Timeline for How to prove the existence of the polytope in $\mathbb{R}^d$ with a given number of faces, minimizing the isoperimetric ratio?
Current License: CC BY-SA 3.0
5 events
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| Jul 4, 2015 at 1:32 | comment | added | student | I interpreted it as a polytope with $f-1$ faces with outer normals $v_1, \dots \, ,v_{f-1}$, but you suggested it should mean counting the multiplicity of the face with $v_1$ vector. Maybe in order to understand the argument, I need to know the expression of the isoperimetric ratio function? | |
| Jul 4, 2015 at 1:26 | comment | added | student | Sorry I still didn't get it. If $\min_{(v_1, \dots \, ,v_f) \in \mathbb{S}^{(n-1) \times f}}\frac{S(v_1, \dots \, ,v_f)^d}{V(v_1, \dots \, ,v_f)^{d-1}}$ is attained at $(v_1, v_1, v_2, \dots \, ,v_{f-1})$, then this means the minimizer is polytope with $f-1$ faces with outer normals $v_1, \dots \, ,v_{f-1}$, but if then how to conclude this is not the minimizer? My main confusion is, I don't know the correspondence of the $(v_1, v_1, \dots \, ,v_{f-1})$ to the geometry of the convex body. It seems that you interpreted it in a different way I explained. | |
| Jul 4, 2015 at 1:08 | comment | added | Igor Rivin | @student The volume is the same as when you would count the face one, but the area is bigger. | |
| Jul 4, 2015 at 0:51 | comment | added | student | Could you explain "counting a face with multiplicity does not help you"? I'm sorry I didn't understand... If $\frac{S^d}{V^{d-1}}=G(v_1, \dots \, , v_f)$ for some continuous function $G$, then what does it mean if the minimum is attained at a point of which two components are the same? | |
| Jul 4, 2015 at 0:27 | history | answered | Igor Rivin | CC BY-SA 3.0 |