I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically."

I've finally sat down to work through this, and -

Okay, yes, it is easy to see that the "canonical isomorphism" from $V$ to $V^{**}$ is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor.

Also, I see that there is no way that the functor $V\mapsto V^*$ could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to:

Is contravariance the whole problem?

To elaborate:

I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism $V\rightarrow V^*$ would feel more like a theorem as opposed to an artifact of the inapplicability of a definition.

Then I tried to create a definition of a natural transformation from a covariant functor $F:\mathscr{A}\rightarrow\mathscr{B}$ to a contravariant functor $G:\mathscr{A}\rightarrow\mathscr{B}$. It seems to me that this definition should be that all objects $A\in\mathscr{A}$ get a morphism $m_A:F(A)\rightarrow G(A)$ such that for all morphisms $f:A\rightarrow A'$ of $\mathscr{A}$, the following diagram (in $\mathscr{B}$) commutes:

$$\require{AMScd}\begin{CD} F(A) @>m_A>> G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @>>m_{A'}> G(A') \end{CD}$$

This is much more stringent a demand on the $m_A$ than the typical definition of a natural transformation. Indeed, it is asking that $m_A=G(f)\circ m_{A'}\circ F(f)$, regardless of how $f$ or $A'$ may vary. Taking $\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k$, $F$ the identity functor and $G$ the dualizing functor, it is clear that this definition can never be satisfied unless $m_V$ is the zero map for all $V\in\text{f.d.Vec}_k$ (because take $f$ to be the zero map). In particular, it cannot be satisfied if $m_V$ is required to be an isomorphism.

Is this the right way to understand (categorically) why there is no natural isomorphism $V\rightarrow V^*$?

As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors?

Note: I have read a number of math.SE answers regarding why $V^*$ is not naturally isomorphic to $V$. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. (This one was closest.) Hence my question here.

Martin Sleziak
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Ben Blum-Smith
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    To have isomorphism $V\rightarrow V^*$ you need to pick base of $V$ if so its hardly canonical. I THINK it depends on the choice. – user52045 Dec 30 '13 at 20:47
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    @user52045 - agreed. It is intuitive to me that any basis is just as good as any other and they lead to different isomorphisms. The question is about how category theory can turn this intuition into a theorem. – Ben Blum-Smith Dec 30 '13 at 20:52
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    In this context, it would make sense to restrict to the non-full subcategory of all f.d. vector spaces and isomorphisms as morphisms (in which you cannot take $f$ to be the zero map!) and the result becomes more interesting. – Mariano Suárez-Álvarez Dec 30 '13 at 22:58
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    Exellent question!! This is similarly discussed in the Eilenberg-MacLane paper where they invent category theory. I completely agree with you that the "proof" of non-naturality is very disappointing (because, IMHO, dinatural transformations are only superficially similar to natural transformations). All this has nothing to do with the usual verbiage about choices of basis. The are natural equivalences, e.g., between categories and their skeletons, which make havy use of the axiom of choice even for classes! – Jochen Oct 22 '20 at 13:27

1 Answers1


Congratulations, you have reinvented the notion of a dinatural transformation (see for instance MacLane's Categories for the working mathematician, section IX.4). And your proof, that every dinatural transformation from the identity functor to the dualization functor is zero, is correct. And I agree that this is one (and perhaps the only) way to make precise that a f.d. vector space is not canonically isomorphic to its dual. By the way, for euclidean vector spaces, there is a canonical isomorphism, given by $V \mapsto V^*, v \mapsto \langle v,- \rangle$.

1st Edit: In the comments, Mariano has suggested to restrict to isomorphisms as morphisms. This comes down to: If $n \in \mathbb{N}$, is there some $M \in \mathrm{GL}_n(K)$, such that for all $A \in \mathrm{GL}_n(K)$ we have $M = A^T \cdot M \cdot A$? By taking $A$ to be some diagonal matrix we immediately see that this is only possible for the trivial case $n=0$ or when $K=\mathbb{F}_2$.

2nd Edit: Let us look more closely at the case $K=\mathbb{F}_2$. For $n=1$ we can take $M=(1)$. As mentioned by ACL (in Mariano's link in the comments), for $n=2$ we can take $M=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$.

Thus, for $2$-dimensional $\mathbb{F}_2$-vector spaces $V$ there is a canonical isomorphism $V \cong V^*$ which is natural with respect to isomorphisms. It is induced by the unique(!) alternating $2$-form on $V$.

For $n=3$ this is not possible: By taking $\small A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ it follows that $M_{11}=M_{13}=0$, and by taking $\small A=\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ it follows $M_{12}=0$, so that $M$ is not invertible. A similar reasoning works for all $n \geq 3$.

Martin Brandenburg
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  • Could you expand on your last sentence, or give a reference? – Michael Albanese Dec 30 '13 at 22:52
  • A good reference is MacLane's book, as Martin mentioned, @MichaelAlbanese – Mariano Suárez-Álvarez Dec 30 '13 at 22:56
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    Regarding your EDIT: see the answer http://mathoverflow.net/a/121989/1409 and the first comment to it. – Mariano Suárez-Álvarez Dec 31 '13 at 16:38
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    There is one case you're missing: $n=1$ and $K=\mathbb{F}_3$ works as well. – Eric Wofsey Dec 18 '17 at 06:17
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    The canonical isomorphism $V → V^*,~x ↦ ⟨v,~⟩$ for finite dimensional inner product spaces only works in the category of finite dimensional inner product spaces – where arrows are given by isometric linear maps. This excludes the zero map, so Bens proof doesn’t apply here. (I was perplexed for a second because how could there be a canonical isomorphism when Ben just proved that there cannot be?) – k.stm Jan 11 '18 at 08:23
  • I think it's a "canonical" isomorphism of vector spaces with an inner product. Once you fix an inner product $\sigma$ there exists a "natural" isomorphism of this Hilbert space to its dual given by [Riesz representation theorem](https://en.wikipedia.org/wiki/Riesz_representation_theorem) $V \rightarrow V^*, x \mapsto \sigma(x, .)$. So I'd consider this a property of Hilbert spaces rather than something unique to euclidean spaces. – chickenNinja123 Jan 26 '22 at 07:50
  • Sure. $\text{~~~}$ – Martin Brandenburg Jan 27 '22 at 19:51