Timeline for Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional
Current License: CC BY-SA 2.5
26 events
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| Apr 16, 2022 at 18:38 | comment | added | Ryan Budney | @darijgrinberg: Perhaps the issue with "very important theorem" is the context. Nobody would argue that the plumbing infrastructure in a home isn't very important. But when you talk about plumbing you've specifically narrowed to an essentially civil-engineering context. This theorem is "important" in that sense. It's a statement that is true, and has to do with basic objects people use in mathematics. It perhaps has more of a relationship to technical issues of computation than anything else. But computability is certainly of some importance. "Important" is a bit of a loaded word. | |
| Apr 15, 2022 at 21:19 | answer | added | Alcides Buss | timeline score: 6 | |
| May 16, 2015 at 18:15 | history | protected | CommunityBot | ||
| Feb 15, 2015 at 19:12 | comment | added | Leon Avery | The proof of this is also Exercise 5 of Section 11.3 of Abstract Algebra, Dummit and Foote. | |
| Nov 11, 2014 at 16:17 | comment | added | Jeremy Rickard | The question is about the dual in the sense of abstract vector spaces (i.e., the space of all functionals with no continuity requirement), not the continuous dual. | |
| Nov 11, 2014 at 16:05 | comment | added | user61643 | I must be misunderstanding the question, because for $1 < p < \infty$, $L(p)$ is a Banach space whose dual is $L(q)$ (where $1/p + 1/q = 1$), and these have the same dimensions. | |
| Dec 9, 2012 at 18:52 | comment | added | Johannes Ebert | +1 for the question and for the answers. Here is an example for an application that I will give in a class, which shows that the theorem in question is not purely a no-go-result: the Poincare duality in de Rham theory states that $H^k (M) \cong (H^{n-k}_{cpt} )^{\ast}$ for an oriented manifold. If $M$ is compact, then $H^k (M) \cong (H^k (M))^{\ast \ast}$, so $H^k (M)$ is finite-dimensional.. | |
| Mar 20, 2010 at 20:59 | vote | accept | Harry Gindi | ||
| Feb 10, 2010 at 5:47 | history | made wiki | Post Made Community Wiki by Harry Gindi | ||
| Jan 29, 2010 at 21:18 | comment | added | Pete L. Clark | Your comments seem disrespectful to Harry and especially to Mariano and Andrea, who have taken time to write out very interesting and instructive answers. Please refrain from making purely negative remarks. | |
| Jan 29, 2010 at 19:43 | comment | added | darij grinberg | My comment was mainly a reply to at least 2 people here overestimating the impact of the original question. If it wouldn't have been called a "very important theorem" by the author or a "'well known' thing that a lot of people seem not to know", I wouldn't have objected. It's not like I wouldn't value actual mathematics over discussions about importance. | |
| Jan 29, 2010 at 18:09 | answer | added | Pace Nielsen | timeline score: 29 | |
| Jan 29, 2010 at 18:01 | comment | added | darij grinberg | It's an exception since it is more or less 1000 negative results in one theorem. Whenever anything has the answer "no" in logic, it is most likely proven by reduction to the halting problem / Gödel incompleteness. But I still consider the positive results (such as Gödel's own completeness theorem, though it is much simpler and less nontrivial than the incompleteness one) way more important. | |
| Jan 29, 2010 at 16:48 | comment | added | Dan Petersen | Is for instance Gödel's incompleteness theorem not important? | |
| Jan 29, 2010 at 16:30 | comment | added | darij grinberg | Yes, but this is basically all it says. I consider a theorem important if it is useful in some proofs, not if it just stands there like a "wrong-way" sign. I don't know what an admissible representation is, but I doubt that the non-topological dual of a vector space is used anywhere in profinite group theory. | |
| Jan 29, 2010 at 13:23 | comment | added | Harry Gindi | This theorem proves that we cannot ever under any circumstances extend the results of finite dimensional linear algebra without considering topology. I think that's pretty important. =p | |
| Jan 29, 2010 at 12:58 | comment | added | Pete L. Clark | @darij: You're entitled to disagree, but saying so, without any justification whatsoever, is not a positive contribution. | |
| Jan 29, 2010 at 12:48 | comment | added | darij grinberg | "very important theorem" - I disagree. | |
| Jan 29, 2010 at 12:45 | answer | added | Andrea Ferretti | timeline score: 142 | |
| Jan 29, 2010 at 8:47 | vote | accept | Harry Gindi | ||
| Mar 20, 2010 at 20:58 | |||||
| Jan 29, 2010 at 6:47 | comment | added | Pete L. Clark | I do recall learning this result in my undergraduate days, so it is not so rarely taught. But you're right that this is one of these "well known" things that a lot of people seem not to know. I think part of the problem is that the proof is -- very unusually for a linear algebra fact -- pretty hard! | |
| Jan 29, 2010 at 4:17 | answer | added | Mariano Suárez-Álvarez | timeline score: 29 | |
| Jan 29, 2010 at 3:23 | comment | added | Harry Gindi | The only one I was able to find was this proof in Jacobson. This result is actually really important. It came to mind earlier tonight, when I used this fact to prove that a smooth representation of a locally profinite group is admissible if and only if the representation is isomorphic to its (smooth) double contragredient. | |
| Jan 29, 2010 at 3:11 | comment | added | Yemon Choi | I've never seen any of the proofs of this, including the one you mention, but given that dimension is - I presume - the cardinality of a Hamel basis, it doesn't seem surprising to me that a proof in full generality requires some mess as opposed to a slick proof. I am tempted to be rash and claim that some kind of diagonal argument should come into play, but that's not based on any serious thought or intuition. | |
| Jan 29, 2010 at 3:03 | history | edited | Harry Gindi | CC BY-SA 2.5 |
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| Jan 29, 2010 at 2:38 | history | asked | Harry Gindi | CC BY-SA 2.5 |