Let $V$ be a vector space over a field $K$. Call a linear map $F : V^* \to K$ representable if there is some $v \in V$ such that $F(w) = \langle w,v \rangle$ for all $w \in V^*$. Here, $\langle w,v \rangle := w(v)$. Remark that this representing vector $v$ is then uniquely determined.

Remark the similarity to category theory: A functor $F : C^{\mathrm{op}} \to \mathrm{Set}$ is representable if there is an object $v \in C$ such that $F \cong C(-,v)$. By the Yoneda Lemma, this object is uniquely determined up to canonical isomorphism. Now there are various theorems which say when a certain functor is representable. Often we require that $F$ preserves limits and satisfies some finiteness condition.

Now back to linear algebra, is there a criterion when a linear map $V^* \to K$ is representable? In other words, is there an intrinsic description of the image of $V \to V^{**}$? If $V$ is finite-dimensional, then we get representability for free. I'm interested in the general case. I would like to mimic somehow the category theoretic conditions.

// Moosbrugger has given below the following nice characterization: Give $K$ the discrete topology, $K^V$ the product topology, and $V^* \subseteq K^V$ the subspace topology. Then a linear map $V^* \to K$ is representable iff it is continuous! However, the proof is rather trivial, except that it uses the result for finite-dimensional vector spaces. So in practise, I doubt that this will be a good criterion. Therefore I would add the following requirement to the criterion: it should be useful in practise (whatever this means) or linear algebraic / geometric: When is a hyperplane in $V^*$ cut out by a single vector $v \in V$?

  • $\begingroup$ The image is contained, at least, in the annihilator of the annihilator of $V$. $\endgroup$ Nov 7, 2011 at 15:03
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    $\begingroup$ Continuity is equivalent to representability (in both cases). $\endgroup$ Nov 7, 2011 at 15:06
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    $\begingroup$ @Moosbrugger: Not every continuous functor $C^{op} \to \mathrm{Set}$ is representable. For example, $\lim_i \hom(-,x_i)$ is continuous, but is representable iff $\lim_i x_i \in C$ exists. However, there many categories with this property, called SAFT by Theo Johnson-Freyd (mathoverflow.net/questions/49175/…). When is a linear map $V^* \to K$ called continuous? Do you assume that $K$ is a topological field, or even $\mathbb{R}$ or alike? $\endgroup$ Nov 7, 2011 at 15:27
  • $\begingroup$ @Mariano: The annihilator of $V$ is the zero subspace of $V^*$. Perhaps I misunderstand your comment. $\endgroup$ Nov 7, 2011 at 15:32
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    $\begingroup$ My answer was meant as a bit of a joke: certainly one requires hypotheses for such representability theorems. The one I know is called "presentable." But anyway, it's a precise statement for vector spaces for any field $K$. You topologize the dual as a subset of $\prod_{v\in{V}}{K}$, where the latter has the usual ("Tychonoff") topology (and the map $V^*\to\prod_{v\in{V}}{K}$ is $\lambda\mapsto (\lambda(v))_{v\in{V}}$). $\endgroup$ Nov 7, 2011 at 18:33

1 Answer 1


Moosbrugger has given an excellent answer to the question on which I'll elaborate a bit.

A similar result holds in functional analysis:
Let $V$ be a Banach space and $V'=hom(V,\mathbb C)$its dual (i.e., the set of continuous linear functionals from $V$ to $\mathbb C$). If you equip $V'$ with the norm topology, then its dual is typically bigger than $V$. However, if you equip it with the weak-star topology, then its dual is exactly $V$. In other words, $V\subset V''$ is the subset of weak-star continuous functionals (and the norm on $V$ is induced by that on $V''$). In that way, you can recover $V$ from $V'$ if you known its weak-star topology.

Now, for the sake of completess, let me also say what the weak-star topology on $V'$ is: it's the topology under which which a net $\{\varphi_i\}_{i\in I}$ converges to some element $\varphi$ if and only if $\{\varphi_i(v)\}_{i\in I}$ converges to $\varphi(v)$ for every $v\in V$.

  • $\begingroup$ Thanks! So the weak-star topology is just the topology of pointwise-convergence. $\endgroup$ Nov 7, 2011 at 20:38
  • $\begingroup$ @Martin: Correct. $\endgroup$ Nov 8, 2011 at 9:33
  • $\begingroup$ @Moosbrugger's comments that you reference: 1 2 3. $\endgroup$
    – LSpice
    Mar 31, 2019 at 20:13

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