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?

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    $\begingroup$ I 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. $\endgroup$ Sep 25, 2018 at 22:09
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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 8:26
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    $\begingroup$ @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. $\endgroup$
    – tst
    Sep 26, 2018 at 11:55
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    $\begingroup$ @JackD'Aurizio In my opinion this kind of question is precisely what the site is for. $\endgroup$ Sep 28, 2018 at 8:20
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    $\begingroup$ @JackD'Aurizio Perhaps you want to take a look at this serious blog post: rjlipton.wordpress.com/2018/09/26/reading-into-atiyahs-proof . $T$ 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) $\endgroup$
    – Jose Brox
    Sep 28, 2018 at 18:35

1 Answer 1


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 $T$:

  1. $T : \mathbb{C} \to \mathbb{C}$ (this is in Section 3.4 of the longer paper "The Fine Structure Constant").

  2. $T$ is "real" (see 2.2 of the shorter paper "The Riemann Hypothesis" - it doesn't mean "real-valued").

  3. $T(1) = 1$ (2.3 of the shorter paper "The Riemann Hypothesis").

  4. 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).

  5. 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.): $$ T\Bigl( e^s(2-e^s))\Bigr) = 1. $$ Now notice that the function $e^s(2-e^s)\rvert_{\mathbb{R}}$ takes any value in $(-\infty,1)$. To see this claim you can solve $$ e^x(2-e^x) = y\ \Leftrightarrow e^{2x} - 2e^x + y = 0 $$ 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.

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    $\begingroup$ What I read about Todd multiplicative sequences (p.122) lets me wonder if $T$ is a function of formal power series instead of complex numbers $\endgroup$
    – reuns
    Sep 26, 2018 at 1:03
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    $\begingroup$ I 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. $\endgroup$
    – tst
    Sep 26, 2018 at 11:53
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    $\begingroup$ @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 $\endgroup$
    – SBK
    Sep 26, 2018 at 15:47
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    $\begingroup$ Unless I do a dumb mistake, property 4 implies that $T$ is a global polynomial. Indeed, pick any two (non-trivial) 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 $$P_1=P_2$$ $\endgroup$
    – N. S.
    Sep 27, 2018 at 0:33
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    $\begingroup$ I guess there are two different $T$, one is a function and one an operator, denoted by different braces: $T(\cdot)$ vs $T\{\cdot\}$. $\endgroup$
    – SampleTime
    Sep 29, 2018 at 8:34

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