# Are maps corresponding to affinoid subdomains flat in the Banach sense?


Let us say that $$B'$$ is Banach-flat over $$B$$ if whenever $$M \to N \to P$$ is an admissible exact sequence of Banach $$B$$-modules then the completed tensor product sequence $$M \comptensor_B B' \to N \comptensor_B B' \to P \comptensor_B B'$$ is also admissible exact.

(A map $$f \colon M \to N$$ of Banach $$B$$-modules is called admissible if there is a constant $$C>0$$ such that any $$n \in f(M)$$ there is a preimage $$m \in M$$ such that $$f(m) = n$$ and $$\abs{m} \le C \abs{n}$$. By Banach's open mapping theorem, this condition is equivalent to $$f(M)$$ being closed in $$N$$. Generally exact sequences of Banach modules can only be expected to behave well, if all mappings are admissible.)

Is it true that if $$V = \Sp(B') \subseteq X = \Sp(B)$$ is an affinoid subdomain, then $$B'$$ is a flat Banach algebra over $$B$$?

Note that for a functor being exact in the Banach sense, it suffices to consider short exact sequences $$0 \to M \to N \to P \to 0$$. Taking the completed tensor products is always Banach-right exact (because being admissible right exact is equivalent to being a cokernel diagram and because completed tensor product is left adjoint to Banach-Hom), so it suffices here to show that tensoring with $$B'$$ preserves admissible injective maps of Banach modules.

One can show that fixing $$M$$, the association $$V = \Sp(B') \mapsto M \comptensor_B B'$$ is a sheaf on $$X$$ (more precisely, its Čech complex is admissible exact). In particular, $$M \comptensor B' \to \prod_i M \comptensor B_i'$$ is admissible injective, if $$\Sp(B') = \bigcup_i \Sp(B_i')$$. By the theorem of Gerritzen and Grauert we can therefore assume that $$V = X(f_1/f, \dots, f_r/f)$$ is a rational subdomain of $$X$$, for which we have a rather explicit description of the algebra $$B'$$. But so far I had no success.

This is not true. Assume that $$X$$ is the closed unit disc (given by $$|T| \le 1$$, with algebra $$B$$) and $$V$$ is a smaller disc (given by $$|T| \le r$$ for some $$r \in (0,1)$$, with algebra $$B_V$$). Consider the annulus $$W$$ defined by $$|T|=1$$ with algebra $$B_W$$. Then the restriction map $$B \to B_W$$ is injective and admissible (because function on the disc reach their maximum on the boundary).
If we do the completed tensor product with $$B_V$$, we get the map $$B_V \to B_W \hat{\otimes}_B B_V$$. But $$B_W \hat{\otimes}_B B_V$$ is the algebra of $$V\cap W = \emptyset$$, that is to say 0, so the map is not injective.
• Yes, I guess this means that flatness in the Banach sense is not such an interesting notion. Let me state a positive result however: Ben-Bassat and Kremnizer (arxiv.org/abs/1312.0338) prove that morphisms of affinoid algebras $A \to B$ corresponding to embedding of affinoid domains may be characterized by the fact that $B \hat{\otimes}^\mathbf{L}_A B \simeq B$. Sep 10, 2019 at 9:15