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Can we say that an isomorphism $V \cong V^*$ is not natural? It seems so intuitively, but formally the notion of a natural transformation between a functor and a cofunctor is not defined (or is it?).

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up vote 12 down vote accepted

Indeed, a natural transformation is defined for two parallel functors, that is, functors with the same domain and codomain. The identity functor I and the single dual functor * are not parallel:

$I:\sf{Vect_K}\to \sf{Vect_K} $ $*:\sf{Vect_K}^{op}\to \sf{Vect_K} $

In other words: I is covariant and * is contravariant.

But let's ignore that. Suppose we have a collection of linear isomorphisms $\{a_V:V\to V^*\}_{V\in\sf{Vect_K}}.$

The naturality square would read

$ \begin{array}{ccccc} & V & \longrightarrow^{a_V} & V^* & \\ L & \downarrow & & \uparrow & L^*\\ & W & \longrightarrow^{a_W} & W^* & \end{array}$

where L* is the usual dual (precomposition) linear map defined by

$L^*f := [v\mapsto f(Lv)]=f\circ L,\ \ \ \ \ f\in W^*$

Now this diagram commutes iff

$\alpha_V = L^*\alpha_W L$

for all linear maps $L:V\to W$. But this cannot be since $\alpha_V$ is an isomorphism, and L not necessarily, e.g. take L=0.

[PS: I copied this answer from my answer at physicsforums ]

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This is more of a long comment on wildildildlife's answer. A natural question after reading that proof is: what happens if we restrict to the groupoid of finite dimensional vector spaces and their isomorphisms? (It may seem that picking $L=0$ is a bit of a "cheat".) The fact is that there is still no equivalence between the identity and the dual functor. The point is that the collection of isomorphisms $\{\alpha_V \colon V \to V^\ast \}$ is only allowed to depend on $V$, i.e. there must be a specific morphism for every $V$ that makes every diagram you can cook up using your category commutative.

In particular we can look at diagrams of the form

$ \begin{array}{ccc} V & \longrightarrow^{\alpha_V} & V^* \\ \downarrow & & \uparrow \\ V & \longrightarrow^{\alpha_V} & V^*. \end{array}$

Now we get $\alpha_V = A^\ast \alpha_V A$ for all $A \in \mathrm{GL}(V)$. Equivalently, if we use $\alpha_V$ to identify two bases of $V$ and $V^\ast$, we must have $\mathrm{id}_V = A^T A$ for all $A \in \mathrm{GL}(V)$, which is absurd.

This point of view also leads directly to what condition we need to impose on the category for the existence of an equivalence like this: we need to find a category of vector spaces where the automorphisms $A \colon V \to V$ are the orthogonal matrices (wrt a basis). The most natural thing to consider is then finite dimensional spaces equipped with an inner product1 and morphisms preserving the inner product. Hence we have rediscovered the principle that you can identify a vector space with its dual precisely when it has an inner product.

1 Remark: By an inner product I really mean an arbitrary nondegenerate bilinear form, so over $\mathbf R$ or $\mathbf C$ this is non-standard usage. Note that this category is automatically a groupoid, since any map of vector spaces preserving the inner product is an isomorphism.

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Interesting answer/comment! – wildildildlife Mar 13 '11 at 13:49
This is great stuff 8) – Alexei Averchenko Mar 13 '11 at 13:55
@Alexei, wildildildlife: thanks! On an unrelated note, what's up with the extra "ild"? – Dan Petersen Mar 13 '11 at 14:34
I am sorry, but what is an "ild"? – wildildildlife Mar 13 '11 at 14:50
Ah! Well, I like the band and their name, but I wanted a nonexisting name; as they already added an ild to 'wildlife', I thought I'd do the same. – wildildildlife Mar 13 '11 at 15:43

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