Timeline for Proof of Rashevskii-Chow theorem
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 6, 2020 at 13:51 | comment | added | Mathsfreak | I think the answer to my question is that is $|\mathcal{F}|<dim(M)$ we can not find enough linear independent vectorfields in $q_0$ | |
| Aug 6, 2020 at 8:32 | comment | added | Mathsfreak | Probably my question is why we have to take always different points $q_i$ in each step and can not stay in $q_0$? | |
| Aug 5, 2020 at 13:03 | comment | added | Mathsfreak | does your argument mean since $f\vert_ {\Sigma_{2}}$ can not be tangent to $\Sigma_{2}$ everywhere on $\Sigma_{2}$ because then the bracket generation condition in $q_{1}$ would not hold we have to find $q_{2}$ where $f(q_{2})$ is not tangent? I'm just wondering why we can use the generation condition in a different point | |
| Aug 5, 2020 at 11:13 | comment | added | Raziel | Assume that, on the contrary, for any $f \in \mathcal{F}$ we have that $f|_{\Sigma_2}$ is tangent to $\Sigma_2$. It follows that any iterated bracket, restricted to $\Sigma_2$, would still be tangent to $\Sigma_2$. This contradicts the bracket generating conditions. | |
| Aug 5, 2020 at 10:54 | comment | added | Mathsfreak | I was working with the proof of Agrachev and I did not get yet the connection between the generated Lie Algebra and the differentials of $\phi_{i}$ could someone help me explain line 3 on page 80 why this is a contradiction? Apart of that I got the proof. Thank you all for your help | |
| Aug 4, 2020 at 13:03 | vote | accept | Mathsfreak | ||
| Aug 3, 2020 at 14:08 | history | edited | Raziel | CC BY-SA 4.0 |
added 1549 characters in body
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| Aug 3, 2020 at 13:05 | history | answered | Raziel | CC BY-SA 4.0 |