Maass Waveforms and Low-Lying Zeros

Abstract

The Katz–Sarnak Density Conjecture states that the behavior of zeros of a family of L-functions near the central point (as the conductors tend to zero) agrees with the behavior of eigenvalues near 1 of a classical compact group (as the matrix size tends to infinity). Using the Petersson formula, Iwaniec, Luo, and Sarnak proved that the behavior of zeros near the central point of holomorphic cusp forms agrees with the behavior of eigenvalues of orthogonal matrices for suitably restricted test functions ϕ. We prove similar results for families of cuspidal Maass forms, the other natural family of $\mathrm{GL}_{2}/\mathbb{Q}$ L-functions. For suitable weight functions on the space of Maass forms, the limiting behavior agrees with the expected orthogonal group. We prove this for $\mathop{\mathrm{supp}}(\hat{\phi }) \subseteq (-3/2,3/2)$ when the level N tends to infinity through the square-free numbers; if the level is fixed the support decreases to being contained in (−1, 1), though we still uniquely specify the symmetry type by computing the 2-level density.

To Professor Helmut Maier on his 60th birthday

Appendices

Appendix 1: Contour Integration

We prove Proposition 3.3 below. We restate it for the reader’s convenience.

Proposition 5.1.

Let T be an odd integer and X ≤ T. Let w _T equal h _T or H _T, where these are the weight functions from Theorems  2.2 to 2.5. Then

$$ \displaystyle\begin{array}{rcl} \int _{\mathbb{R}}J_{2ir}(X)\frac{rw_{T}(r)} {\cosh (\pi r)} dr& =& c_{1}\sum _{k\geq 0}(-1)^{k}J_{ 2k+1}(X)(2k + 1)w_{T}\left (\left (k + \frac{1} {2}\right )i\right ) \\ & & \quad \left [+\ c_{2}T^{2}\sum _{ k\geq 1}(-1)^{k}J_{ 2kT}(X)k^{2}h(k)\right ] {}\end{array}$$

(86)

$$\displaystyle\begin{array}{rcl} & =& \ c_{1}\sum _{k\geq 0}(-1)^{k}J_{ 2k+1}(X)(2k + 1)w_{T}\left (\left (k + \frac{1} {2}\right )i\right ) \\ & & \quad \left [+O\left (Xe^{-c_{3}T}\right )\right ], {}\end{array}$$

(87)

where c ₁ , c ₂ , and c ₃ are constants independent of X and T, and the terms in brackets are included if and only if w _T = h _T.

Proof.

In the proof below bracketed terms are present if and only if w _T  = h _T .

Recall that

$$\displaystyle\begin{array}{rcl} J_{\alpha }(2x)\ =\ \sum _{m\geq 0} \frac{(-1)^{m}x^{2m+\alpha }} {m!\varGamma (m +\alpha +1)}.& &{}\end{array}$$

(88)

By Stirling’s formula, $\varGamma (m + 2ir + 1)\cosh (\pi r) \gg \vert m + 2ir + 1\vert ^{m+\frac{1} {2} }e^{-m}$. Hence by the Lebesgue Dominated Convergence Theorem (remember that w _T (z) is of rapid decay as $\vert \mathfrak{R}z\vert \rightarrow \infty$) we may switch sum and integral to get

$$\displaystyle\begin{array}{rcl} & & \int _{\mathbb{R}}J_{2ir}(X)\frac{rw_{T}(r)} {\cosh (\pi r)} dr \\ & & \quad \quad \ =\ \sum _{m\geq 0}\frac{(-1)^{m}x^{2m}} {m!} \int _{\mathbb{R}} \frac{x^{2ir}rw_{T}(r)} {\varGamma (m + 1 + 2ir)\cosh (\pi r)}dr,{}\end{array}$$

(89)

where X = : 2x.

Now we move the line of integration down to $\mathbb{R} - iR$, $R\not\in \mathbb{Z} + \frac{1} {2}$ (and $\not\in T\mathbb{Z}$ if $w_{T} = h_{T}$). To do this, we note the estimate (for A ≫ 1)

$$\displaystyle\begin{array}{rcl} \int _{\pm A\rightarrow \pm A-iR} \frac{x^{2ir}rw_{T}(r)} {\varGamma (m + 1 + 2ir)\cosh (\pi r)}dr& \ll & \int _{\pm A\rightarrow \pm A-iR}e^{m}x^{2R}\vert r\vert \vert m + 2ir \\ & & +1\vert ^{-m-\frac{1} {2} }\vert w_{T}(r)\vert dr \\ & \ll & _{T,R}A^{-2}, {}\end{array}$$

(90)

where again we have used the rapid decay of w _T along horizontal lines, and B → B − iR denotes the vertical line from $B \in \mathbb{C}$ to $B - iR \in \mathbb{C}$. (Rapid decay also ensures the integral along $\mathbb{R} - iR$ converges absolutely.)

Note that the integrand

$$\displaystyle\begin{array}{rcl} \frac{x^{2ir}rw_{T}(r)} {\varGamma (m + 1 + 2ir)\cosh (\pi r)}& &{}\end{array}$$

(91)

has poles below the real axis precisely at $r \in -i(\mathbb{N} + \frac{1} {2}) =\{ -\frac{1} {2}i,-\frac{3} {2}i,\ldots \}$, and, if w _T  = h _T , poles also at $r \in -iT\mathbb{Z}^{+}$. The residue of the pole at $r = -\frac{2k+1} {2} i$ (k ≥ 0) is, up to an overall constant independent of k,

$$\displaystyle\begin{array}{rcl} \frac{x^{2k+1}} {\varGamma (m + 1 + (2k + 1))}(-1)^{k}(2k + 1)w_{ T}\left (\frac{2k + 1} {2} i\right ).& &{}\end{array}$$

(92)

If w _T  = h _T , the residue of the pole at r = −ikT (k ≥ 1) is, up to another overall constant independent of k,

$$\displaystyle\begin{array}{rcl} \frac{x^{2kT}} {\varGamma (m + 1 + (2kT))}(-1)^{k}k^{2}T^{2} \frac{h(k)} {\cos (\pi kT)} = \frac{x^{2kT}} {\varGamma (m + 1 + (2kT))}k^{2}T^{2}h(k).& &{}\end{array}$$

(93)

Hence the sum of (89) becomes

$$\displaystyle\begin{array}{rcl} & & \sum _{m\geq 0}\frac{(-1)^{m}x^{2m}} {m!} \int _{\mathbb{R}} \frac{x^{2ir}rw_{T}(r)} {\varGamma (m + 1 + 2ir)\cosh (\pi r)}dr \\ & & \quad \quad =\ c_{6}\sum _{m\geq 0}\frac{(-1)^{m}x^{2m}} {m!} \Bigg(\sum _{0\leq k\ll R} \frac{x^{2k+1}} {\varGamma (m + 1 + (2k + 1))}(-1)^{k}(2k + 1)w_{ T} \\ & & \quad \quad \quad \quad \quad \left.\left (\frac{2k + 1} {2} i\right ) +\int _{\mathbb{R}-iR} \frac{x^{2ir}rw_{T}(r)} {\varGamma (m + 1 + 2ir)\cosh (\pi r)}dr\right. \\ & & \quad \quad \quad \quad \quad \Bigg[+c_{7}\sum _{m\geq 0}\frac{(-1)^{m}x^{2m}} {m!} \sum _{0\leq k\ll R} \frac{x^{2kT}} {\varGamma (m + 1 + (2kT))}(-1)^{k}k^{2}T^{2}h(k)\Bigg]\Bigg). {}\end{array}$$

(94)

Now we take R → ∞. Note that

$$\displaystyle\begin{array}{rcl} \int _{\mathbb{R}-iR} \frac{x^{2ir}rw_{T}(r)} {\varGamma (m + 1 + 2ir)\cosh (\pi r)}dr& \,\ll & \,\int _{\mathbb{R}-iR}x^{2R}\vert r\vert \vert w_{ T}(r)\vert \vert m + 2ir + 1\vert ^{-m-\frac{1} {2} -2R}dr \\ & & \ll _{m}\ x^{2R}e^{ \frac{\pi R} {2T} }R^{-m-\frac{1} {2} -2R}, {}\end{array}$$

(95)

again by Stirling and rapid decay of w _T on horizontal lines (that is, $w_{T}(x + iy) \ll (1 + x)^{-4}e^{ \frac{\pi y} {2T} }$ since both h and H have all their derivatives supported in $\left (-\frac{1} {4}, \frac{1} {4}\right )$). This of course vanishes as R → ∞.

Hence we see that

$$\displaystyle\begin{array}{rcl} & & \int _{\mathbb{R}}J_{2ir}(X)\frac{rw_{T}(r)} {\cosh (\pi r)} dr \\ & & \ \ \ \ \ \ =\ c_{8}\sum _{m\geq 0}\frac{(-1)^{m}x^{2m}} {m!} \sum _{k\geq 0} \frac{x^{2k+1}} {\varGamma (m + 1 + (2k + 1))}(-1)^{k}(2k + 1)w_{ T}\left (\frac{2k + 1} {2} i\right ) \\ & & \ \ \ \ \ \ \ \ \ \ \ \left [+c_{9}\sum _{m\geq 0}\frac{(-1)^{m}x^{2m}} {m!} \sum _{k\geq 0} \frac{x^{2kT}} {\varGamma (m + 1 + (2kT))}k^{2}T^{2}h(k)\right ]. {}\end{array}$$

(96)

Switching sums (via the exponential bounds on w _T along the imaginary axis) and applying $J_{n}(X) =\sum _{m\geq 0}\frac{(-1)^{m}x^{2m+n}} {m!(m+n)!}$ gives us the claimed calculation. For the bound on the bracketed term, use

$$\displaystyle\begin{array}{rcl} J_{n}(ny)\ \ll \ \frac{y^{n}e^{n\sqrt{1-y^{2}} }} {(1 + \sqrt{1 - y^{2}})^{n}}& &{}\end{array}$$

(97)

(see [1], p. 362) and bound trivially.

Appendix 2: An Exponential Sum Identity

The following proposition and proof are also used in [2].

Proposition 6.1.

Suppose X ≤ T. Then

$$\displaystyle\begin{array}{rcl} S_{J}(X)\ &:=& \ T\sum _{k\geq 0}(-1)^{k}J_{ 2k+1}(X)\frac{\tilde{\tilde{h}}\left (\frac{2k+1} {2T} \right )} {\sin \left (\frac{2k+1} {2T} \pi \right )} \\ & =& \ c_{10}T\sum _{\vert \alpha \vert <\frac{T} {2} }e\left (Y \sin \left ( \frac{\pi \alpha } {T}\right )\right )\tilde{\tilde{g}}\left (\frac{\pi Y } {T}\cos \left ( \frac{\pi \alpha } {T}\right )\right ) + O(Y ),{}\end{array}$$

(98)

where c ₁₀ is some constant, g(x):= sgn (x)h(x), X =: 2πY, and for any f set $\tilde{f}(x):= xf(x)$ .

Proof.

Observe that $k\mapsto \sin \left (\pi k/2\right )$ is supported only on the odd integers, and maps 2k + 1 to (−1)^k. Hence, rewriting gives

$$\displaystyle\begin{array}{rcl} S_{J}(X)\ =\ T\sum _{{ k\geq 0 \atop k\not\in 2T \mathbb{Z}} }J_{k}(X)\tilde{\tilde{h}}\left ( \frac{k} {2T}\right ) \frac{\sin \left (\frac{\pi k} {2} \right )} {\sin \left ( \frac{\pi k} {2T}\right )}.& &{}\end{array}$$

(99)

Since

$$\displaystyle\begin{array}{rcl} \frac{\sin \left (\frac{\pi k} {2} \right )} {\sin \left ( \frac{\pi k} {2T}\right )}\ =\ \frac{e^{\frac{\pi ik} {2} } - e^{-\frac{\pi ik} {2} }} {e^{ \frac{\pi ik} {2T} } - e^{-\frac{\pi ik} {2T} }}\ =\ \sum _{ \alpha =-\left (\frac{T-1} {2} \right )}^{\frac{T-1} {2} }e^{\frac{\pi ik\alpha } {T} }& &{}\end{array}$$

(100)

when k is not a multiple of 2T, we find that

$$\displaystyle\begin{array}{rcl} S_{J}(X)\ =\ T\sum _{\vert \alpha \vert <\frac{T} {2} }\sum _{{ k\geq 0 \atop k\not\in 2T \mathbb{Z}} }e\left ( \frac{k\alpha } {2T}\right )J_{k}(X)\tilde{\tilde{h}}\left ( \frac{k} {2T}\right ).& &{}\end{array}$$

(101)

Observe that, since the sum over α is invariant under α ↦ −α (and it is non-zero only for k odd!), we may extend the sum over k to the entirety of $\mathbb{Z}$ at the cost of a factor of 2 and of replacing h by

$$\displaystyle{ g(x)\:=\ \mathrm{ sgn}(x)h(x). }$$

(102)

(This is because of the identity J _−k (x) = (−1)^k J _k (x) and the fact that we need only consider odd k.) Note that g is as differentiable as h has zeros at 0, less one. That is to say, $\hat{g}$ decays like the reciprocal of a degree $\mathop{\mathrm{ord}}_{z=0}h(z) - 1$ polynomial at ∞. This will be crucial in what follows.

Next, we add back on the $2T\mathbb{Z}$ terms and obtain

$$\displaystyle\begin{array}{rcl} \frac{1} {2}S_{J}(X)& =& T\sum _{\vert \alpha \vert <\frac{T} {2} }\sum _{k\in \mathbb{Z}}e\left ( \frac{k\alpha } {2T}\right )J_{k}(X)\tilde{\tilde{g}}\left ( \frac{k} {2T}\right ) - T^{2}\sum _{ k\in \mathbb{Z}}J_{2kT}(X)k^{2}h(k) \\ & =& T\sum _{\vert \alpha \vert <\frac{T} {2} }\sum _{k\in \mathbb{Z}}e\left ( \frac{k\alpha } {2T}\right )J_{k}(X)\tilde{\tilde{g}}\left ( \frac{k} {2T}\right ) + O\left (Xe^{-c_{11}T}\right ) \\ & =:& V _{J}(X) + O\left (Xe^{-c_{11}T}\right ), {}\end{array}$$

(103)

where we have bounded the term $T^{2}\sum _{k\in \mathbb{Z}}J_{2kT}(X)k^{2}h(k)$ trivially via $J_{n}(2x) \ll x^{n}/n!$.

Now we move to apply Poisson summation. Write X = : 2π Y. We apply the integral formula (for $k \in \mathbb{Z}$)

$$\displaystyle{ J_{k}(2\pi x)\ =\ \int _{ -\frac{1} {2} }^{ \frac{1} {2} }e\left (kt - x\sin (2\pi t)\right )dt }$$

(104)

and interchange the sum and integral (via rapid decay of g) to get that

$$\displaystyle\begin{array}{rcl} V _{J}(X)\ =\ T\sum _{\vert \alpha \vert <\frac{T} {2} }\int _{-\frac{1} {2} }^{ \frac{1} {2} }\left (\sum _{k\in \mathbb{Z}}e\left ( \frac{k\alpha } {2T} + kt\right )\tilde{\tilde{g}}\left ( \frac{k} {2T}\right )\right )e\left (-Y \sin (2\pi t)\right )dt.& &{}\end{array}$$

(105)

By Poisson summation, (105) is just (interchanging the sum and integral once more)

$$\displaystyle\begin{array}{rcl} V _{J}(X)& =& T^{2}\sum _{ \vert \alpha \vert <\frac{T} {2} }\sum _{k\in \mathbb{Z}}\int _{-\frac{1} {2} }^{ \frac{1} {2} }\hat{g}''\left (2T(t - k)+\alpha \right )e\left (-Y \sin (2\pi t)\right )dt \\ & =& c_{12}T\sum _{\vert \alpha \vert <\frac{T} {2} }\int _{-\infty }^{\infty }\hat{g}''(t)e\left (Y \sin \left ( \frac{\pi t} {T} + \frac{\pi \alpha } {T}\right )\right )dt \\ & =:& c_{12}W_{g}(X). {}\end{array}$$

(106)

As

$$\displaystyle{ \sin \left ( \frac{\pi t} {T} + \frac{\pi \alpha } {T}\right )\ =\ \sin \left ( \frac{\pi \alpha } {T}\right ) + \frac{\pi t} {T}\cos \left ( \frac{\pi \alpha } {T}\right ) - \frac{\pi ^{2}t^{2}} {2T^{2}}\sin \left ( \frac{\pi \alpha } {T}\right ) + O\left ( \frac{t^{3}} {T^{3}}\right ), }$$

(107)

we see that

$$\displaystyle\begin{array}{rcl} & & W_{g}(X) \\ & =& c_{13}T\sum _{\vert \alpha \vert <\frac{T} {2} }e\left (Y \sin \left ( \frac{\pi \alpha } {T}\right )\right )\int _{-\infty }^{\infty }\hat{g}''(t)e\left (\frac{\pi Y t} {T} \cos \left ( \frac{\pi \alpha } {T}\right )\right )dt \\ & & \ \ -\ c_{14}\frac{\pi ^{2}Y } {T} \sum _{\vert \alpha \vert <\frac{T} {2} }e\left (Y \sin \left ( \frac{\pi \alpha } {T}\right )\right )\sin \left ( \frac{\pi \alpha } {T}\right )\int _{-\infty }^{\infty }t^{2}\hat{g}''(t)e\left (\frac{\pi Y t} {T} \cos \left ( \frac{\pi \alpha } {T}\right )\right )dt \\ & & \ \ +\ O\left (\frac{Y } {T} + \frac{Y ^{2}} {T^{2}} \right ) {}\end{array}$$

(108)

$$\displaystyle\begin{array}{rcl} & =& c_{15}T\sum _{\vert \alpha \vert <\frac{T} {2} }e\left (Y \sin \left ( \frac{\pi \alpha } {T}\right )\right )\tilde{\tilde{g}}\left (\frac{\pi Y } {T}\cos \left ( \frac{\pi \alpha } {T}\right )\right ) \\ & & \ \ -\ c_{16}\frac{Y } {T}\sum _{\vert \alpha \vert <\frac{T} {2} }e\left (Y \sin \left ( \frac{\pi \alpha } {T}\right )\right )\sin \left ( \frac{\pi \alpha } {T}\right )\tilde{\tilde{g}}''\left (\frac{\pi Y } {T}\cos \left ( \frac{\pi \alpha } {T}\right )\right ) \\ & & \ \ +\ O\left (\frac{Y } {T} + \frac{Y ^{2}} {T^{2}} \right ). {}\end{array}$$

(109)

Now bound the second term trivially to get the claim.

Appendix 3: 2-Level Calculations

The purpose of this appendix is to provide additional details to the 2-level computation of Sect. 4.4. Letting

$$\displaystyle\begin{array}{rcl} S_{1}(u_{j},\phi _{i})&:=& \sum _{p}\frac{2\lambda _{p}(u_{j})\log p} {p^{1/2}\log R} \hat{\phi }_{i}\left ( \frac{\log p} {\log R}\right ) \\ S_{2}(u_{j},\phi _{i})&:=& \sum _{p}\frac{2\lambda _{p^{2}}(u_{j})\log p} {p\log R} \hat{\phi }_{i}\left (\frac{2\log p} {\log R} \right ){}\end{array}$$

(110)

and summing over the family, a standard calculation reduces the determination of the 2-level density to understanding

$$\displaystyle\begin{array}{rcl} \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \prod _{i=1}^{2}\Bigg[\widehat{\phi _{ i}}(0)\frac{\log (1 + t_{j}^{2})} {\log R} & -& S_{1}^{c(i)}(u_{ j},\phi _{i}) - S_{2}^{c(i)}(u_{ j},\phi _{i}) \\ & +& O\left (\frac{\log \log R} {\log R}\right )\Bigg]^{2} {}\end{array}$$

(111)

(the other terms are straightforward consequences of the combinatorial book-keeping from the inclusion–exclusion argument), where c(1) is the identity map and c(2) denotes complex conjugation.

We can move the factor O(loglogR∕logR) outside the product at the cost of an error of the same size outside all the summations. To see this, since O(loglogR∕logR) is independent of u _j these terms are readily bounded by applying the Cauchy-Schwarz inequality. Letting $\mathcal{S}$ represent any of the factors in the product over i in (111), the product involving this is O(loglogR∕logR):

$$\displaystyle\begin{array}{rcl} & & \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \mathcal{S}\cdot O\left (\frac{\log \log R} {\log R}\right ) \\ & & \ll \left [ \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} O\left (\left (\frac{\log \log R} {\log R}\right )^{2}\right )\right ]^{1/2} \cdot \left [ \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \vert \mathcal{S}\vert ^{2}\right ]^{1/2} \\ & & \ll O\left (\frac{\log \log R} {\log R}\right ) \cdot \left ( \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \vert \mathcal{S}\vert ^{2}\right )^{1/2}. {}\end{array}$$

(112)

We now analyze the four possibilities for the sum involving $\vert \mathcal{S}\vert ^{2}$. If $\mathcal{S}$ is either $\widehat{\phi _{i}}(0)\frac{\log (1+t_{j}^{2})} {\log R}$ or $O\left (\frac{\log \log R} {\log R}\right )$, this sum is trivially O(1), and thus the entire expression is $O\left (\frac{\log \log R} {\log R}\right )$. We are left with the non-trivial cases of $\mathcal{S} = S_{1}$ or $\mathcal{S} = S_{2}$. To handle these cases, we rely on results that we will soon prove in lemmas below: for sufficiently small support | S ₁ | ² = O(1) and | S ₂ | ² = o(1). Note that we use these lemmas for ϕ = ϕ ₁ ϕ ₁ instead of the usual ϕ = ϕ ₁ ϕ ₂, but this does not affect the proofs, and thus we may move the $O\left (\frac{\log \log R} {\log R}\right )$ factor outside the product.

By symmetry, it suffices to analyze the following terms to determine the 2-level density:

$$\displaystyle\begin{array}{rcl} & & \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \widehat{\phi _{1}}(0)\widehat{\phi _{2}}(0)\left (\frac{\log (1 + t_{j}^{2})} {\log R} \right )^{2} \\ & & \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \widehat{\phi _{1}}(0)\frac{\log (1 + t_{j}^{2})} {\log R} S_{k}(u_{j},\phi _{2}),\ \ k \in \{ 1,2\} \\ & & \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} S_{k}(u_{j},\phi _{1})\overline{S_{\ell}(u_{j},\phi _{2})}\ \ k,\ell\in \{ 1,2\}. {}\end{array}$$

(113)

We now analyze these terms. For small support, a similar analysis as performed earlier in the paper shows that the first term in (113) satisfies

$$\displaystyle{ \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \widehat{\phi _{1}}(0)\widehat{\phi _{2}}(0)\left (\frac{\log (1 + t_{j}^{2})} {\log R} \right )^{2} =\widehat{\phi _{ 1}}(0)\widehat{\phi _{2}}(0) + O\left ( \frac{1} {\log \log R}\right ). }$$

(114)

We next handle the terms where we have exactly one S-factor.

Lemma 7.1.

For sufficiently small support, the $\widehat{\phi _{1}}(0)S_{1}(u_{j},\phi _{2})$ and $\widehat{\phi _{1}}(0)S_{2}(u_{j},\phi _{2})$ terms are O(log log R∕log R), and thus do not contribute.

Proof.

The proof is almost identical to the application of the Kuznetsov trace formula to prove similar results for the 1-level density, the only change being that now we have the modified weight function h _T (t _j )log(1 + t _j ²).

We now turn to the more interesting terms in (113). We first handle the second term.

Lemma 7.2.

For sufficiently small support,

$$\displaystyle{ \frac{1} {\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} } \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} S_{1}(u_{j},\phi _{1})\overline{S_{1}(u_{j},\phi _{2})} = 2\int _{-\infty }^{\infty }\vert z\vert \hat{\phi }_{ 1}(z)\hat{\phi }_{2}(z)dz + O\left (\frac{\log \log R} {\log R}\right ). }$$

(115)

Proof.

$$\displaystyle\begin{array}{rcl} & & \sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} S_{1}(u_{j},\phi _{1})\overline{S_{1}(u_{j},\phi _{2})} \\ & & =\ 4\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \sum _{p_{1},p_{2}} \frac{\lambda _{p_{1}}(u_{j})\overline{\lambda _{p_{2}}(u_{j})}} {p_{1}^{1/2}p_{2}^{1/2}} \frac{\log p_{1}\log p_{2}} {\log ^{2}R} \hat{\phi }_{1}\left (\frac{\log p_{1}} {\log R} \right )\hat{\phi }_{2}\left (\frac{\log p_{2}} {\log R} \right ) \\ & & =\ 4\sum _{p_{1},p_{2}} \frac{1} {p_{1}^{1/2}p_{2}^{1/2}} \frac{\log p_{1}\log p_{2}} {\log ^{2}R} \hat{\phi }_{1}\left (\frac{\log p_{1}} {\log R} \right )\hat{\phi }_{2}\left (\frac{\log p_{2}} {\log R} \right )\sum _{j}\frac{h_{T}(t_{j})} {\|u_{j}\|^{2}} \lambda _{p_{1}}(u_{j})\overline{\lambda _{p_{2}}(u_{j})}.{}\end{array}$$

(116)

As before, we apply the Kuznetsov formula to the inner sum. Since the formula has a $\delta _{p_{1},p_{2}}$, we need to split this sum into the case when p ₁ = p ₂ and the case when p ₁ ≠ p ₂. For small support one easily finds the case p ₁ ≠ p ₂ does not contribute; however, the case p ₁ = p ₂ does contribute. Arguing as in the 1-level calculations, after applying the Kuznetsov formula we are left with

$$\displaystyle\begin{array}{rcl} 4\sum _{p}\frac{1} {p} \frac{\log ^{2}p} {\log ^{2}R}\hat{\phi }_{1}\left ( \frac{\log p} {\log R}\right )\hat{\phi }_{2}\left ( \frac{\log p} {\log R}\right ) = 2\int _{-\infty }^{\infty }\vert z\vert \hat{\phi }_{ 1}(z)\hat{\phi }_{2}(z)dz + O\!\left (\frac{\log \log R} {\log R}\right )\!\!.& &{}\end{array}$$

(117)

(the equality follows from partial summation and the Prime Number Theorem, see [42] for a proof).

Lemma 7.3.

For small support, the contribution from the $S_{k}(u_{j},\phi _{1})\overline{S_{\ell}(u_{j},\phi _{2})}$ terms is O(log log R∕log R) if $(k,\ell) = (1,2)$ or (2,2).

Proof.

The proof is similar to the previous lemma, following again by applications of the Kuznetsov trace formula. The support is slightly larger as the power of the primes in the denominator is larger.