In terms of purported proof of Atiyah's Riemann Hypothesis, my question is what is the Todd function that seems to be very important in the proof of Riemann's Hypothesis?

1https://mathoverflow.net/questions/311280/whatisthedefinitionofthefunctiontusedinatiyahsattemptedproofofthe – Randall Sep 25 '18 at 20:29

8I guess this had to come. I would be tempted to close the question to preserve Sir Micheal Atiyah from further embarrassment. On the other hand someone has to stress that, sadly, his alleged proof of RH makes no sense at all. – Jack D'Aurizio Sep 25 '18 at 22:09

4I 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 08:26

3@josebrox I think he needs to define precisely what he means by weak limit. It is a critical part of the proof and he cannot assume that people know what he is talking about. – tst Sep 26 '18 at 11:55

@tst Thanks. Yes, I'm already addressing that. – Jose Brox Sep 26 '18 at 12:44

4@JackD'Aurizio In my opinion this kind of question is precisely what the site is for. – samerivertwice Sep 28 '18 at 08:20

@JoseBrox: even if $T$ makes some sense (which is not the case, in my opinion), "proving" RH without invoking any particular property of the $\zeta(s)$ function (like Euler's product for $\text{Re}(s)>1$ or the reflection formula) is highly suspicious. – Jack D'Aurizio Sep 28 '18 at 17:49

3@JackD'Aurizio Perhaps you want to take a look at this serious blog post: https://rjlipton.wordpress.com/2018/09/26/readingintoatiyahsproof/ . $T$ may make sense, but not as a complex analytic function, of course. But that is not what Atiyah has in mind, since he says so himself (personal communication). Some are trying to make sense of his definitions and terms (which in my opinion are strange in part because they are like coming from the 30's  50's of the past century, not from their usual usage right now) – Jose Brox Sep 28 '18 at 18:35

It is told that Todd dunction has weak analytic Continuation ... any comment on that ? – Anurag Nov 26 '20 at 04:31
1 Answers
While there might be an interesting question here about the older math that Atiyah references, it is worth pointing out what Atiyah actually says about the function $T$:
$T : \mathbb{C} \to \mathbb{C}$ (this is in Section 3.4 of the longer paper "The Fine Structure Constant").
$T$ is "real" (see 2.2 of the shorter paper "The Riemann Hypothesis"  it doesn't mean "realvalued").
$T(1) = 1$ (2.3 of the shorter paper "The Riemann Hypothesis").
On any compact, convex set $K$, $T$ is a polynomial of some degree and the degree is in principle allowed to depend on the set $K$ (this is in the start of Section 2).
If $f$ and $g$ are power series with no constant term then $$ T\Bigl( 1 + f(s) + g(s) + f(s)g(s)\Bigr) = T\Bigl(1 + f(s) + g(s)\Bigr) $$ (this is 2.6 of shorter paper).
The following proof is from a Redditor : Set $f(s) = e^s  1$ and $g(s) = 1  e^s$. Point 5. then implies that $$ T\Bigl( 1 + e^s  1 + 1  e^s + (e^s  1)(1  e^s)\Bigr) = T(1) $$ i.e.(using 3.): $$ T\Bigl( e^s(2e^s))\Bigr) = 1. $$ Now notice that the function $e^s(2e^s)\rvert_{\mathbb{R}}$ takes any value in $(\infty,1)$. To see this claim you can solve $$ e^x(2e^x) = y\ \Leftrightarrow e^{2x}  2e^x + y = 0 $$ by using the quadratic formula and taking logarithms to get a real solution when $y < 1$. This shows that $T\rvert_{(\infty,1)}$ is constant.
OK so now take a compact, convex set $K \subset \{ \mathrm{Re}(z) < 1\}$ that contains an interval on the real line, i.e. $K$ contains a subinterval $I$ of $(\infty,1)$. From the properties 2. and 4. we know that $T\rvert_K$ is a polynomial with real coefficients. But we also know it is constant on $I$ which means $T \rvert_K$ is constant.
Since $K$ was arbitrary we can easily exhaust$ \{ \mathrm{Re}(z) < 1\}$ by compact, convex sets to show that $T$ is constant on $\{ \mathrm{Re}(z) < 1\}$, in particular, this includes the critical strip.

3What I read about [Todd multiplicative sequences](http://120.27.100.167/uploads/soft/all/18729.pdf) (p.122) lets me wonder if $T$ is a function of formal power series instead of complex numbers – reuns Sep 26 '18 at 01:03

When he said T is real, he had T(s*)=(T(s))* which is more general but does mean that real values get mapped to real values. – Kelly Lowder Sep 26 '18 at 04:21

@reuns I don't know but Atiyah clearly says it is from C to C in his paper, so it's unnecessary to give him the benefit of the doubt. – T_M Sep 26 '18 at 04:45

Why not then look at compact convex sets that intersect $\{Re(z)\lt 1\}$ and exhaust the right half of the plane? – theHigherGeometer Sep 26 '18 at 05:12

15I think property 4 by itself kills the proof completely. If a complex function is a polynomial in an open set, it is just a polynomial. There is no way out of this. – tst Sep 26 '18 at 11:53

1@DavidRoberts Yes I think you're right you could add this step. I guess I had in mind that Re < 1 takes care of the critical strip – T_M Sep 26 '18 at 15:47


@DavidRoberts well, that's a tricky bit. This function T may not even be a function at all if we just look at weak limits. If we want to make it a function I cannot see how we can avoid having it constant. – tst Sep 26 '18 at 16:42


@T_M, my previous comment was supposed to be to you, but I got the username wrong :/ Anyway, my point is that the weak limit does not necessarily give you a function (assuming it makes sense at all). Now if we assume it is a function, then we already assume more regularity than what the definition implies. But if we don't then I don't understand what is the meaning of $T(1+g(z))$. – tst Sep 27 '18 at 00:19

5Unless I do a dumb mistake, property 4 implies that $T$ is a global polynomial. Indeed, pick any two (nontrivial) compact convex sets $K_1,K_2$. Then $T=P_1$ on $K_1$ and $T=P_2$ on $K_2$ for some polynomials. I claim that $P_1=P_2$. Indeed, pick a compact convex set $K$ such that $K_{1,2} \subset K$ (for example $K=$ convex hull of $(K_1,K_2)$. Then $T=P$ polynomial on $K$. Next, since $K_1 \subset K$ you have $P_1=P$ on $K_1$, and hence, since $K_1$ is infinite, $P_1=P$. Same was $P_2=P$, and hence $$P_1=P_2$$ – N. S. Sep 27 '18 at 00:33

@N.S. Yes, this basically means that on any neighborhood of any number in the complex plane the function $T$ is the same polynomial. And then it gets worse. Using (5) we have $T((1+z)^2)=T(1+2z)$. Then we need a disk in which the first derivative of $T$ does not vanish. This implies that we can construct a local inverse. This can be done easily since $T$ is a polynomial everywhere. The by using the inverse we get $z^2=0$ for every $z$ in the disk. :/ – tst Sep 27 '18 at 02:16

1@tst $$T((1+z)^2)=T(1+2z) \Rightarrow \deg(T((1+z)^2))=\deg T(1+2z) \Rightarrow 2 \deg(T)= \deg(T) \Rightarrow T \mbox{ is constant}$$ – N. S. Sep 27 '18 at 02:33

@T_M But he also states in the following remark of 2.6 that this formula is just a linear approximation of the power series expansion of $T$ near the origin? So maybe this argument cannot be used for the original $T$? – SampleTime Sep 27 '18 at 17:44

@reuns There are field and ring properties strangely mixed in the definition of $T$ and its use for the alleged proof that invalidate it. – bonif Sep 28 '18 at 23:42

3I guess there are two different $T$, one is a function and one an operator, denoted by different braces: $T(\cdot)$ vs $T\{\cdot\}$. – SampleTime Sep 29 '18 at 08:34