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Restrict the current discussion to finite-dimensional vector spaces, wherein it can be shown that a vector space $V$ and its dual $V^*$ are isomorphic.

In my textbook (Szekeres's A Course in Mathematical Physics) it is claimed (essentially) that there is no way to naturally identify $V$ and its dual $V^*$. By "naturally" (I believe) is meant an isomorphism which does not make reference to an underlying basis.

I have seen a few similar questions but none seem to provide a proof (or point to an authoritative reference). Either-or would be greatly appreciated!

EE18
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    Some discussion of this is in a previous question here. https://math.stackexchange.com/questions/105490/isomorphisms-between-a-finite-dimensional-vector-space-and-its-dual – coffeemath Jul 02 '23 at 22:07
  • It's a bit (a lot?) above the level I am at, but I suppose it is enough to answer my question which is that indeed there is no natural isomorphism, right? @VasilyBolbachan – EE18 Jul 02 '23 at 22:12
  • Yes it is one level of understanding is that we cannot construct any natural(-without using basis) isomorphism of a space with its dual. But we can make a precise statement here: The group $Aut(V)$ acts on the set $Iso(V, V^)$. One can show that there is no element of $Iso(V, V^)$ lying in the fixing points of this action. (Exercise). If you want I can elaborate on this point. – Galois group Jul 02 '23 at 22:23
  • I will have to think about that; I am familiar with an (left) action being a homomorphism from a group (no problem with $Aut(V)$) to the set of transformations (bijections on and to itself) of a set $X$. It's not clear to me what $Iso(V, V^*)$ is doing here (is it $X$, and you are talking about the set of transformations of this set?). From there I think you are saying that (whatever it is), it is a free action? @VasilyBolbachan – EE18 Jul 02 '23 at 22:42
  • $Iso(V,V^)$ is the set of all isomorphisms between these two vector spaces. So if some element $\alpha\in Iso(V,V^)$ is natural then for any $g\in Aut(V)$ we would have $g(\alpha)=\alpha$. However it is easy exercise that if $g$ is multiplication on some scalar $t$ then for any $$\beta\in Iso(V,V^*)$$ we have $g(\beta)=t^2\beta$ and this is different from $\beta$. – Galois group Jul 02 '23 at 22:45

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If you don't have a specific structure (like a prehilbertian space) you have to make a choice of basis to exhibit an isomorphism. That's why it's not canonical or natural.

The iso is given by $$e_i \longmapsto (e_i^* : e_j \mapsto \delta_i^j)$$

With a prehilbertian structure, you don't need a choice of basis, you have an iso: $$ x \longmapsto (y \mapsto \langle x, y \rangle)$$

e.g. for $\mathbb{R}^n$ you take $X \mapsto X^T$.

You have other remarkable spaces for which there exists canonical isomorphisms like $\mathcal{L}(V)$: $$u \in \mathcal{L}(V) \longmapsto (v \mapsto \text{tr}(u\circ v))$$

julio_es_sui_glace
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