Dual Spaces and Natural maps

Question

(I'll explain what I know first and then I'll ask the questions). Given a finite dimensional vector space $V$, it is often remarked that there is no "natural" isomorphism from $V$ to $V^*$ (I guess this means a basis independent isomorphism?). I understand that one typically constructs an isomorphism $V \to V^*$ by fixing a basis for $V$, call this $B = \{v_1, v_2, \ldots, v_n\}$, and then mapping $v_i \mapsto \delta_i$ where $\delta_i$ is the linear functional given by $\delta_i(v_i) = \delta_{ij}$. Here are my questions:

(1) Why is there no natural isomorphism $V \to V^*$?

(2) If $V$ is a finite dimensional inner product space, we can map each $v \mapsto \langle v, \cdot \rangle$. Is this not "natural"? (I know we can consider sesquilinear forms, but let's keep the discussion to this inner product for simplicity).

Answer 1

There are similar posts here:

Why are vector spaces not isomorphic to their duals?

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Isomorphisms Between a Finite-Dimensional Vector Space and its Dual

Dual of a vector space

The moral is for finite dimensional case, the two space $V$ and $V^{*}$ are isomorphic because they have the same dimension and they are over the same field. For infinite dimensional case (for example if $V$ has a countable basis), then $V^{*}$ is larger than $V$ because the cardinality of the two bases differ; the $V$ has cardinality $\mathbb{N}$ while $V^{*}$ has cardinality of all maps from $\mathbb{N}$ to $\mathbb{N}$. We know the space of maps from $\mathbb{N}$ to two points has cardinality $c$, so the second one is strictly larger than the first one. In otherwords they are not isomorphic.

Answer 2

To elaborate on Hagen von Eitzen's answer to question (2), one way to understand an inner product (or, more generally, a nondegenerate bilinear form) on a finite-dimensional vector space is as a choice of isomorphism to the dual!

Indeed, your comment in question (2) shows that fixing a bilinear from $\langle\cdot,\cdot\rangle$ gives us a natural map $V\rightarrow V^*$, and it is straightforward to check that this map is an isomorphism iff the form is nondegenerate.

In the other direction, given any isomorphism $\psi:V\rightarrow V^*$, we can define a nondegenerate bilinear form $$\langle v, w\rangle = \psi(v)[w]$$

Answer 3

Yes, you are right about natural essentially meaning basis-independant here.

(1) There is none. Any isomorphism $V\to V^*$ puts a bias on the basis (though it does not determine a specific base), as it allows us to define an inner product. Since there are different inner products possible (e.g. by choosing different isomorphisms), no choice is natural.

(2) Yes, for spaces with inner product, this is a natural choice. In general there is however no inner product given