$\require{AMScd}$ 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)?

  • $\begingroup$ This might be an answer to my question, but I don't understand it yet: math.stackexchange.com/questions/622589/…. Perhaps it cannot be made any simpler though. $\endgroup$ – A_P Jul 15 '19 at 18:05
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    $\begingroup$ "Canonical" is basically just a synonym for "natural." It's an empirical fact that maps which seem "canonical", intuitively, tend to be components of natural transformations, formally. Incidentally, your definition of $f$ is a bit strange-$v^*$ seems like it should have something to do with $v$, but in fact we should just have $v\mapsto(x\mapsto x(v))$, where $x$ is any element of $V^*$. Maybe that's what you meant, but the notation is unclear. $\endgroup$ – Kevin Arlin Jul 15 '19 at 19:00
  • $\begingroup$ @KevinCarlson Thanks! So we first had to find a map that we could see was canonical (basis-independent), and then confirmed that it was natural. It's not as if we could have first proved the existence of a natural transformation and then use that to find (or infer the existence of) a basis-free map, right? And from your "tend" I infer that there are "canonical" maps that aren't natural? (And thanks, I fixed the $v^*$ thing. I also learned a bit of notational etiquette in the process). $\endgroup$ – A_P Jul 16 '19 at 16:29

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.

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  • $\begingroup$ Thanks very much! If I understood your first half correctly, it's basically saying that there's no natural way to pick an endomorphism on arbitrary $V$ besides scaling (which is intuitively true). It seems like we used the naturality condition to prove this, but for some reason it feels like a sleight of hand: when we say that a choice of basis is equivalent to an isomorphism with $ℝ^n$, we're basically formalizing the intuition that it's meaningless to describe a vector (let alone map) in an arbitrary space without dictating its components. Or maybe I'm way off base here. $\endgroup$ – A_P Jul 16 '19 at 19:28
  • $\begingroup$ I'll try to formalize my feeling. Looking at $E^{-1}\eta_{V}E$, we see that first $E$ converts a $v$ into a basis-free "description," and then $\eta_{V}$ operates on it. But a basis-free description of $v$ is nothing more than a label: $v$ -- and what can you even do to such a thing other than scale it? Alternatively, to describe $\eta_{V}$ (so that we can compare it to the LHS), we have to say what it does to the basis -- but we knew all of this before trying to formalize it in terms of naturality. I'm not entirely sure I'm making sense (and if not, maybe someone can set me straight). $\endgroup$ – A_P Jul 16 '19 at 20:00
  • $\begingroup$ @A_P The point of the first part was that the naturality of η : Id → Id implies that $η_V : V → V$ is coordinate independent in the obvious sense that the matrix of $η_V$ looks the same in every basis of $V$. The rest of the answer suggests how you can think of general natural transformations as also implying coordinate independence in a reasonable sense, with the caveat that naturality is in general significantly stronger. $\endgroup$ – user54748 Jul 16 '19 at 21:37
  • $\begingroup$ Going back to the specific case of $η : V → V$, you can either ask yourself what are the coordinate independent choices of $η$, and you can prove those are exactly the $rI$ without any mention of natural transformations either as an excercise in Linear or general Algebra. Alternatively and more restrictively, you can ask which choices of $η$ extend to a natural transformation η : Id → Id, and show that $η_V$ in this case again has to be $rI$, but you should know that these two notions won't end up agreeing in general. $\endgroup$ – user54748 Jul 16 '19 at 21:39
  • $\begingroup$ Naturality, which respects not only change of basis, but all the ways an object $X$ can be embedded or qotiented or whatevered to any other object $Y$, is a lot more useful of the two, even if the kind of independence and homogeneity it implies is harder to put in words -- words other than "The square commutes.", that is. $\endgroup$ – user54748 Jul 16 '19 at 21:41

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.

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  • $\begingroup$ Thanks! My notation $v^*$ was indeed a mistake -- I didn't mean it to relate to $v$ in any way, but to be an arbitrary element of $V^*$. And as mentioned, I did work through the equivalence of the two paths. I will try to understand the second half of your answer. $\endgroup$ – A_P Jul 15 '19 at 20:57
  • $\begingroup$ Okay, so in the process of proving naturality, we had to use the definition of $(-)^{**}$, which we can plainly see is basis-independent. So is it really true that determining naturality helped us see that such a canonical construction exists? $\endgroup$ – A_P Jul 16 '19 at 16:35
  • $\begingroup$ You're right, we can see that it is basis-independent without even referencing category theory! So perhaps no in this case you don't need to talk about category theory to see this, but it's a good illustration of the broader idea of natural transformations giving rise to canonical constructions. $\endgroup$ – desiigner Jul 16 '19 at 16:39
  • $\begingroup$ Do you know of a simple example where demonstrating the existence of a natural transformation leads to the discovery of a basis-independent (or otherwise canonical, depending on category) map? $\endgroup$ – A_P Jul 16 '19 at 17:18
  • $\begingroup$ The tensor-hom adjunction is a natural isomorphism, and you can show it holds in a lot of different settings. This tells you, for example, that if $U,V,W$ are $k$-vector spaces, we have an isomorphism $\text{Hom}_k(U\otimes_k V, W) \cong \text{Hom}_k(U, \text{Hom}_k(V,W))$ and you can show this without picking bases on any of the vector spaces. $\endgroup$ – desiigner Jul 16 '19 at 17:55

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