Timeline for Question about symmetric bilinear form and convex geometry
Current License: CC BY-SA 4.0
19 events
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| Dec 22, 2022 at 20:56 | comment | added | threeautumn | @RobertBryant Yes, it is unit sphere of Finsler norm and its interior. | |
| Dec 22, 2022 at 14:39 | comment | added | Robert Bryant | Is the convex body $M$ the set of vectors that satisfy $\phi(v)\le 1$? Otherwise, you haven't said how $M$ is related to anything else, so I'm guess that this is what you mean. Let me know if you mean something different. | |
| Dec 22, 2022 at 12:11 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 22, 2022 at 12:08 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 22, 2022 at 11:57 | comment | added | threeautumn | @RobertBryant Also, this problem originates from my thesis of isometric embedding $f$ of second-order flat Finsler metric into Banach-Minkowski space. S(v, w) is an analog of second fundamental form, using projection of $d_p^2 f(v, w)$ into some fixed vector $\tau$. $Q(v, w) = d_p^2 f(v, w) - S(v, w) \tau $. First-order flatness implies that $d_p \varphi(d_p^2 f(v, w)) = 0$. So $d_p \varphi(Q(v, w)) = 0$ if $S(v, w) = 0$. We proved that 3-dim second-order flat Finsler manifold embedded into 4-dim Banach-Minkowski space has degenerate second fundamental form. But not sure for higher dim. | |
| Dec 22, 2022 at 11:51 | comment | added | threeautumn | @RobertBryant Yes, you are correct. I try to reformulate it but it does not work. It should be below: If $S(v, w) = 0$, then $Q(v, w)$ is tangent to level surface of Finsler norm $\varphi$. It does not matter whether basis lies on convex body $M$. Secondly, yes, $Q$ cannot be identically zero. The point is that if it is zero for a zero measurable set, then by continuity, Kakutani still holds. | |
| Dec 22, 2022 at 11:48 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 22, 2022 at 11:45 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 21, 2022 at 15:59 | comment | added | Robert Bryant | There's something missing in your description of your problem. For example, you don't assume any connection between $S$ and $M$ or put any condition on $Q$. If you let $S$ and $M$ be arbitrary and just take $Q(v,w) = 0$ for all $v$ and $w$, then your hypotheses are satisfied but $M$ need not be an ellipsoid. (This is a counterexample to your claim even in dimension $3$.) Moreover, since there is no connection between $S$ and $M$ assumed, for the basis $\{e_1,e_2\}$ that you first construct in your 'proof', there is no reason to suppose that $e_1$ and $e_2$ to belong to $M$. | |
| Dec 19, 2022 at 9:15 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:34 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:28 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:28 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:27 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:26 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:25 | history | edited | threeautumn | CC BY-SA 4.0 |
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| Dec 19, 2022 at 8:22 | history | edited | threeautumn | CC BY-SA 4.0 |
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| S Dec 19, 2022 at 8:21 | review | First questions | |||
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| S Dec 19, 2022 at 8:21 | history | asked | threeautumn | CC BY-SA 4.0 |