Timeline for Why is cotangent more canonical than tangent?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 7, 2010 at 17:49 | comment | added | Deane Yang | @Orbicular: Yes, you are right. | |
| Mar 7, 2010 at 9:07 | comment | added | Orbicular | @Deane Yang: I suppose you mean "exterior differential" instead of covariant derivative! | |
| Mar 7, 2010 at 5:06 | comment | added | Deane Yang | Democracy? Not really. In my answer, I definitely view the tangent bundle as being more fundamental, and the cotangent bundle as arising naturally from differentiating functions. On the other hand, the canonical 1-form and its covariant derivative make the cotangent bundle in many ways much more interesting to study for its own sake than the tangent bundle. | |
| Mar 7, 2010 at 4:52 | comment | added | Eric Zaslow | This nicely captures the duality between the two, but are these answers really honest? Do y'all really think of these two structures with the democracy professed in these answers? I don't (I understand them, but I don't) and I have to twist my mind a bit to find the structure dual to the canonical form. Maybe it's just me. (Sorry, this isn't really a mathematics question.) | |
| Mar 7, 2010 at 4:51 | comment | added | Matt Noonan | No, Terry's answer is right (see also the first paragraph of Deane's answer). A derivative is a linear approximation, so the derivative of a function $\mathbb{R}^n \to \mathbb{R}$ is a linear form on $\mathbb{R}^n$. In the case of a manifold this means you get an element of $T^*M$, which of course turns out to be the differential. | |
| Mar 7, 2010 at 4:42 | vote | accept | Eric Zaslow | ||
| Mar 7, 2010 at 4:13 | comment | added | Petya | Situation is opposite: If I want to differentiate a function on manifold at a point, then I need a tangent vector at that point. | |
| Mar 7, 2010 at 4:08 | history | answered | Terry Tao | CC BY-SA 2.5 |