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.
Are vector spaces and their double duals in fact equal?
2 Answers
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."
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1It really doesn't matter that much. For the sake of not having this question get bumped, just pick one. – 2011-05-08
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 $\text{Spec}(\mathcal{O}_K)$.
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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