Timeline for Is there a version of the Poincaré–Hopf theorem for manifold with corners?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2022 at 19:46 | vote | accept | Ya He | ||
| Feb 7, 2022 at 17:36 | comment | added | Willie Wong | Yes. That would work. | |
| Feb 7, 2022 at 17:12 | comment | added | Ya He | @WillieWong Many thanks! I just checked and your example is correct. So a sufficient condition to make PH theorem still hold for manifolds with corners is: (i) the vector field is non-vanishing at all corner points, and (ii) it is outward-pointing everywhere at the boundary. Is the statement correct now? | |
| Feb 7, 2022 at 5:25 | comment | added | Willie Wong | @YaHe Your specified condition does not rule out the possibility that the vector field vanishes at one (or multiple) of the corners. For an explicit example: Let your domain be $\Omega:= \{(x,y)\in \mathbb{R}^2: |y| \leq \sin(x), x\in [0,\pi]\}$ and $V = (x,y)$. It is not too hard to see that away from the two corners the vector field $V$ points outward through the boundary. | |
| Feb 6, 2022 at 21:18 | comment | added | Ya He | @WillieWong My question is: For a compact manifold with corners (e.g., cubes and simplices), if the inner product of the Gauss map and the vector field is always positive at the boundary (except for the corner points where the Gauss map does not exist), then can we conclude that the PH theorem (i.e., the sum of the indices at the zeros of the vector field equals the Euler characteristic of the manifold) still holds? | |
| Feb 6, 2022 at 21:14 | comment | added | Ya He | @WillieWong Thanks for your comments. I'm working on economics and not very familiar with differential topology. As to me, defining "outward pointing" as a positive inner product of the Gauss map and the vector field is more intuitive and easier to understand. So I want to confirm the following question: | |
| Feb 6, 2022 at 17:22 | comment | added | Willie Wong | @YaHe: You don't have to define inward pointing from the Gauss map. For example, the existence of a curve $\gamma: [0,1)\to A$ such that $\gamma(0)$ is the boundary point, $\gamma'(0)$ equals to vector field at that point, and $\gamma|_{(0,1)}$ sits in the interior of $A$ is another viable definition. | |
| Feb 6, 2022 at 17:13 | comment | added | Willie Wong | @mlk: technically having the v.f. pointing inwards along the boundary away from the corner points is compatible with the v.f. being precisely zero on the corner points. I feel that one has to interpret "pointing inward" to imply "non-vanshing, and continuous on the boundary, and positive inner product against Gauss map where it is well-defined." (Perhaps obvious to you and me, but less so to someone less familiar.) | |
| Feb 6, 2022 at 16:06 | comment | added | mlk | @YaHe Ultimatively you need to be able to smooth the corners without violating this condition, so it needs to point "outwards" in some sense. But this you get automatically because it is already pointing outwards for all the flat pieces of the boundary near the corner. | |
| Feb 6, 2022 at 15:06 | comment | added | Ya He | Thanks for your answer. For manifolds with boundary, the PH theorem assumes that the inner product of the Gauss map and the vector field is always positive at the boundary. For manifold with corners, let's still keep the assumption that there is no singularities at the corners (nor at the flat parts of the boundary). But the Gauss map does not exist at the corners. In order to make PH theorem still hold, how should the vector field behave at the corners? Can you specify in detail the conditions at the corners? | |
| Feb 6, 2022 at 14:35 | history | answered | mlk | CC BY-SA 4.0 |