# Number of facets of the Birkhoff polytopes $B(n)$.

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?

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.

• 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 1 at 13:38