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

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.

Answer 2

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 3

I don’t think the two following statements:

1. There is a canonical map $f: V\to V^{**}$.
   Here “canonical” means $f$ can be determined without resorting to any artifact, such as a choice of the basis of $V$ or $V^{**}$. Such a map $f$ can be described as follows: for any $v∈V$, $f(v)=\text{eval}_v$. $\text{eval}_v$ “evaluates” the value of a linear form $ϕ$ at $v$, i.e. $\text{eval}_v(ϕ)=ϕ(v)$.

2. $m: \text{id} \to(**)$ , whose component $m_V$ is given by the canonical map $f$ for any $V\in \text{Vect}_{\mathbb{K}} $, is a natural isomorphism.
   Here $\text{id}:\text{Vect}_{\mathbb{K}}\to \text{Vect}_{\mathbb{K}} $ is identity functor, $(**): \text{Vect}_{\mathbb{K}} \to \text{Vect}_{\mathbb{K}}^{**} $ is double dual functor.
   “$m$ is a natural isomorphism” means the objects and morphisms (including the composition laws of morphisms) of category $\text{Vect}_{\mathbb{K}}$ and category $\text{Vect}_{\mathbb{K}}^{**}$ are “word by word” identical.

are equivalent. Nor can any of them imply the other. They are intrinsically different. (Don’t be fooled by some sensational slogans of category theory!)

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Nevertheless, the two statements provide us with two different interpretations of the idea “canonical”. Besides, in practice, if we have a canonical map in the same manner with statement 1, we can usually get a natural transformation in the same manner with statement 2.

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In contrast, consider $V$ and $V^*$.

1. Any map $g:V\to V^*$ can’t be determined without using an (artificial) choice of basis of $V$.
   Suppose $v∈V$, then $g(v)$ can’t be determined if we don’t know all the $g(v)(e_i)$, where $e_i$ are elements in the basis of $V$.
   Certainly an inner product structure on $V$ can define a canonical isomorphism between $V$ and $V^*$, but OP’s question doesn’t assume the existence of the inner product structure on V. (In fact, defining an inner product structure needs as many data as specifying all the $g(v)(e_i)$ for all the $v\in V$.)

2. And there is no natural transformation, not to say natural isomorphism, between $\text{id}$ and $(*)$.
   This is because $\text{id}$ is a covariant functor while $(*)$ is a contravariant functor.

Moreover, we can show that extranatural transformation, a reasonable generalization of natural transformation which aims to make sense of the transformation from a covariant functor to a contravariant functor, also does not exist between $\text{id}$ and $(*)$ unless it is $0$ (but the zero extranatural transformation is not invertible): see this post, or page 234 of Eilenberg and MacLane, General theory of natural equivalences.

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Lastly, a helpful reference on this topic is chapter 8.8 of this note (My apologies, not in English).