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Jun 3, 2014 at 12:01 comment added Jochen Wengenroth Since $C^\infty(M)$ is a nuclear Frechet space you can choose more or less any of the usual tensor topologies (like the projective $\pi$ or the injective $\varepsilon$) and you get $C^\infty(M,A)\cong C^\infty(M) \tilde{\otimes} A$ (the completed tensor product) for all Banach (or Frechet) spaces $A$.
Jun 3, 2014 at 11:18 review First posts
Jun 3, 2014 at 11:27
Jun 3, 2014 at 11:12 comment added Igor Khavkine If $A$ is finite dimensional, this is trivial (pick a basis for $A$). If not, then you need to be more specific on the conditions you agree to on $A$ (topology, etc.).
Jun 3, 2014 at 10:59 history asked nielzs CC BY-SA 3.0