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Sep 26, 2019 at 22:52 comment added Simon Henry One can also take this backward: the Notion of C* algebra exists and is so nice essentially because of this coincidence (which by itself is just as you say an example of the law of small number). An evidence for this is that while one can perfectly do Galois theory over p-adic field and p-adic functional analysis with no problem, there are no really nice $p$-adic analogue of the notion C*-algebra. (And I know several people, me included, that have looked for one). I feel like your question is closely related to that problem by the way...
Sep 26, 2019 at 21:35 comment added Yemon Choi This may be slightly orthogonal to what you're seeking, but I just want to say that the depth/importance of Cstar algebras, if one believes in such, is much much much much much much much more to do with the Cstar condition on the norm. Banach star-algebras, let alone ${\mathbb C}$-algebras with involution, have far fewer of the magical properties of ${\rm C}^*$-algebras
Sep 26, 2019 at 19:33 comment added Dmitry Vaintrob Another vague answer is that the E2 operad is canonically a Z/2-equivariant operad via complex conjugation, which a very natural structure to consider for "real homotopy theory" reasons (for example the configuration spaces of n points in the plane are complex points of varieties defined over Q). The (non-equivariant) E1 operad is the fixed point sub-operad. Combining this conjucation equivariance structure with the natural cyclic Z/2 action on E2 should give a Z/2 equivariant E1 operad as some sort of twisted fixed points.
Sep 26, 2019 at 17:47 comment added Noah Snyder You might find Theo’s paper interesting, though I’m not sure it really conclusively gives an answer to your question it has a lot of interesting things to say about it: arxiv.org/abs/1507.06297
Sep 26, 2019 at 17:26 comment added Tim Campion @DenisNardin Thanks! Would you also argue that there's not really a "reason why" $\mathbb C$-algebras with involution are so useful?
Sep 26, 2019 at 17:17 comment added Denis Nardin The Artin-Schreier theorem is actually even stronger. It says that if an algebraically closed field $F$ has a finite index subfield $R$, then $R$ must be a real closed field and $F=R[\sqrt{-1}]$. In particular $\mathrm{char}\,F=0$. That said I would argue that this is a case of the law of small numbers.
Sep 26, 2019 at 16:56 history edited Tim Campion CC BY-SA 4.0
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Sep 26, 2019 at 16:48 history asked Tim Campion CC BY-SA 4.0