Timeline for Are these two structures isometric?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2020 at 20:31 | vote | accept | Ali Taghavi | ||
| Jan 9, 2020 at 6:34 | comment | added | Robert Bryant | @WillieWong: Yes. Because the eigenvalues of Riemann as a linear map are differential invariants of the metric, they are constant on the orbits of the isometry group, which, it turns out, is the group $\mathrm{O}(n)$, and the generic orbit has dimension $\tfrac12n(n{-}1)$. Thus, at most $\tfrac12n(n{+}1)$ differential invariants could be independent functions, and, in fact, that's exactly how many are independent in general. The relations satisfied by any larger number of invariants is an isometry invariant of the metric and may be used to distinguish the two metrics. This is messy, though. | |
| Jan 8, 2020 at 15:43 | comment | added | Willie Wong | Ah, it is the same $P_1$. Now I completely understand; this also explains what you mean by the word "locally". For higher dimensions, is it then the case that thinking of Riemann as a linear map on two forms, the first metric also has non-trivial relation between the (symmetric functions of) eigenvalues? (Thinking about what you mean by the "more involved argument".) | |
| Jan 8, 2020 at 6:49 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Removed relic sentence of the old argument as per Willie Wong's suggestion. Fixed a few typos.
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| Jan 8, 2020 at 6:39 | comment | added | Robert Bryant | @WillieWong: Thanks for your comment. Yes, I will remove the spurious reference to the flat metric in the last paragraph. $P_1$ is just what I described: A nonzero polynomial in 4 variables; it's the same one in both mentions. (It has more than 1000 terms, by the way.) As for the comparison, no, the point is that any (local) diffeomorphism pulling back $ds^2_2$ to $ds^2_1$ (i.e., an isometry), would have to pull back $s_i'$ to $s_i$. (The converse need not be true.) Thus, it would have to pull back the function $P_1(s_1',s_2',s_3',s_4')$ (which vanishes only on a hypersurface) to $0$. | |
| Jan 7, 2020 at 20:56 | comment | added | Willie Wong | (a) Maybe the reference to the flat metric should also be edited in the final paragraph? (b) I am confused about the MAPLE verification: What is the notation $P_1$ in the part about the MAPLE computation? Aren't $s_1'$ etc scalar functions on the manifold? So is the comparison you are making still under the identity map? Then why is it not enough that with respect to the common coordinate systems the two metrics are not equal? | |
| Jan 7, 2020 at 19:15 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Corrected an erroneous argument that attempted to show that the two metrics were not isometric.
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| Jan 5, 2020 at 18:22 | comment | added | Ali Taghavi | Thank you very much Prof. Bryant for your answer. | |
| Jan 5, 2020 at 18:21 | history | bounty ended | Ali Taghavi | ||
| Jan 5, 2020 at 12:36 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Completed the curvature calculation for the first metric and reported that it, indeed, does not vanish.
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| Jan 4, 2020 at 20:49 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added the formula for the second metric and answered most of the OP's questions.
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| Jan 4, 2020 at 16:50 | history | answered | Robert Bryant | CC BY-SA 4.0 |