Is flattening a natural isomorphism?

Question

Consider the following categories:

1. $≔\operatorname{FinVec}_ℝ$ of finite dimensional real vector spaces, whose morphisms are the linear maps
2. $ℍ≔\operatorname{FinHilb}_ℝ$ the category of finite dimensional real Hilbert spaces, with $\hom(ℍ)$ being the linear isometries.
3. (optional) $≔$ The category of finite dimensional real Hilbert spaces which are constructible as a finite combination of direct sums $⊕$, tensor products $⊗$ and dual operations $*$ from $ℝ$, with the morphisms being the linear isometries.

Consider a family $η$ of (isometric) isomorphisms $η_V：V → ℝ^{\dim(V)}$, that map a vector to its flattened representation. For example, if $V=ℝ^n⊗ℝ^m$, then we can consider the classical vectorization mapping.

Question: Is there, for $ ∈ \{, ℍ, \}$, such a family $η$ that is a natural isomorphism? (in the sense that $η$ is a natural isomorphism from the identity functor $\operatorname{id}_$ to some other functor $G：→$)

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My thoughts: First, I am not sure whether $ℍ$ and $$ can be identified as the same here. We know that all Hilbert spaces are isometrically isomorphic to $ℓ^2(B)$. However, it doesn't seem clear to me whether all Hilbert spaces of the same dimension are somehow naturally isomorphic, which of course this very question is a partial to.

Secondly, we need to construct a functor $G：→$ such that $G(f)∈\hom_(ℝ^{\dim(V)}, ℝ^{\dim(W)})$ if $f∈\hom_(V, W)$. I really am not sure what to do here without taking a basis. Of course once we take some basis $B=\{b_1, …, b_{\dim(V)}\}$ in $V$, we can simply map the basis elements to some $e_1, …, e_{\dim(V)}$ in $ℝ^{\dim(V)}$. But it seems essentially impossible to construct $G$ without referencing a basis. So how can we prove such a functor does not exist?

In the case of $$, assuming all spaces are equipped with the standard induced inner product and standard basis, we can even give a complete description of the family $η$, by choosing a consistent ordering: given $U⊕V$, map $∑_{i} α_i |(u_i,0)⟩ + ∑_j β_j |(0, v_j)⟩$ to $∑_i α_i |e_i⟩ + ∑_j β_j |e_{\dim(U)+j}⟩$, and for $U⊗V$ map $∑_{ij} α_{ij} |u_i⊗ v_j⟩$ to $∑_{k=1}^{\dim(U)⋅\dim(V)} α_{k//\dim(U), k \% \dim(U)} |e_k⟩$.

Answer 1

Yes, assuming the axiom of global choice. Let $\mathbb{W}$ be the full subcategory of $\mathbb{V}$ consisting only of the objects $\{\mathbb{R}^n \mid n \in \mathbb{N}\}$. The inclusion functor $\mathbb{W} \to \mathbb{V}$ is fully faithful and essentially surjective. Therefore, it gives rise to an equivalence of categories, which produces the natural isomorphism $\eta$ you seek.

The analogous fact holds for $\mathbb{U}$ and $\mathbb{H}$, since all finite-dimensional real Hilbert spaces are isomorphic to $\mathbb{R}^n$ for some $n$ as Hilbert spaces.

Without choice, I expect this to be impossible.

Edit: indeed, it is impossible. In fact, this principle (for either general real vector spaces or for real Hilbert spaces) implies that for all families $\{S_i \mid i \in I\}$ of finite, non-empty sets, $\prod\limits_{i \in I} S_i$ is non-empty. This fact cannot be proved without choice.

Since we can’t make flattening natural without choice, you should not look for any sort of explicit description of the flattening functor.