# Category Archives: Mathematics

## Update to Second Moments in the Generalized Gauss Circle Problem

Last year, my coauthors Tom Hulse, Chan Ieong Kuan, and Alex Walker posted a paper to the arXiv called “Second Moments in the Generalized Gauss Circle Problem”. I’ve briefly described its contents before.

This paper has been accepted and will appear in Forum of Mathematics: Sigma.

This is the first time I’ve submitted to the Forum of Mathematics, and I must say that this has been a very good journal experience. One interesting aspect about FoM: Sigma is that they are immediate (gold) open access, and they don’t release in issues. Instead, articles become available (for free) from them once the submission process is done. I was reviewing a publication-proof of the paper yesterday, and they appear to be very quick with regards to editing. Perhaps the paper will appear before the end of the year.

An updated version (the version from before the handling of proofs at the journal, so there will be a number of mostly aesthetic differences with the published version) of the paper will appear on the arXiv on Monday 10 December.1

## A new appendix has appeared

There is one major addition to the paper that didn’t appear in the original preprint. At one of the referee’s suggestions, Chan and I wrote an appendix. The major content of this appendix concerns a technical detail about Rankin-Selberg convolutions.

If $f$ and $g$ are weight $k$ cusp forms on $\mathrm{SL}(2, \mathbb{Z})$ with expansions $$f(z) = \sum_ {n \geq 1} a(n) e(nz), \quad g(z) = \sum_ {n \geq 1} b(n) e(nz),$$ then one can use a (real analytic) Eisenstein series $$E(s, z) = \sum_ {\gamma \in \mathrm{SL}(2, \mathbb{Z})_ \infty \backslash \mathrm{SL}(2, \mathbb{Q})} \mathrm{Im}(\gamma z)^s$$ to recognize the Rankin-Selberg $L$-function $$\label{RS} L(s, f \otimes g) := \zeta(s) \sum_ {n \geq 1} \frac{a(n)b(n)}{n^{s + k – 1}} = h(s) \langle f g y^k, E(s, z) \rangle,$$ where $h(s)$ is an easily-understandable function of $s$ and where $\langle \cdot, \cdot \rangle$ denotes the Petersson inner product.

When $f$ and $g$ are not cusp forms, or when $f$ and $g$ are modular with respect to a congruence subgroup of $\mathrm{SL}(2, \mathbb{Z})$, then there are adjustments that must be made to the typical construction of $L(s, f \otimes g)$.

When $f$ and $g$ are not cusp forms, then Zagier2 provided a way to recognize $L(s, f \otimes g)$ when $f$ and $g$ are modular on the full modular group $\mathrm{SL}(2, \mathbb{Z})$. And under certain conditions that he describes, he shows that one can still recognize $L(s, f \otimes g)$ as an inner product with an Eisenstein series as in \eqref{RS}.

In principle, his method of proof would apply for non-cuspidal forms defined on congruence subgroups, but in practice this becomes too annoying and bogged down with details to work with. Fortunately, in 2000, Gupta3 gave a different construction of $L(s, f \otimes g)$ that generalizes more readily to non-cuspidal forms on congruence subgroups. His construction is very convenient, and it shows that $L(s, f \otimes g)$ has all of the properties expected of it.

However Gupta does not show that there are certain conditions under which one can recognize $L(s, f \otimes g)$ as an inner product against an Eisenstein series.4 For this paper, we need to deal very explicitly and concretely with $L(s, \theta^2 \otimes \overline{\theta^2})$, which is formed from the modular form $\theta^2$, non-cuspidal on a congruence subgroup.

The Appendix to the paper can be thought of as an extension of Gupta’s paper: it uses Gupta’s ideas and techniques to prove a result analogous to \eqref{RS}. We then use this to get the explicit understanding necessary to tackle the Gauss Sphere problem.

There is more to this story. I’ll return to it in a later note.

## Other submission details for FoM: Sigma

I should say that there are many other revisions between the original preprint and the final one. These are mainly due to the extraordinary efforts of two Referees. One Referee was kind enough to give us approximately 10 pages of itemized suggestions and comments.

When I first opened these comments, I was a bit afraid. Having so many comments was daunting. But this Referee really took his or her time to point us in the right direction, and the resulting paper is vastly improved (and in many cases shortened, although the appendix has hidden the simplified arguments cut in length).

More broadly, the Referee acted as a sort of mentor with respect to my technical writing. I have a lot of opinions on technical writing,5 but this process changed and helped sharpen my ideas concerning good technical math writing.

I sometimes hear lots of negative aspects about peer review, but this particular pair of Referees turned the publication process into an opportunity to learn about good mathematical exposition — I didn’t expect this.

I was also surprised by the infrastructure that existed at the University of Warwick for handling a gold open access submission. As part of their open access funding, Forum of Math: Sigma has an author-pays model. Or rather, the author’s institution pays. It took essentially no time at all for Warwick to arrange the payment (about 500 pounds).

This is a not-inconsequential amount of money, but it is much less than the 1500 dollars that PLoS One uses. The comparison with PLoS One is perhaps apt. PLoS is older, and perhaps paved the way for modern gold open access journals like FoM. PLoS was started by group of established biologists and chemists, including a Nobel prize winner; FoM was started by a group of established mathematicians, including multiple Fields medalists.6

I will certainly consider Forum of Mathematics in the future.

## The wrong way to compute a sum: addendum

In my previous note, I looked at an amusing but inefficient way to compute the sum $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1}$$ using Mellin and inverse Mellin transforms. This was great fun, but the amount of work required was more intense than the more straightforward approach offered immediately by using Lambert series.

However, Adam Harper suggested that there is a nice shortcut that we can use (although coming up with this shortcut requires either a lot of familiarity with Mellin transforms or knowledge of the answer).

In the Lambert series approach, one shows quickly that $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} = \sum_{n \geq 1} \frac{n}{2^n},$$ and then evaluates this last sum directly. For the Mellin transform approach, we might ask: do the two functions $$f(x) = \sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1}$$ and $$g(x) = \sum_{n \geq 1} \frac{n}{2^{nx}}$$ have the same Mellin transforms? From the previous note, we know that they have the same values at $1$.

We also showed very quickly that $$\mathcal{M} [f] = \frac{1}{(\log 2)^2} \Gamma(s) \zeta(s-1).$$ The more difficult parts from the previous note arose in the evaluation of the inverse Mellin transform at $x=1$.

Let us compute the Mellin transform of $g$. We find that \begin{align} \mathcal{M}[g] &= \sum_{n \geq 1} n \int_0^\infty \frac{1}{2^{nx}} x^s \frac{dx}{x} \notag \\ &= \sum_{n \geq 1} n \int_0^\infty \frac{1}{e^{nx \log 2}} x^s \frac{dx}{x} \notag \\ &= \sum_{n \geq 1} \frac{n}{(n \log 2)^s} \int_0^\infty x^s e^{-x} \frac{dx}{x} \notag \\ &= \frac{1}{(\log 2)^2} \zeta(s-1)\Gamma(s). \notag \end{align} To go from the second line to the third line, we did the change of variables $x \mapsto x/(n \log 2)$, yielding an integral which is precisely the definition of the Gamma function.

Thus we see that $$\mathcal{M}[g] = \frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1) = \mathcal{M}[f],$$ and thus $f(x) = g(x)$. (“Nice” functions with the same “nice” Mellin transforms are also the same, exactly as with Fourier transforms).

This shows that not only is $$\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} = \sum_{n \geq 1} \frac{n}{2^n},$$ but in fact $$\sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1} = \sum_{n \geq 1} \frac{n}{2^{nx}}$$ for all $x > 1$.

I think that’s sort of slick.

## The wrong way to compute a sum

At a recent colloquium at the University of Warwick, the fact that
\label{question}
\sum_ {n \geq 1} \frac{\varphi(n)}{2^n – 1} = 2.

Although this was mentioned in passing, John Cremona asked — How do you prove that?

It almost fails a heuristic check, as one can quickly check that
\label{similar}
\sum_ {n \geq 1} \frac{n}{2^n} = 2,

which is surprisingly similar to \eqref{question}. I wish I knew more examples of pairs with a similar flavor.

[Edit: Note that an addendum to this note has been added here. In it, we see that there is a way to shortcut the “hard part” of the long computation.]

## The right way

Shortly afterwards, Adam Harper and Samir Siksek pointed out that this can be determined from Lambert series, and in fact that Hardy and Wright include a similar exercise in their book. This proof is delightful and short.

The idea is that, by expanding the denominator in power series, one has that

\sum_{n \geq 1} a(n) \frac{x^n}{1 – x^n} \notag
= \sum_ {n \geq 1} a(n) \sum_{m \geq 1} x^{mn}
= \sum_ {n \geq 1} \Big( \sum_{d \mid n} a(d) \Big) x^n,

where the inner sum is a sum over the divisors of $d$. This all converges beautifully for $\lvert x \rvert < 1$.

Applied to \eqref{question}, we find that

\sum_{n \geq 1} \frac{\varphi(n)}{2^n – 1} \notag
= \sum_ {n \geq 1} \varphi(n) \frac{2^{-n}}{1 – 2^{-n}}
= \sum_ {n \geq 1} 2^{-n} \sum_{d \mid n} \varphi(d),

and as

\sum_ {d \mid n} \varphi(d) = n, \notag

we see that \eqref{question} can be rewritten as \eqref{similar} after all, and thus both evaluate to $2$.

That’s a nice derivation using a series that I hadn’t come across before. But that’s not what this short note is about. This note is about evaluating \eqref{question} in a different way, arguably the wrong way. But it’s a wrong way that works out in a nice way that at least one person1 finds appealing.

## The wrong way

We will use Mellin inversion — this is essentially Fourier inversion, but in a change of coordinates.

Let $f$ denote the function

f(x) = \frac{1}{2^x – 1}. \notag

Denote by $f^ *$ the Mellin transform of $f$,

f * (s):= \mathcal{M} [f(x)] (s) := \int_ 0^\infty f(x) x^s \frac{dx}{x}
= \frac{1}{(\log 2)^2} \Gamma(s)\zeta(s),\notag

where $\Gamma(s)$ and $\zeta(s)$ are the Gamma function and Riemann zeta functions.2

For a general nice function $g(x)$, its Mellin transform satisfies

\mathcal{M}[f(nx)] (s)
= \int_0^\infty g(nx) x^s \frac{dx}{x}
= \frac{1}{n^s} \int_0^\infty g(x) x^s \frac{dx}{x}
= \frac{1}{n^s} g^ * (s).\notag

Further, the Mellin transform is linear. Thus
\label{mellinbase}
\mathcal{M}[\sum_{n \geq 1} \varphi(n) f(nx)] (s)
= \sum_ {n \geq 1} \frac{\varphi(n)}{n^s} f^ * (s)
= \sum_ {n \geq 1} \frac{\varphi(n)}{n^s} \frac{\Gamma(s) \zeta(s)}{(\log 2)^s}.

The Euler phi function $\varphi(n)$ is multiplicative and nice, and its Dirichlet series can be rewritten as

\sum_{n \geq 1} \frac{\varphi(n)}{n^s} \notag
= \frac{\zeta(s-1)}{\zeta(s)}.

Thus the Mellin transform in \eqref{mellinbase} can be written as

\frac{1}{(\log 2)^s} \Gamma(s) \zeta(s-1). \notag

By the fundamental theorem of Mellin inversion (which is analogous to Fourier inversion, but again in different coordinates), the inverse Mellin transform will return the original function. The inverse Mellin transform of a function $h(s)$ is defined to be

\mathcal{M}^{-1}[h(s)] (x) \notag
:=
\frac{1}{2\pi i} \int_ {c – i \infty}^{c + i\infty} x^s h(s) ds,

where $c$ is taken so that the integral converges beautifully, and the integral is over the vertical line with real part $c$. I’ll write $(c)$ as a shorthand for the limits of integration. Thus
\label{mellininverse}
\sum_{n \geq 1} \frac{\varphi(n)}{2^{nx} – 1}
= \frac{1}{2\pi i} \int_ {(3)} \frac{1}{(\log 2)^s}
\Gamma(s) \zeta(s-1) x^{-s} ds.

We can now describe the end goal: evaluate \eqref{mellininverse} at $x=1$, which will recover the value of the original sum in \eqref{question}.

How can we hope to do that? The idea is to shift the line of integration arbitrarily far to the left, pick up the infinitely many residues guaranteed by Cauchy’s residue theorem, and to recognize the infinite sum as a classical series.

The integrand has residues at $s = 2, 0, -2, -4, \ldots$, coming from the zeta function ($s = 2$) and the Gamma function (all the others). Note that there aren’t poles at negative odd integers, since the zeta function has zeroes at negative even integers.

Recall, $\zeta(s)$ has residue $1$ at $s = 1$ and $\Gamma(s)$ has residue $(-1)^n/{n!}$ at $s = -n$. Then shifting the line of integration and picking up all the residues reveals that

\sum_{n \geq 1} \frac{\varphi(n)}{2^{n} – 1} \notag
=\frac{1}{\log^2 2} + \zeta(-1) + \frac{\zeta(-3)}{2!} \log^2 2 +
\frac{\zeta(-5)}{4!} \log^4 2 + \cdots

The zeta function at negative integers has a very well-known relation to the Bernoulli numbers,
\label{zeta_bern}
\zeta(-n) = – \frac{B_ {n+1}}{n+1},

where Bernoulli numbers are the coefficients in the expansion
\label{bern_gen}
\frac{t}{1 – e^{-t}} = \sum_{m \geq 0} B_m \frac{t^m}{m!}.

Many general proofs for the values of $\zeta(2n)$ use this relation and the functional equation, as well as a computation of the Bernoulli numbers themselves. Another important aspect of Bernoulli numbers that is apparent through \eqref{zeta_bern} is that $B_{2n+1} = 0$ for $n \geq 1$, lining up with the trivial zeroes of the zeta function.

Translating the zeta values into Bernoulli numbers, we find that
\eqref{question} is equal to
\begin{align}
&\frac{1}{\log^2 2} – \frac{B_2}{2} – \frac{B_4}{2! \cdot 4} \log^2 2 –
\frac{B_6}{4! \cdot 6} \log^4 2 – \frac{B_8}{6! \cdot 8} \cdots \notag \\
&=
-\sum_{m \geq 0} (m-1) B_m \frac{(\log 2)^{m-2}}{m!}. \label{recog}
\end{align}
This last sum is excellent, and can be recognized.

For a general exponential generating series

F(t) = \sum_{m \geq 0} a(m) \frac{t^m}{m!},\notag

we see that

\frac{d}{dt} \frac{1}{t} F(t) \notag
=\sum_{m \geq 0} (m-1) a(m) \frac{t^{m-2}}{m!}.

Applying this to the series defining the Bernoulli numbers from \eqref{bern_gen}, we find that

\frac{d}{dt} \frac{1}{t} \frac{t}{1 – e^{-t}} \notag
=- \frac{e^{-t}}{(1 – e^{-t})^2},

and also that

\frac{d}{dt} \frac{1}{t} \frac{t}{1 – e^{-t}} \notag
=\sum_{m \geq 0} (m-1) B_m \frac{(t)^{m-2}}{m!}.

This is exactly the sum that appears in \eqref{recog}, with $t = \log 2$.

Putting this together, we find that

\sum_{m \geq 0} (m-1) B_m \frac{(\log 2)^{m-2}}{m!} \notag
=\frac{e^{-\log 2}}{(1 – e^{-\log 2})^2}
= \frac{1/2}{(1/2)^2} = 2.

Thus we find that \eqref{question} really is equal to $2$, as we had sought to show.

## Using lcalc to compute half-integral weight L-functions

This is a brief note intended primarily for my collaborators interested in using Rubinstein’s lcalc to compute the values of half-integral weight $L$-functions.

We will be using lcalc through sage. Unfortunately, we are going to be using some functionality which sage doesn’t expose particularly nicely, so it will feel a bit silly. Nonetheless, using sage’s distribution will prevent us from needing to compile it on our own (and there are a few bugfixes present in sage’s version).

Some $L$-functions are inbuilt into lcalc, but not half-integral weight $L$-functions. So it will be necessary to create a datafile containing the data that lcalc will use to generate its approximations. In short, this datafile will describe the shape of the functional equation and give a list of coefficients for lcalc to use.

## Building the datafile

It is assumed that the $L$-function is normalized in such a way that
$$\Lambda(s) = Q^s L(s) \prod_{j = 1}^{A} \Gamma(\gamma_j s + \lambda_j) = \omega \overline{\Lambda(1 – \overline{s})}.$$
This involves normalizing the functional equation to be of shape $s \mapsto 1-s$. Also note that $Q$ will be given as a real number.

An annotated version of the datafile you should create looks like this

2                  # 2 means the Dirichlet coefficients are reals
0                  # 0 means the L-function isn't a "nice" one
10000              # 10000 coefficients will be provided
0                  # 0 means the coefficients are not periodic
1                  # num Gamma factors of form \Gamma(\gamma s + \lambda)
1                  # the \gamma in the Gamma factor
1.75 0             # \lambda in Gamma factor; complex valued, space delimited
0.318309886183790  # Q. In this case, 1/pi
1 0                # real and imaginary parts of omega, sign of func. eq.
0                  # number of poles
1.000000000000000  # a(1)
-1.78381067250408  # a(2)
...                # ...
-0.622124724090625 # a(10000)


If there is an error, lcalc will usually fail silently. (Bummer). Note that in practice, datafiles should only contain numbers and should not contain comments. This annotated version is for convenience, not for use.

You can find a copy of the datafile for the unique half-integral weight cusp form of weight $9/2$ on $\Gamma_0(4)$ here. This uses the first 10000 coefficients — it’s surely possible to use more, but this was the test-setup that I first set up.

## Generating the coefficients for this example

In order to create datafiles for other cuspforms, it is necessary to compute the coefficients (presumably using magma or sage) and then to populate a datafile. A good exercise would be to recreate this datafile using sage-like methods.

One way to create this datafile is to explicitly create the q-expansion of the modular form, if we happen to know a convenient expression. For us, we happen to know that $f = \eta(2z)^{12} \theta(z)^{-3}$. Thus one way to create the coefficients is to do something like the following.

num_coeffs = 10**5 + 1
weight     = 9.0 / 2.0

R.<q> = PowerSeriesRing(ZZ)
theta_expansion = theta_qexp(num_coeffs)
# Note that qexp_eta omits the q^(1/24) factor
eta_expansion = qexp_eta(ZZ[['q']], num_coeffs + 1)
eta2_coeffs = []
for i in range(num_coeffs):
if i % 2 == 1:
eta2_coeffs.append(0)
else:
eta2_coeffs.append(eta_expansion[i//2])
eta2 = R(eta2_coeffs)
g = q * ( (eta2)**4 / (theta_expansion) )**3

coefficients = g.list()[1:] # skip the 0 coeff
print(coefficients[:10])

normalized_coefficients = []
for idx, elem in enumerate(coefficients, 1):
normalized_coeff = 1.0 * elem / (idx ** (.5 * (weight - 1)))
normalized_coefficients.append(normalized_coeff)
print(normalized_coefficients[:10])


## Using lcalc now

Suppose that you have a datafile, called g1_lcalcfile.txt (for example). Then to use this from sage, you point lcalc within sage to this file. This can be done through calls such as

# Computes L(0.5 + 0i, f)
lcalc('-v -x0.5 -y0 -Fg1_lcalcfile.txt')

# Computes L(s, f) from 0.5 to (2 + 7i) at 1000 equally spaced samples
lcalc('--value-line-segment -x0.5 -y0 -X2 -Y7 --number-samples=1000 -Fg1_lcalcfile.txt')

# See lcalc.help() for more on calling lcalc.


The key in these is to pass along the datafile through the -F argument.

## Speed talks should be at every conference

I recently attended Building Bridges 4, an automorphic forms summer school and workshop. A major goal of the conference is to foster communication and relationships between researchers from North America and Europe, especially junior researchers and graduate students.

It was a great conference, and definitely one of the better conferences that I’ve attended. What made it so good? For one thing, it was in Budapest, and I love Budapest. Many of the main topics were close to my heart, which is a big plus.

But what I think really set it apart was that there were lots of relatively short talks, and almost everyone attended almost every talk.1

The amount of time allotted to a talk carries extreme power in deciding what sort of talk it will be. A typical hour-long seminar talk is long enough to give context, describe a line of research leading to a set of results, discuss how these results fit into the literature, and even to give a non-rushed description of how something is proved. Sometimes a good speaker will even distill a few major ideas and discuss how they are related. A long talk can have multiple major ideas (although just one presented very well can make a good talk too).

In comparison, 50, 40, and 30 minute talks require much more discipline. As the amount of time decreases, the number of ideas that can be inserted into a talk decreases. And this relationship is not linear! Thirty minutes is just about long enough to describe one idea pretty well, and to do anything more is very hard.2

Something interesting happens for shorter talks. For 20 minute, 15 minute, and 10 minute talks, the limitation almost serves as a source of inspiration.3 Being forced to focus on what’s important is a powerful organizing force.

The median talk length was 20 minutes, which is a very comfortable number. This is long enough to state a result and give context. It’s also long enough to tempt speakers into describing methodology behind a proof, but not long enough to effectively teach someone how the proof works.

An extraordinary aspect of a 20 minute talk is also that it’s short enough to pay attention to, even if it’s only an okay talk. It is perhaps not a surprise to most conference goers that most talks are not so great. To be a skilled orator is to be exceptional.

At Building Bridges, I was introduced to math speed talks. These are two minute talks. I’ve seen many programming lightning talks (often used to plug a particular product or solution to a common programming problem), but these math speed talks were different.

People used their two minutes to introduce an idea, or a result. And they either chose to give the broadest possible context, or a singular idea in the proof.

People were talking about real mathematics in two minutes. And I loved it.

Simply having a task where you distill some real mathematics into a two minute coherent description is worthwhile. What’s important? What do you really want to say? Why?

Two minutes is so short that it feels silly. And silly means that it doesn’t feel dangerous or scary, and thus many people felt willing to give it a try. At Building Bridges, the organizers gamified the speed talks, so that getting the closest to 2 minutes was rewarded with applause and going over two minutes resulted in a buzzer going off. It was a game, and it was fun. It was encouraging.

I firmly support any activity that encourages people who normally don’t speak so much, especially students and junior researchers. You learn a lot by giving a talk, even if it’s only a two minute talk.4

This conference had 19 (I think) speed talks over a three day stretch. They were given in clumps after the last regular talk each day. Since people were there for the big talk, everyone attended the speed talks. This is also important! In conferences like the Joint Math Meetings, where there might even be something like speed talks, it’s essentially impossible to pay attention since there are too many people in too many places and you never can step in the same river twice. Here, speed talks were given on the same stage as long talks, to the same audience, and with the same equipment.

Every conference should have speed talks. And they should be treated as first-class talks, with the exception that they are irrefutably silly.

Go forth and spread the speed talk gospel.

## Notes from a Talk at Building Bridges 4

On 18 July 2018 I gave a talk at the 4th Building Bridges Automorphic Forms Workshop, which is hosted at the Renyi Institute in Budapest, Hungary this year. In this talk, I spoke about counting points on hyperboloids, with a certain focus on counting points on the three dimensional hyperboloid

$$$$X^2 + Y^2 = Z^2 + h$$$$

for any fixed integer $h$.

I gave a similar talk at the 32nd Automorphic Forms Workshop in Tufts in March. I don’t say this during my talk, but a big reason for giving these talks is to continue to inspire me to finish the corresponding paper. (There are still a couple of rough edges that need some attention).

The methodology for the result relies on the spectral expansion of half-integral weight modular forms. This is unfriendly to those unfamiliar with the subject, and particularly mysterious to students. But there is a nice connection to a topic discussed by Arpad Toth during the previous week’s associated summer school.

Arpad sketched a proof of the spectral decomposition of holomorphic modular cusp forms on $\Gamma = \mathrm{SL}(2, \mathbb{Z})$. He showed that
$$L^2(\Gamma \backslash \mathcal{H}) = \textrm{cuspidal} \oplus \textrm{Eisenstein}, \tag{1}$$
where the cuspidal contribution comes from Maass forms and the Eisenstein contribution comes from line integrals against Eisenstein series.

The typical Eisenstein series $$$$E(z, s) = \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \textrm{Im}(\gamma z)^s$$$$ only converges for $\mathrm{Re}(s) > 1$, and the initial decomposition in $(1)$ implicitly has $s$ in this range.

To write down the integrals appearing in the Eisenstein spectrum explicitly, one normally shifts the line of integration to $1/2$. As Arpad explained, classically this produces a pole at $s = 1$ (which is the constant function).

In half-integral weight, the Eisenstein series has a pole at $s = 3/4$, with the standard theta function

$$$$\theta(z) = \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z}$$$$

as the residue. (More precisely, it’s a constant times $y^{1/4} \theta(z)$, or a related theta function for $\Gamma_0(N)$). I refer to this portion of the spectrum as the residual spectrum, since it comes from often-forgotten residues of Eisenstein series. Thus the spectral decomposition for half-integral weight objects is a bit more complicated than the normal case.

When giving talks involving half-integral weight spectral expansions to audiences including non-experts, I usually omit description of this. But for those who attended the summer school, it’s possible to at least recognize where these additional terms come from.

The slides for this talk are available here.

## (But there are some positive changes that can be made)

This is the final chapter in my series about the state of internet fora, and Math.SE and StackOverflow in particular. The previous chapters are Challenges Facing Community Cohesion and Ghosts of Forums Past. Unlike the previous entries, this also sits on Meta.Math.SE (and was posted there a week before here). (As I write this as a moderator of Math.SE, I refer to the Math.SE community as “we”, “us”, and “our” community).

A couple of weeks ago, there was a proposal on meta.Math.SE to introduce a third level of math site to the SE network. Many members of the the MathSE community have reacted very positively to this proposal, to the extent that even some of the moderators have considered throwing their weight behind it.

But a NoviceMathSE site would be doomed to fail, and such a separation would not solve the underlying problems facing the site.

To explain my point of view, we need to examine more closely the arguments in favor of NoviceMathSE.

## Ghosts of Forums Past

This is the second in a miniseries of posts on internet fora, and Math.SE and StackOverflow in particular. In the previous entry in the miniseries, I described some of the common major problems facing community cohesion. I claimed that when communities get large, they tend to fracture and the ratio of meaningful communication to noise plummets. To combat this tendency, communities use some mixture of core moderation, peer moderation, membership requirements, or creating subcommunities/splitting off in to other communities.

In this chapter I focus more on Math.SE and StackOverflow. Math.SE is now experiencing growing pains and looking for solutions. But many users of Math.SE have little involvement in the rest of the StackExchange network and are mostly unaware of the fact that StackOverflow has already encountered and passed many of the same trials and tribulations (with varying degrees of success).

Thinking more broadly, many communities have faced these same challenges. Viewed from the point of view from the last chapter, it may appear that there are only a handful of tools a community might use to try to retain group cohesion. Yet it is possible to craft clever mixtures of these tools synergistically. The major reason the StackExchange model has succeeded where other fora have stalled (or failed outright) is through its innovations on the implementation of communition cohesion strategies while still allowing essentially anyone to visit the site.

## Imaginary Internet Points

Slashdot1 popularized the idea of associating imaginary internet points to different users. It was called karma. You got karma if other users rated your comments or submissions well, and lost karma if they rated your posts as poor. But perhaps most importantly, each user can set a threshold for minimum scores of content to see. Thus if people have reasonable thresholds and you post crap, then most people won’t even see it after it’s scored badly.

Posted in Programming, SE | | 1 Comment

## Challenges facing community cohesion (and Math.StackExchange in particular)

In about a month, Math.StackExchange will turn 8. Way back when I was an undergrad, I joined the site. This was 7 years ago, during the site’s first year.

Now with some perspective as a frequent contributor/user/moderator of various online newsgroups and fora, I want to take a moment to examine the current state of Math.SE.

To a certain extent, this is inspired by Joel Spolsky’s series of posts on StackOverflow (which he is currently writing and sending out). But this is also inspired by recent discussion on Meta.Math.SE. As I began to collect my thoughts to make a coherent reply, I realized that I have a lot of thoughts, and a lot to say.

So this is chapter one of a miniseries of writings on internet fora, and Math.SE and StackOverflow in particular.

Posted in Programming, SE | | 2 Comments

## Paper Announcement: A Shifted Sum for the Congruent Number Problem

Tom Hulse, Chan Ieong Kuan, Alex Walker, and I have just uploaded a new paper to the arXiv titled A Shifted Sum for the Congruent Number Problem. In this charming, short paper, we investigate a particular sum of terms which are products of square-indicator functions and show that its asymptotics are deeply connected to congruent numbers. This note serves to describe and provide additional context for these results. (This note is also available as a pdf).

## Congruent Numbers

We consider some triangles. There are many right triangles, such as the triangle with sides $(3, 4, 5)$ or the triangle with sides $(1, 1, \sqrt{2})$. We call a right triangle rational when all its side lengths are rational numbers. For illustration, $(3, 4, 5)$ is rational, while $(1, 1, \sqrt{2})$ is not. $\DeclareMathOperator{\sqfree}{sqfree}$

There is mythology surrounding rational right triangles. According to legend, the ancient Greeks, led both philosophcally and mathematically by Pythagoras (who was the first person to call himself a philosopher and essentially the first to begin to distill and codify mathematics), believed all numbers and quantities were ratios of integers (rational). When a disciple of Pythagoras named Hippasus found that the side lengths of the right triangle $(1, 1, \sqrt{2})$ were not rational multiples of each other, the other followers of Pythagoras killed him by casting him overboard while at sea for having produced an element which contradicted the gods. (It with some irony that we now attribute this as a simple consequence of the Pythagorean Theorem).

This mythology is uncertain, but what is certain is that even the ancient Greeks were interested in studying rational right triangles, and they began to investigate what we now call the Congruent Number Problem. By the year 972 the CNP appears in Arabic manuscripts in (essentially) its modern formulation. The Congruent Number Problem (CNP) may be the oldest unresolved math problem.

We call a positive rational number $t$ congruent if there is a rational right triangle with area $t$. The triangle $(3,4,5)$ shows that $6 = 3 \cdot 4 / 2$ is congruent. The CNP is to describe all congruent numbers. Alternately, the CNP asks whether there is an algorithm to show definitively whether or not $t$ is a congruent number for any $t$.

We can reduce the problem to a statement about integers. If the rational number $t = p/q$ is the area of a triangle with legs $a$ and $b$, then the triangle $aq$ and $bq$ has area $tq^2 = pq$. It follows that to every rational number there is an associated squarefree integer for which either both are congruent or neither are congruent. Further, if $t$ is congruent, then $ty^2$ and $t/y^2$ are congruent for any integer $y$.

We may also restrict to integer-sided triangles if we allow ourselves to look for those triangles with squarefree area $t$. That is, if $t$ is the area of a triangle with rational sides $a/A$ and $b/B$, then $tA^2 B^2$ is the area of the triangle with integer sides $aB$ and $bA$.

It is in this form that we consider the CNP today.

Congruent Number Problem

Given a squarefree integer $t$, does there exist a triangle with integer side lengths such that the squarefree part of the area of the triangle is $t$?

We will write this description a lot, so for a triangle $T$ we introduce the notation

\sqfree(T) = \text{The squarefree part of the area of } T.

For example, the area of the triangle $T = (6, 8, 10)$ is $24 = 6 \cdot 2^2$, and so $\sqfree(T) = 6$. We should expect this, as $T$ is exactly a doubled-in-size $(3,4,5)$ triangle, which also corresponds to the congruent number $6$. Note that this allows us to only consider primitive right triangles.

## Main Result

Let $\tau(n)$ denote the square-indicator function. That is, $\tau(n)$ is $1$ if $n$ is a square, and is $0$ otherwise. Then the main result of the paper is that the sum

S_t(X) := \sum_{m = 1}^X \sum_{n = 1}^X \tau(m-n)\tau(m)\tau(nt)\tau(m+n)

is related to congruent numbers through the asymptotic

S_t(X) = C_t \sqrt X + O_t\Big( \log^{r/2} X\Big),

where

C_t = \sum_{h_i \in \mathcal{H}(t)} \frac{1}{h_i}.

Each $h_i$ is a hypotenuse of a primitive integer right triangle $T$ with $\sqfree(T) = t$. Each hypotnesue will occur in a pair of similar triangles $(a,b, h_i)$ and $(b, a, h_i)$; $\mathcal{H}(t)$ is the family of these triangles, choosing only one triangle from each similar pair. The exponent $r$ in the error term is the rank of the elliptic curve

E_t(\mathbb{Q}): y^2 = x^3 – t^2 x.

What this says is that $S_t(X)$ will have a main term if and only if $t$ is a congruent number, so that computing $S_t(X)$ for sufficiently large $X$ will show whether $t$ is congruent. (In fact, it’s easy to show that $S_t(X) \neq 0$ if and only if $t$ is congruent, so the added value here is the nature of the asymptotic).

We should be careful to note that this does not solve the CNP, since the error term depends in an inexplicit way on the desired number $t$. What this really means is that we do not have a good way of recognizing when the first nonzero term should occur in the double sum. We can only guarantee that for any $t$, understanding $S_t(X)$ for sufficiently large $X$ will allow one to understand whether $t$ is congruent or not.

## Intuition and Methodology

There are four primary components to this result:

1. There is a bijection between primitive integer right triangles $T$ with
$\sqfree(T) = t$ and arithmetic progressions of squares $m^2 – tn^2, m^2, m^2 + tn^2$ (where each term is itself a square).
2. There is a bijection between primitive integer right triangles $T$ with
$\sqfree(T) = t$ and points on the elliptic curve $E_t(\mathbb{Q}): y^2 = x^3 – t x$ with $y \neq 0$.
3. If the triangle $T$ corresponds to a point $P$ on the curve $E_t$, then
the size of the hypotenuse of $T$ can be bounded below by $H(P)$, the
(naive) height of the point on the elliptic curve.
4. Néron (and perhaps Mordell, but I’m not quite fluent in the initial
history of the theory of elliptic curves) proved strong (upper) bounds on
the number of points on an elliptic curve up to a given height. (In fact,
they proved asymptotics which are much stronger than we use).

In this paper, we use $(1)$ to relate triangles $T$ to the sum $S_t(X)$ and we use $(2)$ to relate these triangles to points on the elliptic curve. Tracking the exact nature of the hypotenuses through these bijections allows us to relate the sum to certain points on elliptic curves. In order to facilitate the tracking of these hypotenuses, we phrase these bijections in slightly different ways than have appeared in the literature. By $(3)$ and $(4)$, we can bound the number and size of the hypotenuses which appear in terms of numbers of points on the elliptic curve up to a certain height. Intuitively this is why the higher the rank of the elliptic curve (corresponding roughly to the existence of many more points on the curve), the worse the error term in our asymptotic.

I would further conjecture that the error term in our asymptotic is essentially best-possible, even though we have thrown away some information in our proof.

We are not the first to note either the bijection between triangles $T$ and arithmetic progressions of squares or between triangles $T$ and points on a particular elliptic curve. The first is surely an ancient observation, but I don’t know who first considered the relation to elliptic curves. But it’s certain that this was a fundamental aspect in Tunnell’s famous work A Classical Diophantine Problem and Modular Forms of Weight 3/2 from 1983, where he used the properties of the elliptic curve $E_t$ to relate the CNP to the Birch and Swinnerton-Dyer Conjecture.

One statement following from the Birch and Swinnerton-Dyer conjecture (BSD) is that if an elliptic curve $E$ has rank $r$, then the $L$-function $L(s, E)$ has a zero of order $r$ at $1$. The relation between lots of points on the curve and the existence of a zero is intuitive from the approximate relation that

L(1, E) \approx \lim_{X} \prod_{p \leq X} \frac{p}{\#E(\mathbb{F}_p)},

so if $E$ has lots and lots of points then we should expect the multiplicands to be very small.

On the other hand, the elliptic curve $E_t: y^2 = x^3 – t^2 x$ has the interesting property that any point with $y \neq 0$ generates a free group of points on the curve. From the bijections alluded to above, a primitive right integer triangle $T$ with $\sqfree(T) = t$ corresponds to a point on $E_t$ with $y \neq 0$, and thus guarantees that there are lots of points on the curve. Tunnell showed that what I described as “lots of points” is actually enough points that $L(1, E)$ must be zero (assuming the relation between the rank of the curve and the value of $L(1, E)$ from BSD).

Tunnell proved that if BSD is true, then $L(1, E) = 0$ if and only if $n$ is a congruent number.

Yet for any elliptic curve we know how to compute $L(1, E)$ to guaranteed accuracy (for instance by using Dokchitser’s algorithm). Thus a corollary of Tunnell’s theorem is that BSD implies that there is an algorithm which can be used to determine definitively whether or not any particular integer $t$ is congruent.

This is the state of the art on the congruent number problem. Unfortunately, BSD (or even the somewhat weaker between BSD and mere nonzero rank of elliptic curves as is necessary for Tunnell’s result for the CNP) is quite far from being proven.

In this context, the main result of this paper is not as effective at actually determining whether a number is congruent or not. But it does have the benefit of not relying on any unknown conjecture.

And there is some potential follow-up questions. The sum $S_t(X)$ appears as an integral transform of the multiple Dirichlet series

\sum_{m,n} \frac{\tau(m-n)\tau(m)\tau(nt)\tau(m+n)}{m^s n^w}
\approx
\sum_{m,n} \frac{r_1(m-n)r_1(m)r_1(nt)r_1(m+n)}{m^s n^w},

where $r_1(n)$ is $1$ if $n = 0$ or $2$ if $n$ is a positive square, and $0$ otherwise. Then $r_1(n)$ appears as the Fourier coefficients of the half-integral weight standard theta function

\theta(z)
= \sum_{n \in \mathbb{Z}} e^{2 \pi i n^2 z}
= \sum_{n \geq 0} r_1(n) e^{2 \pi i n z},

and $S_t(X)$ is a shifted convolution sum coming from some products of modular forms related to $\theta(z)$.

It may be possible to gain further understanding of the behavior of $S_t(X)$ (and therefore the congruent number problem) by studying the shifted convolution as coming from theta functions.

I would guess that there is a deep relation to Tunnell’s analysis in his 1983 paper, as in some sense he constructs appropriate products of three theta functions and uses them centrally in his proof. But I do not understand this relationship well enough yet to know whether it is possible to deepen our understanding of the CNP, BSD, or Tunnell’s proof. That is something to explore in the future.