How do I determine whether or not an isomorphism $T:V\to W$ is a canonical isomorphism?

Question

Roughly speaking, an isomorphism $T:V\to W$ between vectors spaces $V$ and $W$ is canonical if it can be defined without reference to a base. Facts that are widely known are that if $V$ is an arbitrary finite-dimensional vector space, then there's no canonical isomorphism between $V$ and its dual $V^*$ and that there's a canonical isomorphism between $V$ and its bidual $V^{**}$ (these facts are in the following site: What is a Natural Transformation?).

Next are two canonical isomorphisms:

1. $\mathbb{R}^n\to (\mathbb{R}^n)^*,\, v\mapsto \left(u\mapsto\langle u,v\rangle\right)$
2. $V^*\otimes W\to \mathcal{L}(V,W),\, \sum_{i=1}^kf^i\otimes w_i\mapsto \left(v\mapsto \sum_{i=1}^kf^i(v)w_i\right)$

I've given the above examples to show that I know some canonical isomorphisms. In fact, I even understand intuitively what a canonical isomorphism is. However I would like to know how I can formally define what a canonical isomorphism is.

My questions is: Given an isomorphism $T:V\to W$ between finite-dimensional vector spaces, how do I determine whether or not the isomorphism $T$ is in fact a canonical isomorphism?

I saw in the book "An Infinitely Large Napkin" (in the page 600) that it is possible to formalize the concept of canonical isomorphism using the concept of natural transformation. However I think it is not a simple task to use the concept of natural transformation to prove, for example, that $V^*\otimes W\to \mathcal{L}(V,W),\, \sum_{i=1}^kf^i\otimes w_i\mapsto \left(v\mapsto \sum_{i=1}^kf^i(v)w_i\right)$ is a canonical isomorphism.

I am asking this question because I want a rigorous way to prove whether or not an isomorphism is a canonical isomorphism. I read several answers on this site but I couldn't find a way to rigorously prove whether or not an isomorphism is a canonical isomorphism.

Thanks for your attention!

Answer 1

If you did not choose a basis (or an inner product), chances are the map you defined is natural. I'm going to try and show your latter example slightly more rigorously.

The hardest part initially is to figure out the setup to show that this is a natural transformation. I'm going to tackle that here and leave the details to you.

I'm going to assume based off of your question that you know that $V^\ast \otimes W \cong \mathcal{L}(V,W)$ in some (as of yet non-canonical) way, through the map you have defined (showing injectivity and that both vector spaces have the same dimension suffices). So we are in Case 4 of the link you provided above https://www.math3ma.com/blog/what-is-a-natural-transformation

My initial attempt at this problem is in the box below

Let me clean up some terminology so that I can ride off of the diagram in that link. Let $C$ be the category of pairs of vector spaces over $K$. Objects here are pairs $(V,W)$ and morphisms $(f)$ are pairs of linear maps $(V,W) \to (V',W')$ in $C$. Let $D$ be the category of vector spaces over $K$. Let $F$ be the functor that sends $(V,W) \to V^\ast \otimes W$, $G$ be the functor that sends $(V,W) \to \mathcal{L}(V,W)$, and finally let $\eta$ be the transformation from $V^\ast \otimes W \to \mathcal{L}(V,W)$ as you defined above. $F(f)$ sends a map $(V,W) \to (V',W')$ to a map $(V')^\ast \otimes W \to V^\ast \otimes W'$ (note the different directions the maps go in for the $V$ and $W$ factors). $G(f)$ sends a map $(V,W) \to (V',W')$ to a map $\mathcal{L}(V,W) \to \mathcal{L}(V',W')$. Now we just need to check commutativity of the diagram. Informally, $G(f) \circ \eta_x$ sends $V^\ast \otimes W$ to $\mathcal{L}(V,W)$ to $\mathcal{L}(V',W')$, whereas $\eta_y \circ F(f)$ sends $V^\ast \otimes W$ to $V \otimes W'$ to $\mathcal{L}(V',W')$. You can fill in the details to show this actually commutes, by showing that the explicit maps used in going down either path yield the same answer.

I wrote this at $1$ am and if you carefully follow the details, there are a lot of issues. These issues arise because the functors $F$ and $G$ are covariant in one component but contravariant in another. It is not too hard to see using the ideas above that individually, the transformation is natural in $V$ and $W$ separately. However, as a whole, this is the fix I came up with.

Let $C$ be the category of vector spaces over $K$. Objects are again pairs $(V,W)$, but this time a morphism, $f: (V,W) \to (V',W')$ is actually a pair of maps $f_1: V' \to V$ and $f_2: W \to W'$. The reason for this odd choice will become clear once we work through the details. Now, for any morphism $f$, we have $F(f)$ sends a morphism $(V,W) \to (V',W')$ to a map $V^\ast \otimes W \to (V')^\ast \otimes W$. Similarly, for any morphism $f$, we have $G(f)$ sends a morphism $(V,W) \to (V',W')$ to a map $\mathcal{L}(V,W) \to \mathcal{L}(V',W')$. Because of the way the morphisms in $C$ were set up, these maps exist and are well defined. Now you can go around the diagram to see that it commutes.

Answer 2

The key to formalize this is the concept of a natural transformation. As Osama Ghani showed in their answer, the isomorphism $V^*\otimes W\simeq \mathcal L(V,W)$ is said to be natural (or canonical) because there is an isomorphism $$(-)^*\otimes(-) \simeq \mathcal L(-,-)$$ of functors $\mathrm{finVec}^{\text{op}}\times\mathrm{finVec}\to\mathrm{finVec}$, where $\mathrm{finVec}$ is the category of finite-dimensional vector spaces over some field. This means that, for every two pairs of f.d. vector spaces $(V,W)$ and $(V',W')$ and every morphism $(f,g):(V,W)\to(V',W')$ in $\mathrm{finVec}^{\text{op}}\times\mathrm{finVec}$ (which means $f:V'\to V$ and $g:W\to W'$), then the diagram $$\begin{array}{ccc} V^*\otimes W & \longrightarrow & \mathcal L(V,W) \\[3mm] \downarrow{f^*\otimes g} & & \downarrow{g\circ-\circ f} \\[3mm] V'^*\otimes W' & \longrightarrow & \mathcal L(V',W') \end{array} $$ is commutative.

Now, the isomorphism you mentioned $$v\mapsto \langle v,-\rangle,$$ which I call $\flat:V\to V^*$, is not canonical if your context is the category of finite dimensional vector spaces, $\mathrm{finVec}$, in the first place because the isomorphism $$V\simeq V^*$$ means an isomorphism between the identity functor $\mathrm{id}$ and the dualization functor $(-)^*$, but this doesn't even make sense because $$\mathrm{id}:\mathrm{finVec}\to\mathrm{finVec},$$ while $$(-)^*:\mathrm{finVec}^{\mathrm{op}}\to\mathrm{finVec}.$$ To regard this isomorphism as "canonical" you need to switch to the category of finite dimensional inner product spaces with linear maps as morphisms, $\mathrm{finInnProdSp}$. In that case every vector space comes equipped with its isomorphisms $\flat:V\to V^*$ and $\sharp:V^*\to V$, which allow us to define a covariant dualization functor $$(-)^*:\mathrm{finInnProdSp}\to\mathrm{finInnProdSp}$$ which sends a morphism $f:V\to W$ to $\tilde f:V^*\to W^*$ given by $\tilde f(\alpha)=\flat(f(\sharp(\alpha)))$ and in this case the diagram $$\begin{array}{ccc} V & \xrightarrow f & W \\[3mm] \downarrow{\flat} & & \downarrow{\flat} \\[3mm] V^* & \xrightarrow{\tilde f} & W^* \end{array} $$ commutes basically by definition. This is a way to say that the isomorphism is canonical (or natural) once you have chosen an inner product for every finite dimensional vector space.