What is the definition of the function T used in Atiyah's attempted proof of the Riemann Hypothesis?

Question

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?

https://meta.mathoverflow.net/questions/3894/is-there-a-way-to-discuss-the-correctness-of-the-proof-of-the-rh-by-atiyah-in-mo — mme, Sep 24 '18 at 04:42
Can I ask why you want to know the definition? The consensus on the claim is settled, see the meta link in Mike's comment. — David Roberts, Sep 24 '18 at 04:52
Why do people ask why OP wants to know this? Does it matter? It's a good mathematical question. — Manuel Bärenz, Sep 24 '18 at 07:48
It's a question that many people have wanted to know since the papers got leaked, probably out of morbid interest. The other paper the OP seeks is linked to on Reddit, which is linked to from the meta question. — David Roberts, Sep 24 '18 at 07:51
@DavidRoberts, that's just an assumption. And links are not answers. Someone who understands weakly analytic functions should just step up and define $T(s)$. — Manuel Bärenz, Sep 24 '18 at 07:57
@ManuelBärenz You are presuming that the notion of "weakly analytic" in that document is well-defined and makes sense. Certainly the version stated there does not. Whether or not this is merely a defect of exposition or anything more, is probably not for me to make snap judgments on — Yemon Choi, Sep 24 '18 at 07:59
@YemonChoi, I had no idea it wasn't well defined, thanks for this insight. I'm not active in this field. But I'd still like to learn. It would be incredibly valuable if someone could point out how it's not well-defined, and whether that issue could be salvaged. — Manuel Bärenz, Sep 24 '18 at 08:02
@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... — David Roberts, Sep 24 '18 at 08:05
@DavidRoberts every factor (in the sense of von Neumann algebras) has trivial centre by definition, and the hyperfinite II-1 has a unique trace if one requires the trace to have certain reasonable properties. But I think I am starting to bear out your original point, which (apologies if I have misread) is that the documents provided simply aren't worth speculating about until some of the dust has settled, because as exposition they really are not adequate, even before we start trying to decide "is this a proof of RH or not?" — Yemon Choi, Sep 24 '18 at 08:09
@YemonChoi thanks, I thought something was odd (there's discussion of trace there, but then $T$ doesn't use the trace). Also: eigenvalues can be at least as bad as merely piecewise $C^1$ for a 2-parameter family of 2x2 matrices, so it's not going to help when one is playing with analytic functions. — David Roberts, Sep 24 '18 at 08:11
@ManuelBärenz It is difficult to put this diplomatically, but when you express a belief that " I'd still like to learn. It would be incredibly valuable if..." is putting quite a lot of implicit faith in there being something extractable from these documents. Based on my first impressions, I certainly won't be the one to try and do it — Yemon Choi, Sep 24 '18 at 08:13
@DavidRoberts Gah, since I have started I may as well say: for an operator algebraist (or even someone like me who just drinks beer with operator algebraists) a big red light comes on when anyone starts talking about flow of weights in the context of a finite von Neumann algebra - weights are what you do for infinite von Neumann algebras, in the absence of finite traces. The $\alpha$ preprint seems to be using the words "flow of weights" to mean something different from what the words usually mean — Yemon Choi, Sep 24 '18 at 08:24
@YemonChoi already hearing that probably most constructions there are beyond hope is valuable, because then we know that there is not much to learn there. Sometimes (as in Wiles' first proof attempt to Fermat's theorem) very interesting mathematics can be extracted from a failed proof attempt. — Manuel Bärenz, Sep 24 '18 at 08:25
@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 $\tau\circ\phi=\tau$. — Yemon Choi, Sep 24 '18 at 08:30
To reiterate Yemon Choi's comments, and hopefully to prevent further misinformation, please see subsection 2.1 of the preprint here: https://drive.google.com/file/d/1WPsVhtBQmdgQl25_evlGQ1mmTQE0Ww4a/view. The $\mathbb{C}$--linear trace on the hyperfinite $II_{1}$ factor (with center $\mathbb{C}$) is unique, so if $\varphi$ is any automorphism, $\tau \circ \varphi$ is a trace and so must equal $\tau$. Hence the underlying idea needed to give sense to $T$ is shaky. — Jon Bannon, Sep 24 '18 at 16:55
What is also a little confusing is that supposedly $T$ is a weakly analytic function from $\mathbb{C}$ to $\mathbb{C}$, but according to Weyl's lemma, every weakly analytic function is analytic. Furthermore, Atiyah seems to say that $T$ is polynomial on every compact set, but that directly implies, even without Weyl's lemma, that $T$ is just a polynomial. — Lukas Geyer, Sep 24 '18 at 20:56
@LukasGeyer Are people agreed then that T is identically equal to 1? Which would make the "F" of Section 3 identically equal to zero, without any contradiction. In particular 2.4 does not hold (it clearly doesn't follow from the parenthetical remark, which allows still allows T being identically 1) — SBK, Sep 24 '18 at 22:04
https://www.youtube.com/watch?v=jXugkzFW5qY&feature=youtu.be&t=1476 — Count Iblis, Sep 24 '18 at 22:10
@LukasGeyer Not an expert, but does not the analytic and polynomial problem go away if the domain of the function is not all $\mathbb{C}$? That is what happens with the example by analogy that Atiyah gives: the step function over the real line (constant for each of the compact sets contained in its domain, yet not constant) — Jose Brox, Sep 25 '18 at 08:37
Something I don't understand: didn't Atiyah say that the Todd function was already defined in some older paper? Supposing it is indeed ill-defined or useless, is that extendable to said paper, is it that Atiyah has changed the definition...? — Jose Brox, Sep 25 '18 at 09:08
@Jose T is supposed to be an isomorphism of C with itself... — David Roberts, Sep 25 '18 at 12:19
Step functions aren't constant...therefore if I take an interval that contains the step, its clearly not constant on that interval. I think people should be able to agree that there are numerous errors surrounding the definition of this function T. — SBK, Sep 25 '18 at 13:10
@T_M What I meant is that if you take the step function as nondefined in the origin and consider only compact sets inside its domain, then Atiyah's claim about the step function is true (and it seems to me like the natural way of looking at the claim) — Jose Brox, Sep 25 '18 at 13:23
@JoseBrox: I said it is confusing (not that it is wrong) particularly because Atiyah is very unclear about where $T$ exactly is defined and what it is. Also, his definition of weak analyticity seems to differ from the standard one (distribution with locally integrable derivatives for which Cauchy-Riemann holds) where the Weyl lemma applies. I am not sure if there is a way to make Atiyah's idea of the function $T$ rigorous, but it would need some more details. — Lukas Geyer, Sep 25 '18 at 16:22
@JoseBrox Isn't it clear that you are contorting to make what he says true? i.e. if this were from an unknown mathematician you would just say that it was obviously nonsense. Re: the step function He clearly says "any closed interval K on the line". And re: the Todd function itself, he says in the longer paper its domain is C and then he claims its a real-valued polynomial on any ball and that T(1) = 1. This is obviously just the constant function. — SBK, Sep 25 '18 at 20:28
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 — Jose Brox, Sep 26 '18 at 08:18
My earlier critique of the argument regarding the uniqueness of the trace is too blunt. The point is that two realizations of the hyperfinite II_1 factor, with explicitly calculable traces that are a priori different, are actually equal via an isomorphism. This provides identities. This is the meaning of "isomorphic traces". — Jon Bannon, Sep 27 '18 at 14:52
https://rjlipton.wordpress.com is a further place to look for a serious discussion of the issues here, and trying to unpack some of the possible ambiguities — Mark Bennet, Sep 27 '18 at 18:43
@JoseBrox While you're at it, could you ask him to check the formula (7.1) for correctness? Over on reddit, several people tried to calculate the sum, and, instead of slowly converging to 137*0.577/3.14 ~ 25, it rapidly converges to about 0.16. — user11325, Sep 28 '18 at 05:27
@user11325 Since I don't know how much of Atiyah's attention I can hope to get (I don't know if he'll even answer my second email), for the moment being I prefer to stick to essential things. My opinion on those computations: the Todd function is needed to do them and we are all having a hard time trying to understand what its definition is, so I doubt them; it is clearly not "just a complex function", and even when Atiyah "does the computation" himself in the paper, he already starts from a value close to the actual one,so this could simply be a matter of local convergence to different points — Jose Brox, Sep 28 '18 at 18:42
@JoseBrox The problem is that 7.1 does not need a starting value. It is a straightforward infinite sum with real-valued terms that can be calculated exactly. And 8.1-8.5, defining the alpha, don't need a starting value of ж either. Todd function is only used to map standard formulas for gamma and pi into 7.1 and 8.1-8.5. So the question of what he means by "specify the initial data" (p.8) / "start from ж=..." (pp.13-14) is rather essential to understanding that paper. — user11325, Sep 29 '18 at 20:20

Wilem2 · Answer 1 · 2020-01-11T03:28:31.583

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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.

edited Jan 11 '20 at 03:28

answered Sep 24 '18 at 11:04

Wilem2

452

23

you might want to check the comments by David Roberts in the OP for why this is not really a "definition" – Carlo Beenakker Sep 24 '18 at 11:11
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Moreover, someone has shown that (some of) the properties assumed of $T$ in the RH notes make it the constant function at $1$, certainly not an isomorphism as 'defined' in the fine structure constant notes. – David Roberts Sep 25 '18 at 07:30
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@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 – Jose Brox Sep 25 '18 at 08:33
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@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. – David Roberts Sep 25 '18 at 21:46
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If anyone is interested I checked again and wrote up the Redditor's proof on Math.SE as an answer to this question https://math.stackexchange.com/questions/2930742/what-is-the-todds-function-in-atiyahs-paper/2930796#2930796 – SBK Sep 26 '18 at 00:12
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@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. – David Roberts Sep 26 '18 at 05:07

score 1 · Answer 2 · answered Nov 27 '23 at 22:03

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:

We start with some $v\in \mathbb C$
Then write $v=\sum p_iz^i$ in some specific way
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'.

What is the definition of the function T used in Atiyah's attempted proof of the Riemann Hypothesis?

2 Answers2