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Is a vector space naturally isomorphic to its dual?

The answer is no: you know it, I know it, everyone who has spent years becoming familiar with this family of spaces knows through intuition that the answer is no: a vector space is not naturally isomorphic to its dual. But pinning down exactly why not proves challenging.

I originally asked this question on MathOverflow on 3rd November 2019 but sadly the question was closed. I repeat it here verbatim before proceeding.

The question

This question may not be as easy to answer as you think! Some tangentially-related questions have appeared on math.stackexchange but I’m not really convinced by the answers.

In what follows I will assume all vector spaces under discussion are finite dimensional.

A vector space is naturally isomorphic to its double dual

In an early linear algebra course we are told that “a finite dimensional vector space is naturally isomorphic to its double dual”. The isomorphism in question is \({**}_V : V \to V^{**}\), \(v^{**}(\phi) = \phi(v)\). We are told that this isomorphism is “natural” because it doesn’t depend on any arbitrary choices. The notion of “natural”, or “independent or arbitrary choice”, is made precise via the concept of a category theoretical “natural transformation”. Specifically, the operation \(**\) on vector spaces gives rise to a functor whose action on maps is \(f^{**} : V^{**} \to W^{**}\), \(f^{**}(v^{**}) = f(v)^{**}\). In fact this is exactly the condition for the naturality square to commute and so \({**}_V\) is indeed a natural transformation (between the identity functor and \(**\)) which is an isomorphism.

A vector space is naturally isomorphic to its dual!

So far, so familiar. But there’s something that doesn’t quite hold up about all this. Let’s adapt the above to show that \(V\) and \(V^*\) are “naturally isomorphic”. We do this by following exactly the same procedure, replacing \(**\) everywhere with \(*\). The only change we have to make is to come up with an arbitrary isomorphism \(*_V\) for each \(V\). Other than that, the whole construction goes through unchanged. Specifically, once we have chosen \(*_V\) we define the functorial action on morphisms to be \(f^{*} : V^{*} \to W^{*}\), \(f^{*}(v^{*}) = f(v)^{*}\).

In particular I have a natural isomorphism between the identity functor and \(*\)!

Objections to the construction

One could make a few objections to this construction, but they seem to be circular.


All attempts to explain why I haven’t really shown that a finite dimensional vector space is naturally isomorphic to its dual seem to invoke circular reasoning.

I can only conclude that if the notion of natural transformation is going to be used to formalise the concept of “independent of arbitrary choice” then something needs to be tightened up. My choice of \(*_V\) was indeed arbitrary but it is not ruled out by the notion of natural transformation.

How could we proceed? Could the absence of choice be used to rule out the construction of \(*_V\)? Answers to an earlier question seem to suggest that is an irrelevant issue but in light of the above I’m not convinced. More generally, does this kind condition perhaps only make sense in a constructive or intuitionistic setting? I have a clue about how to formalise this condition in type theory via parametricity, so perhaps that is the key!

(This question is similar to an earlier one. I’m not convinced by the answer. It seems to be making an objection of the third form above, which seems circular to me.)

Proceeding to the answer

The aim of the question is twofold: firstly to demonstrate that there is something fishy about the standard justification for the statement that “a finite-dimensional vector space is naturally isomorphic to its double dual” and secondly to ask for a less fishy definition of “natural” that can be used to show whether or not there is a natural mapping from a vector space to its dual (presumably there is not).

[Terminological note: when talking about the technical and specific notions of natural transformation and natural isomorphism from category theory I will use italics; when talking about the informal and intuitive notion of “naturalness” I will use normal font.]

The statement that “a finite dimensional vector space is naturally isomorphic to its double dual” appears in various places1. We have an intuitive sense that the isomorphism \(v \mapsto \phi \mapsto \phi(v)\) is natural but formally pinning down what “natural” means is hard. The standard category theoretical explanation in terms of a natural transformation between functors offers one attempt. Indeed the statement holds first position in Tom Leinster’s list of “phenomena, facts and concepts that category theory sheds light on”. But appealing to natural transformation as an explanation of naturalness falls down on two counts: it is too restrictive in one sense and insufficiently restrictive in another.

The sense in which it is too restrictive is that many natural things cannot be expressed as a natural transformation between functors at all! For example, suppose we want to know whether it is natural to contract a vector with a dual vector \(V \otimes V^* \to \mathbb{R}\). The dual vector space functor is contravariant, so \(V \mapsto V \otimes V^*\) is not a functor at all and the notion of natural transformation cannot shed light on this question! But what operation could be more natural than contraction? If the concept of natural transformation cannot speak about whether contraction is natural then there must be a vast range of natural constructions that it cannot speak about. So, there are some candidates for naturality which the concept of natural transformation gives no answer to.

By contrast, there is a sense in which appealing to natural transformations to define naturalness is insufficiently restrictive: there is a functor which on objects takes a space to its dual space and is naturally isomorphic to the identity! That functor is the one described above in my original MathOverflow question above. Yet the components of the natural transformation in question are not natural. They are utterly arbitrary! We all feel it in our bones that there is nothing “natural” about that transformation yet it is a natural transformation. So, there are some candidates for naturality which category theory gives a false positive answer to.

A elementary attempt at an answer

The former difficulty, the one in which the notion of natural transformation has insufficient expressivity, is the first hurdle that we encounter when asking whether “a vector space is naturally isomorphic to its dual”. To reiterate, the usual dual vector space functor is contravariant, so it doesn’t even make sense to ask whether it is naturally isomorphic to the identity functor. On the other hand, this difficulty can be finessed.

Here is a elementary route to the answer. Let us take as axiomatic that the following operations are natural and that the composition of natural maps is natural2.

Next let \(g\) be an arbitrary natural map \(V \to V^*\). Our goal is to determine what \(g\) could possibly be. We start by noticing that given an element of \(V \otimes V\) we can apply \(g\) to the right component to obtain an element of \(V \otimes V^*\). Subsequently we can contract to obtain an element of \(\mathbb{R}\). If our hypothesis holds then we have produced a natural linear map \(b : V \otimes V \to \mathbb{R}\) or in other words, a natural bilinear form on \(V\). Our intuition is even stronger about this that it was about dual spaces: vector spaces do not come with a natural non-zero bilinear form!

This simple change of perspective brings the question into a covariant form: both \(V \otimes V\) and \(\mathbb{R}\) are functors3 covariant in \(V\). Therefore we can ask whether \(b\) is a natural transformation \(V \otimes V \to \mathbb{R}\). If so its naturality square must commute, that is \(b_{V'}(f(v) \otimes f(v')) = b_{V}(v \otimes v')\). We deduce that \(b\) is zero by simply choosing \(f = 0\), hence \(g : V \to V^{*}\) is zero.

If we accept (more on this later) that category theory never gives false negative answers to questions of naturalness then we have our answer: there is no non-zero natural bilinear form and therefore no non-zero natural map \(V \to V^*\).

Hidden in this approach is the answer to why my earlier attempt to construct of a natural map \(V \to V^*\) should be ruled out. Specifically, it shows why my definition of \(*\) is not “the real” dual space functor: it does not respect the structure of \(V^*\) as a function space because under it contraction \(V \otimes V^* \to \mathbb{R}\) is not natural. The concept of natural transformation cannot directly show the non-naturalness of my construction because it cannot be applied to contraction. It was only by taking contraction as axiomatically natural that we were able to change our perspective to one that the concept of natural transformation could talk about.

That is, we can answer the following questions

Naturality of contraction

We took as axiomatic that the following operations are natural. Can category theory provide us with justification for that assumption?

Let’s look at contraction. The category of vector spaces is closed monoidal, that is, its tensor operation \(\otimes\) and its linear map former \(\multimap\) satisfy

\[ \textrm{Hom}(A, B \multimap C) \cong \textrm{Hom}(A \otimes B, C) \]

where this bijection (which arises from a particular adjunction) is natural in \(A\) and \(C\)4. Let’s look at what happens if we choose \(A = B \multimap \mathbb{R}\), and \(C = \mathbb{R}\). The bijection becomes

\[ \textrm{Hom}(B \multimap \mathbb{R}, B \multimap \mathbb{R}) \cong \textrm{Hom}((B \multimap \mathbb{R}) \otimes B, \mathbb{R}) \]

or, since \(B \multimap \mathbb{R}\) is typically written \(B^*\),

\[ \textrm{Hom}(B^{*}, B^{*}) \cong \textrm{Hom}(B^{*} \otimes B, \mathbb{R}) \]

Now, applying a natural bijection to a natural element ought to result in another natural element, so if we hypothesise that \(\mathrm{id} \in \mathrm{Hom}(B^*, B^*)\) is natural, then we conclude that its image in \(\mathrm{Hom}(B^{*} \otimes B, \mathbb{R})\) is natural. That image is contraction, thus contraction is natural! We can replace our axiom about naturality of contraction with a simpler axioms about the naturality of \(\mathrm{id}\) and the naturality of bijections arising from adjunctions.

Presumably a similar line of reasoning can conclude that \(v \otimes w \mapsto v \otimes g(w)\) is natural, based on the naturality properties of monoidal categories.

Remaining questions

  1. for example in Jason Polak’s writing↩︎

  2. If they were not natural then we would be surprised if anything was!↩︎

  3. As a functor \(\mathbb{R}\) is the zero-fold tensor product of \(V\). It takes all morphisms \(f : V \to V'\) to the identity function \(I : \mathbb{R} \to \mathbb{R}\).↩︎

  4. It would probably be required to be natural in \(B\) too, if it could be!↩︎