Timeline for Poincaré-Hopf Theorem for domains with a point of vanishing curvature
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 9 at 22:48 | comment | added | Ryan Budney | The answer to your question is nothing happens. The formula is the same, and it does not depend at all on the curvature of the boundary curve. | |
| Jan 9 at 12:18 | comment | added | Tom Goodwillie | You seem to think that the outward normal vector at a point is determined by the curvature of the boundary at that point. But any non-zero vector orthogonal to the tangent line is a normal vector (outward or inward). | |
| Jan 9 at 9:17 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor formatting and Math Jaxing (embedded link+`\operatorname`)
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| Jan 8 at 20:52 | history | edited | DrHAL | CC BY-SA 4.0 |
added 475 characters in body
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| Jan 8 at 20:20 | history | edited | DrHAL | CC BY-SA 4.0 |
added 262 characters in body
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| Jan 8 at 18:50 | comment | added | Ryan Budney | I'm a little confused by your question. Most statements of the Poincare-Hopf index theorem don't mention curvature. What definition of Euler characteristic are you using? It sounds like you might be using the Gauss-Bonnet theorem? Do look up the various alternative definitions. | |
| S Jan 8 at 10:08 | review | First questions | |||
| Jan 8 at 10:31 | |||||
| S Jan 8 at 10:08 | history | asked | DrHAL | CC BY-SA 4.0 |