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In Michael Atiyah's paper purportedly proving the Riemann hypothesis, he relies heavily on the properties of a certain function $T(s)$, known as the Todd function. My question is, what is the definition of $T(s)$?

Atiyah states that this function is defined in his paper "The Fine Structure Constant", but I can't seem to find a copy of the paper. So can anyone tell me how Atiyah defined $T$ in that paper?

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    $\begingroup$ meta.mathoverflow.net/questions/3894/… $\endgroup$
    – mme
    Sep 24, 2018 at 4:42
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    $\begingroup$ Why do people ask why OP wants to know this? Does it matter? It's a good mathematical question. $\endgroup$ Sep 24, 2018 at 7:48
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    $\begingroup$ @ManuelBärenz I was trying to avoid discussion, but... T is defined as a composite of isomorphisms $\mathbb{C} \stackrel{t_+}{\to} Z(A) \stackrel{t_-}{\to}\mathbb{C}$ where $Z(A)$ is apparently the centre of the hyperfinite type II von Neumann factor, and each $t_\pm$ is induced (somehow, it's not clear) by the map sending a 2x2 complex matrix to its eigenvalues (and recalling that $A$ is an infinite tensor product of such 2x2 matrix algebras). I'm not sure these maps $t_\pm$ are well-defined, or if only $T$ is supposed to be, but even then I'm suspicious. I'm not sure it's continuous, even... $\endgroup$
    – David Roberts
    Sep 24, 2018 at 8:05
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    $\begingroup$ @ManuelBärenz I hesitate to say "most constructions" but I can say (perhaps demonstrating less tact than others have been doing) that there are problems even in the explanations of how one is supposed to get started. Here's another instance from the $\alpha$ preprint: it is claimed that a finite von Neumann algebras always has a trace (true) and then it is claimed that inner automorphisms give different but isomorphic traces. However, if $\tau$ is a trace on any algebra and $\phi$ is an inner automorphism then one will find rather quickly that $\tau\circ\phi=\tau$. $\endgroup$
    – Yemon Choi
    Sep 24, 2018 at 8:30
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    $\begingroup$ I just decided to email Atiyah asking for clarifications, and he has answered. If I figure something worthy out of the conversation, I will post it here (of course, since I'm not an expert in analysis, I may fail to understand subtle ideas). For starters, the preprints are from him (although he didn't know they had leaked, and is going to circulate a paper), and address the "T would be constant" issue: since it is defined as a weak limit (which is not unique), it has no analytic continuation. It is uniquely determined by Hirzebruch theory. If you want to help me, write to josebrox at mat.uc.pt $\endgroup$
    – Jose Brox
    Sep 26, 2018 at 8:18

2 Answers 2

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Here's a public paper of the "the fine structure constant" by Atiyah.

It doesn't seem to be the original, but a copy: https://drive.google.com/file/d/1WPsVhtBQmdgQl25_evlGQ1mmTQE0Ww4a/view

See the section 3.4, the Todd function is defined there.

[Edit] As pointed by @T_M in the comments, here you would find the main properties of the T function.

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    $\begingroup$ you might want to check the comments by David Roberts in the OP for why this is not really a "definition" $\endgroup$ Sep 24, 2018 at 11:11
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    $\begingroup$ @DavidRoberts As I understand that particular issue, from the step function analogy given in the preprint, I think that Atiyah is asking for analiticity on every compact set of its domain, not all compact sets of $\mathbb{C}$. I don't know if the analogy is good enough, but the step function is indeed polynomial in each of the compact sets contained in its domain, without being polynomial $\endgroup$
    – Jose Brox
    Sep 25, 2018 at 8:33
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    $\begingroup$ @Jose to reiterate, the function T is defined in the fine structure constant paper as being as isomorphism from C to itself, hence the domain is all of C. $\endgroup$
    – David Roberts
    Sep 25, 2018 at 21:46
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    $\begingroup$ If anyone is interested I checked again and wrote up the Redditor's proof on Math.SE as an answer to this question math.stackexchange.com/questions/2930742/… $\endgroup$
    – SBK
    Sep 26, 2018 at 0:12
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    $\begingroup$ @T_M would you add an answer here linking to your answer on M.SE? Then the OP can accept it and we can all do something else. $\endgroup$
    – David Roberts
    Sep 26, 2018 at 5:07
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In his article "The Fine Structure Constant", Atiyah says the following about T on page 6:

For the case of inverse-integer weights, Hirzebruch formalized the notion of exponential maps and found the most general solutions 1-Chapter 3. His motivation came from the behaviour of certain manifold invariants, in algebraic geometry, under taking Cartesian products. But the problem was purely formal and applied to all manifolds, giving topological invariants [especially after Atiyah-Singer index theory]. He showed that the basic example was the Todd genus, named after J.A. Todd, whose generating function, due to Bernoulli, is $\frac{x}{ 1−e^{-x}}$, already exploited by Euler. It is crucial that this function, which underpins Hirzebruch’s formalism, is analytic in the closed interval [0, 1/2] (see section 6). This ensures that the normalized trace of 4.7 extends to the weak closure A. Following Hirzebruch, we have named it after Todd and denoted it by T .

Reference 1 is the book "Topological methods in algebraic geometry (with appendices by R.L.E.Schwarzenberger, and A.Borel)" by F. Hirzebruch. Here is a link to the book.

I don't understand the definition of Todd genus in chapter 3. However, in the first section of the book it defines a multiplicative sequence $T$. Such a sequence can be viewed as a function $T:\mathbb C\times \mathbb C^{\infty}\rightarrow \mathbb C$. The definition is then: $$T(z,p_1,p_2,\dots)=\sum_{i=0}^{\infty} T_i(p_1,\dots,p_i)\cdot z^i$$ or $$T(x,c_1,c_2,\dots)=\sum_{i=0}^{\infty} T_i(c_1,\dots,c_i)\cdot x^i.$$ Here the $T_i$ are the Todd polynomials. This $T$ might be related to the $T$ that Atiyah is using.

At page 9 of the book, it states about general multiplicative sequences $K$:

In abbreviated notation we write $$K(\sum_{i=0}^{\infty} p_iz^i)=\sum_{i=0}^{\infty} K_j(p_1,\dots,p_j)z^j$$

In this notation, it seems as if $K$ is a function from $\mathbb C$ to $\mathbb C$, but that is not technically accurate. The left-hand side should actually be interpreted as $K(z,p_1,p_2,\dots)$.

This might be the reason that Atiyah presents $T$ as a function from $\mathbb C$ to $\mathbb C$. That could also be 'abbreviated notation'. Maybe, this is what actually happens:

  1. We start with some $v\in \mathbb C$
  2. Then write $v=\sum p_iz^i$ in some specific way
  3. Now define $T(v)$ as $T(z,p_1,p_2\dots)$ for these specific values of $z,p_1,p_2,\dots$

However, I do not know how the $z,p_1,p_2,\dots$ have to be chosen. There is an infinite number of way to get the equation $v=\sum p_iz^i$.

If Atiyah was really on to something, then I think his articles contain a great lot of abbreviated notation. Therefore, most logical steps in his article are incomprehensible to us 'mere mortals'.

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