Timeline for Mappings between 2-manifolds with symmetries with fixed singular values
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2022 at 10:30 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed a typo and added clarifying language
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| Feb 3, 2022 at 10:21 | comment | added | Robert Bryant | @DanielCastro: Since $f^*(\mathrm{d}\phi)$ and $\mathrm{d}\theta$ are closed, we have $\mathrm{d}(M\,\mathrm{d}r) = 0$, so $M$ is a function of $r$. Since $A(0)=B(0)=0$ while $B'(0)>0$, by shrinking $\epsilon$, we can guarantee that $c_2\,A(r)$ is in $\bigl(B(-\delta),B(\delta)\bigr)$ when $|r|<\epsilon$. Yes, that $V$ should be a $W$. I'll fix those points. | |
| Feb 2, 2022 at 19:54 | comment | added | Daniel Castro | I guess that there should be a typing error, so that in the first paragraph it should read $(s,\phi):W$ ... on an open $q$-neighborhood $W\subset N$, isn't it ? | |
| Feb 2, 2022 at 19:51 | comment | added | Daniel Castro | Thank you so much for such a great answer. I wish to clarify just a couple of details. In constructing $ f^*\left( d \phi\right)$ we use the fact that $ f^*\left( d \phi\right)\left(\frac{\boldsymbol{\partial}\:\:}{\boldsymbol{\partial}\theta} \right)=d \phi \left(d f\left(\frac{\boldsymbol{\partial}\:\:}{\boldsymbol{\partial}\theta} \right) \right)=1$, then $f^*\left(d \phi\right)=M(r,\theta)d r+d \theta$, that is shouldn't be $M$ in general a function of $r,\theta$ ? Also, why is it guaranteed that $c_2A(r)$ is in the image of $B(r)$, when solving for $R(r)$ ? Thank you. | |
| Jan 30, 2022 at 19:03 | history | bounty ended | Daniel Castro | ||
| Jan 30, 2022 at 19:03 | vote | accept | Daniel Castro | ||
| Jan 24, 2022 at 10:59 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a remark about 'globalizing' the adapted coordinates.
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| Jan 23, 2022 at 16:25 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Edited to make it clearer how the functions M and R determine a local solution f.
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| Jan 23, 2022 at 14:28 | history | answered | Robert Bryant | CC BY-SA 4.0 |