There are quite a few questions both on this site and math.SE related to this topic as well as what we mean when we say "natural" or "canonical". For the purposes of this question, I'm going to consider canonical and natural to be synonyms, and use wikipedia's definition of an unnatural isomorphism:

A particular map between particular objects may be called an unnatural isomorphism (or "this isomorphism is not natural") if the map cannot be extended to a natural transformation on the entire category.

From here: https://en.wikipedia.org/wiki/Natural_transformation

There's an excellent question here: https://math.stackexchange.com/q/622589/816960 which raises the issue that it seems to be an artefact of the definition of a natural transformation that there is no canonical isomorphism between $V$ and $V^*$, since the dual functor is contravariant. One of the answers suggests that the only way to resolve this is by showing that every dinatural transformation from the identity functor to the dual functor must be zero.

Personally I feel this doesn't get around the issue, because it seems to require us to redefine a canonical isomorphism between objects in a category as one which can be extended to a nontrivial dinatural transformation, and abandon the definition given above. Otherwise we're still left with a "definition not applicable"-based proof.

Another candidate is given here: https://mathoverflow.net/a/345148/175537 in which we change the definition of the dual functor.

One way to get around this is by working instead with the core groupoid $\mathbf{Vect}_{core}$, consisting of vector spaces and invertible linear transformations, and defining $*:\mathbf{Vect}_{core} \to \mathbf{Vect}_{core}$ to be the functor taking $f:V \to W$ to $(f^{-1})^{\ast}: V^\ast \to W^\ast$, the linear adjoint of its inverse. Then one can ask whether the identity is naturally isomorphic to the covariant dual functor $\ast$. It is not.

But it seems like in both cases, we need a different definition of something to prove the nonexistence of a canonical isomorphism.

My questions are:

Can we prove that any isomorphism between a finite-dimensional vector space and its dual is unnatural using the definition above, without appealing to the fact that the definition of a natural transformation doesn't allow comparisons of covariant functors with contravariant ones?

Since I suspect the answer to the above is "no", does this mean the definition of "unnatural isomorphism" isn't quite right? what is the right definition of "canonical isomorphism" to use in order to do this properly with category-theoretic machinery?

  • 3
    $\begingroup$ Okay, I misunderstood your question at first. But now I'm not sure I get it. You say "cannot be extended to a natural transformation". But a natural transformation is something between functors. So are you really asking if the map cannot be extended into a triple (functor, functor, natural transformation)? Do you have any requirement on the functors? For example, between what categories...? $\endgroup$ Mar 9, 2021 at 14:12
  • 2
    $\begingroup$ @TimCampion It does come up (e.g. in the MSE question), and the result is that any dinatural transformation between the identity and the dual functors is zero. $\endgroup$ Mar 9, 2021 at 14:14
  • 2
    $\begingroup$ What do you call the dual functor? The contravariant one? Then there can't be a natural isomorphism by definition, or a dinatural isomorphism by the MSE answer you linked. The inverse of the dual, restricted to the core groupoid? Then it's also done in the MSE answer you've linked. $\endgroup$ Mar 9, 2021 at 15:03
  • 2
    $\begingroup$ I don't see any future in attempting to define an "unnatural isomorphism". Consider the positive question instead: any isomorphism (indeed homomorphism) $V\to V^*$ is a structure, otherwise written $\mu:V\otimes V\to K$. Then $(V,\mu)$ and $(V,\mu')$ are different mathematical objects. $\endgroup$ Mar 9, 2021 at 15:55
  • 8
    $\begingroup$ You could consider the category "Pair" of pairs $(V, \mu)$ where $V$ is a vector space and $\mu: V \otimes V \to K$ is a non-degenerate pairing (i.e. an isom $V \cong V^*$. A morphism is a map of vector spaces compatible with pairings. There is a forgetful functor from Pair to Vect which forgets $\mu$. You can ask: Does the forgetful functor admit a section? This would be a functorial choice of isom $V \cong V^*$ for each $V$. The answer is no, there is no such section. $\endgroup$ Mar 9, 2021 at 18:00

2 Answers 2


Some people (including me) think that "canonical" should be synonymous with "natural on isomorphisms" or "Functorial on isomorphisms" (depending on if you are talking of a "canonical object" or a "canonical arrow"). Doing so solves the problem of variance in the definition.

To be clear, by "Functorial on isomorphism" I justs means that we have a functor $F:Core(C) \to D$ where $Core(C)$ is the subcategory of $C$ containing all objects but only isomorphisms as arrows. And by "Natural on isomorphisms", I'm asking for things that are natural transformation between such functors, i.e. that satisfies the naturality condition only with respect to isomorphisms.

If you look at the category of finite dimensional vector spaces and linear isomorphisms between them, here $V \mapsto V^*$ can be made into an actual (covariant) endofunctor of this category as you can fix the contravariance by inverting the morphisms, so that an invertible arrow $f:V \to W$ induces $(f^*)^{-1} : V^* \to W^*$.

And there you can concretely show that there is no natural isomorphism between this functor and the identity.

Indeed, chosing an isomorphism $V \simeq V^*$ gives you a non-degenerate bilinear form on $V$ and you can always find an automorphism of $V$ that does not preserves this bilinear form, which is exactly what the naturality on isomorphisms would mean! So at the end of they day, one recovers exactly the argument that Paul Taylor or Chris Schommer-Pries made in the comments, but starting with a concrete categorical definition of "canonical".

  • 1
    $\begingroup$ I think my answer is canonically isomorphic to yours. :) $\endgroup$ Mar 9, 2021 at 19:35
  • 3
    $\begingroup$ I think they are actually quite different. You are looking at automorphism of k-Vect, I'm not. (But I also like your answer, in fact I'd love to see a situation where both leads to different conclusion) $\endgroup$ Mar 9, 2021 at 19:36
  • 1
    $\begingroup$ I think an automorphism of k-Vect (which preserves the dual space functor) can be constructed from any automorphism $\alpha$ of a single vector space $V$. Precompose anything from $V$ with $\alpha^-1$; postcompose anything to $V$ with $\alpha$; similarly with $V^*$ and $\alpha^*$,$(\alpha^*)^{-1}$. $\endgroup$ Mar 9, 2021 at 19:39
  • 3
    $\begingroup$ Here is an example where I think the two differs: Let $P$ be the category of posets with all supremum and infimum and all order preserving maps between them. Now the sentence "each object $X$ of P admits a canonical morphism $1 \to X$". Is true in my definition of caninical as I can send the unique element of $1$ to the top elements and this is functorial on isomorphisms, but it is not true for you definition as $P$ admit an automorphisms that "reverse the order relation" on each object which preserve $1$ but do not preserve my "canonical" choice. $\endgroup$ Mar 9, 2021 at 19:44
  • 1
    $\begingroup$ OK, sorry, I see your point. You set things up so that we do not have the data of which relation is $\ge$ and which is $\le$ for each poset. $\endgroup$ Mar 9, 2021 at 20:00

$\newcommand\kVect{k\text{-Vect}}$When people say ‘canonical’ what they mean in this context is something like ‘definable without parameters’ (i.e., without choosing bases; actually, without choosing anything at all). See for instance the entry for “Definable Set” in Wikipedia. The important point is that canonical objects are invariant under automorphisms that preserve the relevant structure.

This means our isomorphism $V\to V^*$ should be invariant under all automorphisms of $\kVect$ that preserve the dualizing functor, considered as a contravariant functor from $\kVect$ to $\kVect$. This is not possible to obtain even for one finite-dimensional $V$. The reason is that with respect to some bases $B$, $C$ on $V$, $V^*$, a given isomorphism has the form of an identity matrix. But for any matrix $A$ representing an isomorphism $V\to V$, there is an automorphism of $\kVect$ which preserves the dual space functor and which transforms the representation of the given automorphism (with respect to the same bases $B$, $C$) to $A^T A^{-1}$. We can find a matrix $A$ for which this expression is different than $I$ (unless $\dim(V)=1$ or $\lvert k\rvert=2=\dim(V)$, as I learned here). Thus the isomorphism $V \to V^*$ is not preserved by this automorphism.

  • $\begingroup$ Of course "unless $\dim(V) = 1$ or $\lvert k\rvert = 2 = \dim(V)$" should be "unless $\dim(V) \le 1$ …". (You could also be cutesy and re-phrase as "$\dim(V) \le 1$ or $\lvert V\rvert = 4$".) $\endgroup$
    – LSpice
    Mar 9, 2021 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.