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Jan 11, 2020 at 3:28 history edited Wilem2 CC BY-SA 4.0
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Sep 26, 2018 at 5:07 comment added David Roberts @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.
Sep 26, 2018 at 0:12 comment added SBK 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/…
Sep 25, 2018 at 21:46 comment added David Roberts @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.
Sep 25, 2018 at 8:33 comment added Jose Brox @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
Sep 25, 2018 at 7:30 comment added David Roberts 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.
Sep 24, 2018 at 11:11 comment added Carlo Beenakker you might want to check the comments by David Roberts in the OP for why this is not really a "definition"
Sep 24, 2018 at 11:04 history answered Wilem2 CC BY-SA 4.0