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I've read in several sources that the number of facets of the Birkhoff polytope $\mathcal{B}(n)$ is $n^2$.

Is this supposed to hold when $n=2$? Since $\mathcal{B}(2)$ has dimension $1$, the facets would be the two $0$-dimensional vertices, which are the two permutation matrices below: $$\begin{pmatrix} 1 & 0 \\ 0 &1 \end{pmatrix} \text{ and } \begin{pmatrix} 0 & 1 \\ 1 &0 \end{pmatrix}$$ However, the claim is that there should be $2^2 = 4$ facets. None of my sources have given any restriction on $n$. What am I missing?

Rodrigo de Azevedo
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M47145
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1 Answers1

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No, it doesn't apply to $n=2$; your sources (like this one) apparently failed to treat this special case.

The generally $n^2$ facets correspond to the non-negativity constraints for the $n^2$ entries of the matrix. But for $n=2$, the $4$ non-negativity constraints form two pairs of identical constraints if you restrict them to the space defined by the row and column sum constraints: The row and column sum constraints span a $3$-dimensional space and thus leave only a $1$-dimensional space of doubly stochastic matrixes of the form

$$ \pmatrix{x&1-x\\1-x&x}\;, $$

in which the non-negativity constraints are pairwise identical on the diagonal and off the diagonal.

So you're right; there are only two facets in this case, defined by $x=0$ and $x=1$, which corresponds to the two matrices you gave.

joriki
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  • This is from a different post but I received no answers from there, so I'd like to ask it here: could you explain the claim that the $n^2$ facets correspond to the non-negativity constraints for the $n^2$ entries? I tried different things but can't see how to deduce ones from the others. – ensbana Mar 21 '19 at 09:36