Skip to main content
13 events
when toggle format what by license comment
Nov 7, 2011 at 20:34 comment added B. Bischof I like this question a lot.
Nov 7, 2011 at 20:20 answer added André Henriques timeline score: 9
Nov 7, 2011 at 19:57 history edited Martin Brandenburg CC BY-SA 3.0
added 636 characters in body
Nov 7, 2011 at 19:01 comment added Moosbrugger Yes, it carries the discrete topology.
Nov 7, 2011 at 18:49 comment added Martin Brandenburg @Gerald: Sure, the maps $hom(W,V) \to \hom(V^*,W^*) \to \hom(V^**,W^**)$ are injective, but not surjective in general. By intrinsic I mean a condition on a linear map $V^* \to K$ which is free from uncanonical choices. @Moosbrugger: So $K$ carries the discrete topology? Feel free to add this as an answer :)
Nov 7, 2011 at 18:42 comment added Gerald Edgar If $V^{**}$ is isomorphic to $W^{**}$, I see no reason to suspect that the isomorphism takes the image of $V$ to the image of $W$. But is that what you mean by "intrinsic"?
Nov 7, 2011 at 18:33 comment added Moosbrugger 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 $\lambda\mapsto (\lambda(v))_{v\in{V}}$).
Nov 7, 2011 at 15:32 comment added Martin Brandenburg @Mariano: The annihilator of $V$ is the zero subspace of $V^*$. Perhaps I misunderstand your comment.
Nov 7, 2011 at 15:27 comment added Martin Brandenburg @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?
Nov 7, 2011 at 15:19 history edited Martin Brandenburg CC BY-SA 3.0
deleted 2 characters in body
Nov 7, 2011 at 15:06 comment added Moosbrugger Continuity is equivalent to representability (in both cases).
Nov 7, 2011 at 15:03 comment added Mariano Suárez-Álvarez The image is contained, at least, in the annihilator of the annihilator of $V$.
Nov 7, 2011 at 14:56 history asked Martin Brandenburg CC BY-SA 3.0