6
$\begingroup$

Excuse me, in a course of linear algebra, our assistant stated that, if $\mathbb{V}$ is a finite-dimensional vector space, and $\mathbb{W}$ its double dual, $\mathbb{V}$ and $\mathbb{W}$ are actually equal to each other; I am wondering if this has anything to do with the viewpoint in algebraic number theory that realizes elements, in algebraic number fields, as functions?
In any case, thank you very much.

2 Answers 2

8

Regarding your second question, it is true in some informal sense that when we view elements of a commutative ring $R$ as functions on $\text{Spec } R$, we are also viewing the points $\text{Spec } R$ as functions on $R$; in fact they are precisely the morphisms $R \to k$ where $k$ is a field, up to a certain equivalence relation. So I would say that this is not completely unrelated to double duals of vector spaces, although there isn't a direct formal connection since in this case the dual of an object is a different kind of object. This is sometimes summarized in the slogan "algebra is dual to geometry."

  • 1
    It really doesn't matter that much. For the sake of not having this question get bumped, just pick one.2011-05-08
6

Firstly, a vector space $V$ and its double dual are never equal. They may or may not be isomorphic, depending on whether $V$ is finite dimensional. Also, the answer to your original question is no; this is not related to viewing elements of a number field $K$ as rational functions on $\text{Spec}(\mathcal{O}_K)$.

  • 2
    @awllower: I see. You have canonical identifications $U \cong U^{\ast\ast} \cong (U^{\perp})^{\perp} \subset V^{\ast\ast}$, so Zev's answer and some of the comments explain why the first and the second isomorphisms aren't *equalities*. However, in the notation of my previous comment, if $U$ is a finite dimensional subspace of $V$ then $(U^{\perp})_{\perp} = U$ (really equality, this time). I fully second Zev's and Joel's comments.2011-05-07