Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f$ and $g$ be two cuspforms on $\Gamma \backslash \mathbb{H}$. They could be Maass cuspforms, or holomorphic modular forms. Let us say that they are holomorphic and also that $\Gamma = \operatorname{SL}_2(\mathbb{Z})$ for simplicity. The product $f \overline g$ is not necessarily a cuspidal function on $\Gamma \backslash \mathbb{H}$ although it decays even faster than $f$ or $g$ as it approaches a cusp. One can see that the function is not cuspidal by taking $f = g$ and noting that, $$ \int_0^1 |f|^2(x+iy) dx \neq 0. $$

I guess this is not a question but rather a surprised statement about cuspidality not being synonymous with vanishing at cusps.

It seems like the two statements are equivalent only when the function in question is also an eigenfunction of the Laplacian.

Could you elaborate on this issue, for example could you find a cuspidal function, which does not vanish at the cusps? Of course the function has to be not an eigenfunction.

share|improve this question

2 Answers 2

up vote 5 down vote accepted

I don't think it's possible to find a "nice" (say, smooth) function $f \in L_2(\Gamma \backslash \mathbb{H})$ such that $(1) \int_0^{1} f(x+iy) dx = 0$ for all $y > 0$ and $\lim_{y\rightarrow \infty} f(x+iy) \neq 0$. This may be total overkill, but consider the spectral decomposition of such an $f$, namely $$(2) \qquad f(z) = \sum_{j} \langle f, u_j \rangle u_j(z) + \frac{1}{4\pi } \int_{\mathbb{R}} \langle E(\cdot, 1/2 + it), f\rangle E(z, 1/2 + it) dt.$$ By unfolding, the inner product of $f$ with the Eisenstein series $E(z,s)$ is zero by the assumption (1); initially this is easy for the real part of $s$ large but then follows by analytic continuation. By inserting (2) into (1) we see that $\langle f, u_0 \rangle = 0$, that is $f$ is orthogonal to the constant eigenfunction. Now in (2) take $z= x+iy$ with $y$ large. Each term in the sum is very small since all the Maass forms vanish at the cusp, and the projections of $f$ onto the constant eigenfunction and the Eisenstein series are zero.

share|improve this answer
    
I think that the proof you have given in fact shows that no such $f$ could exist in $L^2(\Gamma \backslash \mathbb{H})$, not only smooth ones. I also realize that I can find an automorphic function $f \in L^2(\Gamma \backslash \mathbb{H})$ such that $\int_0^1 f(x+iy) dx = 0$ for $y >2$ say, but having $\lim_{y\to \infty}f(x+iy) \neq 0$. That would be obtained by averaging the function $\phi(z) = e^{2\pi i x} \mathbf{1}_{[-1/2,1/2]\times [2,\infty]}(z)$ with respect to the group. $$f(z) = \sum_{\gamma \in \Gamma} \phi(\gamma z).$$ With some more work I think $f$ can be made smooth. –  Eren Mehmet Kiral May 2 '12 at 21:49
    
At the last stage of my proof one has to interchange a limit and an infinite sum, so there needs to be some kind of continuity/smoothness assumption. As a counterexample, you can change $f$ on a set of measure zero, say making it be $1$ on some vertical line. –  Matt Young May 3 '12 at 0:41

This sort of issue is significant when we're trying to get a grip on the analysis of automorphic forms, but/and, already non-automorphic situations provide much insight. I can't help but comment that adding "automorphic" in any discussion creates enough cognitive dissonance (at some level) that otherwise-classical examples and counterexamples often get lost.

Yes, one should think in terms of automorphic spectral decompositions, and note that eigen-cuspforms are of rapid decay, and that a not-vanishing-at-infinity but vanishing-constant-term function must necessarily have unpleasantly-behaving spectral decomposition coefficients.

That a (very nicely convergent, at least uniformly pointwise on compacts) sum of rapidly-decreasing functions need not be decreasing at all, etc., (yes, I know this is a somewhat different issue) is illustrated by $\sum x^n e^{-nx}/n! = 1$. This suggests something about the spectral analysis.

Similarly, more directly, (but, yes, I know, somewhat differently), an $L^2$ function on the real line can have ever-narrower spikes parading out to infinity, so not go to $0$ at infinity. (Of course, if it had any limit at all, it would have to be 0, on the real line... though this is not true for automorphic forms, because of finite volume at infinity.)

In the automorphic case, take $f(x+iy)$ to be $0$ for $y<2$, and for $y\ge 2$ let $f$ be $e^{2\pi ix}h(y)$ with $h(y)=y^{1/3}$. Can smooth this out, too. And can make the lim sup be $+\infty$ by spikes.

In particular, here, note that dropping the eigenfunction condition gives us license to look at the much simpler "tapering cone" that is the image of a high-up Siegel set under the quotient map.

In summary, it is (with hindsight, sure) boringly easy to make a "cuspidal" automorphic form, in $L^2$, that does not go to $0$ at infinity. But, when we see what the possibilities are, they do not disprove other important principles.

Edit: should have said that the lim sup can grow arbitrarily fast by (narrow) spikes.

Edit: That is, a natural argument based on a spectral expansion is incomplete without further information on the rapidly decreasing functions (here, cuspforms), since a sum of rapidly decreasing functions need not be decreasing at all. Of course, some sums of rapidly-decreasing are rapidly decreasing... but if the sups occur further and further out, this need not be so, as in $\sum {2^n\over n!}\,x^ne^{-x} = e^x$.

Although, for example, being in $L^2(\mathbb R)$ does not imply a function goes to $0$ at infinity pointwise, if it is in a Sobolev space (has an $L^2$ derivative) then by the fundamental theorem of calculus it has a bound something like $\sqrt{x}$. To actually have decay requires more.

Sums of cuspforms that are in $L^2$ can easily map to non-$L^2$ "cuspidal" things under $\Delta$, or even under "first derivatives" coming from the Lie algebra acting on the right. This does not necessarily contravene "going to $0$", but it shows that a simple argument fails, as in the sum of $x^n e^{-x}$'s.

In fact, I think Iwaniec' "spectral theory of afms" book has some remarks in it about the sups of cuspforms occurring further and further out, which creates the danger alluded to above. There was a paper of Iwaniec-Sarnak in which an infelicity about something of this sort occurred, remarked upon in Sarnak's letter to Morawetz. The extent to which this enables wild-ish growth at infinity of (smooth?) $L^2$ "cuspidal" things would require computation, at least, and sharp answers may depend on serious unproven things, now that I think about it...

Edit again: yes, some sort of smoothness hypothesis presumably makes a spectral argument work, even with "relatively easy" estimates on sups of an orthonormal basis of cuspforms. An easy sort of smoothness assumption is not merely smoothness, but that $\Delta f$ is in $L^2$, and/or $\Delta^\ell f$ is in $L^2$ for some sufficient $\ell$. A Sobolev-ish condition. Already on the real line the analogous issue is present: smooth functions in $L^2$ without constraints on integrability of derivatives need not decay.

share|improve this answer
    
What are the important principles that you say? In the case of vanishing at cusps but not being cuspidal case, the function I had found had good properties in terms of its spectral decomposition. (I know this doesn't violate any theorem, but did violate my intuition about cuspidal functions). –  Eren Mehmet Kiral May 4 '12 at 1:41
    
@Paul Yes there is a potential issue with reversing the limit as $y\rightarrow \infty$, and the sum over $j$. If $f$ is sufficiently smooth then using the fact that the Laplacian is self-adjoint, one can show that the spectral coefficients decay as the reciprocal of some polynomial of the Laplace eigenvalue. The sup norm of a Maass form grows like a fixed polynomial in the Laplace eigenvalue, and also $u_j(x+iy)$ is exponentially small once $y$ is a little larger than the square-root of the Laplace eigenvalue. Maybe I'll go back and edit my answer when I get some time... –  Matt Young May 4 '12 at 14:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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