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It is well known that finite dimensional vector spaces (over any field $K$) canonically satisfy $V''\otimes W''\simeq (V\otimes W)''$. My question is, roughly, if this canonical map can somehow be salvaged for infinite dimensional spaces. (A more precise version is at the end).

As far as I see, there are (at least) two ways to go about constructing this isomorphism:

  1. If we let $j_V:V\hookrightarrow V''$ denote the double dual embedding, we obtain the two natural maps $$j_{V\otimes W}:V\otimes W \to (V\otimes W)''$$ and $$j_V\otimes j_W:V\otimes W \to V''\otimes W''.$$ If $V$ and $W$ are finite dimensional, both of these maps are isomorphisms and combine to give an isomorphism $V''\otimes W'' \simeq (V\otimes W)''$ that is natural in both $V$ and $W$. If at least one space is infinite dimensional, however, we just conclude that both $V''\otimes W''$ and $(V\otimes W)''$ receive an embedding from $V\otimes W$.

  2. The second way to go about this is to use the evaluation pairing $$\eta_{V,W}: V'\otimes W'\to (V\otimes W)', f\otimes g\mapsto (v\otimes w\mapsto f(v)g(w))$$ In particular, it also follows that $V''\otimes W''\simeq (V\otimes W)''$ canonically by composing $$V''\otimes W'' \stackrel{\eta_{V',W'}}{\to}(V'\otimes W')' \stackrel{(\eta_{V,W}')^{-1}}{\to} (V\otimes W)''.$$ Here, if the spaces are infinite dimensional, we can just conclude that $(V'\otimes W')$ receives a map from both $V''\otimes W''$ and $(V\otimes W)''$.

Moreover, these two constructions actually give the same isomorphism, as is easily verified by observing that the diagram

\begin{array} 0&&V\otimes W & \stackrel{j_V\otimes j_W}{\longrightarrow} & V''\otimes W''\newline (D)&&\downarrow j_{V\otimes W} & & \downarrow \eta_{V',W'}\newline &&(V\otimes W)'' & \stackrel{\eta_{V,W}'}{\longrightarrow} & (V'\otimes W')' \end{array} is in fact commutative for all spaces.

In this situation, note that $\eta_{V',W'}$ is injective and $\eta_{V,W}'$ is surjective, so that there must be some (unnatural) embedding of $V''\otimes W'' \hookrightarrow (V\otimes W)''$ (and, likewise, quotient map going the other way). But so far I had no luck coming up with any natural (nonzero) map in any direction.

Does there exist such a canonical map? To make this precise: Are there any natural transformations between the functors $(V,W)\mapsto V''\otimes W''$ and $(V,W)\mapsto (V\otimes W)''$ that restrict to the above isomorphisms for finite dimensional spaces? If so, are there any examples that constitute a lift of the diagram $(D)$? If there are no such maps (or even consistently none, if this turns out to be sensitive to the underlying set theory), I would also be grateful for any information on that matter.

Tim Seifert
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  • To confirm, you want to work with the algebraic dual spaces? Also would perhaps a duality (pairing) between relevant spaces work for your applications versus map restricting to iso? – blargoner Mar 01 '24 at 01:53
  • Yes, this is about algebraic duals. I don't really understand your second quesion however, can you perhaps clarify? What I'm after is really some natural map between these spaces - if that turns out to be an embedding or a quotient map (or even iso, even though I doubt that), all the better. – Tim Seifert Mar 01 '24 at 09:30
  • If arbitrary spaces $V,V^$ are dual in the sense of having a non-degenerate bilinear form $V\times V^\to k$, and similarly for $W,W^$, then the same is naturally true for $V\otimes W$ and $V^\otimes W^*$. In the infinite-dimensional case this is sufficient for many uses of (algebraic) duality, although you won't have induced isos to the spaces of linear forms. I was just thinking that might be of use to you. See here for more detail. – blargoner Mar 01 '24 at 16:36
  • Okay, thanks for the clarification! Yes, any natural pairing in this setting would also be of interest. – Tim Seifert Mar 01 '24 at 16:57

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