Naturality vs canonicality of double dual

Question

$\require{AMScd}$ I'm learning some linear algebra and basic category theory, and have been asked to show two things.

1. There is a canonical map $f: V \rightarrow V^{**}$
2. The following diagram commutes:

$$ \begin{CD} V @>\phi>> W\\ @V{m_V}VV @VV{m_W}V\\ V^{**} @>>\phi^{**}> W^{**} \end{CD} $$

For the first problem, we exhibit the map $v \mapsto x \mapsto x(v)$ (note for posterity: originally I had written $v \mapsto v^* \mapsto v^*(v)$, making it seem like $v$ and $v^*$ were related).

For the second, I was able to perform the requisite symbol manipulation (though I had to do it mindlessly, as I find each expression nearly impossible to interpret. For example, $w^* \mapsto eval(\phi^*(w^*), v)$ is "the function that takes a function from $W$ to $k$ to a function that takes (the function from $W^*$ to $V^*$ (given by ...) and applies it to $w^*$) and applies it to v").

Anyway, I've inferred that the canonicality of the first map (i.e., the fact that it is basis-independent) is somehow related to the naturality indicated by the second diagram, but I don't understand why.

Is my inference correct, and is there a simple way to understand it (without having to understand much category theory beyond what a natural transformation is)?

Answer 1

As Kevin Carlson noted in the comments, you should not be defining the map $V \to V^{\ast\ast}$ by passing through the dual space $V^\ast$, since it is entirely unnecessary.

Instead we define $V\to V^{\ast\ast}$ by, as he remarked, $v \mapsto (x \mapsto x(v))$. This is canonical since it does not depend on any selection of basis (compared to any isomorphism $V\cong V^\ast$ which necessarily requires you to pick a basis).

To show that the diagram commutes, you should check that, given a $v\in V$, we get the same thing going along the top/right as along the left/bottom. Along the top/right, we get an element in $W^{\ast\ast}$ which is determined by $\phi(v) \mapsto (x\mapsto x(\phi(v))$, where $x\in W^\ast$. Along the left/bottom, we get an element in $W^{\ast\ast}$ which is $$\phi^{\ast\ast} \left( v\mapsto (y\mapsto y(v)) \right),$$ where $y\in V^\ast$.

You should check that these are the same by figuring out what $\phi^{\ast\ast}$ does.

Categorically: what is happening is that we have a natural isomorphism $\eta: \text{id} \Rightarrow (-)^{\ast\ast}$ between endofunctors on the category of vector spaces over some field $k$. What this means is that each of the components $V \to V^{\ast\ast}$ is an isomorphism. The reason this is interesting is that it ensures us that there is a canonical way to construct the isomorphism $V\cong V^{\ast\ast}$ without picking a basis.

We should compare this to the fact that there is no natural isomorphism $\text{id}\Rightarrow (-)^\ast$ from the identity to the dual functor, meaning that there is no canonical isomorphism $V\cong V^\ast$. This isomorphism does exist, it just means that we are forced to select a basis in order to show it.

Answer 2

I'll expand a bit on the relationship between naturality and independence on the choice of basis.

The connection is most obvious when looking at natural transformation $η : \mathrm{Id} ⇒ \mathrm{Id}$. In that case, naturality means that for every $A : V → W$ we have $η_WA = Aη_V$.

Notice first that a choice of basis for $V$ is the same thing as an isomorphism $E : ℝ^n → V$, and given an operator $A : V → W$ and a basis $F : ℝ^m → W$, $F^{-1}AE : ℝ^n → ℝ^m$ is exactly the matrix of the operator $A$ in the given pair of bases, under the canonical identification of $L(R^n, R^m)$ and $M_{mn}$.

Now by applying the naturality condition of $η$ to a choice of basis $E : ℝ^n → V$, you get that $η_VE = Eη_{ℝ^n}$, and consequently $η_{ℝ^n} = E^{-1}η_VE$. This literally says that the matrix of $η_V$ is equal to (the matrix of) $η_{ℝ^n}$ in every basis $E$ of $V$ (which incidentally means that $η$ has to equal $rI$ for $r ∈ ℝ)$.

In your case you have $η : V → V^{**}$, which means that to talk about coordinate independence you need bases for both $V$ and $V^{**}$, and you can't ask for $η$ to be the same for an arbitrary choice of these bases. It doesn't make sense intuitively because you still want the entire thing to only depend on the choice of basis for $V$, and it would force $η$ to be $0$ anyway.

Fortunately, by the functoriality of $(-)^{**}$ a choice of basis $E$ induces a choice of basis $E' = (ℝ^n ≅ (ℝ^n)^{**} \stackrel{E^{**}}{\longrightarrow} V^{**})$ for $V^{**}$, and we have that $η_V$ always looks the same in any pair of bases $(E, E')$ (it's the identity matrix if you choose the canonical isomorphism for $ℝ^n ≅ (ℝ^n)^{**}$), so $η$ is again coordinate independent in the appropriate (and only reasonable, really) sense.

In general this talk about bases becomes cumbersome, and it's better to think about automorphisms $φ : V → V$ directly as changes of coordinates on $V$, or symmetries of $V$. Then you can just think of natural morphisms $η : FX → GX$ as of being invariant under the change of coordinates $(Fφ, Gφ)$ of the pair $(FX, GX)$, which is induced by the change of coordinates of the base object $X$, which makes the slogan that natural transformations don't depend on the choice of coordinates formally true.

Just keep in mind that the converse isn't true. Not every transformation between functors that is coordinate independent in the sense above is natural, because coordinate independence only gives you naturality squares for isomorphisms, and not for every morphism in the category.