Timeline for Is there a global obstruction for a diffeomorphism to be an isometry?
Current License: CC BY-SA 3.0
10 events
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| Apr 20, 2015 at 18:35 | comment | added | Vladimir S Matveev | I can but may be instead of my example you take the one which appeared in the comment of Will Sawin: as the diffeomorphism $R\to R$ you take $x\mapsto x+ x^3$. It has only one fixed or periodic point, this is the point $x=0$; its derivative at $x=0$ is $1$, and it can not be an isometry since isometry preserves volume and therefore can not map the interval [-1, 1] on the interval $[-2,2]$ as the diffeomorphism does. | |
| Apr 20, 2015 at 13:49 | comment | added | Asaf Shachar | Can you please elaborate about how to see that the differential of $F_1$ at $0$ is $1$ (The identity). I am not sure I understand why the fact that all the derivatives of $v$ are $0$ implies this. (In particular I also do not understand why it works with only two derivatives equal to zero). | |
| Apr 20, 2015 at 10:51 | comment | added | Ben McKay | To ensure complete flow, maybe try $v=\arctan(x)^3 \partial_x$. That vector field $v$ generates a flow on the real line, preserving orientation. If its flow $F_t$ after some positive time $t$ preserved a metric, we would have $F_t'(0)=I$. But then $F_t$ preserves the exponential map, and acts trivially on one tangent space, so $F_t$ is the identity map. But clearly that is not possible. | |
| Apr 20, 2015 at 7:03 | comment | added | Vladimir S Matveev | I agree that the diffeomorphism $x\mapsto x+ x^3$ is possibly the simples counterexample. The observation that the flow of $ x^3 \frac{\partial}{\partial x} $ diverges does not really make any problems since it is well defined on small intervals around $0$ | |
| Apr 20, 2015 at 6:53 | history | edited | Vladimir S Matveev | CC BY-SA 3.0 |
fixed misprint
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| Apr 19, 2015 at 22:18 | comment | added | Will Sawin | Actually $v(x)=x^3$ doesn't quoite work because the flow along that vector field diverges to $\infty$ in finite time. So $F_1$ is not a well defined function (it's something like $f(x) = x/ \sqrt{1-x^2}$). But just adding $x^3$ works fine, or taking a slower-growing vector field. | |
| Apr 19, 2015 at 22:16 | comment | added | Will Sawin | @RickyDemer He means $x \to x+x^3$. | |
| Apr 19, 2015 at 22:12 | comment | added | user5810 | $x\mapsto x^3 \;$ is not a diffeomorphism from $\mathbf{R}$ to itself, since the inverse of that map is not differentiable at zero. $\;\;\;\;$ | |
| Apr 19, 2015 at 17:49 | history | edited | Vladimir S Matveev | CC BY-SA 3.0 |
added 31 characters in body
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| Apr 19, 2015 at 17:44 | history | answered | Vladimir S Matveev | CC BY-SA 3.0 |