Timeline for Is the tensor product of polyhedra a polyhedron?
Current License: CC BY-SA 3.0
6 events
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| Dec 19, 2012 at 6:20 | comment | added | Will Sawin |
Clearly it contains $P_a \otimes Q_b$, for $a,b\in \{p,c,l\}$. We need to prove it contains the sum of these objects. Write $x+y = \lim _\lambda \to 0 \lambda (x/\lambda) + (1-\lambda) y$. So we get the sum as long as one of the objects is closed under multiplication of positive reals, and the convex hull is closed. Cones and linear subspaces both have this property, and tensor product preserves it. So we only need to check closure. Unfortunately, closure is not preserved by taking the convex hull, so this is a bit tricky.
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| Dec 8, 2012 at 7:30 | comment | added | Dima Pasechnik | and once there are no $P_\ell$ and $Q_\ell$, a seemingly natural finite set of generators (?) of conv($P\otimes Q$) would be pairs of vertices and extreme rays. | |
| Dec 7, 2012 at 18:41 | comment | added | Dima Pasechnik | One wants to separate $P_c$ from $P_\ell$ to be able to say that $P_p+P_c$ are in the orthogonal complement to $P_\ell$. Whether it's useful here, is another question, of course. | |
| Dec 7, 2012 at 17:23 | comment | added | darij grinberg | Yes, this is the approach I have tried too. Unfortunately, while proving that my convex hull is a subset of your $(*)$ is easy, the other inclusion is not, and I am not even sure that it is true. (By the way, is there really much use in separating $P_c$ from $P_{\ell}$ ?) | |
| Dec 7, 2012 at 12:46 | history | edited | Dima Pasechnik | CC BY-SA 3.0 |
rectified the decomposition (*) into a more meaningful one
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| Dec 7, 2012 at 11:45 | history | answered | Dima Pasechnik | CC BY-SA 3.0 |