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The wikipedia's page for Birkhoff polytope states that the polytope has $n^2$ facets, determined by the inequalities $x_{ij} \geq 0$, for $1 \leq i,j \leq n$. I've tried different things but can't see how this claim makes sense. How should I look at this?

Also is this only true for $n > 2$? According to the Birkhoff - von Neumann theorem, $B(2)$ would have dimension $(n-1)^2 = 1$, $n! = 2$ vertices, but also $n^2 = 4$ facets, which are also vertices?

Rodrigo de Azevedo
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ensbana
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    I believe you are correct. It may be helpful to refer to the attached link https://math.stackexchange.com/questions/2858882/what-are-the-facets-of-the-birkhoff-polytope-when-n-2 – mxnoqwerty Mar 21 '19 at 02:29

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The penultimate sentence of the Wikipedia article is referring to half-spaces of the $(n-1)^2$ dimensional space containing $B_n$. $B_n$ is the intersection of these half-spaces. The formula for the facets themselves is $a_{i,j} = 0$ for each of the $n^2$ entries.

And, yes, the number of facets of $B_n$ is $n^2$ only for $n \gt 2$. The Wikipedia article is not stated correctly. As you have deduced, $B_2$ is a line segment.

Dan Moore
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    I have made my first Wikipedia edit to the Birkhoff polytope page, clarifying the number of facets in the case n = 2. – Dan Moore Apr 01 '19 at 13:38