Natural isomorphism in linear algebra: is the naturalness asymmetric?

Question

Let $V$ be a finite-dimensional vector space. It is well known that there is a natural isomorphism between $V$ and its double dual $V^{\ast\ast}$ defined by $T(x)(f)=f(x)$ for every $x\in V$ and $f\in V^\ast$.

However, I am unable to write down a definition of $T^{-1}$ directly without any reference to $T$ and without picking any basis. Thus it seems to me that the isomorphism between the two vector spaces is not really so natural if we look at the direction from $V^{\ast\ast}$ to $V$.

Is it possible to define $T^{-1}$ in a direct and natural way? If not, does this asymmetry in the easiness of definition have any significance in linear algebra or other branches of mathematics?

Answer 1

There is no issue with naturality here when interpreted in the categorical sense (for other interpretations, see the 2nd paragraph). If $\alpha : F \to G$ is a natural transformation between functors whose components $\alpha(x) : F(x) \to G(x)$ are invertible, then $\alpha^{-1} : G \to F$ defined by $(\alpha^{-1})(x) := \alpha(x)^{-1}$ is also a natural transformation (the proof is easy to write down in one line).

But also in the more informal interpretation of "natural", it is clear that $T^{-1}$ is natural when you remember that functions are actually special relations, and the concept of a relation is symmetric. So if you believe that the function $T : V \to V^{**}$ is natural, this means that its graph $\Gamma_T \subseteq V \times V^{**}$ is natural, and hence its opposite $(\Gamma_T)^{\mathrm{op}} \subseteq V^{**} \times V$ is natural. Now one has to prove that, if $V$ is finite-dimensional, this relation is induced by a function $T^{-1} : V^{**} \to V$, but this is really not a matter of naturality and of course one has to uses finite bases (or something equivalent, like direct sums of copies of $K$) here since this does not hold when $V$ is infinite-dimensional. Since the graph of $T^{-1} : V^{**} \to V$ is natural in $V$ (it just the set $\{(w,v) : \forall f \in V^* (w(f)=f(v)\}$ after all) the function $T^{-1}$ must also be interpreted as natural. The natural definition is: $T^{-1}(w)$ is the uniquely determined $v \in V$ such that $w(f)=f(v)$ for all $f \in V^*$ (see also my PPS). Again, being a function is not part of being natural or not.

PS: If $D : \mathbf{Vect} \to \mathbf{Vect}^{\mathrm{op}}$ is the dual space functor, so that $B := D^{\mathrm{op}} \circ D : \mathbf{Vect} \to \mathbf{Vect}$ is the bidual space functor, there is exactly one natural transformation $B \to \mathrm{id}$, namely the zero transformation. So without finite dimensionality you cannot write down any non-trivial natural maps $V^{**} \to V$.

Proof: Let $\alpha : B \to \mathrm{id}$ be a natural transformation. Let $K$ be the ground field and consider $\alpha_K : K^{**} \to K$. It is a multiple of the canonical isomorphism, say with the factor $u \in K$. Then for every $w \in V^{**}$ and $l \in V^{*}$ naturality with respect to $l : V \to K$ shows that $l(\alpha_V(w)) = u \cdot w(l)$. This determines $\alpha_V(w)$ completely. In particular, if $u=0$ we get $\alpha=0$, and we are done. If $u \neq 0$, we can work with $u^{-1} \alpha$ instead and assume $u=1$. So we have $l(\alpha_V(w))=w(l)$. Thus, if $\{b_i\}$ is a basis of $V$ and $l_i \in V^*$ extracts the coefficient of $b_i$, we see that $\alpha_V(w) = \sum_i w(l_i) \cdot b_i$. This is required to be a finite sum, so almost all $w(l_i)$ are zero. But since the $l_i$ are linearly independent, there is some $w \in V^{**}$ with $w(l_i)=1$ for all $i$, and for infinite-dimensional $V$ this leads to the desired contradiction.

PPS: Let me explain more why I think that also definitions like "$f(x)$ is the unique $y$ such that ..." do not prevent naturality in the informal sense. Consider one of the most natural functions out there, $f : X \to P(X)$, $f(x) := \{x\}$ (where $P(X)$ denotes the power set of a set $X$). What is $\{x\}$? Well, it is the unique set which contains $x$ as an element and no other element. When working in ZFC for example, it is an axiom that for all sets $x,y$ there is a set $z$ which contains $x,y$ and nothing else, and then we denote this set $z$ by $\{x,y\}$. The axiom of extensionality implies that $\{x,y\}$ is uniquely determined. The notation $\{\dotsc\}$ is not built into our language of ZFC, which only has first-order logic and $\in$. So in some sense, $f$ is really defined as the natural relation $\{(x,A) \in X \times P(X) : x \text{ is the only element of } A\}$ and it is a (very basic) theorem of ZFC that this relation is a function. Back to linear algebra: If $V$ is a finite-dimensional vector space, we have the natural relation $\{(\omega,v) \in V^{**} \times V : \omega(f)=f(v) \text{ for all } f \in V^*\}$, and it is a theorem of ZFC that it is a function. Concerning naturality, there is absolutely no difference here.

Answer 2

It is a common fallacy to attach to mathematical concepts meanings that pertain to the meaning of their name in the natural language.

For instance, a number can be irrational, but this doesn't mean it is not logic or not reasonable (the meanings I get from a dictionary); “real”, “imaginary” and “transcendental” are other examples. Don't attach to the adjectives used for classifying these numbers more than the mathematical definition says.

The name “natural isomorphism” was initially chosen based on examples from algebraic topology and the similarity with the already known case of the double dual was noticed, where the isomorphism $V\to V^{**}$ (for a finite dimensional vector space $V$) can be defined in “a natural way”.

When other instances of “natural (iso)morphisms” came around, it became apparent that a precise definition had to be found, not based on the intuition that the “definition is easy and obvious”, because in some cases it was definitely not easy and obvious, although it fitted the framework. And category theory was born.

If $F,G\colon \mathbf{A}\to\mathbf{B}$ are (covariant) functors, a natural morphism $\eta\colon F\to G$ is a family of morphisms $\eta_X\colon F(X)\to G(X)$, one for each object $X$ in $\mathbf{A}$ in such a way that, given any morphism $f\colon X\to Y$ in $\mathbf{A}$, we have the commutative diagram $$\require{AMScd} \begin{CD} F(X) @>\eta_X>> G(X) \\ @V{F(f)}VV @VV{G(f)}V \\ G(Y) @>\eta_Y>> G(Y) \end{CD} $$ that is, $\eta_Y\circ F(f)=G(f)\circ\eta_X$. If each $\eta_X$ is an isomorphism, we say that $\eta$ is a natural isomorphism.

It easily follows from the definition that if $\eta$ is a natural isomorphism $\eta\colon F\to G$, then the collection $(\eta_X^{-1})$ defines a natural isomorphism $\eta^{-1}\colon G\to F$.

In the particular case when $\mathbf{A}=\mathbf{B}$ is the category of finite dimensional vector spaces over some fixed field, $F$ is the identity functor and $G$ the double-dual functor, then there exists a natural isomorphism $F\to G$, which is the well-known one.

Thus also the inverse is a natural isomorphism. But nothing in the definition suggests that this inverse can be “computed in a natural way”. And indeed it cannot: if it could, then the “natural” definition would also apply to infinite dimensional spaces, but this is not possible.

Answer 3

Let $w \in V^{\ast \ast} \setminus \{0\}$. Let $$V_1 := \bigcap_{f \in \ker(w)} \ker(f).$$

Because $w \neq 0$, there exists $\ell \in V^\ast$ with $w(\ell) = a \neq 0$. Note that $V^\ast = \langle \ell \rangle + \ker(w)$.

Choose $0\neq v \in V_1$. Then $b := \ell(v) \neq 0$. (If $\ell(v)=0$, then because also $f(v)=0$ for all $f \in \ker(w)$, we would have $l(v)=0$ for all $l \in V^\ast$ which implies $v=0$.)

Set $T^{-1}:w \mapsto \frac{a}{b} v$.

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Edit: Commenters point out the issue I poorly tried to hide: Somewhere we need to invoke the fact that $V$ has finite dimension. It goes into proving that $V_1$ contains non-zero elements, and then into the fact that my definition of $T^{-1}$ does not depend on the choice of such a non-zero $v \in V_1$.

Of course one would also need that to prove that $T$ is surjective.

Also, look at https://math.stackexchange.com/a/3781591/96384 for a different way to interpret the "naturalness" of $V \simeq V^{\ast \ast}$, and see how close it comes to parts of Martin's answer. (And again, the finite dimension enters almost secretly in a step to prove surjectivity.)

Answer 4

It seems to me that you do not use the phrase natural isomorphism in the sense of category theory (as one expects when one reads the title), but in a more motivational sense:

The definition of $T : V \to V^{**}$ looks so obvious and compelling - so natural - that one should expect that $T^{-1}$ also has such a nice description.

In my opinion this is a fallacy. In fact, $T$ is defined for each (finite or infinite dimensional) vector space $V$ and one can show that it is always injective. For a general $V$ the proof requires the axiom of choice (for each $x \in V, x \ne 0$, you have to find a linear map $u \in V^*$ such that $u(x) \ne 0$), but let us ignore that here. The main issue is that $T$ is an isomorphism only for finite dimensional $V$. To prove surjectivity you must invoke at some point that $V$ has a finite basis.

That is, although $T$ is an isomorphism for finite dimensional $V$, you need to pick a basis to prove it. Viewed in this light it is not surprising that you need a basis of $V$ to describe $T^{-1}$.

Update:

You correctly observe that there is an asymmetry between the "easiness" to define of $T$ and the "trouble" to describe $T^{-1}$. As we have seen, this is due to the the fact that it easy to define $T$, but non-trivial to prove that it is an isomorphism (which is true only for finite dimensional $V$ and has a serious impact on finding the inverse).

Does this have any significance in linear algebra or other branches of mathematics? In this particular case I do not think so. But this is a somewhat "philosophical question" which I would reformulate as follows:

If we consider bijections $f : A \to B$ (we shall not be specific; $A,B$ may be sets, groups or vector spaces or something else, and $f$ may be a function or a homomorphism), can we always easily determine their inverses $f^{-1}$?
Of course, the interpretation of easy will depend on the context.

No, neither on a generic level (as for $T : V \to V^{**}$) nor on a concrete level as e.g. for one-way functions. Quotation from Wikipedia:

A one-way function is a function that is easy to compute on every input, but hard to invert given the image of a random input.

Such are used for example in the RSA cryptosystem: It is easy to encrypt a message, but very hard to decrypt the result - unless you have access to secret information.

Remark:

Martin Brandenburg has proved in his answer that there does not exist a non-zero natural transformation from the bidual functor $B$ to the identity functor on the category $\mathbf{Vect}$ of vector spaces over a field $K$. Actually his proof shows that for no full subcategory $\mathbf{Vect'} \subset \mathbf{Vect}$ containg $K$ and an infinite dimensional $V$ there exists a non-zero natural transformation from $B' = B \mid_{\mathbf{Vect'}}$ to the inclusion functor $\mathbf{Vect'} \to \mathbf{Vect}$. Therefore the best possible positive result is available on the category $\mathbf{Vect}_{fin}$ of finite dimensional vector spaces over $K$: The inverse $T^{-1}$ is a non-zero natural transformation $B_{fin} \to id_{fin}$.

However, here "natural" is understood in the formal sense of category theory. Could it be possible that for each $V$ there exists a linear map $S : V^{**} \to V$ such that $S \circ T = id$ which has a natural definition in some more intuitive sense? Again: No. First I would argue that if something has a short and elegant generic definition for all $V$, then it must be natural also in the formal sense. Okay, this is just a heuristic point of view and does not prove anything. But there is a more formal argument supporting the non-existence of an easily defined $S$.

Obviously $S$ can only exist if $T$ is injective. For infinite dimensional $V$ the injectivity of $T$ is proved by the axiom of choice which implies that $V^*$ separates the points of $V$ (i.e. that for each $x \in V, x\ne 0$, there exists $u \in V^*$ such that $u(x) \ne 0$). Indeed, this fact (assuming it for arbitrary $V$) is equivalent to the axiom of choice. See Andreas Blass' answer to Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

Therefore a universal easy definition of $S$ would substitute the axiom of choice in the proof of injectivity of $T$. In other words, $S$ would be something like a universal choice function - which is impossible. Actually any definition of $S$ will require the choice of a suitable basis of $V^{**}$. For finite-dimensional $V$ this can be overlooked since we do not need the full power of the axiom of choice; but even in that case we have to pick a basis $b_1,\ldots, b_n$ of $V$ and can then construct the dual and bidual bases of $V^*$ and $V^{**}$.