# What are the facets of the Birkhoff Polytope when $n=2$?

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?

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
$$\pmatrix{x&1-x\\1-x&x}\;,$$
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.
• 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 9:36