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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.

awllower
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2 Answers2

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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."

Qiaochu Yuan
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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 $\text{Spec}(\mathcal{O}_K)$.

Zev Chonoles
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  • @Zev Chonoles: Yes, that is exactly my point, when I talked to the assistant, but he insisted that they are actually equal; in fact he announced that some books do not explicitly prove that they are equal, but they are actually so; moreover, he said that a subspace of $\mathbb{V}$ and its annihilator are equal!
    Besides, I am sorry about the dimension of the vector space; they all should be finite.     – awllower May 07 '11 at 01:40
  • And thank you for such an elaborate explanation. – awllower May 07 '11 at 01:44
  • @awllower: It is great that you appreciate the subtlety of the difference between "isomorphic" and "equals", and very unfortunate that the assistant does not. I suppose how much you should press this issue depends on how much control the assistant has over your grade. Perhaps you can ask the professor to settle the matter, or write the assistant an email after the course is over. – Zev Chonoles May 07 '11 at 01:46
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    I think "equal" was a bit of an overstatement. But nonetheless, a vector space is always canonically embedded in its bidual by seeing a vector $v$ as the function $|v \rangle : l \in V^* \mapsto l(v) \in K$. In the finite dimensional case, the dimension being equal, you get a canoncial isomorphism (roughly no arbitrary choice involved in the definition). In a categorical point of view, that's as close as you can get to being equal. – Joel Cohen May 07 '11 at 01:49
  • @Zev Chonoles: I will discuss with my professor; thanks very much. @Joel Cohen: this is why I said that they seem alike in algebraic number theory; thanks very much. – awllower May 07 '11 at 01:52
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    @awllower: I'm not sure, but it could be that there was a basic misunderstanding: You can define the annihilator $U^{\perp} \subset V^{\ast}$ of a subspace $U \subset V$ (the elements in $V^{\ast}$ sending $U$ to zero) and you can define the annihilator $W_{\perp} \subset V$ of a subspace $W \subset V^{\ast}$ (the elements of $V$ that are sent to zero by all elements of $W$). Then in fact you always have equality $(V^{\perp})_{\perp} = V$, because $V^{\perp} = {0} \subset V^{\ast}$. – t.b. May 07 '11 at 06:58
  • @Theo Burhler: It was my bad, and he said that the double annihilator of a subspace of a finite-dimensional vector space is equal to that subspace; he proved this by double incusion, and this is my point!! Also, thanks for the explanation. – awllower May 07 '11 at 08:45
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    @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. – t.b. May 07 '11 at 08:52