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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 Information

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.

Post $1$, Post $2$, Post $3$, Post $4$.

Hosein Rahnama
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    See here for a related question (and answer) – rogerl Aug 22 '16 at 16:26
  • The Riesz representation theorem gives a natural isomorphism between a finite-dimensional $\textit{inner product space}$ and its dual. I don't think it's a stretch to say that this is THE distinguishing factor between inner product spaces and arbitrary vector spaces. It is taken advantage of endlessly in, say, Hilbert space theory. – Noah Olander Aug 22 '16 at 17:49
  • @NoahOlander: So do we have a basis free proof for the Reisz representation theorem I mentioned in the question? If Yes, would you please provide a link or reference? :) – Hosein Rahnama Aug 22 '16 at 17:59
  • The usual proof depends on finite dimensionality, that is, the existence of a basis. However, it does not require a $\textit{choice}$ of basis. This is more like how you should think of naturality. Also, the definition of the isomorphism is just $v \mapsto \langle _ , v \rangle$, and this definition is intrinsic - no choice of basis necessary. However, to even $\textit{define}$ an isomorphism $V \cong V^*$ for a general vector space, you need to start by choosing a basis. – Noah Olander Aug 22 '16 at 18:06
  • @NoahOlander: $v \mapsto \langle _ , v \rangle$ is not an isomorphism because it does not have the homogeneity property! Usually they call it anti-isomorphism, I think. :) (I am assuming that the definition of isomorphism is an invertible linear map). – Hosein Rahnama Aug 22 '16 at 18:11
  • Haha, I think it's more standard in mathematics (as opposed to physics) to make an inner product linear with respect to the first variable and conjugate linear with respect to the second, but by your book's convention, yes, it should be $v \mapsto \langle v, \underline{ } \rangle$. – Noah Olander Aug 22 '16 at 18:21
  • It really is Riesz, not Reisz (as it is Riemann, not Reimann). – k.stm Aug 22 '16 at 18:39
  • @k.stm: Thanks, for clarifying. :) – Hosein Rahnama Aug 22 '16 at 18:41
  • @NoahOlander: I am exactly assuming the usual convention in math that you said, i.e., linearity with respect to first variable. :) I think you mixed up the conventions or notations. :) – Hosein Rahnama Aug 22 '16 at 18:48
  • You're totally right! Sorry, don't know what I was thinking! – Noah Olander Aug 22 '16 at 19:34
  • @NoahOlander: No Problem! :) This happens a lot to me! :D – Hosein Rahnama Aug 22 '16 at 19:35
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    We don't need a basis to prove the Riesz theorem. Part of the assumptions in the Riesz theorem is the inner product, and thus the spaces it applies to have more structure than just a plain vector space. This additional structure gives you a "natural" (in the sense that one doesn't need a basis) isomorphism (or anti-isomorphism, in case of complex inner product spaces; if the space has a natural conjugation - as e.g. spaces of complex-valued functions often have - then composing this with the Riesz map gives a natural isomorphism for such complex inner product spaces). – Daniel Fischer Aug 23 '16 at 13:07
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    But if you have only an abstract finite-dimensional vector space over an arbitrary field, you don't have enough structure to define a nontrivial natural map $V \to V'$. – Daniel Fischer Aug 23 '16 at 13:08

1 Answers1

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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:

For every $f:V \to W$, we have $\eta_W \circ \id(f) = F(f) \circ \eta_V$

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

for any $f:V \to W$, $\eta_W \circ f \circ \eta_V^{-1} = f''$

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.

Hosein Rahnama
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Ben Grossmann
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