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

1 Answer 1


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/… $\endgroup$
    – T_M
    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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