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For ease of exposition, I will stick to the simplest case: consider the Eisenstein series for $SL_2(\bf R)$ $$E(z,s)=\sum_{\gamma\in P_{\bf Z}\backslash SL_2(\bf Z)}\text{Im}(\gamma z)^s=\sum_{(c,d)\in{\bf Z}\backslash (0,0)}\frac{y^s}{|cz+d|^{2s}}$$ and its truncation used to compute the inner product of Eisenstein series (called the Maaß-Selberg relation) over $SL_2(\bf Z)\backslash H$, $$\Lambda^TE(z,s)=\begin{cases} E(z,s) & y<T\\E(z,s)-y^s-\varphi(s)y^{1-s}&y>T\end{cases}$$ (I am being sloppy with the difference between the 'naive' truncation and Arthur's truncation, but I think in this case it does not make a difference; please correct me otherwise.)

My question concerns the parameter $T$. The first condition is that $T>0$, large enough that the truncated fundamental domain is such that the curve $y=T$ cuts the fundamental domain only on the two sides that pass through the cusp, and such that the truncated function has rapid decay at cusps to be automorphic.

What is the smallest value that $T$ can take so that the inner product formula remains valid?

Why I am asking: from what I understand, Selberg in Harmonic Analysis (Collected works, p.633) says that $T=1$ will suffice, while Garrett in his note Simplest Example of Truncation and Maaß-Selberg Relations claims $T=\sqrt3/2$ (p.2). Unfortunately, a computation I am making seems to fail for such small values, but holds for some other given $T\gg0$. I would like to know if there is a computational error, or I should expect the Maass-Selberg relation to fail or alter at some limit.

EDIT: I am starting a bounty, as after making certain computations with Hejhal's calculation and my own formulas I am sure something is fishy. From Hejhal in The Selberg Trace Formula and the Riemann Zeta Function one sees that $$\frac{1}{4\pi}\int^{\infty}_{-\infty}h(r)\int_{\Gamma\backslash\bf H}|\Lambda^T E(z,\frac{1}{2}+ir)|^2dz\ dr$$ is equal to the sum of $$\frac{1}{4}m(\frac{1}{2})h(0)-g(0)\log\pi+\frac{1}{2\pi}\int^\infty_{-\infty}h(r)\frac{\Gamma'}{\Gamma}(1+ir)dr-2\sum_{n=1}^\infty\frac{\Lambda(n)}{n}g(2\log n)$$ where $\Lambda(n)$ is the von Mangoldt function, and $$+\frac{1}{4\pi i}\int^\infty_{-\infty}m(\frac{1}{2}-ir)h(r)e^{2ir \log T}\frac{dr}{r}+g(0)\log T$$ where $m(s)=\xi(2-2s)/\xi(2s),$ the quotient of completed Riemann zeta functions. Then taking as a test function $h(r)=e^{-r^2}$ (which is allowable), and $T$ very close to 1, one finds that the second line is negligible, the digamma function is negative, and the resulting expression is negative, whereas the original integral is clearly positive. More generally, if one chooses $g(0)$ or $h(0)$ to be zero, the expression is even more likely negative.

Certainly for $T$ large enough, the expression is positive for positive test functions. I have been unable to find accurate information on what $T$ is allowable for the inner product formula to hold.

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    $\begingroup$ It is not clear to me what you are asking, i.e. what identity you need precisely (note that I am familiar with the Maass-Selberg relations). From the classical fundamental domain for $\mathrm{SL_2}(\mathbf{Z})\backslash\mathbf{H}$ it is clear that $T\geq\sqrt{3}/2$ is necessary if you want the natural projection from $\{z:\ |\Re(z)|<1/2,\ \Im(z)>T\}$ to $\mathrm{SL_2}(\mathbf{Z})\backslash\mathbf{H}$ to be injective. $\endgroup$ – GH from MO Oct 25 '15 at 23:57
  • $\begingroup$ A little more info: I would like to integrate this against a test function, where one gets terms involving $\log T$, so that taking some small $T$ the terms in fact can be made to be zero or negative. $\endgroup$ – TA Wong Oct 26 '15 at 0:35
  • $\begingroup$ For $T\geq\sqrt{3}/2$, the usual fundamental domain decomposes into the strip where $y>T$ and the so-called central polygon where $y\leq T$. So if you integrate on the fundamental domain, you do obtain the Maass-Selberg relations. See Theorem 6.14 and the subsequent Remarks in Iwaniec: Spectral methods of automorphic forms. The proof uses that the central polygon is bounded by part of the boundary of the fundamental domain and a single horocycle with $y=T$, so I don't think the statement holds for smaller values $T$ than $\sqrt{3}/2$. $\endgroup$ – GH from MO Oct 26 '15 at 1:38
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    $\begingroup$ As @GHfromMO observed, that the cut-off be above $\sqrt{3}/2$ is certainly necessary, as I should have said clearly in my note, whatever I did happen to say. As Peter Humphries quite correctly observes in his answer, a naive form of truncation can go that low, ... but it's dangerous. So, no, one ought not try to truncate below the bottom of a Siegel set that behaves well. No doubt. No contest. What is the underlying issue? $\endgroup$ – paul garrett Oct 30 '15 at 23:09
  • $\begingroup$ @paulgarrett Thank you for the comment. To be clear, using Hejhal's expression it appears that the Maass-Selberg does not hold unconditionally for some range of $T>1$. (Already seems to exceed the condition posed by Selberg, surprisingly.) I require an estimate for the lowest allowable $T$, and perhaps as you suggest looking at the Siegel set more closely may lead to something, I will certainly take that hint. $\endgroup$ – TA Wong Oct 30 '15 at 23:39
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I think the answer truly does depend on whether you use the naive truncation, as in Iwaniec's book, or the Arthur truncation, as in Paul Garrett's note. In particular, the method of proof for the Maaß-Selberg relation with the naive truncation is proved in Iwaniec's book using Green's identity, and as GH from MO mentioned in his comment, the geometric picture here suggests that the proof only requires that $T \geq \sqrt{3}/2$.

On the other hand, I believe that with the Arthur truncation, one really does require that $T \geq 1$, with the proof of the Maaß-Selberg relation now instead via an unfolding argument as in Paul Garrett's notes. Here the Arthur truncation is \[\Lambda^T E(z,s) = E(z,s) - \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} c_0 E(\gamma z,s), \] where for $z = x + iy \in \mathbb{H}$, \[c_0 E(\gamma z,s) = \int_{0}^{1} E(\gamma z,s) \, dx,\] so that $c_0 E(z,s) = y^s + \varphi(s) y^{1-s}$.

One can easily show that if $\gamma = \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$ with $\gamma \notin \Gamma_{\infty}$, so that $c \neq 0$, then \[\Im(z) \Im(\gamma z) = \frac{y^2}{(cx + d)^2 + c^2 y^2} \leq 1,\] while $\Im(\gamma z) = \Im(z)$ for $\gamma \in \Gamma_{\infty}$.

So for $T \geq 1$, the Arthur truncation is equal to \[\Lambda^T E(z,s) = \begin{cases} E(z,s) & \text{if $1/T \leq y \leq T$,} \\\ E(z,s) - y^s - \varphi(s) y^{1-s} & \text{if $y > T$.} \end{cases}\] If $y < 1/T$, then there may be more terms (and which terms are also present will now depend on $x$ as well), as there may be more coset representatives $\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z})$ other than just the identity for which $\Im(\gamma z) > T$.

If $T < 1$, then $\Lambda^T E(z,s) = E(z,s) - y^s - \varphi(s) y^{1-s}$ if $y \geq 1/T$, but there may be more terms if $y < 1/T$.

In particular, the Arthur truncation coincides with the naive truncation on the standard fundamental domain if $T \geq 2/\sqrt{3}$, but if $T < 2/\sqrt{3}$ then this is no longer the case.

Now a key step in proving the Maaß-Selberg relation is showing that \[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0.\] One shows this by unfolding to find that this is \[-\int_{T}^{\infty} \overline{c_0 E(z,r)} \int_{0}^{1} \Lambda^T E(z,s) \, dx \, dy,\] using the fact that $c_0 E(z,r) = y^r + \varphi(r) y^{1-r}$ does not depend on $x$.

If $T \geq 1$, then $\Lambda^T E(z,s) = E(z,s) - c_0 E(z,s)$ for $T < y < \infty$ and $0 < x < 1$, and so the inner integral vanishes. But if $T < 1$, then there may be other terms, and so this is no longer the case. So the condition $T \geq 1$ truly is necessary here.

EDIT: With regards to your comment, note that the Maaß-Selberg relation states that \[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{\Lambda^T E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}. \] Here \[\varphi(s) = \frac{\Lambda(2 - 2s)}{\Lambda(2s)}, \qquad \Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s),\] so that $|\varphi(1/2 + it)|^2 = 1$ for all $t \neq 0$, and taking logarithmic derivatives shows that \[\frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) = -2\Re\left(\frac{\Gamma'}{\Gamma}\left(\frac{1}{2} + it\right)\right) - 4\Re\left(\frac{\zeta'}{\zeta}(1 + 2it)\right) + 2 \log \pi.\]

Setting $s = r = 1/2 + it + \varepsilon$ for $t \neq 0$ and $\varepsilon > 0$, taking the limit as $\varepsilon \to 0$, and using the Laurent expansions of each term, we find that \[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + it\right)\right|^2 \, d\mu(z) = 2 \log T - \frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) - \Im \left( \varphi\left(\frac{1}{2} + it\right) \frac{T^{-2it}}{t}\right). \] Now it's not obvious to me either that this is nonnegative, and of course if $T$ is very close to zero then you wouldn't expect it to be. But for $T$ near $1$ (because as mentioned earlier, even using the naive truncation we can't have $T < \sqrt{3}/2$) it's not obvious to me either whether this is negative, so I'm not sure if there's truly a problem here; certainly for large $t$, the digamma function should dominate. Of course, I could be missing something here.

SECOND EDIT: I still don't understand your objection; as I have already stated, one requires that $T \geq 1$ for the Arthur truncation, and this is also sufficient (i.e. with the Arthur truncation, the Maaß-Selberg relation holds for all $T \geq 1$). Indeed, I already showed that \[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0\] when $T \geq 1$, so it remains to show that \[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}. \]

By the definition of the Arthur truncation, the left-hand side is \[ \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_ {\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) \leq T}} \Im(\gamma z)^s \overline{E(z,r)} \, d\mu(z) - \varphi(s) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} \Im(\gamma z)^{1-s} \overline{E(z,r)} \, d\mu(z), \] which, by the $\mathrm{SL}_2(\mathbb{Z})$-invariance of $E(z,r)$, unfolds to \[ \int_{0}^{T} y^s \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2} - \varphi(s) \int_{T}^{\infty} y^{1-s} \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2}. \] By the definition of the constant term of $E(z,r)$ (namely that it is $y^r + \varphi(r) y^{1-r}$), one can evaluate these integrals, with the result being the Maaß-Selberg relation.

Of course, to ensure convergence of all the integrals involved, this is initially only valid for $\Re(s), \Re(r) > 1$ with $\Re(s) - \Re(r) > 1$, but then by analytic continuation this extends to all $s,r \in \mathbb{C}$ with $s \neq \overline{r}$ and $s + \overline{r} \neq 1$ provided $\varphi(s), \varphi(r)$ are well-defined. As before, one can extend this to $s = r = 1/2 + it$ with $t \neq 0$ by setting $s = r = 1/2 + it + \varepsilon$ and taking the limit as $\varepsilon \to 0$.

All this is valid if $T \geq 1$, so one certainly should be able to take $T = 1$ in the Maaß-Selberg relation, unless I'm missing something.

With regards to the integral \[\int_{-\infty}^{\infty} h(r) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + ir\right)\right|^2 \, d\mu(z) \, dr,\] to me it looks like there may be an issue at $r = 0$, and one must be careful when breaking up the integral and blindly using Weil's explicit formula. (In particular, I don't see why you say the second line is negligible - doesn't the integrand blow up at $r = 0$?) But I haven't looked too closely at this.

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  • $\begingroup$ Thanks for the detailed answer. Here's my problem: I'm looking at the inner product $|\Lambda^TE(z,1/2+ir)|^2$ for all real $r$, it does not seem clear that the result is nonnegative. $\endgroup$ – TA Wong Oct 26 '15 at 16:57
  • $\begingroup$ Thanks also for the addendum: I agree with you, that when $T$ grows large the real part of digamma is basically log $T$, so these two should produce the main term, whereas when $T$ is close to $1^+$ the positivity isn't clear to me from the RHS, but I'm wondering what that says about the behaviour of the LHS, say, how it depends on $T$. $\endgroup$ – TA Wong Oct 27 '15 at 3:22
  • $\begingroup$ Hi Peter, let me try explicate the issue: I've gone through several derivations of the formula the conditions are more or less as stated, the issue is that when I choose a particular test function $e^{-r^2}$, the integral appears to be negative for $T$ close to 1. That the second integrand is small (and regular at 0) takes a little work, but you can see that $m(1/2+ir)$ is equal to $r^{2i\theta}$, $\theta$ being the argument of $\xi(1/2+ir)$, hence has norm one. If you're willing I could write you an email if you'd like to check the work. $\endgroup$ – TA Wong Nov 3 '15 at 19:55
  • $\begingroup$ But what about the $dr/r$? $\endgroup$ – Peter Humphries Nov 3 '15 at 20:04
  • $\begingroup$ So $m(1/2-ir)$ is $e^{2i\theta}$ where $\theta$ is the argument of $\xi(1+2ir)$, then splitting it into the $\cos$ and $\sin$ part one sees that only the $\sin(2\theta+2Tr)$ part remains. Then as $r$ approaches $0$, this $\sin$ term approaches 0, and matches with the $1/r$ term. $\endgroup$ – TA Wong Nov 3 '15 at 20:25

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