I am not a mathematician by training, so please feel free to correct my logic or descriptions when necessary:

Let $P$ denote a $\textit{permutation matrix}$: $$ \begin{equation} P := \{X \in \{0,1\}^{n\times n} : X \mathbf{1}_n = \mathbf{1}_n\,,\,\mathbf{1}_n^T X = \mathbf{1}_n^T\}. \end{equation} $$ where $\mathbf{1}_n$ denotes an $n$-dimensional ones vector.

When doing optimizations, to circumvent the discrete nature, such matrices are generally relaxed using the doubly-stochastic matrices $\mathcal{DP}_n$, that live on the Birkhoff Polytope, an $(n - 1)^2$-dimensional convex submanifold of the ambient $\mathbb{R}^{n\times n}$, defined as: $$ \begin{align} \mathcal{DP}_n = \{\,\,X :& X_{ij}>0 \,,\, i,j \in\{1,...,n\} \,\,\, \wedge \sum\limits_{i=1}^n X_{ij}=1 \,\wedge\, \sum\limits_{j=1}^n X_{ij}=1 \,\,\} \end{align} $$ In other words, row and column sums equate to $1$ and entries can take real values between $0$ and $1$.

The permutation matrices live on the vertices of Birkhoff Polytope. One can also describe the discrete set of permutation matrices as the intersection of Birkhoff Polytope $\mathcal{DP}_n$ with the orthogonal group (special case of Stiefel manifold when $m=n$): $$ P=\{X \in \mathcal{DP}_n : X X^T=\mathbf{I}\} $$ The orthogonal group $O(n)$ can be regarded as an $(n(n—1)/2)$-dimensional surface in the $n^2$-dimensional Euclidean space. This surface is a subset of the sphere of radius $n^{1/2}$.

So, the orthogonal group touches the polytope exactly on the permutation matrices. In low-dimensions, one can probably think this as a tessellated shape, approximating the sphere. Or, in the other way around, we can speak of an encapsulating sphere of the polytope (though for higher dimensions this might not be true).

Now comes my question: As $n \rightarrow \infty$, does the sphere become a good approximation of the polytope, or does the gap grow? In other words, instead of using the Birkhoff polytope itself as a relaxation of the discrete permutations, can we live with using just the sphere? And, if possible, would it be possible to bound the error of approximation?


  • $\begingroup$ By "error of approximation", do you mean the Frobenius norm of the difference between a doubly schoastic matrix and the nearest point on the sphere? And if so, isn't the maximum distance attained at the matrix with constant entries $\frac1n$? $\endgroup$ – joriki Jul 21 '18 at 23:24
  • $\begingroup$ I meant the largest (or mean) error one attains by walking on the sphere rather than on the polytope (assuming that Birkhoff is the correct relaxation of course). $\endgroup$ – Tolga Birdal Jul 21 '18 at 23:29
  • $\begingroup$ "largest (or mean)" in the sense that you'd be interested in knowing either? What do you want to average over? The sphere and/or the polytope? And what do you mean by "the correct relaxation" -- correct in what sense? $\endgroup$ – joriki Jul 21 '18 at 23:39
  • $\begingroup$ Let's say we walk from $P_0$ to $P_1$. On Birkhoff the path would be a straight line, an edge rather, if they are neighbors. On $O(n)$, that would be a geodesic of the Stiefel, and takes longer to travel. Therefore, I am assuming that Birkhoff is a better relaxation. However, how better? As dimension grows, do we gain or do we lose? Any measure would be fine in that case, as soon as it can give a hint on the quality of approximation. Does that make sense? You might also be right that $1/n$ becomes the farthest point of $\mathcal{DP}_n$ to $\mathcal{S}_n$. Maybe it is possible to show this? $\endgroup$ – Tolga Birdal Jul 21 '18 at 23:44

The answer to the question literally asked is $\infty$. I will argue that the answer to the question the OP probably meant to ask is $(n-1)$. I'll write $J$ for the $n \times n$ matrix of all $1$'s, and $\vec{j}$ for the length $n$ vector of all $1$'s.

Let $D$ be the affine linear space of $(n-1) \times (n-1)$ matrices whose rows and columns sum to $1$. (So, doubly stochastic but with no condition on the signs.) The Birkhoff polytope $P$ is contained in $D$, whereas the orthogonal group $O(n)$ is not. So, if $\lambda$ is a linear functional which is $0$ on $D$ but not on $O(n)$, then $\lambda$ is $0$ on $P$ but not on $O(n)$, and the ratio of $\max_{X \in O(n)} |\lambda(X)|$ to $\max_{X \in P} |\lambda(X)|$ is infinite. An example of such a $\lambda$ is $\lambda(X) = \sum_{j=1}^n (X_{1j} - X_{2j})$. Another way to put this is that, if we travel on a ray starting at $\tfrac{1}{n} J$ not in the plane $D$, we leave $P$ immediately but stay in the convex hull of $O(n)$ for a nonzero amount of time.

Let $DO(n) = D \cap O(n)$. It seems more natural to ask about the difference between optimizing a linear functional on $P$ and on $DO(n)$. However, $D$ is an affine linear space so we can only speak of affine linear functionals on it, not linear functionals. It seems natural to me to solve this issue by normalizing our functionals to be $0$ on $\tfrac{1}{n} J$, the center of $P$. Another way to phrase our question is, if we travel on a ray from $\tfrac{1}{n} J$ within $D$, how large can the ratio be between the time we leave $P$ and the time we leave $DO(n)$?

Let us see that this ratio can be as large as $n-1$. Consider the line through $\tfrac{1}{n} J$ and $\mathrm{Id}$, and extend it in the opposite direction through $\tfrac{1}{n} J$. So a general point on this line is of the form $\tfrac{1+x}{n} J - x \mathrm{Id}$. This ray leaves $P$ when $x=\tfrac{1}{n-1}$ and the diagonal entries become $0$. However, when $x=1$, this matrix is orthgonal (it is the transformation which fixes $\vec{j}$ and negates $\vec{j}^{\perp}$). So, for all $x \in [-1,1]$, the matrix $\tfrac{1+x}{n} J - x \mathrm{Id}$ is in the convex hull of $DO(n)$. So the ray can be in $DO(n)$ for $n-1$ times as long as in $P$. In terms of optimizing linear functionals, the linear functional $\mathrm{Tr}(JX-nX)$ can only be as large as $n$ on $P$ (achieved when $X$ is a permutation with no fixed points), but is as large as $n^2-n$ at $2\tfrac{1}{n} J - \mathrm{Id}$.

We now prove this ratio is optimal. Suppose that $Z$ is in the convex hull of $DO(n)$. We must show that $\tfrac{n-2}{n-1}J+\tfrac{1}{n-1} Z$ is in $P$. In other words, we must show that each entry of $\tfrac{n-2}{n-1}J+\tfrac{1}{n-1} Z$ is nonnegative. Write $e_i$ for the $i$-th standard basis vector; we must show that $(\tfrac{n-2}{n-1}J+\tfrac{1}{n-1} Z) e_i$ has nonnegative entries. We will use two key properties of $Z$: We have $Z \vec{j} = \vec{j}$, and we have $|Z \vec{w}| \leq |\vec{w}|$ for any vector $\vec{w}$. Both properties are clearly true on $DO(n)$, and preserved by convex combinations.

Write $e_i = \tfrac{1}{n} \vec{j} + \vec{w}$ and put $\vec{x} = Z \vec{w}$. We note that $\vec{w}$ is perpendicular to $\vec{j}$ and has length $\sqrt{(1-1/n)^2 + (n-1) (1/n)^2} = \sqrt{1-1/n}$, so $\vec{x}$ is perpendicular to $\vec{j}$ and has length $\leq \sqrt{1-1/n}$. We have $$\begin{array}{rcl} (\tfrac{n-2}{n-1}J+\tfrac{1}{n-1} Z) e_i &=& (\tfrac{n-2}{n-1}J+\tfrac{1}{n-1} Z) (\tfrac{1}{n} \vec{j}+\vec{w}) \\ &=& \tfrac{1}{n} \vec{j} + \tfrac{1}{n-1} Z \vec{w} \\ &=& \tfrac{1}{n} \vec{j} + \tfrac{1}{n-1} \vec{x} . \\ \end{array}$$ We have used that $J \vec{j} = Z \vec{j} = \vec{j}$ and $J \vec{w}=0$ since $\vec{w}$ is in $\vec{j}^{\perp}$. We want to show that every element of $\tfrac{1}{n} \vec{j} + \tfrac{1}{n-1} \vec{x}$ is nonnegative or, in other words, that each entry of $\vec{x}$ is $\geq -1+1/n$.

Now, $|\vec{x}| \leq \sqrt{1-1/n}$, so each entry of $\vec{x}$ is at most $-\sqrt{1-1/n}$, but we need a better bound. We can get one by using that $\vec{x}$ and $\vec{j}$ are perpindicular. Let $\vec{x} = (x_1, \ldots, x_n)$. So we know $\sum x_j = 0$ and $\sum x_j^2 \leq 1-1/n$. So we have $$x_n^2 \leq 1-\tfrac{1}{n} - \sum_{j=1}^{n-1} x_j^2 \leq 1-\tfrac{1}{n} - \tfrac{1}{n-1} \left( \sum_{j=1}^{n-1} x_j \right)^2 = 1-\tfrac{1}{n} - \tfrac{x_n^2}{n-1}$$ where we have used the root-mean-square inequality in the middle step. Rearranging, $x_n^2 \tfrac{n}{n-1} \leq \tfrac{n-1}{n}$ or $|x_n| \leq \tfrac{n-1}{n}$. So in fact each entry of $\vec{x}$ is $\geq -1+1/n$, as desired.

As a side note, if we instead focus on the intersection of $D$ with the sphere of radius $\sqrt{n}$, the approximation ratio is much worse. Look at the matrix $$\tfrac{1}{n} J + x \begin{bmatrix} -1 & 1/(n-1) & 1/(n-1) & \cdots & 1/(n-1) \\ 1/(n-1) & -1/(n-1)^2 & -1/(n-1)^2 & \cdots & -1/(n-1)^2 \\ 1/(n-1) & -1/(n-1)^2 & -1/(n-1)^2 & \cdots & -1/(n-1)^2 \\ \vdots & & & \ddots & \\ 1/(n-1) & -1/(n-1)^2 & -1/(n-1)^2 & \cdots & -1/(n-1)^2 \\ \end{bmatrix}.$$ This exits the Birkhoff polytope at $x=1/n$ (when the upper left entry becomes $0$) but stays in the sphere of radius $\sqrt{n}$ until $x = \pm \tfrac{(n-1)^{3/2}}{n}$. So the ratio is as bad as $(n-1)^{3/2}$ if we use the sphere instead of the orthogonal group. I also get that this ray exits the convex hull of $DO(n)$ at $x = (n-1)/n$, so this is another case of the $n-1$ ratio.

  • $\begingroup$ Good thought, I have two questions though: 1. How can one prove that we are not shooting the ray at a location close to one of the vertices, but to the center of the face? And would that ray be orthogonal to the face? Maybe this was encoded in the second part but wasn't clear to me. $\endgroup$ – Tolga Birdal Aug 2 '18 at 8:18
  • $\begingroup$ There is a catch here though. The matrix you described : $\frac{1}{\sqrt{n}}J$ is not on the sphere (does not belong to the orthogonal group). So don't think it exits the sphere at that point. $\endgroup$ – Tolga Birdal Oct 10 '18 at 22:23
  • $\begingroup$ It is on the sphere of radius $\sqrt{n}$. It is not in the orthogonal group. Which set did you want to optimize over? (That said, this answer has other flaws; I am working on an updated version now.) $\endgroup$ – David E Speyer Oct 11 '18 at 10:22
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    $\begingroup$ Answer completely rewritten. Thank you for making me think this through. I was confused by the (1) the fact that the Brikhoff polytope and the convex hull of $O(n)$ have different dimensions and (2) the difference between $O(n)$ and the sphere of radius $\sqrt{n}$. $\endgroup$ – David E Speyer Oct 15 '18 at 14:26
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    $\begingroup$ Some facts which I noticed while working on this answer, but didn't wind up being relevant to the final solution: (1) $DO(n)$ is the group of orthogonal matrices preserving the vector $\vec{j}$, and hence isomorphic to $O(n-1)$. (2) The convex hull of $O(n)$ is the matrices with operator norm $\leq 1$, see math.stackexchange.com/questions/434720 . (3) For an $n \times n$ matrix $L$, the maximum of $\mathrm{Tr}(LX)$ as $X$ ranges over $O(n)$ is the sum of the singular values of $L$. $\endgroup$ – David E Speyer Oct 15 '18 at 14:29

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