Timeline for "A gentleman never chooses a basis."
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2012 at 16:41 | history | edited | Todd Trimble | CC BY-SA 3.0 |
the old answer had not been properly latexed
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| Nov 23, 2009 at 15:26 | comment | added | Qiaochu Yuan | I guess Andrew also means that, for example, Hilbert spaces <em>are</em> isomorphic to their continuous double duals. | |
| Nov 23, 2009 at 15:24 | comment | added | Qiaochu Yuan | Presumably Andrew means that one almost never talks about unadorned infinite-dimensional vector spaces. An analyst naturally thinks of the dual of a finite-dimensional vector space as a special case of the continuous dual of a topological vector space, and in this situation spaces are rarely isomorphic to their double duals. | |
| Nov 23, 2009 at 3:22 | comment | added | Mariano Suárez-Álvarez | @Andrew: what other duals are there in this context? | |
| Nov 8, 2009 at 21:26 | comment | added | Andrew Stacey | When you say "isomorphic to its double dual" you presumably mean its algebraic double dual. | |
| Oct 23, 2009 at 23:37 | comment | added | Todd Trimble | Defining? No... you asked me to prove that the conditions I gave imply that V is finite-dimensional. So, I did. My proof shows that there exists a finite-dimensional subspace of V that contains V. It uses the fact that any vector space is the union of its finite-dimensional subspaces. | |
| Oct 23, 2009 at 22:18 | comment | added | Eric Chopin | But Todd, you're defining a finite dimensional vector space using finite dimensional subspaces? I must have missed something? Rgds, Eric | |
| Oct 23, 2009 at 14:54 | comment | added | Todd Trimble | Here's one way: the condition says that the functor V \otimes - is left adjoint to the functor W \otimes -. But it's also left adjoint to hom(V, -), hence we have a canonical iso W \otimes - \cong hom(V, -). In particular, hom(V, -) preserves all colimits. Now V is the union, i.e., the filtered colimit of the system of inclusions V_i \to V_j of its finite-dimensional subspaces. So hom(V, V) is the filtered colimit (union) of subspaces hom(V, V_i). In particular, the identity map 1_V must factor through one of the V_i, hence is finite-dimensional. | |
| Oct 23, 2009 at 11:10 | comment | added | Eric Chopin | Hi all, For Todd Trimble: how do you prove that a vector space V fulfilling your conditon "there exists a space W together with maps e: W \otimes V \to k, f: k \to V \otimes W making triangular equations hold" implies that V is finite dimensional? Thanks and best regards, Eric | |
| Oct 23, 2009 at 4:24 | vote | accept | Richard Dore | ||
| Oct 22, 2009 at 4:15 | comment | added | Todd Trimble | Yes, there a number of ways one might think of characterizing finite-dimensionality (including being isomorphic to its double dual!), Noetherian/Artinian hypotheses, etc. But some of these characterizations don't port so well to modules over other commutative rings. The present characterization is equivalent to being finitely generated and projective, for any commutative ring. | |
| Oct 21, 2009 at 22:54 | comment | added | Tom Leinster | OK, great! So you can define finite-dimensionality without mentioning bases (or chains of subspaces). The answer to the question is then easy. But this recasting of the definition of finite-dimensionality is, I think, much the most interesting thing. | |
| Oct 16, 2009 at 6:40 | history | answered | Todd Trimble | CC BY-SA 2.5 |