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Let $p$ be a prime and $\mathbb{C}_{p}$ the completion of the algebraic closure of the $p$-adic number field $\mathbb{Q}_p$.

Let $M$ be a $\mathbb{Q}_p$-Banach space.

We denote by $M\mathbin{\widehat{\otimes}_{\mathbb{Q}_p}}\mathbb{C}_{p}$ the complete tensor product over $\mathbb{Q}_p$.

Is the natural map $M\rightarrow M\mathbin{\widehat{\otimes}_{\mathbb{Q}_p}}\mathbb{C}_{p}$ an isometry ?

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  • $\begingroup$ What is the complete tensor product? $\endgroup$
    – Mikael de la Salle
    Commented Feb 9 at 12:40
  • $\begingroup$ TeX notes: I think that the hat in a completed tensor product is usually placed only over $\otimes$, not the ground ring. But, either way, it needs a \mathbin for spacing: see, e.g., $M \widehat\otimes \mathbb C_p$ M \widehat\otimes \mathbb C_p vs. $M \mathbin{\widehat\otimes} \mathbb C_p$ M \mathbin{\widehat\otimes} \mathbb C_p. I edited accordingly. $\endgroup$
    – LSpice
    Commented Feb 9 at 14:08
  • $\begingroup$ Thank you for your comments. The complete tensor product is the one defined in §2.1.7 of "Non Archimedean Analysis" (S. Bosch, U. Guntzer, R. Remmert). $\endgroup$
    – user521844
    Commented Feb 9 at 16:07
  • $\begingroup$ By Schneider, Nonarchimedean Functional Analysis, Coro 17.5 on page 106, the map is a homeomorphism on its image. Maybe studying the proof of Prop 17.4 and Coro 17.5 will help shed some light on your question? $\endgroup$
    – Laurent Berger
    Commented Feb 27 at 10:05

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