w(\lambda^{-1}\boldsymbol{y})\psi(\boldsymbol{w}\cdot \boldsymbol{y})\int_{|\gamma|<\widehat{\Gamma}}\psi(\gamma \lambda^{-3}F(\boldsymbol{y}))\text{d}\gamma \text{d}\boldsymbol{y}, \end{align*} $$

where we applied the change of variables $\boldsymbol {y}=\lambda \boldsymbol {x}$ . It follows from (2.3) that

$$\begin{align*}\int_{|\gamma|<\widehat{\Gamma}}\psi(\gamma \lambda^{-3}F(\boldsymbol{y}))\text{d}\gamma = \begin{cases} \widehat{\Gamma} &\text{if }|F(\boldsymbol{y})|< \widehat{\Gamma}^{-1}|\lambda|^3,\\ 0&\text{else.} \end{cases} \end{align*}$$

This formula clearly only depends on $|\lambda |$ . Moreover, $w(\lambda ^{-1}\boldsymbol {y})$ depends only on $\lambda \bmod {1+t^{-L}\mathbb {T}}$ by (5.1). Since we assume $L\geqslant 0$ , any two elements of $K_\infty ^\times $ whose image in $K_\infty ^\times / (1+t^{-L}\mathbb {T})$ coincides must have the same absolute value, from which we deduce that $J^\Gamma _{F,w}(\lambda \boldsymbol {w})$ only depends on $\lambda \bmod {1+t^{-L}\mathbb {T}}$ .

Proof. Let $\boldsymbol {x}\in R$ . Since $\det H(\boldsymbol {x}) \neq 0$ , it follows from the inverse function theorem for local fields [Reference Serre48, Part II, § III.9, Theorem 2] that there exist open neighbourhoods $U, U'\subset K_\infty ^n$ , with $\det H(\boldsymbol {y})\neq 0$ for any $\boldsymbol {y}$ in the closure of U, such that $\boldsymbol {x}\in U$ , $\nabla F(\boldsymbol {x})\in U'$ and $\nabla F\colon U \to U'$ is an isomorphism. In addition, after shrinking $U'$ if necessary, we may assume that it is an open ball of radius $\widehat C_0^{-1}$ around $\nabla F(\boldsymbol {x})$ , for some real $C_0=C_0(\boldsymbol {x})\geqslant 0$ . It follows that if $\boldsymbol {w} \in U'$ , there exists $\boldsymbol {s}\in U$ such that $\nabla F(\boldsymbol {s})= \boldsymbol {w}$ . Therefore, we have

$$ \begin{align*} \boldsymbol{w}= \nabla F (\boldsymbol{s})= \nabla F(\boldsymbol{s}-\boldsymbol{x}) + H(\boldsymbol{x})(\boldsymbol{s}-\boldsymbol{x})+\nabla F(\boldsymbol{x}). \end{align*} $$

As $\boldsymbol {x}\in R$ , we have $\det H(\boldsymbol {x})\neq 0$ , and hence

$$\begin{align*}\boldsymbol{s}-\boldsymbol{x}= H(\boldsymbol{x})^{-1}(\boldsymbol{w}-\nabla F(\boldsymbol{x}) - \nabla F(\boldsymbol{s}-\boldsymbol{x})). \end{align*}$$

Moreover, as R is compact, the entries of $H(\boldsymbol {x})^{-1}$ are $O_R(1)$ . It follows that there exist constants $C_1, C_2\geqslant 0$ depending only on R and F such that

$$\begin{align*}|\boldsymbol{s}-\boldsymbol{x}|\leqslant \widehat{C}_1|\boldsymbol{w}-\nabla F(\boldsymbol{x})| +\widehat{C}_2 |\boldsymbol{s}-\boldsymbol{x}|^2. \end{align*}$$

Upon shrinking U if necessary, we may assume that $|\boldsymbol {s}-\boldsymbol {x}|< (2\widehat {C}_2)^{-1}$ , so that we obtain

$$\begin{align*}|\boldsymbol{s}-\boldsymbol{x}|\leqslant 2\widehat{C}_1|\boldsymbol{w}-\nabla F(\boldsymbol{x})|. \end{align*}$$

Since R is compact, we can cover it with finitely many open sets U and then take C to be the maximum of the constants that arise in this process.

Proof. Note that if $|\boldsymbol {w}|\ll 1$ , then $|F^*({\boldsymbol {w}})|\ll 1+|\boldsymbol {w}|^{\deg F^*-1}$ holds trivially, so that we may assume $|\boldsymbol {w}|\gg 1$ from now on. Similarly we can assume $1\ll |\boldsymbol {w}|\ll \widehat {\Gamma }$ , since otherwise $J^\Gamma _{F,w}(\boldsymbol {w})=0$ by Lemma 5.5. Using Lemma 5.5 again, it follows that

$$\begin{align*}J^\Gamma_{F,w}(\boldsymbol{w})=\int_{|\gamma|\asymp |\boldsymbol{w}|} J_{F,w}(\gamma,\boldsymbol{w}) \text{d}\gamma. \end{align*}$$

Let $Z\in \mathbb {Z}_{>0}$ be such that $\widehat {Z}\asymp |\boldsymbol {w}|$ . We will now consider the contribution to $J^\Gamma _{F,w}(\boldsymbol {w})$ from those $\gamma $ with $|\gamma |=\widehat {Z}$ . To do so, write $\gamma = a t^Z(1+\gamma ')$ , where $a\in \mathbb {F}_q^\times $ and $\gamma '\in \mathbb {T}$ . Let $\boldsymbol {y} := \boldsymbol {x}-\boldsymbol {x}_0$ and $G(\boldsymbol {y}) := F(\boldsymbol {y}+\boldsymbol {x}_0)$ , as in the proof of Lemma 5.5. Then let

$$ \begin{align*} \boldsymbol{x}' := (1+\gamma')^{-1/2} \boldsymbol{x} = (1+\gamma')^{-1/2} (\boldsymbol{y}+\boldsymbol{x}_0). \end{align*} $$

Note that since $(1+\gamma ')^{-1/2} - 1 \in \mathbb {T}$ and $L=0$ , property (5.1) of the weight function w implies $w(\boldsymbol {x})=w(\boldsymbol {x}')$ . In particular, after the change of variables $\boldsymbol {x}\mapsto (1+\gamma ')^{-1/2}\boldsymbol {x}$ by condition (ii) on w, we get

$$ \begin{align*} \int_{\mathbb{T}^n}w(\boldsymbol{x}')\psi(\gamma F(\boldsymbol{x}')+\boldsymbol{w}\cdot\boldsymbol{x}')\text{d}\boldsymbol{x} &=\psi((1+\gamma')^{-1/2}\boldsymbol{x}_0\cdot\boldsymbol{w})\int_{\mathbb{T}^n}\psi((1+\gamma')^{-1/2}(at^ZG(\boldsymbol{y})+\boldsymbol{y}\cdot\boldsymbol{w}))\text{d}\boldsymbol{y}\\ &= \psi((1+\gamma')^{-1/2}\boldsymbol{x}_0\cdot\boldsymbol{w})\int_{\Omega}\psi((1+\gamma')^{-1/2}(at^ZG(\boldsymbol{y})+\boldsymbol{y}\cdot\boldsymbol{w}))\text{d}\boldsymbol{y}, \end{align*} $$

by Lemma 5.2, where

$$\begin{align*}\Omega := \{\boldsymbol{y}\in\mathbb{T}^n\colon |at^Z\nabla G(\boldsymbol{y})+\boldsymbol{w}|\ll \widehat{Z}^{1/2}\} \end{align*}$$

and we used that $\gamma F(\boldsymbol {x}')=at^Z (1+\gamma ')^{-1/2}F(\boldsymbol {y}+\boldsymbol {x}_0)$ since F is homogeneous of degree $3$ . In summary, we have proven the identity

$$ \begin{align*}J_{F,w}(\gamma,\boldsymbol{w}) = J_{F,w}((1+\gamma')^{-3/2}\gamma,(1+\gamma')^{-1/2}\boldsymbol{w}),\end{align*} $$

and applied stationary phase to the latter integral to get an integral over $\boldsymbol {y}\in \Omega $ .

Upon expressing the integral over $\boldsymbol {y}\in \Omega $ in terms of $\boldsymbol {x} = \boldsymbol {y}+\boldsymbol {x}_0$ , we find that

$$\begin{align*}J_{F,w}(\gamma,\boldsymbol{w}) = \int_{\mathbb{T}^n}w(\boldsymbol{x}')\psi(\gamma F(\boldsymbol{x}')+\boldsymbol{w}\cdot\boldsymbol{x}')\text{d}\boldsymbol{x}' = \int_{\Omega'}\psi((1+\gamma')^{-1/2}\Phi(\boldsymbol{x}))\text{d}\boldsymbol{x}, \end{align*}$$

where $\Phi (\boldsymbol {x}) := at^ZF(\boldsymbol {x})+\boldsymbol {w}\cdot \boldsymbol {x}$ and

$$\begin{align*}\Omega' := \{\boldsymbol{x}\in\operatorname{\mathrm{supp}}(w)\colon |a\nabla F(\boldsymbol{x})+t^{-Z}\boldsymbol{w}|\ll \widehat{Z}^{-1/2}\}. \end{align*}$$

The crucial observation here is that $\Omega '$ is independent of $\gamma '$ . In particular, when considering the total contribution to $J^\Gamma _{F,w}(\boldsymbol {w})$ from the set $\{\gamma = at^Z(1+\gamma '): \gamma '\in \mathbb {T}\}$ , we may exchange the order of integration and see that it is given by

$$\begin{align*}\int_{\Omega'}\int_{\mathbb{T}}\psi((1+\gamma')^{-1/2}\Phi(\boldsymbol{x}))\text{d}\gamma' \text{d}{\boldsymbol{x}}. \end{align*}$$

Here $\Omega '$ , $\Phi $ depend on Z and a, but all estimates below will be uniform over Z and a. Let us now take the Taylor expansion

$$\begin{align*}(1+\gamma')^{-1/2}=\sum_{k\geqslant 0}a_k \gamma^{\prime k}, \end{align*}$$

where $a_k=\binom {-1/2}{k}\in \mathbb {F}_q$ . This is well defined, because $|\gamma '|<1$ and $\operatorname {\mathrm {char}}(\mathbb {F}_q)\neq 2$ . Since $|\gamma '|<1$ , the value of $\psi ((1+\gamma ')^{-1/2}\Phi (\boldsymbol {x}))$ only depends on $f(\gamma ')=\sum _{k=0}^{k_0}a_k \gamma ^{\prime k}$ if we choose $k_0$ sufficiently large. We may now apply Lemma 5.2 again with $n=1$ to infer that the inner integral is given by

$$\begin{align*}\int_{\Theta} \psi(f(\gamma')\Phi(\boldsymbol{x}))\text{d}\gamma', \end{align*}$$

where

$$\begin{align*}\Theta = \{\gamma'\in\mathbb{T}\colon |f'(\gamma')\Phi(\boldsymbol{x})|\leqslant \text{max}\{1, |\Phi(\boldsymbol{x})|^{1/2}\}\}. \end{align*}$$

It is easy to see that $|f'(\gamma ')|=1$ , so that $\Theta $ is nonempty only if $|\Phi (\boldsymbol {x})|\leqslant 1$ , which we shall assume for the rest of the proof.

Since $\boldsymbol {x}\in \Omega '$ , we have $\boldsymbol {x}\in \operatorname {\mathrm {supp}}(w)$ and $|a\nabla F(\boldsymbol {x})+t^{-Z}\boldsymbol {w}|\ll \widehat {Z}^{-1/2}\ll |\boldsymbol {w}|^{-1/2}$ . Since we assume $|\boldsymbol {w}|\gg 1$ for a sufficiently large implied constant, we may invoke Lemma 5.7 to deduce the existence of $\boldsymbol {s}\in K_\infty ^n$ such that $|\boldsymbol {s}-\boldsymbol {x}|\ll |a \nabla F(\boldsymbol {x})+t^{-Z}\boldsymbol {w}|\ll |\boldsymbol {w}|^{-1/2}$ and $\nabla F(\boldsymbol {s})=-a^{-1}t^{-Z}\boldsymbol {w}$ . Using Taylor expansion at $\boldsymbol {s}$ it follows that

$$ \begin{align*} |aF(\boldsymbol{x})+t^{-Z}\boldsymbol{w}\cdot \boldsymbol{x}|&= |aF(\boldsymbol{s})+t^{-Z}\boldsymbol{w}\cdot\boldsymbol{s}|+O(|\boldsymbol{x}-\boldsymbol{s}|^2)\\ &=|aF(\boldsymbol{s})+t^{-Z}\boldsymbol{w}\cdot\boldsymbol{s}|+O(|\boldsymbol{w}|^{-1}). \end{align*} $$

Moreover, as $a\nabla F(\boldsymbol {s})=-t^{-Z}\boldsymbol {w}$ , we have $3a F(\boldsymbol {s})=-t^{-Z}\boldsymbol {w}\cdot \boldsymbol {s}$ and hence

$$ \begin{align*}{|a F(\boldsymbol{s})+t^{-Z}\boldsymbol{w}\cdot \boldsymbol{s}|=|F(\boldsymbol{s})|}.\end{align*} $$

By (2.12), the ratio $F^*(\nabla F(\boldsymbol {x})) / F(\boldsymbol {x})$ is a homogeneous polynomial in $\mathcal {O}[x_1,\dots , x_n]$ of degree $2\deg F^*-3$ , which immediately implies

$$\begin{align*}|F^*(\nabla F(\boldsymbol{s}))|\ll |F(\boldsymbol{s})||\boldsymbol{s}|^{2\deg F^*-3}. \end{align*}$$

In addition, as $|\boldsymbol {s}-\boldsymbol {x}|\ll |\boldsymbol {w}|^{-1/2}$ , we have $|\boldsymbol {s}|\asymp 1$ . Together with $-a^{-1}t^{-Z}\boldsymbol {w}=\nabla F(\boldsymbol {s})$ , this implies $|F^*(t^{-Z}\boldsymbol {w})|=|F^*(\nabla F(\boldsymbol {s}))|\ll |F(\boldsymbol {s})|$ and hence

$$\begin{align*}|F^*(\boldsymbol{w})|\ll \widehat{Z}^{\deg F^*}(\widehat{Z}^{-1}|\Phi(\boldsymbol{x})|+|\boldsymbol{w}|^{-1})\ll |\boldsymbol{w}|^{\deg F^*-1}, \end{align*}$$

as desired.

Proof. Part (1) is due to Hooley [Reference Hooley32, Lemma 60]. Part (2) is straightforward and is recorded in the proof of [Reference Glas and Hochfilzer26, Eq. (4.7)], although we take the opportunity to correct the definition of $\{\varpi ^e,c_i\}$ to be equal to $1$ when $e=2$ . To see part (3) we invoke [Reference Glas and Hochfilzer26, Eq. (4.7)] to deduce that

$$ \begin{align*}S_{\varpi^e}^\natural(\boldsymbol{c})\ll \frac{1}{|\varpi^e|^{(n+1)/2}} \cdot |\varpi^e| \prod_{1\leqslant i\leqslant n} |\varpi^e|^{1/2} \{\varpi^e,c_i\}^{1/4}, \end{align*} $$

where $\{\varpi ^e,c_i\}=|\varpi |^{-1}$ if $\varpi \| c_i$ , with $\{\varpi ^e,c_i\}=|\gcd (\varpi ^e,c_i)|$ otherwise. The statement of the lemma follows on noting that $\{\varpi ^e,c_i\}\leqslant |\gcd (\varpi ^e,\operatorname {\mathrm {sq}}(c_i))|$ .

Proof. By multiplicativity, it suffices to show that $ S_{\varpi ^e}^\natural ({\boldsymbol {c}}) \ll |\varpi ^e|^{1/2+\theta _e}$ , for any prime power $\varpi ^e\in \mathcal {O}$ , where

$$ \begin{align*}\theta_e= \begin{cases} 0 & \text{ if } e\leqslant 2,\\ n/6 & \text{ if } e\geqslant 3. \end{cases} \end{align*} $$

For $e\leqslant 2$ this follows from parts (1) and (2) of Lemma 6.1. For $e\geqslant 3$ , we apply the $1$ -dimensional Hua type estimate recorded in [Reference Glas and Hochfilzer26, § 4.2].

Proof. After expressing $S_{\varpi ^e}(\boldsymbol {c})$ in terms of one-dimensional sums, this is immediate from the function field analogues of [Reference Hooley29, Eq. (44)–(45)], which hold since $\operatorname {\mathrm {char}}(\mathbb {F}_q)\ne 3$ .

Proof. Here $\Delta $ is defined by a specialisation process explained in [Reference Busé and Jouanolou14, paragraph after Eq. (4.2.1)]. Write $\boldsymbol {c}=(c_1,\dots ,c_{n-1},1)$ and let $\operatorname {\mathrm {disc}}(F,\boldsymbol {c})$ be the discriminant associated in [Reference Terakado50, § 1.1] to the pair $(F(\boldsymbol {x}), \boldsymbol {c}\cdot \boldsymbol {x})$ .

As explained in [Reference Wang53, Proposition 4.4], for example, we have $\operatorname {\mathrm {disc}}(F,\boldsymbol {c}) / F^\ast (\boldsymbol {c}) \in \mathbb {F}_q^\times $ . (This is the only step where we use $\operatorname {\mathrm {char}}(\mathbb {F}_q)> 3$ .) Also

$$ \begin{align*} \operatorname{\mathrm{disc}}(F,\boldsymbol{c}) = \pm \operatorname{\mathrm{disc}}(F(y_1,\dots,y_{n-1},-c_1y_1-\dots-c_{n-1}y_{n-1})), \end{align*} $$

by [Reference Wang53, Proposition 3.2]. It therefore follows that $\Delta / F^\ast \in \mathbb {F}_q^\times $ , which suffices.

Proof. Let $e>1$ and assume that $\varpi \nmid {\boldsymbol {c}}$ . We may appeal to an argument of Hooley [Reference Hooley34, § 6], which carries over to the function field setting to give

$$ \begin{align*}S_{\varpi^e}(\boldsymbol{c})=\frac{|\varpi|}{|\varpi|-1}\left\{|\varpi|^e\nu_0(\varpi^e,{\boldsymbol{c}})- |\varpi|^{e-1}\nu_1(\varpi^e,{\boldsymbol{c}})\right\}. \end{align*} $$

Here, for $i\in \{0,1\}$ , $\nu _i(\varpi ^e,{\boldsymbol {c}})$ is the number of $\boldsymbol {y}\bmod {\varpi ^e}$ such that $F(\boldsymbol {y})\equiv 0\bmod {\varpi ^e}$ , with $\varpi \nmid \boldsymbol {y}$ and ${\boldsymbol {c}}.\boldsymbol {y}\equiv 0\bmod {\varpi ^{e-i}}$ . Since F is a nonsingular cubic form with coefficients in $\mathbb {F}_q$ , it follows that $\varpi \nmid \nabla F(\boldsymbol {y})$ for any $\boldsymbol {y}$ such that $\varpi \mid F(\boldsymbol {y})$ but $\varpi \nmid \boldsymbol {y}$ . Hence $\nu _1(\varpi ^e,{\boldsymbol {c}})=|\varpi |^{n-1}\nu _0(\varpi ^{e-1},{\boldsymbol {c}})$ , so that

$$ \begin{align*}S_{\varpi^e}(\boldsymbol{c})=\frac{|\varpi|^{e}}{1-|\varpi|^{-1}}\left\{\nu_0(\varpi^e,{\boldsymbol{c}})- |\varpi|^{n-2}\nu_0(\varpi^{e-1},{\boldsymbol{c}})\right\}. \end{align*} $$

Since $\varpi \nmid {\boldsymbol {c}}$ , we can assume without loss of generality that $\varpi \nmid c_n$ . Eliminating $y_n$ , we are left with the expression

$$ \begin{align*}S_{\varpi^e}(\boldsymbol{c})=\frac{|\varpi|^{e}}{1-|\varpi|^{-1}}\left\{\nu(\varpi^e,{\boldsymbol{c}})- |\varpi|^{n-2}\nu(\varpi^{e-1},{\boldsymbol{c}})\right\}, \end{align*} $$

where $\nu (\varpi ^e,{\boldsymbol {c}})$ is the number of $\boldsymbol {y}'=(y_1,\dots ,y_{n-1})\bmod {\varpi ^e}$ such that $G(\boldsymbol {y}')\equiv 0\bmod {\varpi ^e}$ and $\varpi \nmid \boldsymbol {y}'$ , where $ G(\boldsymbol {y}')=F(c_ny_1,\dots ,c_ny_{n-1},-c_1y_1-\dots -c_{n-1}y_{n-1}). $ Part (2) for $\varpi \nmid \boldsymbol {c}$ now follows from taking $m=e-1$ , $\delta \in [0, \tfrac {m-1}{2}]$ and $\ell =0$ in the multivariate form of the Hensel-type lemma [Reference Browning and Matthiesen11, Lemma 3.3], carried over to $\mathcal {O}$ . Indeed, for any $\boldsymbol {u}'$ counted by $\nu (\varpi ^{e-1},{\boldsymbol {c}})$ , we must have

$$ \begin{align*}v_\varpi(\nabla G(\boldsymbol{u}')) \leqslant (e-2)/2 = (m-1)/2,\end{align*} $$

or else $v_\varpi (F^\ast (\boldsymbol {c})) \geqslant e-1$ by the $\beta = \lceil \frac {e-1}{2} \rceil $ case of Lemma 6.7. Using Lemma 6.4 and arguing by induction on $v_\varpi (\boldsymbol {c})$ , we can also conclude that part (2) holds if $\varpi \mid \boldsymbol {c}$ .

Turning to part (1), we take $e=2$ and note that $\varpi \nmid {\boldsymbol {c}}$ if $v_\varpi (F^*({\boldsymbol {c}}))\leqslant 1$ , since $\deg F^*>1$ . But then it follows that

$$ \begin{align*}S_{\varpi^2}(\boldsymbol{c})=\frac{|\varpi|^{2}}{1-|\varpi|^{-1}}\left\{\nu(\varpi^2,{\boldsymbol{c}})- |\varpi|^{n-2} \nu(\varpi,{\boldsymbol{c}})\right\}. \end{align*} $$

Pick $\boldsymbol {u}'$ counted by $\nu (\varpi ,{\boldsymbol {c}})$ . If $\varpi \nmid \nabla G(\boldsymbol {u}')$ then the contribution to the right hand side vanishes, by Hensel lifting. On the other hand, if $\varpi \mid \nabla G(\boldsymbol {u}')$ then Lemma 6.7 implies that the contribution to $\nu (\varpi ^2,{\boldsymbol {c}})$ vanishes. Hence

$$ \begin{align*}S_{\varpi^2}^\natural(\boldsymbol{c})\ll \frac{|\varpi|^{2}}{ |\varpi^2|^{(n+1)/2} }\cdot |\varpi|^{n-2} \#\{\boldsymbol{u}'\bmod{\varpi}: \varpi\mid \nabla G(\boldsymbol{u}'), ~\varpi\nmid \boldsymbol{u}'\}. \end{align*} $$

It follows from a result of Zak [Reference Hooley31, Katz’s Appendix, Theorem 2] that the hypersurface $G=0$ in $\mathbb {P}^{n-2}$ has only $O(1)$ singular points, whence the number of $\boldsymbol {u}'$ is $O(|\varpi |)$ . It follows that

$$ \begin{align*}S_{\varpi^2}^\natural(\boldsymbol{c})\ll \frac{|\varpi|^{2}}{ |\varpi^2|^{(n+1)/2} }\cdot |\varpi|^{n-1}\ll 1, \end{align*} $$

which is what we wanted.

Proof. Since $\varpi \nmid \boldsymbol {u}'$ , we have $\varpi \nmid u_a$ , say. Then by Proposition 6.5 and [Reference Busé and Jouanolou14, Corollary 4.30; see the beginning of § 4 for the definition of $\tilde{f}$ there], $F^\ast (\boldsymbol {c})$ is an $\mathcal {O}[\boldsymbol {u}'][1/u_a]$ -linear combination of G and the pairwise products $(\partial G/\partial u_i)(\partial G/\partial u_j)$ . Since $\varpi ^{2\beta }$ divides each pairwise product, the desired divisibility of $F^\ast (\boldsymbol {c})$ follows.