Timeline for Quick definition of the tangent space
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2022 at 8:14 | comment | added | painday | See also the question mathoverflow.net/questions/88880/… and the answers therein. | |
| Jun 8, 2012 at 22:32 | answer | added | Tom Goodwillie | timeline score: 5 | |
| Jun 8, 2012 at 18:16 | answer | added | Dan Lee | timeline score: 1 | |
| May 22, 2012 at 2:36 | comment | added | Mariano Suárez-Álvarez | @Micheal-grade83: you can easily show that if $v$ is a derivation at $x$ and $f$ and $g$ are functions which vanish at $x$ then $v(fg)=0$. Use that. | |
| May 21, 2012 at 21:00 | comment | added | Bruno Martelli | @Johannes: on any chart, it is immediate to prove that $T_pM$ is a vector space with the old-style definition (and that the vector structure does not depend on the chart). It is immediate to prove (1) and (2). | |
| May 21, 2012 at 19:44 | answer | added | JHM | timeline score: 18 | |
| May 21, 2012 at 17:37 | comment | added | user21706 | @Mariano Suárez-Alvarez So if $v$ is a derivation we have $v(f) = \sum_i \frac{\partial f}{\partial x_i}|_p v(x_i) + v\left(\sum_{i,j} f_{i,j} x_i x_j\right)$ but how I can conclude that the last term vanishes? @Angelo If you define $T_p M$ by equivalence classes of curves which passes through $p$ then $T_p M$ is only a set, not a vector space. You need to define the operations of vector space passing through some chart and prove that they are independend of the choice of chart... is more convoluted. | |
| May 21, 2012 at 17:35 | comment | added | Johannes Ebert | And the argument that Mariano sketches shows that each derivation $X$ equals $\sum_{i=1}^{n} (X x^i) \frac{\partial}{\partial x^i}$. There does not seem to be a quicker definition. | |
| May 21, 2012 at 17:31 | comment | added | Johannes Ebert | @Angelo: I think it is inadequate because it is not immediately clear that $T_p M$, viewed as equivalence classes of curves, is a vector space. | |
| May 21, 2012 at 17:04 | comment | added | Angelo | Why is the old-style definition inadequate? | |
| May 21, 2012 at 17:02 | comment | added | Mariano Suárez-Álvarez | You can prove (1) and (2) in one go using the Taylor theorem using the definition in terms of derivtions, if you write that theorem in the form: every smooth function $f$ can be written locally as $f(x)+\sum_i\frac{\partial f}{\partial x_i}x_i+\sum_{i,j}f_{i,j}x_ix_j$ for some smooth functions $r_{i,j}$. | |
| May 21, 2012 at 16:50 | history | asked | user21706 | CC BY-SA 3.0 |