Timeline for Is the intersection of boundaries of convex bodies a topological sphere?
Current License: CC BY-SA 3.0
20 events
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| Jun 15, 2012 at 20:27 | history | edited | Alfredo Hubard | CC BY-SA 3.0 |
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| Jun 15, 2012 at 12:52 | answer | added | Anton Petrunin | timeline score: 5 | |
| Jun 15, 2012 at 3:03 | answer | added | Tom Goodwillie | timeline score: 1 | |
| Jun 14, 2012 at 20:32 | vote | accept | Alfredo Hubard | ||
| Jun 14, 2012 at 20:31 | answer | added | André Henriques | timeline score: 8 | |
| Jun 14, 2012 at 20:26 | vote | accept | Alfredo Hubard | ||
| Jun 14, 2012 at 20:32 | |||||
| Jun 14, 2012 at 20:11 | answer | added | Will Sawin | timeline score: 8 | |
| Jun 14, 2012 at 20:07 | comment | added | Autumn Kent | That's nice, Will! You should post that as an answer. | |
| Jun 14, 2012 at 20:03 | comment | added | Will Sawin | A topologically transverse counterexample: $x^2+y^2\leq 1$ and $z^2+w^2\leq 1$. Intersection of boundaries is a torus. | |
| Jun 14, 2012 at 20:02 | history | edited | Alfredo Hubard | CC BY-SA 3.0 |
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| Jun 14, 2012 at 20:00 | comment | added | Will Sawin | Or: Take a cube and a sphere with the same center of mass, but the radius of the sphere equal to the distance from the center of the cube to the center of an edge. Then the intersection of the boundary will be a union of 6 circles on the surface of the cube, but the circles will not be disjoint - they will be tangent to each other, forming a fairly complex graph. | |
| Jun 14, 2012 at 19:56 | comment | added | Will Sawin | Here's another counterexample: Take a big square and a little square that share a corner. Then the intersection of their boundaries will be two adjacent edges of the little square, forming a topological line, not a topological sphere. | |
| Jun 14, 2012 at 19:52 | history | edited | Alfredo Hubard | CC BY-SA 3.0 |
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| Jun 14, 2012 at 19:51 | comment | added | Autumn Kent | So, what is the generality condition? That the boundaries are topologically transverse? | |
| Jun 14, 2012 at 19:51 | comment | added | Alfredo Hubard | @ Patricia, that would be nice, at least I don't see a counterexample. I'll edit the question. Thanks! | |
| Jun 14, 2012 at 19:47 | history | edited | Alfredo Hubard | CC BY-SA 3.0 |
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| Jun 14, 2012 at 19:45 | comment | added | Patricia Hersh | Maybe for your generality condition you could use that the intersection of the relative interiors of the convex bodies is nonempty? | |
| Jun 14, 2012 at 19:44 | comment | added | Autumn Kent | The generality condition you give holds for two cubes intersecting in a single edge. (@Robert: The intersection of the boundaries isn't the boundary of the intersection.) | |
| Jun 14, 2012 at 19:36 | comment | added | Robert Israel | The intersection of the bodies is convex, and the intersection of the boundaries is the boundary of the intersection of the (closed) bodies. This is a sphere of whatever dimension it has. | |
| Jun 14, 2012 at 19:13 | history | asked | Alfredo Hubard | CC BY-SA 3.0 |