Local average in hyperbolic lattice point counting, with an Appendix by Niko Laaksonen


The hyperbolic lattice point problem asks to estimate the size of the orbit \(\Gamma z\) inside a hyperbolic disk of radius \(\cosh ^{-1}(X/2)\) for \(\Gamma \) a discrete subgroup of \({\hbox {PSL}_2( {{\mathbb {R}}})} \). Selberg proved the estimate \(O(X^{2/3})\) for the error term for cofinite or cocompact groups. This has not been improved for any group and any center. In this paper local averaging over the center is investigated for \({\hbox {PSL}_2( {{\mathbb {Z}}})} \). The result is that the error term can be improved to \(O(X^{7/12+{\varepsilon }})\). The proof uses surprisingly strong input e.g. results on the quantum ergodicity of Maaß cusp forms and estimates on spectral exponential sums. We also prove omega results for this averaging, consistent with the conjectural best error bound \(O(X^{1/2+{\varepsilon }})\). In the appendix the relevant exponential sum over the spectral parameters is investigated.

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The author of the appendix would like to thank Peter Sarnak for useful discussions and for providing notes for the co-compact case.

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Correspondence to Morten S. Risager.

Additional information

The Morten S. Risager was supported by a Sapere Aude grant from The Danish Council for Independent Research.

Appendix: Numerical investigation of exponential sums over eigenvalues in \({\hbox {PSL}_2( {{\mathbb {Z}}})} \backslash {\mathbb {H}}\)

Appendix: Numerical investigation of exponential sums over eigenvalues in \({\hbox {PSL}_2( {{\mathbb {Z}}})} \backslash {\mathbb {H}}\)

Let \(\lambda _{j}=\frac{1}{4}+t_{j}^{2}\) be the eigenvalues of \(\Delta \) in \({\Gamma \backslash {{\mathbb {H}}}}\), where \(\Gamma ={\hbox {PSL}_2( {{\mathbb {Z}}})} \). For \(X>1\), we define the following sum

$$\begin{aligned} S(T,X)=\sum _{|t_{j}|\le T} X^{it_{j}}, \end{aligned}$$

which is symmetrised by including both \(t_{j}\) and \(-t_{j}\). Petridis and Risager (Conjecture 2.2) conjecture that up to a factor of the order of \(X^{\epsilon }\), the sum has square root cancellation in T.

Conjecture 1

For every \(\epsilon >0\) and \(X>1\) we have

$$\begin{aligned} S(T,X)\ll _{\epsilon } T^{1+\epsilon }X^{\epsilon }. \end{aligned}$$

We report on the numerical investigation of the function S(TX) and prove a theorem about its behaviour as \(T\rightarrow \infty \) and \(X>1\) is fixed. Our investigation resulted in the following observations.

Experimental Observation 1

The growth of S(TX) is consistent with the conjecture.

Experimental Observation 2

For a fixed \(X>1, S(T,X)\) has a peak of order T whenever X is equal to a power of a norm of a primitive hyperbolic class of \(\Gamma \) or an even power of a prime number \(p\in {\mathbb {N}}\).

Experimental Observation 2 is also in agreement with the results of Chazarain [5] that for the wave kernel the singularities occur at the lengths of closed geodesics (or in our case when \(\log X\) is a multiple of a length of a prime geodesic). The peaks at even powers of rational primes are due to the scattering determinant \(\varphi \). Experimental Observation 2 leads us to prove asymptotics for S(TX) for a fixed \(X>1\). Let \(\Lambda (X)\) be the von Mangoldt function extended to \({\mathbb {R}}\) by defining it to be 0 when X is not equal to a power of a prime number. We also define a similar function \(\Lambda _{\Gamma }\) for the norms of hyperbolic classes of \({\hbox {PSL}_2( {{\mathbb {Z}}})} \) given by

$$\begin{aligned} \Lambda _{\Gamma }(X)= {\left\{ \begin{array}{ll} \log (N({\mathfrak {p}})), &{}\text {if}\,X=N({\mathfrak {p}})^{\ell }, \ell \in {\mathbb {N}},\\ 0, &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

Let |F| be the volume of the fundamental domain of \({\Gamma \backslash {{\mathbb {H}}}}\). We prove the following theorem.

Theorem 1

For a fixed \(X>1\), we have

$$\begin{aligned} S(T,X)= & {} \frac{|F|}{\pi }\frac{\sin (T\log X)}{\log X}T+\frac{T}{\pi }(X^{1/2}-X^{-1/2})^{-1}\Lambda _{\Gamma }(X)\\&+\frac{2T}{\pi }X^{-1/2}\Lambda (X^{1/2})+O\left( T/\log T\right) , \end{aligned}$$

as \(T\rightarrow \infty \).


Let \(\psi \) be a positive even test function supported on \([-1,1]\) with \(\int \psi =1\). Then define \(\psi _{\epsilon }(x)=\epsilon ^{-1}\psi (x/\epsilon )\). So \(\psi _{\epsilon }\) is supported on \([-\epsilon , \epsilon ]\) and \(\int \psi _{\epsilon }=1\). Also, let G be the convolution \(G(r)=(\chi _{[-T,T]}*\psi _{\epsilon })(r)\) for some \(\epsilon >0\) to be chosen later. Define a function h, depending on TX and \(\epsilon \), given by \(h(r)=G(r)(X^{ir}+X^{-ir})\). Let g be the Fourier transform of h and denote the determinant of the scattering matrix \(\Phi \) by \(\varphi \). Then the Selberg Trace Formula [18, Theorem 10.2] says that

$$\begin{aligned} {\mathscr {S}}(T,X) + \frac{1}{4\pi }\int _{-\infty }^{\infty }h(r)\frac{-\varphi '}{\varphi } \left( \frac{1}{2}\!+\!ir\right) dr\!=\!I(T,X)\!+\!H(T,X)+E(T,X)+L(T,X), \end{aligned}$$


$$\begin{aligned} {\mathscr {S}}(T,X)&=\sum _{t_{j}>0}h(t_{j}),\\ I(T,X)&= \frac{|F|}{4\pi }\int _{-\infty }^{\infty }h(r)r\tanh (\pi r)dr,\\ H(T,X)&= \sum _{{\mathfrak {p}}}\sum _{\ell =1}^{\infty }\left( N({\mathfrak {p}})^{\ell /2}-N({\mathfrak {p}})^{-\ell /2}\right) ^{-1}g(\ell \log N({\mathfrak {p}}))\log N({\mathfrak {p}}),\\ E(T,X)&= \sum _{{\mathcal {R}}}\sum _{0<\ell <m}\left( 2m\sin \frac{\pi \ell }{m}\right) ^{-1}\int _{-\infty }^{\infty } h(r)\frac{\cosh \pi \left( 1-\frac{2\ell }{m}\right) r}{\cosh \pi r}dr,\\ L(T,X)&= \frac{h(0)}{4}{{\mathrm{Tr}}}\left( I-\Phi \left( \tfrac{1}{2}\right) \right) -h_{\Gamma }g(0)\log 2-\frac{h_{\Gamma }}{2\pi }\int _{-\infty }^{\infty }h(r)\frac{\Gamma '(1+ir)}{\Gamma (1+ir)}dr, \end{aligned}$$

where \({\mathfrak {p}}\) and \({\mathcal {R}}\) range over the primitive hyperbolic and elliptic classes of \({\hbox {PSL}_2( {{\mathbb {Z}}})} \), respectively, and \(h_{\Gamma }\) is the number of cusps of \(\Gamma \). First observe that \(S(T,X)={\mathscr {S}}(T,X)+O(T\epsilon )\), so we can work with \({\mathscr {S}}\). For the identity motion we have

$$\begin{aligned} I(T,X)&= \frac{|F|}{2\pi }\int _{-\infty }^{\infty }G(r)\cos (r\log X)r\tanh \pi r\,dr\\&=\frac{|F|}{\pi }\int _{0}^{\infty }G(r)\cos (r\log X)r\left( 1-\frac{2}{e^{2\pi r}+1}\right) \,dr\\&=\frac{|F|}{\pi }(I_{1}(T,X)+I_{2}(T,X)). \end{aligned}$$

From \(I_{1}\) we obtain a part of the main term:

$$\begin{aligned} I_{1}(T,X)&=\int _{0}^{\infty }G(r)\cos (r\log X)r\,dr\\&=\left( \int _{0}^{T-\epsilon }+\int _{T-\epsilon }^{T+\epsilon }\right) G(r)\cos (r\log X)r\,dr\\&=I_{11}+I_{12}, \end{aligned}$$

since G is even and supported on \([-T-\epsilon ,T+\epsilon ]\). Then

$$\begin{aligned} I_{11}&= \int _{0}^{T-\epsilon }\cos (r\log X)r\,dr=\frac{\sin ((T-\epsilon )\log X)}{\log X}(T-\epsilon )+O(1),\\ I_{12}&\ll \int _{T-\epsilon }^{T+\epsilon }r\,dr=O(T\epsilon ). \end{aligned}$$


$$\begin{aligned} I_{2}(T,X)=-\int _{0}^{\infty }G(r)\cos (r\log X)r\frac{2}{e^{2\pi r}+1}\,dr\ll \int _{0}^{\infty }re^{-2\pi r}\,dr=O(1). \end{aligned}$$

For g(r) we compute

$$\begin{aligned} g(r)&=\frac{1}{2\pi }\int _{-\infty }^{\infty }e^{-irt}h(t)\,dt\\&=\frac{1}{2\pi }\int _{-\infty }^{\infty }G(t)e^{-irt}(e^{it\log X}+e^{-it\log X})\,dt\\&=\frac{1}{2\pi }\left( \widehat{G}\left( \frac{r-\log X}{2\pi }\right) +\widehat{G}\left( \frac{r+\log X}{2\pi }\right) \right) . \end{aligned}$$

So in particular \(g(\ell \log N({\mathfrak {p}}))\sim T/\pi \) if \(X=N({\mathfrak {p}})^{\ell }\) and decays as \(O((\ell \log N({\mathfrak {p}}))^{-k-1}\epsilon ^{-k})\) otherwise, for any \(k\in {\mathbb {N}}\). For the elliptic terms we need to evaluate

$$\begin{aligned} \int _{-\infty }^{\infty }h(r)\frac{\cosh \pi (1-\frac{2\ell }{m})r}{\cosh \pi r}\,dr\ll \int _{0}^{\infty }\frac{e^{-2\pi r\ell /m}+e^{-2\pi r}}{1+e^{-2\pi r}}\,dr=O(1). \end{aligned}$$

Hence E(TX) is bounded. By the explicit formula of \(\varphi '/\varphi \) for \({\hbox {PSL}_2( {{\mathbb {Z}}})} \), [18, 3.24], we have

$$\begin{aligned} \int _{-\infty }^{\infty }h(r)\frac{-\varphi '}{\varphi }\left( \frac{1}{2}+ir\right) dr&=\int _{-\infty }^{\infty }h(r)\left( -2\log \pi +\frac{\Gamma '(\frac{1}{2}\pm ir)}{\Gamma (\frac{1}{2}\pm ir)}+2\frac{\zeta '(1\pm 2ir)}{\zeta (1\pm 2ir)}\right) \,dr.\\&= C_{1}+C_{2}+C_{3}. \end{aligned}$$

The integral \(C_{1}\) is the Fourier transform of G and is thus bounded. For \(C_{2}\) we use Stirling asymptotics to get

$$\begin{aligned} C_{2}=\int _{-\infty }^{\infty }h(r)\log \left( \frac{1}{4}+r^{2} \right) \,dr+O(1). \end{aligned}$$

This is \(O(\log T)\). The same computation shows that \(L(T,X)=O(\log T)\). The remaining part of the main term comes from \(C_{3}\). We first expand h and isolate the important terms:

$$\begin{aligned} C_{3}=2\left( \int _{-T-\epsilon }^{-T+\epsilon }+\int _{-T+\epsilon }^{T-\epsilon }+\int _{T-\epsilon }^{T+\epsilon }\right) (X^{ir}+X^{-ir})G(r)\frac{\zeta '(1\pm 2ir)}{\zeta (1\pm 2ir)}\,dr. \end{aligned}$$

The first and third integrals are bounded by \(O(\epsilon \log T)\). Notice that \(G(r)=1\) in the range of the second integral, hence we can write it as

$$\begin{aligned} 2\int _{\frac{1}{2}-(T-\epsilon )i}^{\frac{1}{2}+(T-\epsilon )i}(X^{s-1/2}+X^{1/2-s})\left( \frac{\zeta '(2s)}{\zeta (2s)}+\frac{\zeta '(2-2s)}{\zeta (2-2s)}\right) \,ds. \end{aligned}$$

We separate this into two integrals by adding and subtracting the singular part:

$$\begin{aligned} C_{3}&=2\int _{\frac{1}{2}-(T-\epsilon )i}^{\frac{1}{2}+(T-\epsilon )i}(X^{s-1/2}+X^{1/2-s})\left( \frac{\zeta '(2s)}{\zeta (2s)}-\frac{1}{2s-1}\right) \,ds\\&\quad +\,2\int _{\frac{1}{2}-(T-\epsilon )i}^{\frac{1}{2}+(T-\epsilon )i}(X^{s-1/2}+X^{1/2-s})\left( \frac{\zeta '(2-2s)}{\zeta (2-2s)}-\frac{1}{(2-2s)-1}\right) \,ds\\&=2(C_{31}+C_{32}). \end{aligned}$$

For the first integral we move the contour to \(\mathfrak {R}s=1\) and for the second one to \(\mathfrak {R}s=0\). It is easy to see that the top and bottom parts of the contours yield \(O(\log T)\). For the line at \(\mathfrak {R}s=1\) we get

$$\begin{aligned} C_{31}&=\int _{1-(T-\epsilon )i}^{1+(T-\epsilon )i}(X^{s-1/2}+X^{1/2-s})\left( \frac{\zeta '(2s)}{\zeta (2s)}-\frac{1}{2s-1}\right) \,ds\\&= \int _{-T+\epsilon }^{T-\epsilon }(X^{1/2+ir}+X^{-1/2-ir})\left( \frac{\zeta '(2+2ir)}{\zeta (2+2ir)}-\frac{1}{1+2ir}\right) \,dr, \end{aligned}$$

For the rest of the proof we will follow an argument similar to [22, Hilfssatz 2]. We start by writing out the Dirichlet series:

$$\begin{aligned} C_{31}&=-\int _{-T+\epsilon }^{T-\epsilon }X^{1/2+ir}\sum _{n=1}^{\infty }\frac{\Lambda (n)}{n^{2+2ir}}\,dr+O(\log T)\\&=-\sum _{n\ne \sqrt{X}}\frac{\sqrt{X}\Lambda (n)}{n^{2}}\int _{-T+\epsilon }^{T-\epsilon }\left( \frac{X}{n^{2}}\right) ^{ir}\,dr - X^{-1/2}\Lambda (X^{1/2})\int _{-T+\epsilon }^{T-\epsilon }\,dr+O(\log T). \end{aligned}$$

Since \(X>1\), the term in \(C_{31}\) with the negative exponent gets absorbed into the error term. Hence,

$$\begin{aligned} \Bigg |\int _{-T+\epsilon }^{T-\epsilon }X^{1/2+ir}\frac{\zeta '(2+2ir)}{\zeta (2+2ir)}\,dr&+2X^{-1/2}\Lambda (X^{1/2})(T-\epsilon )\Bigg |\\&\le \sum _{n\ne \sqrt{X}}\frac{\sqrt{X}\Lambda (n)}{n^{2}}\left|\frac{(\frac{X}{n^{2}})^{i(T-\epsilon )}-(\frac{X}{n^{2}})^{-i(T-\epsilon )}}{\log \frac{X}{n^{2}}}\right|\\&\ll 2\sqrt{X}\left| \frac{\zeta '(2)}{\zeta (2)}\right| . \end{aligned}$$

So we see that \(C_{31}=-2X^{-1/2}\Lambda (X^{1/2})(T-\epsilon )+O(\log T)\). A similar argument shows that \(C_{32}\) has the same asymptotics. Letting \(\epsilon =1/\log T\) concludes the proof. \(\square \)

Fig. 1

\(\Sigma (T,X)\) in terms of X for \(X\in [3,10]\)

Fig. 2

\(\Sigma (T,X)\) in terms of X for \(X\in [13,20]\)

We will now present plots of S(TX) in terms of both T and X. In Figs. 12 and 3 we have fixed \(T=800\) with \(X\rightarrow \infty \), while in Fig. 4 we are fixing X with \(T\rightarrow \infty \). Taking into account the conjecture, we plot the normalised sum \(\Sigma (T,X)=S(T,X)T^{-1}\). In Fig. 4 we present a comparison for different powers of T, which suggests that \(1+\epsilon \) is the correct exponent. The programs used for the plots are available on the website [21]. We used \(53\,000\) eigenvalues from the data of Then [33] with 13 decimal digit precision. We have also used the data of Booker and Strömbergsson related to [1], which has a much higher precision of 53 decimal digits for 2280 eigenvalues. We verify that the computations are robust, that is, the number of eigenvalues or their precision has no significant impact on our calculations. More details are available on the website [21].

Recall that we expect a peak of order T at all even prime powers as well as powers of the norms of the primitive hyperbolic classes. The first few norms (up to 8 decimals) are given by

$$\begin{aligned} g_{1}&= 6.85410196&g_{5}&= 46.97871376\\ g_{2}&= 13.92820323&g_{6}&= 61.98386677\\ g_{3}&= 22.95643924&g_{7}&= 78.98733975\\ g_{4}&= 33.97056274&g_{8}&= 97.98979486. \end{aligned}$$

These can be computed by expressing the norm in terms of the trace (see e.g. [18, p. 146]).

We start by considering \(\Sigma (T,X)\) in terms of X.

Figure 1 clearly shows peak points at \(X=4=2^{2}, X=g_{1}\) and \(X=9=3^{2}\). In Fig. 2 we can see the peak points \(X=g_{2}\) and \(X=16=2^{4}\):

Figures 1 and 2 verify Theorem 1 numerically in accordance with Experimental Observation 2. In Fig. 3 we look at \(\Sigma (T,X)\) for X in a much larger interval. The graph agrees with Experimental Observation 1. On the other hand we cannot dispose of \(X^{\epsilon }\) in the conjecture. The frequencies \(t_{j}\) are conjecturally linearly independent over \({\mathbb {Q}}\), which makes S(TX) the partial sums of an almost periodic function. Therefore, for a choice of arbitrarily large X, compared to TS(TX) will be of size \(T^{2}\).

Fig. 3

\(\Sigma (T,X)\) in terms of X for \(X\in [100,10\,000]\)

Fig. 4

Different normalisations of \({\tilde{\Sigma }}(T,X)\) at \(X=49\) (A) \({\tilde{\Sigma }}(T,X)T^{-1}\), (B) \({\tilde{\Sigma }}(T,X)T^{-3/2}\)

In Theorem 1 the asymptotics show an oscillatory term with an amplitude of order T coming from the identity motion. We subtract it from S(TX) and define

$$\begin{aligned} {\tilde{\Sigma }}(T,X)=S(T,X)-\frac{|F|}{\pi }\frac{\sin (T\log X)}{\log X}T. \end{aligned}$$

We plot \({\tilde{\Sigma }}(T,X)\) in terms of T at \(X=49\), which is one of the peak points.

Notice that clearly the normalisation \(T^{-1}\) seems to be closer to the correct one, which is evidence towards our Experimental Observation 1.

It is of interest to compare the behaviour of S(TX) with a similar sum over the Riemann zeros. Landau [22, Satz 1] showed that for a fixed \(x>1\), if \(\rho =\beta +i\gamma \) is a non-trivial zero of \(\zeta (s)\), we have the formula

$$\begin{aligned} \sum _{0<\gamma <T}x^{\rho }=-\frac{T}{2\pi }\Lambda (x)+O(\log T). \end{aligned}$$

We call the left-hand side of (7.20) Z(TX). We used our program with \(10\,000\) zeros of \(\zeta (s)\) to 9 decimal places, provided by Odlyzko [26]. With our program we obtain the plot for the normalized sum \(T^{-1}Z(T,X)\). Here blue denotes the real part and green the imaginary part of the sum) (Fig. 5).

Fig. 5

\(x\in [1.5,30]\)

Notice that S(TX) is the analogue of the real part of Z(TX) only. Since the Selberg Trace Formula demands that the test function is even, we cannot analyse the imaginary part directly. For numerical study of this we again refer the reader to the website [21].

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Petridis, Y.N., Risager, M.S. Local average in hyperbolic lattice point counting, with an Appendix by Niko Laaksonen. Math. Z. 285, 1319–1344 (2017). https://doi.org/10.1007/s00209-016-1749-z

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