The Burgess bound via a trivial delta method

Abstract

Let g be a fixed Hecke cusp form for $\mathrm SL(2,{\mathbb {Z}})$ and $\chi$ be a primitive Dirichlet character of conductor M. The best known subconvex bound for $L(1/2,g\otimes \chi )$ is of Burgess strength. The bound was proved by a couple of methods: shifted convolution sums and the Petersson/Kuznetsov formula analysis. It is natural to ask what inputs are really needed to prove a Burgess-type bound on $\mathrm GL(2)$. In this paper, we give a new proof of the Burgess-type bounds ${L(1/2,g\otimes \chi )\ll _{g,\varepsilon } M^{1/2-1/8+\varepsilon }}$ and $L(1/2,\chi )\ll _{\varepsilon } M^{1/4-1/16+\varepsilon }$ that does not require the basic tools of the previous proofs and instead uses a trivial delta method.

Appendix A: Shifted character sums and counting lemmas

For this section, let $M>3$ be a prime and define

$$\begin{aligned} \mathfrak {K}=\sum _{z\in \mathbb {F}_M^{\times }}\bar{\chi }(r+\ell z) e\left( \frac{n\overline{z}}{M}\right) ,\qquad (\ell ,M)=1,\quad n,r,\ell \in \mathbb {Z}. \end{aligned}$$

Lemma A.1

Suppose that $(r,M)=1$. If M|n, then $\mathfrak {K}=-\bar{\chi }(r).$ If $M\not \mid n$, then

$$\begin{aligned} \mathfrak {K}\ll M^{1/2}. \end{aligned}$$

Proof

For M|n, trivial. If $M\not \mid n$, by the Fourier expansion of $\chi$ in terms of additive characters

$$\begin{aligned} \chi (a)=g_{\bar{\chi }}^{-1}\sum _{y\bmod M}\bar{\chi }(y)e\left( \frac{a y}{M}\right) , \end{aligned}$$

(26)

we have

$$\begin{aligned} \mathfrak {K}=g_{\chi }^{-1}\sum _{y,z\in \mathbb {F}_M^{\times }}\chi (y) e\left( \frac{ry+\ell yz+n\overline{z}}{M}\right) . \end{aligned}$$

Then the bound follows from [1, Corollary 4.3]. $\square$

We define

$$\begin{aligned} {\mathfrak {C}}=\mathop {\sum _{z\in \mathbb {F}_M^{\times }}}_{ (n+\beta \overline{z},M)=1}\bar{\chi }(r_1+z) \chi \bigg (r_2+\alpha (\overline{n+\beta \overline{z}})\bigg ), \qquad (\alpha \beta ,M)=1,\quad n,r_1,r_2,\alpha ,\beta \in \mathbb {Z}. \end{aligned}$$

Lemma A.2

Suppose that $(r_1r_2,M)=1$. If M|n, we have

$$\begin{aligned} {\mathfrak {C}}=\chi (\alpha \overline{\beta }) R_M(r_2-r_1\alpha \overline{\beta })-\chi (r_2\overline{r_1}), \end{aligned}$$

where $R_M(a)=\sum _{z\in \mathbb {F}_M^{\times }}e(az/M)$ is the Ramanujan sum. If $M\not \mid n$ and at least one of $r_1-\overline{n}\beta$ and $r_2+\overline{n}\alpha$ is nonzero in $\mathbb {F}_M$, then

$$\begin{aligned} {\mathfrak {C}}\ll M^{1/2}. \end{aligned}$$

Finally, if $n\ne 0$ and $r_1-\overline{n}\beta =r_2+\overline{n}\alpha =0$ in ${\mathbb {F}}_M$, then

$$\begin{aligned} {\mathfrak {C}}= {\left\{ \begin{array}{ll} -\chi (nr_2\overline{\beta }) \quad &{}\text { if } \chi \text { is not a quadratic character}, \\ \chi (\overline{n}r_2\beta )(M-1) \quad &{}\text { if } \chi \text { is a quadratic character}. \end{array}\right. } \end{aligned}$$

Proof

For M|n,

$$\begin{aligned} {\mathfrak {C}} =\sum _{z\in \mathbb {F}_M}\bar{\chi }(r_1+z) \chi (r_2+\alpha \overline{\beta }z)-\chi (r_2\overline{r_1}). \end{aligned}$$

Then the first statement follows from (26). If $M\not \mid n$, by making change of variables $z\rightarrow \overline{n}\beta z$ and $z+1\rightarrow z$, we have

$$\begin{aligned} {\mathfrak {C}} =\sum _{z\in \mathbb {F}_M^{\times }} \bar{\chi }(r_1+\overline{n}\beta (z-1)) \chi (r_2+\overline{n}\alpha (1-\overline{z}))-\chi (r_2\overline{r_1}). \end{aligned}$$

Applying (26) again we get

$$\begin{aligned} {\mathfrak {C}}= & {} g_{\chi }^{-1}g_{\bar{\chi }}^{-1}\sum _{x,y,z\in \mathbb {F}_M^{\times }}\chi (x)\bar{\chi }(y) e\left( \frac{(r_1-\overline{n}\beta )x+ (r_2+\overline{n}\alpha )y+\overline{n}\beta xz-\overline{n} \alpha y\overline{z}}{M}\right) \\&-\chi (r_2\overline{r_1}). \end{aligned}$$

Consider the Newton polyhedron $\Delta (f)$ of

$$\begin{aligned} f(x,y,z)=(r_1-\overline{n}\beta )x+ (r_2+\overline{n}\alpha )y+\overline{n}\beta xz-\overline{n} \alpha yz^{-1}\in \mathbb {F}_M^{\times }[x,y,z,(x y z)^{-1}]. \end{aligned}$$

We separate into two cases.

• (1) If $r_1-\overline{n}\beta =0$ or $r_2+\overline{n}\alpha =0$ in $\mathbb {F}_M$, then $\Delta (f)$ is the tetrahedron in $\mathbb {R}^3$ with vertices (0, 0, 0), (0, 1, 0), (1, 0, 1), $(0,1,-1)$ or (0, 0, 0), (1, 0, 0), (1, 0, 1), $(0,1,-1)$. It is easy to check that f is nondegenerate with respect to $\Delta (f)$. By [1] (see also [9]), we have ${\mathfrak {C}}\ll M^{1/2}.$

• (2) If both $r_1-\overline{n}\beta$ and $r_2+\overline{n}\alpha$ are nonzero in $\mathbb {F}_M$, then $\Delta (f)$ is the pentahedron in $\mathbb {R}^3$ with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 0, 1) and $(0,1,-1)$. The only face which fails to meet the criterion for nondegeneracy is the quadrilateral one with polynomial $f_{\sigma }(x,y,z)=(r_1-\overline{n}\beta )x+ (r_2+\overline{n}\alpha )y+\overline{n}\beta xz-\overline{n} \alpha yz^{-1}$ for which the locus of $\partial f_{\sigma }/\partial x=\partial f_{\sigma }/\partial y= \partial f_{\sigma }/\partial z=0$ is empty in $(\mathbb {F}_M^{\times })^3$ only if $n\not \equiv \beta \overline{r_1}-\alpha \overline{r_2}\bmod M$. Therefore, for $n\not \equiv \beta \overline{r_1}-\alpha \overline{r_2}\bmod M$, we can apply the square-root cancellation result in [1] or [9] to get ${\mathfrak {C}}\ll M^{1/2}$.

For $n\equiv \beta \overline{r_1}-\alpha \overline{r_2}\bmod M$, we have $r_1-\overline{n}\beta \equiv -\alpha r_1\overline{nr_2}\bmod M$ and $r_2+\overline{n}\alpha \equiv \beta r_2\overline{nr_1}\bmod M.$ By changing variables $x\rightarrow x\overline{z}$ and $\overline{n}x\rightarrow x$, $\overline{n}y\rightarrow y$ we obtain

$$\begin{aligned} \begin{aligned} {\mathfrak {C}} =&g_{\chi }^{-1}g_{\bar{\chi }}^{-1}\sum _{x,y,z\in \mathbb {F}_M^{\times }}\chi (x)\bar{\chi }(y)\bar{\chi }(z) e\left( \frac{ (r_1\overline{r_2}x+y)(\beta r_2\overline{r_1}-\alpha \overline{z})}{M}\right) -\chi (r_2\overline{r_1}). \end{aligned} \end{aligned}$$

Since $\chi$ is primitive, the sum over x and y vanishes if $z\equiv \alpha r_1\overline{\beta r_2}\bmod M$. Thus we can make change of variables $(\beta r_2\overline{r_1}-\alpha \overline{z})x\rightarrow x$ and $(\beta r_2\overline{r_1}-\alpha \overline{z})y\rightarrow y$ to get

$$\begin{aligned} \begin{aligned} {\mathfrak {C}} =&g_{\chi }^{-1}g_{\bar{\chi }}^{-1}\mathop {\sum _{z\in \mathbb {F}_M^{\times }}}_{z\not \equiv \alpha r_1\overline{\beta r_2}\bmod M} \bar{\chi }(z) \sum _{x,y\in \mathbb {F}_M^{\times }} \chi (x)\bar{\chi }(y) e\left( \frac{ r_1\overline{r_2}x+y}{M}\right) -\chi (r_2\overline{r_1})\\ =&-\chi ^2(r_2\overline{r_1})\chi (\beta \overline{\alpha })-\chi (r_2\overline{r_1}). \end{aligned} \end{aligned}$$

Finally, if $n\ne 0$ and $r_1-\overline{n}\beta =r_2+\overline{n}\alpha =0$ in ${\mathbb {F}}_M$,

$$\begin{aligned} \begin{aligned} {\mathfrak {C}}&= \chi (n)\sum _{z\in {\mathbb {F}}_M^\times }\bar{\chi }(r_1+z)\bar{\chi }(n+\beta \overline{z})\chi (r_2+\overline{n}\alpha +r_2\beta \overline{zn})\\&= \chi (r_2\beta )\sum _{z\in {\mathbb {F}}_M^\times }\bar{\chi }(r_1+z)\bar{\chi }(nz+\beta ) =\chi (r_2\beta )\sum _{\begin{array}{c} z\in {\mathbb {F}}_M \\ z\ne \overline{n}\beta \end{array}}\bar{\chi }(r_1-\overline{n}\beta +z)\bar{\chi }(nz)\\&= \chi (\overline{n}r_2\beta )\sum _{\begin{array}{c} z\in {\mathbb {F}}_M \\ z\ne \overline{n}\beta \end{array}}\bar{\chi }^2(z). \end{aligned} \end{aligned}$$

If $\chi$ is not a quadratic character, then by orthogonality of characters, ${\mathfrak {C}}= -\chi (nr_2\overline{\beta })$. If $\chi$ is a quadratic character, then ${\mathfrak {C}}=\chi (\overline{n}r_2\beta )(M-1)$. $\square$

Lemma A.3

$\Sigma _{10}\ll (PML)^{\varepsilon }\frac{L^6M^4}{P^4N^2}$ and $\Delta _{10}\ll (PML)^{\varepsilon }\frac{L^3M^2}{P^2N}$.

Proof

We recall $|r_2|<R:=N^{\varepsilon }PM/N$. Suppose

$$\begin{aligned} \begin{aligned} P^2L\ll N^{1-\varepsilon }. \end{aligned} \end{aligned}$$

(27)

Then the congruence $r_2\ell _1p_2\equiv r_1\ell _2 p_1\bmod {M}$ implies that $r_2\ell _1p_2= r_1\ell _2 p_1$. Similarly, $r_2'\ell _1p_2'= r_1'\ell _2 p_1'$. Therefore fixing $\ell _1,p_2,r_2$ fixes $\ell _2,p_1,r_1$ up to factors of $\log M$. If $\ell _1\ne \ell _2$, then the equality $r_2'\ell _1p_2'= r_1'\ell _2 p_1'$ implies $\ell _2|r_2'$. That saves a factor of L in $r_2'$ sum. Further, for fixed $p_2',r_2'$, there are only $\log M$ many $p_1', r_1'$. In the case $\ell _1=\ell _2$, the previous identities become $r_2p_2=r_1p_1$. Therefore fixing $r_2, p_2$ fixes $r_1, p_1$ up to factors of $\log M$.

Finally, the congruence conditions on $r_1, r_2$ and n can be combined to write

$$\begin{aligned} \begin{aligned} -\overline{r_1}\ell _1 p_2+\overline{r_2}\ell _2p_1+n\equiv 0\bmod p_1p_2M. \end{aligned} \end{aligned}$$

From (22), the n satisfies $|n|\ll N^{1+\varepsilon }L/M$, which is smaller than the size of the modulus $p_1p_2M$. Therefore for fixed $r_i, \ell _i, p_i$, the n sum is at most singleton. Similarly, $n'$ is at most a singleton. Therefore up to a factor of $(PML)^\varepsilon$,

$$\begin{aligned} \begin{aligned} \Sigma _{10}\ll L^5L^2\frac{1}{P^3}R\frac{1}{P^3}\frac{R}{L}M^2\ll \frac{L^6M^4}{P^4N^2} \quad \text { and } \quad \Delta _{10}\ll L^3\frac{1}{P^3}RM \ll \frac{L^3M^2}{P^2N}. \end{aligned} \end{aligned}$$

$\square$

Lemma A.4

$\Sigma _{11}\ll (PML)^\varepsilon \frac{L^8M^4}{N^4P^4}\left( 1+\frac{P^2}{{\mathcal {N}}_0}\right) ^2$ and $\Delta _{11}\ll (PML)^\varepsilon \frac{L^3M^2}{N^2P^2}\left( 1+\frac{P^2}{{\mathcal {N}}_0}\right)$.

Proof

We let the variables of summation $\ell _i, p_i, r_i, n$ to be the same as before. The expressions for $\Delta _{11}$ and $\Sigma _{11}$ are the same as $\Delta _{10}$ and $\Sigma _{10}$ with the condition $r_2\ell _1p_2\equiv r_1\ell _2p_1\bmod M$ replaced by $r_2\ell _1p_2\not \equiv r_1\ell _2p_1\bmod M$. First let $\ell _1\ne \ell _2$. If $p_1\ne p_2$, then $(n,p_1p_2)=1$, $r_1\equiv \overline{nM}\ell _1p_2\bmod {p_1}$ and $r_2\equiv - \overline{nM} \ell _2 p_1 \bmod {p_2}$. Since $R\gg P$, these congruence conditions therefore save a factor of O(P) in each $r_i$-sum. Similarly we save a factor of O(P) in each $r_i'$ sum. The congruence $n\equiv 0\bmod M$ (resp $n'\equiv 0\bmod M$) saves a factor of at most M in the n-sum (resp $n'$-sum). If $p_1=p_2=p$, then the congruence conditions imply p|n. We already have M|n. Recall from (22) the n-sum satisfies $|n|\ll N^{1+\varepsilon }L/M$, which is smaller than pM by our choice of P and L. Hence we have $n=0$. The remaining congruence condition $r_1\ell _2\equiv r_2\ell _1\bmod p$ shows that fixing $r_1,\ell _2,\ell _1$ saves a factor of O(P) in the $r_2$-sum. The exact same savings follow for $n'$ and $r_2'$ sums. Also, the exact same analysis as done for $p_i,r_i,n$-sums follows for the case $\ell _1=\ell _2$. Therefore up to a factor of $(PML)^\varepsilon$,

$$\begin{aligned} \begin{aligned} \Sigma _{11}&\ll L^8\bigg [\frac{1}{P^2}\frac{R^2}{P^2}\left( 1+\frac{P^2}{{\mathcal {N}}_0}\right) \bigg ]^2 + L^8\bigg [\frac{1}{P^3}\frac{R^2}{P}\bigg ]^2 \ll \frac{L^8M^4}{N^4P^4}\left( 1+\frac{P^2}{{\mathcal {N}}_0}\right) ^2,\\ \Delta _{11}&\ll L^3\bigg [\frac{1}{P^2}\frac{R^2}{P^2}\left( 1+\frac{P^2}{{\mathcal {N}}_0}\right) \bigg ] + L^3\bigg [\frac{1}{P^3}\frac{R^2}{P}\bigg ] \ll \frac{L^3M^2}{N^2P^2}\left( 1+\frac{P^2}{{\mathcal {N}}_0}\right) . \end{aligned} \end{aligned}$$

$\square$

Lemma A.5

$\Sigma _{20}\ll (PML)^\varepsilon \frac{L^6M^4}{P^4N^2}$ and $\Delta _{20}\ll (PML)^\varepsilon \frac{L^3M^2}{P^2N}$.

Proof

The congruence conditions on $r_1, r_2$ and n can be combined to write

$$\begin{aligned} -\ell _1p_2+nr_1\equiv 0\bmod p_1M \quad \text { and } \quad \ell _2p_1+nr_2\equiv 0\bmod p_2M. \end{aligned}$$

By (22), we have $|nR|\ll N^{\varepsilon }PL<PM^{1-\varepsilon }$. The congruence conditions therefore give equalities

$$\begin{aligned} \begin{aligned}&n r_1 = \ell _1 p_2 \quad \text { and } \quad n r_2 = -\ell _2 p_1, \\&n' r_1' = \ell _1 p_2' \quad \text { and } \quad n' r_2' = -\ell _2 p_1'. \end{aligned} \end{aligned}$$

Note that $n=\ell _1p_2/r_1 = -\ell _2 p_1/r_2$ implies $\ell _1p_2r_2=-\ell _2p_1r_1$. Therefore fixing $\ell _1,p_2,r_2$ fixes $\ell _2,p_1,r_1$ up to factors of $\log M$. Similarly, $n'=\ell _1p_2'/r_1' = -\ell _2 p_1'/r_2'$, so that $\ell _1p_2'r_2'=-\ell _2p_1'r_1'$. If $\ell _1\ne \ell _2$, then $\ell _2|r_2'$. That saves a factor of L in $r_2'$ sum. Moreover for fixed $\ell _1,p_2',r_2'$, there are only $\log M$ many $p_1', r_1'$. Finally the identities $n r_1 = \ell _1 p_2$ and $n' r_1' = \ell _1 p_2'$ fix n and $n'$.

In the case $\ell _1=\ell _2$, the previous identities become $r_2p_2=-r_1p_1$. Therefore fixing $r_2, p_2$ fixes $r_1, p_1$ up to factors of $\log M$. Finally the identity $n r_1 = \ell p_2$ fixes n. Therefore up to a factor of $(PML)^\varepsilon$,

$$\begin{aligned} \begin{aligned} \Sigma _{20}\ll L^5L^2\frac{1}{P^3}R\frac{1}{P^3}\frac{R}{L}M^2\ll \frac{L^6M^4}{P^4N^2}, \quad \text { and } \quad \Delta _{20} \ll L^3\frac{1}{P^3}RM \ll \frac{L^3M^2}{P^2N}. \end{aligned} \end{aligned}$$

$\square$

Lemma A.6

$\Sigma _{21}\ll (PML)^\varepsilon \frac{L^8M^5}{N^4P^4}\left( 1+\frac{P^2M}{{\mathcal {N}}_0}\right) ^2$ and $\Delta _{21}\ll (PML)^\varepsilon \frac{L^3M^{5/2}}{N^2P^2}\left( 1+\frac{P^2M}{{\mathcal {N}}_0}\right)$.

Proof

When $p_1\ne p_2$, the congruence $-\overline{r_1}\ell _1 p_2+\overline{r_2}\ell _2p_1+n\equiv 0(p_1p_2)$ implies that $(n,p_1p_2)=1$. Moreover, for fixed n, $p_i$ and $\ell _i$, $i=1,2$,

$$\begin{aligned} \begin{aligned} r_1\equiv \overline{n}\ell _1p_2\bmod {p_1} \quad \text { and } \quad r_2\equiv -\overline{n}\ell _2p_1\bmod {p_2}. \end{aligned} \end{aligned}$$

These congruence conditions save a factor of P in each $r_i$-sum. In case $p_1=p_2=p$, the congruence condition shows p|n. Moreover, $-\overline{r_1}\ell _1+\overline{r_2}\ell _2+n/p \equiv 0 \bmod p$. Therefore fixing $r_1, \ell _1, \ell _2, n$ saves P in $r_2$-sum. Similarly we get saving of P for each of the $n'$ and $r_2'$ sums. Also, the exact same analysis as done for $p_i,r_i,n$-sums follows for the case $\ell _1=\ell _2$. Therefore up to a factor of $(PML)^\varepsilon$,

$$\begin{aligned} \Sigma _{21}\ll&L^8\bigg [\frac{1}{P^2}\frac{R^2}{P^2}\left( 1+\frac{P^2M}{{\mathcal {N}}_0}\right) M^{1/2}\bigg ]^2\\&+ L^8\bigg [\frac{1}{P^2}\frac{R^2}{P^2}\left( 1+\frac{PM}{{\mathcal {N}}_0}\right) M^{1/2}\bigg ]^2 \ll \frac{L^8M^5}{N^4P^4}\left( 1+\frac{P^2M}{{\mathcal {N}}_0}\right) ^2, \nonumber \\ \Delta _{21} \ll&L^3\bigg [\frac{1}{P^2}\frac{R^2}{P^2}\left( 1+\frac{P^2M}{{\mathcal {N}}_0}\right) M^{1/2}\bigg ]\\&+ L^3\bigg [\frac{1}{P^2}\frac{R^2}{P^2}\left( 1+\frac{PM}{{\mathcal {N}}_0}\right) M^{1/2}\bigg ] \ll \frac{L^3M^{5/2}}{N^2P^2}\left( 1+\frac{P^2M}{{\mathcal {N}}_0}\right) . \end{aligned}$$

$\square$