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Let $V$ be a topological linear space, and let $\operatorname{Hom}(V,V)$ be the space of continuous linear maps from $V$ back to $V$, equipped with a suitable topology.

Is there a non-trivial example which satisfies the equivalence $V \cong \operatorname{Hom}(V,V)$?

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  • $\begingroup$ Is a topological linear space the same as a topological vector space? $\endgroup$ – Andrej Bauer Jan 27 '14 at 6:40
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    $\begingroup$ Also, why is this question tagged with type-theory? $\endgroup$ – Andrej Bauer Jan 27 '14 at 6:40
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    $\begingroup$ @Andrej: yes, "topological linear space" and "topological vector space" are synonyms. The type-theory tag is present since types of the form $T \cong T \to T$ are of particular importance. e.g., that is the single type of the untyped lambda calculus. I am looking for a functional-analytic analogue. $\endgroup$ – Tom LaGatta Jan 27 '14 at 8:08
  • $\begingroup$ @alpha, I'm afraid I don't see it. Could you write that up as an answer? Does $s$ therefore satisfy $s \cong \operatorname{Hom}(s,s)$? $\endgroup$ – Tom LaGatta Jan 27 '14 at 8:11
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    $\begingroup$ So in terminology of $\lambda$-calculus, you are looking for a $C$-monoid in the category of topological vector spaces which in addition satisfies the $\eta$-rule? (Also: there is a lambda-calculus tag, I retagged, I hope you're ok with that.) $\endgroup$ – Andrej Bauer Jan 27 '14 at 8:15
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If you drop the requirement on linearity, domain theory gives a nice example. (You can think of domains as topological spaces that don't have nice separation properties, but that do model "approximation of information" in computer science.) The construction of a domain $D$ with $D \cong [D \to D]$ was Dana Scott's breakthrough that pretty much started domain theory. See e.g. here.

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  • $\begingroup$ Chris: thanks! Is there a nice way to represent domains as topological linear spaces? i.e., do you know of a functor $F : \operatorname{Domains} \to \operatorname{TopLinSpaces}$? A necessary property would then be that $V := F(D)$ satisfy the equivalence $V \cong \operatorname{Hom}(V,V)$. There are (at least) two trivial cases: $V = $the zero space, and $V = $the one-dimensional space. If @alpha's conjecture is true, then we might also be able to find a functor with $V = L(s)$. That would be very cool. $\endgroup$ – Tom LaGatta Jan 27 '14 at 18:57
  • $\begingroup$ Tom: I can't think of a direct way. But: the forgetful functor $\mathbf{TopLinSpaces} \to \mathbf{Top}$ has a left adjoint. We could compose that with the functor $\mathbf{Domains} \to \mathbf{Top}$. But I don't know whether the resulting functor preserves "reflexive objects" like $V \cong \mathrm{Hom}(V,V)$ ... $\endgroup$ – Chris Heunen Jan 27 '14 at 22:00
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    $\begingroup$ Mike Mislove proved there is no $T_2$ space homeomorphic to its own function space. $\endgroup$ – Andrej Bauer Jan 28 '14 at 6:09
  • $\begingroup$ @Andrej, that's a great fact to know. Can you provide a reference? $\endgroup$ – Tom LaGatta Jan 28 '14 at 23:14
  • $\begingroup$ @AndrejBauer, sorry for the bump. Do you have a reference for Mislove's result? $\endgroup$ – Tom LaGatta Mar 10 '14 at 20:05
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A simple and natural example of such a space is that of the row-finite matrices. These play a central row in the theory of summation of divergent series since many of the classical methods are implemented by such matrices (prominent example---Cesaro summation and its higher degree variants). The above space has a natural locally convex structure and it enjoys the requested property.

EDIT after Tom's comment. Am typing on my iPad and so can't use many math symbols. The standard reference on infinite matrices and summability is "Infinite matrices and sequence spaces" by R. Cooke (now a bit dated). On nuclear spaces, spaces of operators and tensor products it is Grothendieck's thesis. If you are interested in such topics, I recommend its study, a potentially life-changing experience.

The basic spaces involved are the nuclear Fréchet space $\omega$ of all sequences and its dual $\phi$, the nuclear Silva space of all finite sequences. The row finite matrices map the former into itself and it is easy to see that all such continuous linear mappings arise in this way. The rest is just formal manipulation of tensor products and the corresponding operator spaces, one further ingredient being the fact that $\omega$ is isomorphic to $\omega \otimes \omega$ (but, notabene, not in any natural way).

Now if $E$ is a nuclear Fréchet space which is isomorphic to $E\times E$ and there are many such and we put $V=E\otimes E'$, then $$L(V,V)=V'\otimes V=(E\otimes E')\otimes(E\otimes E')'=(E\otimes E')\otimes(E'\otimes E'')=(E\otimes E')\otimes(E'\otimes E)=(E\otimes E) \otimes(E'\otimes E')=E\otimes E'=V$$ (equality denote "is isomorphic to", $\otimes$ any tensor product since everything in sight is nuclear).

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  • $\begingroup$ This is a great observation, @blackburne. Could you provide a reference, please? I'd love to learn more about row-finite matrices, particularly from a functional-analytic point of view. $\endgroup$ – Tom LaGatta Jul 31 '14 at 1:09
  • $\begingroup$ Also, can you demonstrate this property in your answer? I'll be happy to accept it in that case. $\endgroup$ – Tom LaGatta Jul 31 '14 at 1:10
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    $\begingroup$ @blackburne I think one has to be careful with the "formal manipulations". Isn't your example similar to Andrej Bauer's attempt? If I understand correctly, the space of row-finite matrices is (isomorphic to) $\phi^{\mathbb N}$ and then you have $L(\phi^{\mathbb N},\phi^{\mathbb N})= L(\phi^{\mathbb N},\phi)^{\mathbb N}$. But why is $L(\phi^{\mathbb N},\phi)$ isomorphic to $\phi^{\mathbb N}$? $\endgroup$ – Jochen Wengenroth Aug 1 '14 at 7:26
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    $\begingroup$ I still have doubts. In particular concerning the isomorphism $(E\otimes E')' = E'\otimes E''$. I think that Remarque 1 on page 47 of Grothendieck's thesis might be relevant here. $\endgroup$ – Jochen Wengenroth Aug 1 '14 at 13:15
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    $\begingroup$ This argument seems to indicate that $V=\text{Hom}(E,E)$, so there is a canonical element $1_E\in V$. It also indicates that $V'=E'\otimes E''=E'\otimes E=V$. This much does not use $E=E\otimes E$, so it is canonical. The element $1_E\in V$ corresponds to a canonical element $t\in V'$, which looks like it has to be some kind of trace. Thus $\langle t,1_E\rangle$ wants to be $\text{dim}(E)$, which seems to be a problem. $\endgroup$ – Neil Strickland Aug 1 '14 at 13:50
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[Disclaimer: this answer does not hold water but Tom likes its spirit.]

Let me live dangerously and attempt to answer a question in functional analysis, slightly generalising @alpha's suggestion, I think.

Consider any topological (complex) vector space $W$ such that $W \otimes W \cong W$ and $\mathrm{Hom}(W, \mathbb{C}) \cong W$. Then take $V = \mathrm{Hom}(W, \mathbb{C})$ and calculate (noting that $V$ is isomorphic to $W$): $$\mathrm{Hom}(V,V) \cong \mathrm{Hom}(V, \mathrm{Hom}(W, \mathbb{C})) \cong \mathrm{Hom}(V \otimes W, \mathbb{C}) \cong \mathrm{Hom}(W \otimes W, \mathbb{C}) \cong \mathrm{Hom}(W, \mathbb{C}) \cong V. $$ There ought to be many examples of such spaces. For instance $\ell^2$ seems to satisfy the conditions.

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    $\begingroup$ Unfortunately, the adjunction between tensor product and ``hom'' no longer hold for (infinite dimensional) topological vector spaces. For example, in the case of Hilbert space, one has to replace Hom(A,B) by the space of Hilbert-Schmidt operator from $A$ to $B$ in order to have such an adjunction. Also , $l^2$ is extremely different from its space of endomorphism. $\endgroup$ – Simon Henry Jan 27 '14 at 9:20
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    $\begingroup$ Someone just has to find the correct definitions of $\mathrm{Hom}$ and $\otimes$ so that my "proof" can work. $\endgroup$ – Andrej Bauer Jan 27 '14 at 21:03
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    $\begingroup$ Change "doesn't obviously" to "obviously doesn't" since the former is separable and the latter is not (or try using the attribute "reflexive"). $\endgroup$ – blackburne Aug 1 '14 at 6:56
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    $\begingroup$ Andrej, it seems to me that this could unfortunately be like finding the correct definition of morphism for which $X$ and $2^X$ are isomorphic sets $\endgroup$ – Yemon Choi Aug 1 '14 at 10:21
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    $\begingroup$ I have no idea, that's why the first word in my "answer" is "disclaimer". I am just saying that Cantor's diagonal argument does not necessarily always go through in an arbitrary category. $\endgroup$ – Andrej Bauer Aug 1 '14 at 10:31

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