Question
While looking over the exercise $3.F-34$ in Linear Algebra Done Right, I encountered the following paragraph
Suppose $V$ is finite dimensional. Then $V$ and $V'$ are isomorphic, but finding an isomorphism from $V$ onto $V'$, generally, requires choosing a basis of $V$. In contrast, the isomorphism from $V$ to $V''$ does not require a choice of basis and thus is considered more natural.
and these questions showed up in my mind:
$1$. Does the word natural just means that we don't need to choose a basis? I have seen the word canonical is used in the similar manner too. Is there a more precise definition for natural or canonical?
$2$. Assuming the answer to question $1$ is Yes, then why there is no natural isomorphism from $V$ onto $V'$?
$3$. I think that there is a relation between the answer to question $2$ and the proof of Riesz representation theorem. So, if we cannot find a natural isomorphism between $V$ and $V'$ then it means that we cannot prove Riesz representation theorem without choosing a basis of $V$. Is this true?
Complementary Informations
The Isomorphism from $V$ onto $V^{''}$.
Suppose $V$ is a finite dimensional vector space. Consider the following map
$$
\Lambda(v)(\phi)=\phi(v), \qquad \forall v \in V, \,\, \forall \phi \in V^{'}
$$
then $\Lambda$ is an isomorphism from $V$ onto $V''$.
Riesz Representation Theorem.
Suppose $V$ is a finite dimensional linear space equipped with an inner product and $\phi$ is a linear functional on $V$. Then there is a unique vector $v_0 \in V$ such that
$$\phi(v) = {\langle v,v_0 \rangle}_{V}, \qquad \forall v \in V$$
Other Related Posts
I found the following posts related to this question on MSE and MO.
Best Answer
There is an interpretation of natural within category theory that allows us to rigorously state that while there is a natural isomorphism from $V$ to $V''$, there is no natural isomorphism between $V$ and $V'$. This interpretation is explained here and here. The second bit is a little too elaborate for me to wrap my head around, but I'll explain what it is about the $V \to V''$ map which is "natural".
Let $\mathcal C$ denote the category whose objects are finite dimensional vector spaces. The morphisms of this category are the linear maps between vector spaces. We define a functor $F:\mathcal C \to \mathcal C$ by $F(V) = V''$ and $F([V \overset{f}{\to}W]) = [V'' \overset{f''}{\to}W'']$. What makes this a functor is that for any $f:V \to W$ and $g:U \to V$, we have $$ F(f \circ g) = F(f) \circ F(g) $$ We define the much simpler identity functor by $$ \DeclareMathOperator{\id}{id} \id(V) = V; \qquad \id([V \overset{f}{\to}W]) = V \overset{f}{\to}W $$ When we say that $V$ is naturally isomorphic to $V''$, we mean that there is a natural isomorphism between the functors $\id$ and $F$. In this case, what this means is that we can assign an isomorphism (invertible morphism) $\eta_V:\id(V) \to F(V)$ to every vector space $V$ in such a way that:
Or, as we can rephrase it in this context (noting $\id$ is just the identity), we need an $\eta_V:V \to V''$ for every $V$ such that
Now, what is this $\eta_X$? Well, it suffices to take $$ \eta_V:V \to V''\\ [\eta(x)](\alpha) = \alpha(x) $$ You know that this map is an isomorphism from the text. Now, we note that for any $\beta \in V''$, there is an $x_\beta$ for which $\alpha(x_{\beta}) = \beta(\alpha)$ for any $\alpha \in V'$, and we have $\eta^{-1}(\beta) = x_{\beta}$. With that in mind, we can see that for any $f:V \to W$ and for any $\beta \in V''$ and $\alpha \in V'$, we have
$$\begin{align} [[\eta_W\circ f \circ \eta_V^{-1}](\beta)](\alpha) &= [[\eta_W\circ f](x_{\beta})](\alpha) \\ &= [\eta_W(f(x_{\beta}))](\alpha) \\ &= \alpha(f(x_{\beta})) \\ &= [\alpha \circ f](x_{\beta}) \\ &= \beta (\alpha \circ f) \\ &= \beta (f'(\alpha)) \\ &= [\beta \circ f'](\alpha) \\ &= [f'' (\beta)](\alpha) \end{align}$$ as required.