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What, precisely, do we mean when we say that a f.d. vector space is canonically isomorphic to its double dual?

I’ve been reading the Xena Project blog, which has been loads of fun. In the linked post Kevin gives the natural isomorphism $V to V^{ast ast}$ from a f.d. vector space to its dual as an example of a "canonical isomorphism":

People say that the obvious map from a vector space to its double dual is “canonical” but they could instead say that it is a natural transformation.

I think this is not a complete description. When we say that $V$ is canonically isomorphic to its double dual we don’t just mean that there exists some natural isomorphism from the identity functor to the double dual functor – what we mean, and what we use in practice, is that a particularly canonical natural transformation $V to V^{ast ast}$ is an isomorphism. We can multiply any such natural isomorphism by a scalar $c neq 0, 1$ in the ground field $k$ and obtain a different natural isomorphism between these two functors, and these are not canonical!

But reading the linked post showed me I don’t know precisely what I mean by "canonical" in the paragraph above. So:

Question: What, precisely, do we mean by "canonical" when we say that the usual natural transformation $V to V^{ast ast}$ is the "canonical" one we want? What other "canonical" maps are canonical in the same way?

What follows are some off-the-cuff thoughts about this.


One thought is that when we write this map down we’re "using as little as possible"; we’re not even really using that we’re working in vector spaces. If we define the dual to be the internal hom $[V, k]$ then ultimately all we’re using is the evaluation map $V otimes [V, k] to k$ together with currying. In other words, one way to make precise what we’re using is the closed monoidal structure on $text{Vect}$. (We don’t even need a closed structure if we define the dual to be the monoidal dual but I expect the internal hom to be more familiar to more mathematicians.)

So, here’s what I think ought to be true: in the free closed monoidal category on an object $V$ (blithely assuming that such a thing exists – maybe I should have used monoidal duals after all, because I’m much more confident that the free monoidal category on a dualizable object exists), there ought to be a literally unique morphism $V to [[V, 1], 1]$ where $1$ denotes the unit object. This is the "walking double dual map" and, by the universal property, reproduces all the other ones, and I think this is a candidate for what we might mean by "canonical map."

This trick of taking free categories is a really good trick actually, since it also cleanly describes the nonexistence of canonical maps. For example there is no canonical map $V to V otimes V$, and you could use representation theory to show that the only $GL(V)$-equivariant map is zero, but actually it’s just true that in the free monoidal category on $V$ there are no maps $V to V otimes V$ whatsoever. Similarly there is no canonical map $V to V^{ast}$, and similarly you could use representation theory to show that the only $GL(V)$-equivariant map is zero, but actually it’s just true that in the free closed monoidal category on $V$ (if it exists; take the free monoidal category with duals if not) there are no such maps whatsoever.

But I don’t think this sort of reasoning is enough to capture other examples where two functors are naturally isomorphic and there’s a particularly canonical natural isomorphism that we want in practice. For example, it’s also true that de Rham cohomology $H^{bullet}_{dR}(X, mathbb{R})$ on smooth manifolds is canonically isomorphic to singular cohomology $H^{bullet}(X, mathbb{R})$ with real coefficients, and by this we don’t mean just that there exists some natural isomorphism but that the particularly canonical natural transformation given by integrating differential forms over simplices is an isomorphism. We can multiply any such natural isomorphism by $c^n$ in degree $n$ where $c in mathbb{R} setminus { 0, 1 }$ and we’ll get a different one, even one that respects cup products, and again these are not canonical (although I wonder if we could easily distinguish the usual one from the one obtained by setting $c = -1$ – do we have to choose how simplices are oriented or something like that somewhere?) But now I really don’t know what I mean by that! It doesn’t seem like I can pull the same trick of zooming out to a more general categorical picture. de Rham cohomology is a pretty specific functor defined in a pretty specific way. Maybe this one is a genuinely different sense of "canonical," closer to "preferred," I don’t know.


Some previous discussion of "canonical" stuff on MO:

MathOverflow Asked by Qiaochu Yuan on January 4, 2021

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