Timeline for Is there a diffeomorphism of the disk with constant sum of singular values?
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11 events
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| May 3, 2020 at 21:13 | comment | added | Connor Mooney | @Asaf: Yes, concavity implies that $c^2 - (\psi' + r^{-1}\psi)^2$ is nonnegative so we can solve for $\phi'$. | |
| May 3, 2020 at 17:15 | comment | added | Asaf Shachar | Oh, after some further thought I think that I understand now. The concavity of $\psi$ is the crucial property which ensures that $\psi' + r^{-1}\psi $ remains below or equal $c^2$ at all times, so the ODE has a solution for as long as we want. | |
| May 3, 2020 at 8:24 | comment | added | Asaf Shachar | It seems rather obvious, but I wonder whether there is something more that needs to be said here... Also, am I right that the reason for choosing $\psi$ concave is that we need to lower its derivative-since we start with derivative $\psi'(0)=\frac{c}{2}>1$ and we need to finish with $\psi(1)=1$, so we have to lower the speed. So, do you think that indeed any such $\psi$ would be OK? | |
| May 3, 2020 at 8:24 | comment | added | Asaf Shachar | Thank you! this is a very nice answer. Are you are implying that one can create such a diffeomorphism for any value of $c>2$? Can you please elaborate on the general scheme you have described in the last paragraph? I understand that taking $\psi$ to be linear near the origin, you get an $f$ that is a simple dilation near the origin, so there is no problem there. Now, given such $\psi$, how do you ensure that the ODE for $\phi(r)$ will be solvable up to $r=1$?... | |
| May 3, 2020 at 8:07 | vote | accept | Asaf Shachar | ||
| May 1, 2020 at 22:25 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| May 1, 2020 at 19:46 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| May 1, 2020 at 19:36 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| May 1, 2020 at 16:56 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| May 1, 2020 at 16:34 | history | edited | Connor Mooney | CC BY-SA 4.0 |
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| May 1, 2020 at 16:17 | history | answered | Connor Mooney | CC BY-SA 4.0 |