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3 votes
0 answers
155 views

Colimits in commutative Banach algebras?

Let $K$ be a complete non-Archimedean field. It is known that the category $\mathrm{Ban}_K$ of $K$-Banach spaces with bounded linear maps does not have infinite colimits. The usual argument for $\...
user577413's user avatar
2 votes
1 answer
291 views

Čech complex of rigid $K$-space - Closed image of boundary maps

Let $(X,\mathcal{O}_X)$ be a rigid $K$-space with a finite affinoid covering $(U_i)_{i\in I}$ such that any intersection of the $U_i$ is affinoid too. Equipping the direct products with the maximum ...
KKD's user avatar
  • 473
9 votes
1 answer
596 views

Why is the Berkovich spectrum of a C*-Algebra the same as the Gelfand spectrum?

Let $A = \mathcal{C}(X)$ be a commutative (unital) C*-Algebra. Let $Spec(A)$ denote its Gelfand spectrum $$ Spec(A) = \{A \rightarrow \mathbb{C} : \text{non-zero *-homomorphism} \} \simeq X. $$ Now ...
Niki's user avatar
  • 335
3 votes
0 answers
275 views

Seminorms on tensor products of affinoid algebras

Let $A \leftarrow C \rightarrow B$ be affinoid $K$-algebras, where $K$ is a non-archimedean field with non-trivial absolute value. Equipping $A$, $B$, $C$ with the supremum seminorms, there is a ...
user114125's user avatar
12 votes
1 answer
885 views

bornological vector spaces over a non-archimedean field

Let $k$ be a complete non-archimedean field. In definitions I have seen of bornological vector spaces over $k$ there are usually some extra assumptions on the non-archimedean field. For instance in '...
Oren Ben-Bassat's user avatar