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Consider the symmetric monoidal category TVS of complete Hausdorff topological vector spaces equipped with the completed projective, injective, or inductive tensor product.

Every finite-dimensional vector space is dualizable in the symmetric monoidal category TVS. Is every dualizable object in TVS finite-dimensional?

A negative answer to the above question (i.e., infinite-dimensional dualizable vector spaces exist) would be far more interesting than a positive one, so even if the answer for the three monoidal products (projective, injective, inductive) is positive, we can still ask whether nontrivial examples exist for other tensor products.

Does TVS admit a subcategory $C$ that contains all finite-dimensional vector spaces and linear maps and at least one topological vector space $V$ that admits endomorphisms with infinitely many distinct eigenvalues such that $C$ admits a symmetric monoidal structure with $\def\C{{\bf C}}\C$ as the monoidal unit and in which $V$ is dualizable?

For example, consider the vector space $\def\sm{{\sf C}^{\sf\infty}}\def\dis{{\sf C}^{\sf-\infty}}V=\sm(M)$ of smooth functions on a compact smooth manifold $M$. Take $V^\ast$ to be the space $\dis(M)$ of distributions on $M$. Take the unit map $V^\ast⊗V→\C$ to be the evaluation map. Naively, we can imagine the following construction of the counit map $\C→V⊗V^\ast$. Suppose $V⊗V^\ast=\sm(M)⊗\dis(M)$ is isomorphic to $\sm(M,\dis(M))$, which could be true for some choices of the tensor product $⊗$. Then we can take the map $m↦δ_m$, where $δ_m$ is the Dirac delta-distribution at $m$. It is far from clear whether there is a tensor product $⊗$ that allows evaluation and coevaluation maps with these properties, but it is also not clear why such a tensor product cannot exist.

Of course, this leaves open the question of what the dimension of $\sm(M)$ (defined as the composition of the counit, the braiding, and the unit) should be, and perhaps one might be able to extract a contradiction here, but it is not immediately clear to me how.

I am also happy to see answers for formalisms other than topological vector spaces, e.g., $p$-liquid vector spaces of Clausen–Scholze.

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    $\begingroup$ This type of question comes up a lot in my work with Clausen; you can ask it over any "analytic ring". Modulo possible technical issues of the specific setup, the answer is "it's true iff Fredholm theory works" in the sense that any trace-class perturbation of an isomorphism has finite-dimensional kernel and cokernel (or, over a general ring, better to say "has perfect cone"). I believe this argument shows that in any of the examples you propose, the answer is Yes. But interestingly, the answer is No for topological $\mathbb C[T]$-modules. $\endgroup$ Commented May 21 at 20:11

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