Let $-D < 0$ be a negative fundamental discriminant and let $y$ range over the values $y = y_Q = \frac{\sqrt{|D|}}{2a}$, as the values $(a,b,c)$ run through the reduced binary quadratic forms $Q = aX^2+bXY+cY^2$ of discriminant $b^2-4ac = -D$ (thus $\mathrm{gcd}(a,b,c) = 1$, $-a < b \leq a \leq c$). Those are the ordinates of the CM points $z_Q = \frac{b + \sqrt{-D}}{2a}$ in the standard fundamental domain for $\mathcal{H} / \mathrm{PSL}(2,\mathbb{Z})$.

By Duke's theorem, the $z_Q$ are equidistributed in the measure $\frac{3}{\pi} \frac{dx \, dy}{y^2}$. The highest lying point has ordinate $k := \sqrt{|D|}/2$, and if $\varphi: [1,\infty) \to \mathbb{R}$ is any compactly supported or fast enough decaying function, it follows that the mean value of $\varphi(\frac{\pi}{3}y_Q)$ (over all reduced forms $Q$ if the given discriminant $-D$) is equal to $\int_1^k \varphi(t) \frac{dt}{t^2} + o_{\varphi}(1)$, as $D \to \infty$. The asymptotic $\mathrm{Avg}_Q \varphi(\frac{\pi}{3}y_Q) \sim \int_1^k \varphi(t) \frac{dt}{t^2}$ turns out to be also satisfied for $\varphi(t) = t$ (assuming $L(s,\chi_D)$ satisfies the Riemann hypothesis), for a very special reason: the Kronecker-Chowla-Selberg limit formula expresses $$ \mathrm{Avg}_Q \big(\frac{\pi}{3} y_Q - \log{y_Q} \big) = \log{k} + \frac{L'}{L}(1,\chi_D) + O(1), $$ with an absolutely bounded $O(1)$ term.

On the other hand it is plain that $\mathrm{Avg}_Q \varphi(\frac{\pi}{3}y_Q) \sim \int_1^k \varphi(t) \frac{dt}{t^2}$ fails to hold with $\varphi(t) = t^{1+\epsilon}$, for any fixed $\epsilon > 0$: already the principal form (with $a = 1$) contributes an $\sim k^{\epsilon} \frac{k}{h}$ to this average ($h$ denoting the class number), a contribution that can be as big as $k^{\epsilon} \log{\log{k}}$.

I am interested in what happens for $\varphi(t) = t\log{t}$.

Question. Assume GRH, and possibly some other standard analytic hypotheses. What can be said about the mean value of $\frac{\pi}{3} y_Q \log{y_Q}$, asymptotically as $D \to \infty$?

Evidently this mean value lies somewhere between $\log{k}$ and $(\log{k})^2$. Should the limit $\lim_{D \to \infty} \mathrm{Avg}_Q(y_{Q}\log{y_Q}) \big/ (\log{k})^2$ even exist?

[Of course, the same question could be asked for the positive discriminants (indefinite binary quadratic forms and closed geodesics on the modular surface). ]

  • 1
    $\begingroup$ I think the limit "should" exist and should be given by the limit of that integral (so $1/2$ if I calculated correctly). In lieu of a proof, how about a lower bound? $\operatorname{Avg}_Q( y_Q \log y_Q) \geq \log C \operatorname{Avg}_Q( y_Q ) - \int_{1}^C t (C - \log t ) \frac{dt}{t^2} - o_C(1)$. If you can make the dependence on $C$ in the $o_C(1)$ (which comes from Duke's theorem) explicit, and optimize in $C$, this gives an asymptotic lower bound beating $\log k$. $\endgroup$ – Will Sawin Jun 27 '17 at 12:29
  • $\begingroup$ @WillSawin: Thanks! Yes, I believe it might be shown that the limit must be $1/2$ if it exists; perhaps this could be done in the averaged sense over $D \leq X$. Here is what made me curious about this question. If $L(s,\chi_D)$ has a Siegel zero $\beta = 1 - c / \log{k}$ with $c \to 0$ (which of course isn't supposed to actually happen), then the basic mean value formula for $y_Q$ modifies as follows: $\lim \mathrm{Avg}_Q (c \frac{\pi}{3} y_Q) = \log{k}$ (note the factor of $c$ inside the average). [continued.] $\endgroup$ – Vesselin Dimitrov Jun 27 '17 at 13:33
  • $\begingroup$ [continued.] But apparently, in this illusory situation, there does not a priori seem to be any `expected' asymptotic of $c \frac{\pi}{3} y_Q \log{y_Q}$. As far as I could tell, this could just as easily turn out to be $\gg (\log{k})^2$ as it could $o((\log{k})^2)$. $\endgroup$ – Vesselin Dimitrov Jun 27 '17 at 13:46
  • $\begingroup$ (And well, in my first comment, I should of course have written that the limit quotient of the two sides is $1$... :) Also, I believe your idea can indeed be made to work to show, without any hypothesis, that the mean value of $y_Q \log{y_Q}$ certainly beats $\log{k}$ by any given factor.) $\endgroup$ – Vesselin Dimitrov Jun 27 '17 at 15:56
  • $\begingroup$ I hope that in addition to beating $\log k$ by any constant factor one can actually give an explicit lower bound that beats $\log k$ by any constant factor, which should just require an explicit bound for the error term in Duke's theorem. $\endgroup$ – Will Sawin Jun 27 '17 at 16:19

Let $g : \Gamma \backslash \mathbb{H} \to \mathbb{C}$ be any bounded continuous function. Duke's theorem states that \[\frac{1}{h(D)} \sum_{A \in \mathrm{Cl}_K} g(z_A) = \frac{1}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} \int_{\Gamma \backslash \mathbb{H}} g(z) \, d\mu(z) + o_g(1)\] as $D \to -\infty$ through negative fundamental discriminants, where $h(D) = \# \mathrm{Cl}_K$ the class number of $K = \mathbb{Q}(\sqrt{D})$, $\mathrm{Cl}_K$ the class group, and $d\mu(z) = \frac{dx \, dy}{y^2}$ the $\mathrm{SL}_2(\mathbb{R})$-invariant measure on the upper half-plane $\mathbb{H}$.

The proof uses the spectral decomposition \[g(z) = \frac{\langle g, 1\rangle}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} + \sum_{f \in \mathcal{B}_0(\Gamma)} \langle g, f \rangle f(z) + \frac{1}{4\pi} \int_{-\infty}^{\infty} \left\langle g, E\left(\cdot, \frac{1}{2} + it\right)\right\rangle E\left(z, \frac{1}{2} + it\right) \, dt,\] where $\mathcal{B}_0(\Gamma)$ denotes an orthonormal basis of Hecke-Maass cusp forms and $E(z,s)$ denotes the real-analytic Eisenstein series.

We want to take $g(z) = y \log y$, but this is of course not square-integrable with respect to $d\mu(z)$. The trick is to let $g_s(z) = y^s \log y$ and $G_s(z) = g_s(z) - \frac{\partial}{\partial s} E(z,s)$ with $\Re(s) > 1/2$ and $s \neq 1$. Then $G_s(z)$ is square-integrable, with \[g_s(z) = \frac{\partial}{\partial s} E(z,s) + \frac{\langle G_s, 1\rangle}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} + \sum_{f \in \mathcal{B}_0(\Gamma)} \langle G_s, f \rangle f(z) + \frac{1}{4\pi} \int_{-\infty}^{\infty} \left\langle G_s, E\left(\cdot, \frac{1}{2} + it\right)\right\rangle E\left(z, \frac{1}{2} + it\right) \, dt.\]

It is clear that $\langle G_s, f\rangle = \langle g_s, f\rangle$. Using Zagier's method of renormalisation of integrals, one can show that \[\langle G_s, 1\rangle = \int_{\sqrt{3}/2}^{1} y^{s - 2} \log y \left(1 - 2\sqrt{1 - y^2}\right) \, dy + \frac{1}{(s - 1)^2},\] and that \[\left\langle G_s, E\left(\cdot, \frac{1}{2} + it\right)\right\rangle = 0.\] (Both of these are somewhat nontrivial to show, however.) It follows by analytic continuation that \[g(z) = \lim_{s \to 1} \left(\frac{\partial}{\partial s} E(z,s) + \frac{1}{\mathrm{vol}(\Gamma \backslash \mathbb{H}) (s - 1)^2}\right) + C + \sum_{f \in \mathcal{B}_0(\Gamma)} \langle g, f \rangle f(z),\]

where \[C = \frac{1}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} \int_{\sqrt{3}/2}^{1} y^{-1} \log y \left(1 - 2\sqrt{1 - y^2}\right) \, dy \approx -0.00181698.\]

Now we let $z = z_A$ and sum over all $A \in \mathrm{Cl}_K$. Then the sum over $f \in \mathcal{B}_0(\Gamma)$ is $O((-D)^{1/2 - \delta})$ for some $\delta > 0$ via the period formula for Hecke-Maass cusp forms at Heegner points and subconvexity bounds for $L(1/2,f \otimes \chi_D)$. Next, we have that \[\sum_{A \in \mathrm{Cl}_K} E(z_A,s) = \frac{w_K}{2} \frac{\Lambda_K(s)}{\Lambda(2s)},\] where $w_K = \# \mathcal{O}_{K,\mathrm{tors}}^{\times}$ (which is equal to $2$ for $D \leq -4$), $\Lambda_K(s) = (2\pi)^{-s} \Gamma(s) \zeta(s) L(s,\chi_D)$, and $\Lambda(2s) = \pi^{-s} \Gamma(s) \zeta(2s)$. We may rewrite this as \[\frac{w_K \sqrt{-D}}{4} \frac{\exp\left((s - 1) \log \frac{\sqrt{-D}}{2}\right) \zeta(s) L(s,\chi_D)}{\zeta(2s)}.\] This has a simple pole at $s = 1$ with residue \[\frac{3}{\pi} \frac{w_K \sqrt{-D} L(1,\chi_D)}{2\pi} = \frac{h(D)}{\mathrm{vol}(\Gamma \backslash \mathbb{H})}\] by the class number formula. It follows that if $C' = C'(D)$ denotes the coefficient of $s - 1$ in the Laurent expansion of this, then \[\sum_{A \in \mathrm{Cl}_K} g(z_A) = C' + C h(D) + O((-D)^{1/2 -\delta}).\] The main term from $C'$ is \[\frac{3}{\pi} \frac{w_K \sqrt{-D} L(1,\chi_D)}{2\pi} \frac{1}{2} \left(\log \frac{\sqrt{-D}}{2}\right)^2 = \frac{h(D)}{\mathrm{vol}(\Gamma \backslash \mathbb{H})} \frac{1}{2} \left(\log \frac{\sqrt{-D}}{2}\right)^2.\] To prove that the remaining terms in $C'$ are smaller requires bounds of the form \[\frac{L'}{L}(1,\chi_D) = o(\log (-D)), \quad \frac{L''}{L}(1,\chi_D) = o((\log (-D))^2).\] Such bounds certainly follow from the generalised Riemann hypothesis, though I don't believe they are known unconditionally.

I should mention that this method works more generally for $g(y) = y (\log y)^m$ with $m$ a nonnegative integer, in which case you need to take $m$-th derivatives with respect to $s$, or even more generally for $g(y) = y^s (\log y)^m$ with $s \neq 1$, in which case you no longer need to take limits, but $h(D)$ will no longer naturally appear in the main term, so the answer may fluctuate wildly depending on the size of $L^{(k)}(s,\chi_D)/L(1,\chi_D)$ for $0 \leq k \leq m + 1$.

| cite | improve this answer | |
  • 1
    $\begingroup$ Many thanks for this detailed and instructive answer! Yes, according to Iwaniec and Friedlander (a recent reference below), not even the trivial bound $L'(1,\chi_D) \ll (\log{|D|})^2$ has been unconditionally improved to a $o(\cdot)$. This also explains my 'observation' that no asymptotic may be given under the extraordinary hypothesis of a Siegel zero, for then, unless one assumes something supplementary for the complex zeros, the derivative $L'(1,\chi_D)$ may a priori be anything in the range $1 \ll R \ll (\log{|D|})^2$ (cf. Iwaniec and Friedlander's Note on Dirichlet $L$-functions). $\endgroup$ – Vesselin Dimitrov Jun 27 '17 at 21:11
  • 1
    $\begingroup$ [Reg. the edit,to rectify my comment: While $L(1) = o(1)$ and $(L'/L)(1) \sim 1/(1-\beta)$ under the hypothesis of a Siegel zero with $(1-\beta) \log{|D|} \to 0$, I suppose it is $(1-\beta)(L''/L)(1)$ whose magnitude rel. $\log{|D|}$ may not 'a priori' be determined - unless e.g. one assumes the remaining non-trivial zeros to lie on the critical line. In contrast to the case $g(z) = y$, this means that the average value for $g(z) = y\log{y}$ may not be asymptotically computed as a function of $D$ and $\beta$ under the sole hypothesis of such a Siegel zero $\beta$.] $\endgroup$ – Vesselin Dimitrov Jun 27 '17 at 22:37

Can we do this by reducing it to a sum over ideals?

By Duke's theorem, we can easily deal with the low $y$ terms. Thus we can adjust the boundary condition to $-a < b \leq a < \sqrt{- D /3 }$.

Instead of CM points, we are now working simply with pairs of an oriented lattice of discriminant $-D$ and an primitive lattice point of norm $a < \sqrt{- D /3 }$ with the bijection that sets the fixed lattice point to $(1,0)$ and sets $(0,1)$ to be the unique other generator with positive orientation where $b$ is in the correct range.

In other words, these are $\mathcal O_{\mathbb Q(\sqrt{-D})}$-modules with a primitive element generating a submodule of index $a < \sqrt{- D /3 }$ or, dually, ideals of $\mathcal O_{\mathbb Q(\sqrt{-D})}$ of norm $a < \sqrt{-D/3}$ that are not divisible by any natural number.

So you are looking at something like the sum of $$\frac{ \sqrt{D}}{2|I|} \log \left( \frac{ \sqrt{D}}{2|I|} \right) = \frac{\sqrt{D}}{2|I|} \log \left( \frac{\sqrt{D}}{2} \right) - \frac{ \sqrt{D}}{2|I|} \log |I| $$ over all indivisible ideals $I$ of $\mathcal O_{\mathbb Q(\sqrt{-D})}$ satisfying $|I| < \sqrt{ -D /3}$.

It should be possible to evaluate these terms by contour-shifting the generating functions $ \frac{ \zeta_{ \mathbb Q(\sqrt{-D})} (s) }{ \zeta(2s)}$ and $\frac{d}{ds} \frac{ \zeta_{ \mathbb Q(\sqrt{-D})} (s) }{ \zeta(2s)}$ respectively, no? The leading terms should come from the pole of the zeta function at $1$, which when divided by the class number should give the desired $(\log k)^2/2$ factor.

| cite | improve this answer | |
  • $\begingroup$ Thanks much for the insightful answer! In a sense, one can repeatedly derive Eisenstein series and look at the full Laurent expansion at $s = 1$ of the partial zeta functions making up $\zeta_{\mathbb{Q}(\sqrt{-D})}(s)$, and use these to establish the expected formula for $\varphi(t) = t (\log{t})^m$, with any fixed $m \in \mathbb{N}_0$, under the Riemann hypothesis for $L(s,\chi_D)$. $\endgroup$ – Vesselin Dimitrov Jun 27 '17 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.