Are business schools intellectually bankrupt? (Part Two)

[I just found this draft of a blog post from 2017, I thought it might be a light diversion.]

There are essentially three types of people who claim proofs of the Riemann Hypothesis.

First, there are the cranks. The crank often writes something which is utterly incoherent — possibly invoking something from physics, possibly also simultaneously proving RH and Fermat’s Last Theorem at the same time. Alternatively, there are cranks who have some basic knowledge of mathematical formalism, and manage to scribble down something which at least shares some of the grammatical structure of a mathematical argument. My mental image of a crank used to be a retired 60 year old engineer, but, like so many other things, the internet has expanded my horizons, and nowadays cranks come in many different flavours.

Second, there are the amateurs with a sufficient amount of hubris that they somehow believe they can make a contribution a problem which — after a century of study — is clearly very deep and almost everyone believes will require fundamentally new ideas to crack. These people are often successful in their own careers — including Nobel prize winners in Chemistry (or was that a proof of Fermat?). This appears to be related to a (no doubt well-known) variation of the Dunning–Kruger effect — people with a high level of competence in one domain mistakenly overrate their competence in another (in this case number theory). This is usually a very bad mistake when it comes to higher mathematics. (Although don’t imagine that the shoe is never on the other foot — try to imagine for a moment otherwise smart mathematicians pontificating about biology.)

Third, there is the mentally ill. (Obviously there is often a non-trivial intersection between some or all of these classes.)

It is really not worth my while bothering to look at any purported proof of RH — it’s fairly clear that attempting to interact with any character of type one, two, or three above is not really worth one’s time. However, recent circumstances have unfortunately brought me to do so. It began when a probabilist was asked to review a short paper for MathSciNet in a Brazilian probability journal. The reviewer then noticed that the main claim of the paper was a proof of the Riemann hypothesis. Naturally he was confused! The author of the paper was a professor at the University of Chicago. Not of the mathematics department, of course, nor of the statistics department, but of the business school! (For more on business schools, see this post.) The reviewer decided to consult Reddit, who suggested forwarding it to me:

Seeing as the author is at U Chicago (but not in the math department), just email Calegari and inform him that a “colleague” of his has somehow managed to get a proof of RH published in a low-tier probability journal. He’ll deal with it (or get someone to), that’s a professional embarrassment for U Chicago.

Well, of course, I do whatever Reddit tells me to do (/s)

I place the argument — to the extent that I can tell — in the second class. Half an hour of study was sufficient to determine a hole big enough to drive a lorry through. It was the type of mistake that I might have made (but didn’t) when I was 15. There’s a certain amount of probabilistic window dressing in the paper which is syntactically related to real mathematics but tangential to RH. Once this window dressing is removed, the argument is essentially as follows:

1. Take the function \( 1/\zeta(s)\), then take its inverse Mellin transform.
2. Assume without comment properties of this inverse Mellin transform which require RH.
3. Deduce RH by considering the Mellin transform again.

In fact, the good news is that the argument can be upgraded in a smaller number of pages so that it proves that all the zeros of the function \((1/4-s)(3/4-s)\) lie on the critical line. Now that would be a spectacular result!

Let’s prove it! We can start at around equation (2.35) in the paper, where the function

\[ \displaystyle{G(x) = \sum_{k=1}^{\infty} \frac{ (-1)^{k+1}}{\Gamma(k)} \frac{\xi(1/2) \xi(k + 3/2)}{\xi(k + 1/2)} x^k,}\]

is defined, with corresponding Laplace transform

\[\displaystyle{m_G(s) = \int_{0}^{\infty} x^{s-1} G(x) dx.}\]

Here \(\xi(s)\) is the part of the zeta function coming from the Hadamard factorization (i.e. without the Gamma factors). Now let’s imagine instead that \(\xi(s)\) is just the function \(3/4 – s\).
A standard calcuation (for either the \(\xi\) coming from Riemann or from \(3/4 – s)\) gives

\[ \displaystyle{m_{G}(s) = \int_{0}^{\infty} x^{s-1} G(x) dx = \frac{\xi(3/2 – s) \Gamma(1+s) \xi(1/2)}{\xi(1/2 – s)}}.\]

In particular, this is a purely “formal” calculation (Ramanujan Master Theorem Style). The crux of the argument, the part where something “gets done” and one gets access to \(\xi\) in the critical strip is where he “eliminates” \(m_{G}(s)\) by using the identity

\[\displaystyle{
\frac{1}{\xi(1/2 – s)} = \frac{m_G(s)}{\xi(1/2)\Gamma(1+s) \xi(3/2 – s)}.}\]

He really wants to use this identity (see the line before (2.45), where variables have been changed slightly) in the range

\[ \displaystyle{s \in (-1/2,0).}\]

After all, he wants to understand \(\xi\) — even if just on the real line — in the critical strip. In order to get this, you need to know something about the growth of \(G(x)\) as \(x\) goes to infinity, because you want the Mellin transform to be well defined. In order to get convergence for real negative s close to zero (which he uses), you certainly want to assume that

\[\displaystyle{G(x) = O(x^{\epsilon})}.\]

So let’s assume exactly this. And now let me prove that \(3/4 – s\) has no zeroes for \(1/2 \le \mathrm{Re}(s)\le 1\) which is nonsense. From our equation above, \(m_G(s)\) is now well defined for \(\mathrm{Re}(s)\) in \((-1/2,0)\). But then the RHS is well defined in this range, so the LHS has no poles, so \(\xi(1/2 – s)\) has no zeroes for \(-1/2 \le \mathrm{Re}(s) \le 0\), or \(\xi(s)\) has no zeroes for \(1/2 \le \mathrm{Re}(s) \le 1\).. Done!

So what is wrong with this argument? Obviously one actually has to understand the growth of \(G(x)\). In the case of \(\xi(s) = r – s\), one can compute \(G(x)\) explicitly, and at least away from \(1/2\) where symmetry forces \(G(x)\) to vanish one can compute it explicitly in terms of incomplete Gamma functions and get

\[\displaystyle{G(x) \sim x^{\mathrm{Re}(s) – 1/2}}.\]

Of course, this is no surprise, since this exactly eliminates the contradiction. So now let us return to the paper. We have the function \(G(x)\), and we need to say something
about its growth at infinity. In order to prove RH we need to show that it grows slower than any power of \(x\). So what is going to happen? Well, we are going to have that

\[\displaystyle{G(x) = O(x^{\rho – 1/2 + \epsilon})},\]

where \(\rho\) is the supremum over the real parts of all the non-trivial zeros of \(\zeta(s)\). So, in order to prove RH, one only needs to prove … the Riemann Hypothesis!

Note that understanding the growth of functions like \(G(x)\) and their link to RH is not new. In fact, already over 100 years ago, Riesz proves the following. Let

\[ \displaystyle{ F(x) = \sum_{n=1}^{\infty} (-1)^n \frac{x^n}{\zeta(2n) \Gamma(n)}.}\]

Then

\[ \displaystyle{F(x) = O(x^{1/4 + \epsilon})}\]

is equivalent to RH. What I have sketched above is basically already a moral explanation of this argument — poles of a function imply growth of the inverse Mellin transform and vice versa. Indeed, Grosswald (in the paper cited by Polson!) proves that the rate of growth (up to \(\epsilon\)) of \(F(x)\) is exactly \(x^{\theta/2}\) where \(\theta\) is the supremum of the real part of zeros of \(\zeta(s)\).

Well, now at least we know the answer to “what does it take to get you to look at my proof of RH?” The answer: you have to have tenure at Chicago, and you have to have a published proof of RH.

One month later: This, at least, was the original story as of a month ago. But there was a twist. Greg Lawler and I actually contacted the author of this paper. Communications via email were not particularly successful. He actually produced a second purported proof (!?) which was worst than the first — basically writing down integrals for a complex parameter s related to \(1/\zeta(s)\) paying no attention to the domains of applicability, and then using (in effect) precisely facts about convergence of these integrals which require being careful about the domain of applicability. But then we met in person, and he was very polite, and seemed to realize that both approaches were flawed. The original published paper was retracted by the author, and balance was restored. Success! Or at least I thought so, until I just found out that he recently updated his second paper with more of the same claims! (A little effort — more than it is worth — shows that simply by changing some 1/2s to 1/3rds or 2/3rds one can prove that \(\zeta(s)\) doesn’t have any zeroes at all.)

So where does this lead us with respect to the question in the title? I guess in the context where producing banal observations about human behavior that have been repeatedly observed by others gets you a (not really a) Nobel Prize, paying someone $400000 a year (Note: guesstimate) to produce quisquilian proofs of the Riemann Hypothesis sounds perfectly sane. C’est la vie.

Nine years later: Apparently Polson is back in the news for authoring 258 papers in 2026. Well, I guess when you can prove the RH, nothing is beyond you!