Analytic functions on spaces over non-Archimedean fields and troubles with totally disconnectedness. Why totally disconnectedness of the spaces causes big troubles if one tries to study the space using naively defined analytic functions on it locally expressible in power series?

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Krish Finley · Accepted Answer

Step 1I would highly suggest reading Andre&#039;s book.Let us say that a sheaf       O   on a topological space X  is non-degenerate if whenever   U  ⊆  X is a non-empty open subset, one has that       O    (  U  ) is non-zero. This assumption is completely reasonable if, for instance, you desire that O(U) should contain &#039;all constant functions on U&#039; (where this is intentionally vague, but is clear in all examples).So, let us make the following trivial observationIf O is non-degenerate and O(X) is an integral domain (or more specifically &#039;connected&#039; -- i.e. that Spec(O(X)) is connected) then X is connected.Proof: Indeed, suppose that   X  =  U  ⊔  V where   U  ,  V  ⊆  X are open. Then,       O    (  X  )  =      O    (  U  )  ×      O    (  V  ) and by our assumption on O(X) we see that O(U) or O(V) is zero, and so one of U or V is empty.So then, where is the issue with disconnectedness in rigid geometry? Let K be a non-archimedean field (maybe algebraically closed for total realism) and let us denote            B        1    (  K  )  :=  {  x  ∈  K  :      |    x      |    ⩽  1  }  ,endowed with the subspace topology of K (where K itself is given the topology associated to       |    ⋅      |   ). One then wants to define a sheaf of analytic functions O on             B        1    (  K  ) and one expects that1. Ome has the equality      O    (            B        1    (  K  )  )  =  K  ⟨  x  ⟩  :=      {          ∑      n              a      n              x      n        :    lim          |              a      n              |        =    0    }    ,the ring of convergent (on the unit ball) power series,2. For all non-empty open subsets U of             B        1    (  K  ) one has that O(U) contains the set of constant functions   U  →  K (and in-particular is non-zero).From 2. we see that O is non-degenerate, but as       O    (            B        1    (  K  )  ) is an integral domain by 1. this is a contradiction as             B        1    (  K  ) is highly disconnected as K is non-archimedean.

Analytic functions on spaces over non-Archimedean fields and troubles with

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