# How is the isomorphism between a vector space and its dual not natural?

Let $$V$$ be a vector space over the field $$K$$ and $$V^*=\mathcal{L}(V,K)$$ its dual space. We can prove that $$V$$ is naturally isomorphic to its double dual $$V^{**}$$, but why does every isomorphism between $$V$$ and its dual $$V^*$$ depend on the choice of basis? We certainly use dual basis, but the number of element of a basis i.e. the dimension is not basis-dependent. For example, in Linear Algebra by Serge Lang, I have found this:

Let $$V$$ be a vector space over $$K$$ with a non-degenerate scalar product, $$\langle\cdot,\cdot\rangle:V\times V\rightarrow K$$. Let $$v\in V$$, the map $$L_v$$ such that $$$$V\ni u\overset{L_v}{\longrightarrow}\langle u,v\rangle$$$$ is a linear functional, thus an element of $$V^*$$.

The map such that $$$$V\ni v\rightarrow L_v$$$$ is an isomorphism (between $$V$$ and its dual). This is proved by showing that this map is linear, injective (because of non-degeneracy) and surjective ($$dimV=dimV^*$$). So, how does this depend on the choice of basis? It is true that we used the dual basis at the beginning, but as I said above, every basis would give us the same answer as for the dimension of the space, that is what the author used in the last proof.

• Yes, if you fix an inner product you get an isomorphism between the space and its dual. However, there is no natural choice of inner product on a vector space.
– lulu
Aug 1, 2020 at 18:29
• So what you're saying is that the fact this isomorphism is not natural is not because of the dual basis but because of the scalar product? Aug 1, 2020 at 18:38
• I didn't mention a basis. I'm just saying that you got your isomorphism by adding structure, so there was nothing natural about it.
– lulu
Aug 1, 2020 at 18:40
• May be it's time to define what is natural... Aug 1, 2020 at 18:45
• Given any basis $\beta = \{ v_1, \cdots, v_n\}$ there is an unique inner product $\langle \cdot, \cdot\rangle_\beta$ which makes that basis orthonormal. The set of inner product is in one-to-one correspondence with the set of basis. Aug 1, 2020 at 19:35

This is not the case for the correspondence between the dual of the dual and the original space, where a basis-independent identification can be made. In this way, although (for finite dim vector spaces) any vector space of the same dimension is equivalent to the original space, the identification $$(V^*)^* \cong V$$ is natural because everyone will agree on the correspondence between vectors regardless of the basis they're using.