Timeline for Choice of metric/topology on $\mathbb{R}^n$ when we say a manifold is locally homeomorphic to it
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2020 at 11:21 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited body
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| Jun 30, 2020 at 10:49 | comment | added | Qmechanic♦ | I updated the answer. | |
| Jun 30, 2020 at 10:49 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
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| Jun 30, 2020 at 8:18 | comment | added | Shirish | Awesome! My main takeaway so far is that I was wrong to assume that there can only be one metric on a set (spacetime in this case). There can be different metrics corresponding to different structures on it. And each structure serves a different purpose. Topology (structure for which Euclidean is used) serves the purpose of us being able to talk about continuity of curves. Inner product space (structure for which Minkowski is used) serves the purpose of us enforcing that spacetime interval is invariant. That's the rough idea I have. Looking forward to your update! | |
| Jun 29, 2020 at 13:54 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
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| Jun 29, 2020 at 11:53 | comment | added | Qmechanic♦ | $\uparrow$ Yes. | |
| Jun 29, 2020 at 11:49 | comment | added | Shirish | Your fifth point seems to tackle the question. Could you elaborate on that, if possible? The metric is (in the case of standard topology) used to define open sets, which in turn are used to specify continuity of maps. So a curve $\gamma:\mathbb{R}\supset I\to\mathbb{R}^4$, which physically corresponds to the worldline of a particle, can be called continuous / non-continuous based on standard topology, which assumes a Euclidean metric. Effectively then, to state whether or not the worldline of a particle is continuous, we're ignoring the Minkowski structure of flat spacetime. Is that correct? | |
| Jun 29, 2020 at 11:09 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
Added explanation
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| Jun 29, 2020 at 10:51 | comment | added | Shirish | I agree that a topological manifold, in general, doesn't necessarily need a metric to be defined. But I'm specifically talking about the standard topology on $\mathbb{R}^n$ here, in which open sets are defined as open balls. An open ball, by definition, needs the underlying $\mathbb{R}^n$ to be equipped with a Euclidean metric. | |
| Jun 29, 2020 at 10:39 | history | answered | Qmechanic♦ | CC BY-SA 4.0 |