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Timeline for What are "perfectoid spaces"?

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Oct 15, 2022 at 15:06 comment added Steffen Jaeschke Everything on the p-adics in is difficult to visualize. Over at multicomputational-irreducibility is an attempt to do this with the p-adics-metric. So the $ln$ shall provide a cursory glimpse.
Oct 15, 2022 at 8:21 comment added Steffen Jaeschke By studying the definition of p-adical convergence I found, Converging sequences of p-adic Galois representations and density theorems in [limits of p adic representations] by J. Bellaiche, G. Chenevier, C. Khare, M. Larsen in (mathoverflow.net/questions/69080/…). I ask whether this is towards operational applicability. The text asks for exists. Ideas about existence: p-adic numbers vs real numbers,
Oct 11, 2022 at 17:42 comment added Peter Scholze @SteffenJaeschke The limit should be taken $p$-adically, not in the usual archimedean metric.
Oct 1, 2022 at 20:01 comment added Steffen Jaeschke I made the effort in calculating what is the limit n to infinity following "In fact it is not difficult to see that. My result is that this limit is $1$ if $ln[p]<0$. This is divergent for $p>1$. I used for reference Mathematica.
Oct 23, 2017 at 19:12 history edited Arrow CC BY-SA 3.0
Introduced \varprojlim and added accents to 'etale'.
S Oct 21, 2017 at 17:39 history suggested Glorfindel CC BY-SA 3.0
typos corrected
Oct 21, 2017 at 17:21 review Suggested edits
S Oct 21, 2017 at 17:39
Apr 19, 2016 at 3:46 comment added S. Li you said 'At least on differentials Ω1Ω1, one can believe this'... I was wondering given perfectoid space over char 0 perfectoid field, is there a notion of sheaf of (relative) differential? I tried to google around yet got nothing...
Feb 13, 2014 at 20:00 comment added fherzig The polynomial $X^2−7p^{1/p^n}X+p^{5/p^n}$ actually splits over $\mathbb{Q}_p(p^{1/p^n})$ (Newton), and similarly for $x^2-7tX+t^5$. :-) For $p > 2$ one could take e.g. $X^2−7p^{1/p^n}X+p^{1/p^n}$ instead (which doesn't split because the roots have valuation outside the value group of the extensions considered here). However, it seems quadratic extensions are too simple to fully illustrate what's going on, because the quadratic extension defined is the same for any $n$.
Nov 4, 2011 at 10:57 history edited Peter Scholze CC BY-SA 3.0
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Jun 1, 2011 at 6:40 vote accept Thomas Riepe
May 31, 2011 at 21:57 comment added SGP @Scholze: Many thanks for this! @Oliver: well put indeed!
May 31, 2011 at 18:17 comment added Olivier Yeah... And so we watch math history in the making.
May 31, 2011 at 15:55 history edited Peter Scholze CC BY-SA 3.0
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May 31, 2011 at 15:41 history answered Peter Scholze CC BY-SA 3.0