Timeline for Problem with definition of tangent vectors as derivations
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 23 at 19:03 | comment | added | Moishe Kohan | @NateRiver: No, just sense of an unsatisfactory answer. The actual question remains unanswered. | |
| Oct 23 at 17:27 | comment | added | Nate River | @MoisheKohan I sense some sour grapes… | |
| Oct 22 at 11:03 | comment | added | Moishe Kohan | @OrangeMushroom: The original question asked for a "description" of accesible derivation. Your answer only proves existence; not a single exotic derivation is described. What you described is a linearly independent subset of the dual space of the space of derivations. But, I see that even this is enough to OP, so be it. | |
| Oct 22 at 6:32 | comment | added | Sebastian Goette | @MoisheKohan You are of course right, the axiom of choice is used here to construct actual derivations. Anyway, the answer below halfway solves my original problem: it explains why there is a rich predual of the space of derivations. For me, that is "accessible" enough, even though a concrete construction of one such thing would have been nicer (it is still not clear that this is impossible). | |
| Oct 22 at 0:42 | history | became hot network question | |||
| Oct 22 at 0:34 | history | edited | LSpice | CC BY-SA 4.0 |
Name of "this question"; link to book
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| Oct 21 at 22:53 | comment | added | user113407 | @MoisheKohan my answer does not answer your question of "do there exist exotic derivations independent of choice," so I don't see why you call it "quite wrong". It answers the OP's question of whether or not there is an accessible way of describing an exotic derivation. I gave one example and the OP seemed to find it helpful. If you think using choice makes it not accessible then that would be a matter of opinion. | |
| Oct 21 at 22:29 | comment | added | Moishe Kohan | Take a look at my answer here where this is spelled out in the case $r=1$. When one says "extend the given linearly independent subset to a basis" one uses AC. | |
| Oct 21 at 22:19 | comment | added | Moishe Kohan | If you look more closely, then you see how the axiom of choice shows up. In particular, the accepted answer to your question is quite wrong. | |
| Oct 21 at 21:38 | vote | accept | Sebastian Goette | ||
| Oct 21 at 19:52 | comment | added | Sebastian Goette | @MoisheKohan I would be surprised if this was a logics problem. The only proof that I know rewrites a $C^k$ function as a sum of products $g_i\cdot x^i$, where the $g_i$ are $C^{k-1}$ functions and the $x^i$ are coordinates. But then $g_i$ is no longer in the domain of the derivation, unless $k=\infty$. So for finite $k$, you cannot apply the product rule anymore. | |
| Oct 21 at 19:06 | answer | added | user113407 | timeline score: 5 | |
| Oct 21 at 17:41 | answer | added | Iosif Pinelis | timeline score: 0 | |
| Oct 21 at 17:05 | comment | added | Moishe Kohan | I do not see how it would be possible since it would yield a "description" of the algebraic dual of some infinite dimensional vector space. Then you get into issues related to the axiom of choice... I am actually unsure if the existence of exotic derivations is independent of the axiom of choice. Maybe logicians here can help with this... | |
| Oct 21 at 16:41 | history | asked | Sebastian Goette | CC BY-SA 4.0 |