Timeline for The group of isometries of Shahshahani metric
Current License: CC BY-SA 4.0
12 events
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| Jul 8, 2018 at 12:24 | comment | added | Ali Taghavi | @RobertBryant Yes I understand. by sharp dimension(may be not a good terminology) I meant "Maximum possible dimension". after your explanation I realize that the isometric group does not have maximum possible dimension.Thank you. | |
| Jul 8, 2018 at 10:45 | comment | added | Robert Bryant | @AliTaghavi: I'm not sure what you mean by 'sharp dimension'. When $n>2$, the isometry group of the metric considered to be on $\mathbb{R}^n\setminus\{0\}$ is exactly $\mathrm{O}(n)$, and it is not flat (there are no 'translations'). In particular, the 'global isometry group' has dimension strictly less than the (maximum possible) $\tfrac12n(n{+}1)$. Another way to see this is that the sphere $|u|=1$ under the metric $g$ is isometric to the standard Euclidean sphere of radius $2$ (instead of radius $1$), even though the $g$-distance from $0\in\mathbb{R}^n$ is only $1$. | |
| Jul 8, 2018 at 6:23 | comment | added | Ali Taghavi | I am sorry if my question is elementary: What is my mistake in the following: We consider the metric $g$ in the third line of your answer. We consider $g$ as a metric on $\mathbb{R}^n\setminus \{0\}$. Then $g$ is invariant under $O(n)$ action.So the isometric group of this Riemannian manifold has sharp dimension. Does not imply this that the metric is flat? So the restriction of metric to M is flat too. What is my mistake? | |
| Jun 20, 2018 at 13:46 | comment | added | Robert Bryant | @AliTaghavi: Yes, there are metrics with arbitrarily large finite group of isometries. For example, for any $n\ge 1$ you can have a metric on $\mathbb{R}^2$ whose only isometries are the rotations by $2k\pi/n$ about the origin, a cyclic group of order $n$. | |
| Jun 20, 2018 at 7:31 | comment | added | Ali Taghavi | @RobertBryant Is there a uniform upper bound for the cardinality of the isometric group of a Riemannian metric on the plane whose isometric group is a finite group. According to your answer the Shahshani metric, which can be counted as a metric on the whole plane(after a diffeomorphism), has an order 2 isometric group. Now, can one imagine metrics whith arbitrary large order of finite isometry group? | |
| Jun 15, 2018 at 13:54 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Rearranged the answer for clarity
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| Jun 15, 2018 at 8:44 | comment | added | Ali Taghavi | @RobertBryant Thank you for extending your comment to this very helpful answer. | |
| Jun 15, 2018 at 8:39 | vote | accept | Ali Taghavi | ||
| Jun 14, 2018 at 10:23 | comment | added | Robert Bryant | @jarauh: I have added a note to address your concerns. You are right that the case $n=2$ is special. | |
| Jun 14, 2018 at 10:22 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a note about the exceptional case n=2
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| Jun 14, 2018 at 9:24 | comment | added | jarauh | I wanted to write something similar, but I was missing one argument. You write, correctly, that any rotation preserves the metric. But the question remains: are there additional transformations that preserve the metric? Without the factor $(u_1^2 + \cdots + u_n^2)$, in the Euclidean case, the obvious answer would be no. | |
| Jun 13, 2018 at 21:56 | history | answered | Robert Bryant | CC BY-SA 4.0 |