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There are several questions on this already, and my question arose after reading the explanation given in the following:

In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual?

We know a natural isomorphism $\eta:F \to G$ is defined as a natural transformation whose components are all isomorphisms in $\mathbf{D}$, where $F,G: \mathbf{C} \to \mathbf{D}$ are functors. What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

Also, in the answer to the quoted question, the author states that there is a natural isomorphism between a Euclidean vector space and its dual, and gives it as $V \mapsto V^*, v \mapsto \langle v,- \rangle$. What does this notation mean? And also, if there are natural isomorphisms between some finite-dimensional vector spaces, how can we "prove" that there are none in general, as the OP wants? It seems like the best we could do is a counterexample? Or am I missing a key distinction between Euclidean spaces and finite-dimensional spaces?

Brendan Langfield
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    The point is that a Euclidean vector space comes equipped with a distinguished inner product while, in general, a finite dimensional vector space has many inner products, but no way to pick one of them from the others. – Somos Mar 03 '21 at 18:36
  • The key difference is that the Euclidean spaces have an inner product, or what amount to the same thing, the Euclidean distance. In vector spaces over fields other than reals, or over infinite-dimensional real vector spaces, there is no such way of turning a vector into a functional. – Jyrki Lahtonen Mar 03 '21 at 18:37

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A Euclidean space is not just a finite-dimensional vector space. It is a finite-dimensional vector space over the reals together with an inner product space. So it is an object of a very different category than mere vector spaces (over the reals). The extra structure gives in this case a particular choice for a mapping $V\to V^*$, namely using the inner product. In the absence of inner products, that is if all you have is a finite-dimensional vector space over the reals, then there is no canonical choice for a mapping $V\to V^*$ even though the two spaces have the same dimension. So, they are isomorphic but there is no canonical choice for any particular isomorphism.

Further, the construction $V\mapsto V^*$ is contravariant, so it does not even make sense to ask for a natural transformation between the identity functor on $\mathrm {Vect}$ and the (contravariant!) functor $-^*$. The double dual is covariant and so the question becomes meaningful and there is then a natural isomorphism between the functors.

You could account for the reversal of arrows and try to compare the identity functor and the contravariant dualisation. Once this is given precise meaning it can be shown that any assumed such transformation leads to a contradiction, so it does not exist.

Ittay Weiss
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  • The issue of contravariance appears to be treated in the OP in the linked question, using the notion of a dinatural transformation. So would it be more clear to say that there's no natural transformation between a finite-dimensional vector space and its dual given that it has no additional structure? – Brendan Langfield Mar 03 '21 at 18:44
  • Yes, and there is also no dinatural transformation either. – Ittay Weiss Mar 03 '21 at 19:37
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What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

I assume that the most puzzling aspect of this is the fact that the term "isomorphism" is used for two different kinds of objects; one is a map between vector spaces, and the other is a family of such maps. In the most general sense, "isomorphism" can refer to any reversible structure preserving map. An invertible linear map between vector spaces is called an isomorphism because it preserves the "linear" structure of a vector space. In particular, the essential thing that makes an object a vector space is the ability to take linear combinations of elements, i.e. to ("nicely" and "meaningfully") refer to an object $ax + by$ for scalars $a,b$ and vectors $x,y$. $f:V \to W$ is an isomorphism, then this essential structure is preserved in the sense that $$ f(ax + by) = af(x) + bf(y). $$ In other words, taking a linear combination in the first space and then applying an isomorphism yields the same result as applying the isomorphism and then taking (the same) linear combination in the second space.

On the other hand, the essential thing that makes a map $F:\mathbf C \to \mathbf D$ a functor is the fact that $F(f \circ g) = F(f) \circ F(g)$ holds for all compatible morphisms $f,g \in \mathbf C$. A natural isomorphism $\eta:F \to G$ preserves this structure in the sense that for any morphism $f:X \to Y$ in $\mathbf C$, we have $$ \eta_Y \circ F(f) = G(f) \circ \eta_X. $$ In other words, applying the first functor to a morphism and then applying the natural isomorphism yields the same result as applying the natural isomorphism and then applying the second functor to that same morphism.

Ben Grossmann
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  • You are correct in thinking that that is the puzzling bit. I think I'm still confused as to the direct relation. Suppose we had a situation where there was a natural isomorphism, i.e. that of the double dual. If we have the natural isomorphism, i.e. collection of arrows, how could we explicitly construct the isomorphism between vector spaces, i.e. the linear transformation, from this? If $F,G$ are the identity, dual functors, it seems like we want to somehow get a morphism in their source/domain category, but I don't see how it could be done. – Brendan Langfield Mar 03 '21 at 18:52
  • In your question "If we have the natural isomorphism, i.e. collection of arrows, how could we explicitly construct the isomorphism between vector spaces, i.e. the linear transformation, from this?" it's unclear what you mean by "construct the isomorphism between vector spaces". By definition, a natural isomorphism (between functors) is a collection of isomorphisms (in this case between vector spaces). For (one choice of) the natural isomorphism between the identity functor and the double dual functor, each map $\eta_X:X \to X^{**}$ has the form $\eta_X(v)(f) = f(v)$. – Ben Grossmann Mar 03 '21 at 19:02
  • @bxw You might find my earlier post here to be useful. – Ben Grossmann Mar 03 '21 at 19:03
  • Is is true that the functors we're talking about, identity, dual, double dual, all map from $\mathbf{Vect} \to \mathbf{Vect}$? So their source and target categories are identical? – Brendan Langfield Mar 04 '21 at 15:49
  • I think I understand. What I was trying to ask in the bit that was unclear was this: "if we have a natural transformation between the identity and double dual functors, and we choose a specific vector space $V$, how do we recover the specific natural isomorphism (as a morphism in $\mathbf{Vect}$) between $V$ and it's double dual?" But I believe the answer is just "take the component $\eta_V$ of your natural transformation $\eta$ associated with $V$." – Brendan Langfield Mar 04 '21 at 15:56
  • So this I think is a satisfying answer to the first question in the OP. The "natural" maps between vector spaces are just the components of the natural transformation (if one exists). – Brendan Langfield Mar 04 '21 at 15:58
  • @bxw That's all correct. I'm not sure if $\eta_V$ is referred to as a "component" of $\eta$, but yes the "natural map" between $V$ and its double dual in that context would be the linear isomorphism $\eta_V:V \to V^{**}$. – Ben Grossmann Mar 04 '21 at 16:47
  • Thanks for the confirmation! As far as that terminology, just following the wikipedia definition of a component here: https://en.wikipedia.org/wiki/Natural_transformation – Brendan Langfield Mar 04 '21 at 19:57
  • @bxw Oh, thanks for pointing that out – Ben Grossmann Mar 04 '21 at 21:14
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$ \newcommand{\A}{\mathscr A} \newcommand{\B}{\mathscr B} \newcommand{\F}{\mathscr F} \newcommand{\G}{\mathscr G} \newcommand{\V}{\mathscr V} \newcommand{\I}{\mathscr I} \newcommand{\D}{\mathscr D} $In order to better discuss this problem it is useful to have the precise categorical definition of a natural transformation in mind.

So, suppose that $\A$ and $\B$ are categories, and that $\F$ and $\G$ are functors from $\A$ to $\B$.

Definition. A natural transformation from $\F$ to $\G$ is a correspondence that, to each object $X$ in $\A$, assigns a morphism $$ \Phi_X: \F(X)\to \G(X), $$ satisfying the following crucial property: if $X$ and $Y$ are objects in $\A$, and if $f\in \text{Hom}(X, Y)$, then the diagram $\require{AMScd}$ \begin{CD} \mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\ @V\mathscr{F}(f)VV @VV\mathscr{G}(f)V\\ \mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y) \end{CD} commutes.

This is, of course, presuming that both $\F$ and $\G$ are covariant functors. If, on the other hand, both $\F$ and $\G$ are contravariant then the above diagram must be replaced by

\begin{CD} \mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\ @A\mathscr{F}(f)AA @AA\mathscr{G}(f)A\\ \mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y) \end{CD}

The question as to whether there is a natural isomorphism from a vector space to its dual, if phrased in the context of category theory, should therefore be:

Question. Considering the category $\V$ of vector spaces over a given field, is there a natural transformation $\Phi$ from the identity functor $$ \I: \V\to \V, $$ to the dual functor $$ \D: \V\to \V, $$ namely the functor sending a vector space to its dual, and such that $\Phi_X:X\to X^*$ is an isomorphism for every vector space $X$?

The first major difficulty one encounters is that the identity functor $\I$ is covariant, while $\D$ is contravariant, so one should first attempt to adapt the notion of a natural transformation for this situation. Although this will soon be understood to be a somewhat foolish pursuit, in the general case where our functor $\F$ above is covariant and $\G$, contravariant, we could ask for the commutativity of the diagram \begin{CD} \mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\ @V\mathscr{F}(f)VV @AA\mathscr{G}(f)A\\ \mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y) \end{CD}

Applying this to the functors $\I$ and $\D$, we would be asking for an a priori choice of isomorphism $$ \Phi_X:X\to X^*, $$ for every vector space $X$, such that

\begin{CD} X @>{\Phi_X}>> X^*\\ @VTVV @AAT^*A\\ Y @>{\Phi_Y}>> Y^* \end{CD}

commutes for every linear map $T$ from $X$ to $Y$. However this is impossible for many reasons, e.g. we may apply it twice, once for $T$ and another time for $2T$, leading up to $$ \Phi_X = T^*\Phi_YT = (2T)^*\Phi_Y(2T) = 4 T^*\Phi_YT = 4\Phi_X. $$

Of course this cannot hold!

Ruy
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