Angle-restricted sets and zero-free regions for the permanent

TL;DR

This paper develops a systematic geometric approach to identify zero-free regions for the permanent of matrices, refining Barvinok's method by introducing angle-restricted sets, leading to improved bounds and explicit solutions.

Contribution

It introduces angle-restricted sets and a geometric framework to construct zero-free regions for the permanent, improving upon Barvinok's previous results.

Findings

Derived explicit zero-free regions for the permanent.

Improved bounds over previous results.

Reduced the problem to low-dimensional geometry.

Abstract

The goal of this note is to give a systematic method of constructing zero-free regions for the permanent in the sense of A. Barvinok, i.e. regions in the complex plane such that the permanent of a square matrix of any size with entries from this region is nonzero. We do so by refining the approach of Barvinok, which is based on his clever observation that a certain restriction on a set S involving angles implies zero-freeness; we call sets satisfying this requirement angle-restricted. This allows us to reduce the question to a low-dimensional geometry problem (notably, independent of the size of the matrix!), which can then be solved more or less explicitly. We give a number of examples, improving some results of Barvinok.

Equations

$$|\arg (a_j)|<\frac{1}{2}(\pi -\theta )$$

$$v:=\sum _{i,j,k}{a}_{ij}{u}_{ijk}=\sum _{i,j,k}{a}_{ij}{r}_{ij}{s}_{ik}{u}_i=\sum _{i,k}{a}_i{s}_{ik}{u}_i=\sum _i{a}_i{u}_i$$

$$w:=\sum _{i,j,k}{b}_{ik}{u}_{ijk}=\sum _{i,j,k}{b}_{ik}{r}_{ij}{s}_{ik}{u}_i=\sum _{i,j}{b}_i{r}_{ij}{u}_i=\sum _i{b}_i{u}_i.$$

$$f(u_1,...,u_n)=\mathrm{Im}\log \frac{{\sum }_j{a}_j{u}_j}{{\sum }_j{b}_j{u}_j}$$

$$v=\sum _j{a}_j{u}_j=ar{e}^{i\theta /2}+b{e}^{-i\theta /2},w=\sum _j{b}_j{u}_j=cr{e}^{i\theta /2}+d{e}^{-i\theta /2},$$

$$a=\sum _{j=1}^m{a}_j{r}_j/r,b=\sum _{j=m+1}^n{a}_j{r}_j,c=\sum _{j=1}^m{b}_j{r}_j/r,d=\sum _{j=m+1}^n{b}_j{r}_j.$$

$$|y|\le \frac{2\sqrt{x}+(x+1)\cos \theta }{\sin \theta },$$

$${\left(x+\frac{1}{\cos \theta }\right)}^2+y^2\le {\tan }^2\theta .$$

$$\frac{1-\sin \theta }{|\cos \theta |}\le x\le \frac{1+\sin \theta }{|\cos \theta |},$$

$$\mathrm{Re}\left(\frac{r{e}^{\pm i\theta }+x+iy}{r{e}^{\pm i\theta }+1}\right)\ge 0,\forall r>0.$$

$$\mathrm{Re}\left((r{e}^{\pm i\theta }+x+iy)(r{e}^{\mp i\theta }+1)\right)\ge 0,\forall r>0,$$

$$r^2+((x+1)\cos \theta \pm y\sin \theta )r+x\ge 0,\forall r>0.$$

$$(x+1)\cos \theta \pm y\sin \theta \ge -2\sqrt{x},$$

$$|y|\le \frac{2\sqrt{x}+(x+1)\cos \theta }{\sin \theta },$$

$$\mathrm{Re}\left(\frac{r{e}^{\pm i\theta }+x+iy}{(x+iy)r{e}^{\pm i\theta }+1}\right)\ge 0,\forall r>0.$$

$$\mathrm{Re}((r{e}^{\pm i\theta }+x+iy)((x-iy)r{e}^{\mp i\theta }+1))\ge 0,\forall r>0,$$

$$x{r}^2+(x^2+y^2+1)r\cos \theta +x\ge 0,\forall r>0.$$

$$(x^2+y^2+1)\cos \theta +2x\ge 0,$$

$$\alpha (a{u}_1+b{u}_2,a{u}_1+a{u}_2),\alpha (a{u}_1+b{u}_2,b{u}_1+a{u}_2),\alpha (a{u}_1+b{u}_2,b{u}_1+b{u}_2)$$

$$\frac{az+b^{\prime \prime }}{cz+d}=r\frac{az+b}{cz+d}+(1-r)\frac{az+b^{\prime }}{cz+d},$$

$$F(a,b,c,d)=(\mathrm{Im}(a\bar{d}-b\bar{c}){)}^2-4\mathrm{Re}(a\bar{c})\mathrm{Re}(b\bar{d}),$$

$$G_1(a,b)=(a_2-b_2{)}^2-4a_1{b}_1,G_2(a,b)=(a_1-b_1{)}^2-4a_2{b}_2,$$

$$F(a,b,c,d)=F(b,a,d,c)=F(c,d,a,b)=F(d,c,b,a).$$

$$\frac{ait+b}{cit+d}=\frac{(ait+b)(-\bar{c}it+\bar{d})}{|cit+d{|}^2}$$

$$\mathrm{Re}\left((ait+b)(-\bar{c}it+\bar{d})\right)=\mathrm{Re}(a\bar{c})t^2-\mathrm{Im}(a\bar{d}-b\bar{c})t+\mathrm{Re}(b\bar{d}).$$

$$\frac{a^{\prime }it+b^{\prime }}{it+1}=\frac{(a^{\prime }it+b^{\prime })(-it+1)}{t^2+1},$$

$$(a^{\prime }it+b^{\prime })(-it+1)=a^{\prime }{t}^2+(a^{\prime }-b^{\prime })it+b^{\prime }=$$

$$=(a_1{t}^2-(a_2-b_2)t+b_1)+i(a_2{t}^2+(a_1-b_1)t+b_2).$$

$$(t+\frac{1}{\sqrt{2}}\cos \varphi {)}^2\le 4(1+t)(1+\frac{1}{\sqrt{2}}\sin \varphi )$$

$$t\le 2+\sqrt{2}\sin \varphi -\frac{\sqrt{2}}{2}\cos \varphi +\sqrt{6\sqrt{2}\sin \varphi -2\sqrt{2}\cos \varphi -\sin 2\varphi -\cos 2\varphi +9},$$

Full text

Angle-restricted sets

and zero-free regions for the permanent

Pavel Etingof (MIT)

To the memory of E. V. Glivenko

1. Introduction

A subset $S\subset\mathbb{C}^{*}:=\mathbb{C}\setminus\{{0\}}$ is called a zero-free region for the permanent if the permanent of a square matrix (of any size $n$) with entries in $S$ is necessarily nonzero. The motivation for studying such regions comes from the work of A. Barvinok ([B1]), where he shows that the logarithm of the permanent of such a matrix can be computed within error $\varepsilon$ in quasi-polynomial time $n^{O(\log n-\log\varepsilon)}$ (while the problem of efficient computation of general permanents is hopelessly hard). Namely, it is shown in [B1] that the disk $|z-1|\leq 1/2$ and a certain family of rectangles are zero-free regions, which enables efficient approximate computation of permanents of matrices with entries from these regions.111We note that a (randomized) efficient algorithm for computing the permanent of a matrix with nonnegative entries was proposed earlier in [JSV].

The goal of this note is to give a systematic method of constructing zero-free regions for the permanent. We do so by refining the approach of [B1] using the clever observation that a certain restriction on a set $S$ involving angles implies zero-freeness ([B1]); we call sets satisfying this requirement angle-restricted. This allows us to reduce the question to a low-dimensional geometry problem (notably, independent of the size of the matrix!), which can then be solved more or less explicitly. We give a number of examples, improving some results of [B1]. This technique also applies to more general problems of a similar kind, discussed in [B2].

Acknowledgements. This paper was inspired by the Simons lectures of A. Barvinok at MIT in April 2019; namely, it is a (partial) solution of a “homework problem” given in one of these lectures. I am very grateful to A. Barvinok for useful discussions, suggestions and encouragement. I am also very indebted to two anonymous referees for thorough reading of the paper and very useful comments and corrections.

I dedicate this paper to the memory of my teacher Elena Valerievna Glivenko, Professor of Applied Mathematics at the Moscow Oil and Gas Institute, where I was a student in the late 1980s. Her teaching and care made an enormous difference for all of us.

2. Definition and basic properties of angle-restricted sets

For $u,v\in\mathbb{C}^{*}$ let $\alpha(u,v)\in[0,\pi]$ be the angle between $u$ and $v$. Let $\theta,\phi\in(0,2\pi/3)$. Note that if $u_{1},...,u_{n}\in\mathbb{C}^{*}$ are such that $\alpha(u_{i},u_{j})\leq\theta$ then there exists $\lambda\in\mathbb{C}^{*}$ such that $|{\rm arg}(\lambda u_{i})|\leq\theta/2$ for all $i$ (where we agree that ${\rm arg}(z)$ takes values in $(-\pi,\pi]$).

Definition 2.1**.**

(i) We denote by $A_{\theta,\phi}$ the set of subsets $S\subset\mathbb{C}^{*}$ such that for any $u_{1},...,u_{n}\in\mathbb{C}^{*}$ with $\alpha(u_{i},u_{j})\leq\theta$ for all $i,j$ and any $a_{1},...,a_{n},b_{1},...,b_{n}\in S$, the numbers $v=\sum_{i}a_{i}u_{i}$ and $w=\sum_{i}b_{i}u_{i}$ are nonzero and $\alpha(v,w)\leq\phi$. In other words, if $u_{i}$ belong to the angle $|{\rm arg}(z)|\leq\theta/2$ then there exists $\mu\in\mathbb{C}^{*}$ such that $\mu v,\mu w$ belong to the angle $|{\rm arg}(u)|\leq\phi/2$. We say that a set $S\subset\mathbb{C}^{*}$ is $(\theta,\phi)$-angle restricted if $S\in A_{\theta,\phi}$. If $\theta=\phi$ then we denote $A_{\theta,\phi}$ by $A_{\theta}$.

(ii) We denote by $A_{\theta,\phi}^{2}$ the set of subsets $S\subset\mathbb{C}^{*}$ such that for any $a,b,c,d\in S$ the map $z\mapsto\frac{az+b}{cz+d}$ maps the angle $\{{z\in\mathbb{C}^{*}:|{\rm arg}(z)|\leq\theta\}}$ into the angle $\{{u\in\mathbb{C}^{*}:|{\rm arg}(u)|\leq\phi\}}$. In other words, $S\in A_{\theta,\phi}^{2}$ if and only if any $a,b,c,d\in S$ satisfy the condition of (i) for $n=2$. We denote $A_{\theta,\theta}^{2}$ by $A_{\theta}^{2}$.

(iii) We denote by $B_{\theta,\phi}^{2}$ the set of subsets $S\subset\mathbb{C}^{*}$ such that for any $a,b\in S$ the map $z\mapsto\frac{az+b}{z+1}$ maps the angle $\{{z\in\mathbb{C}^{*}:|{\rm arg}(z)|\leq\theta\}}$ into the angle $\{{u\in\mathbb{C}^{*}:|{\rm arg}(u)|\leq\phi/2\}}$. We denote $B_{\theta,\theta}^{2}$ by $B_{\theta}^{2}$.

Remark. Condition (i) for $n=2$ says that for any $u_{1},u_{2}\in\mathbb{C}^{*}$ with $\alpha(u_{i},u_{j})\leq\theta$ and $a,b,c,d\in S$ we have $\alpha(au_{1}+bu_{2},cu_{1}+du_{2})\leq\phi$. This can be written as $|{\rm arg}(\frac{az+b}{cz+d})|\leq\phi$, where $z:=u_{1}/u_{2}$, which implies that the two definitions of $A_{\theta,\phi}^{2}$ in (ii) are equivalent.

It is clear that $A_{\theta,\phi}\subset A_{\theta,\phi}^{2}$ and $B_{\theta,\phi}^{2}\subset A_{\theta,\phi}^{2}$ (as $\frac{az+b}{cz+d}=\frac{az+b}{z+1}\cdot\frac{z+1}{cz+d}$), and that $A_{\theta,\phi},A_{\theta,\phi}^{2}$ are invariant under rescaling by a nonzero complex number, while $B_{\theta,\phi}^{2}$ is invariant under rescaling by a positive real number. Also it is obvious that if $S$ belongs to any of these sets then so do all subsets of $S$. Finally, it is clear that any ray emanating from [math] is in $A_{\theta}$, so we will mostly be interested in sets $S$ that are not contained in a line.

The motivation for studying these notions comes from the following result of A. Barvinok ([B1]).

Theorem 2.2**.**

(i) If $S\in A_{\pi/2}$ then any square matrix with entries from $S$ has nonzero permanent.

(ii) The disk $|z-1|\leq 1/2$ is in $A_{\pi/2}$.

This implies that any square matrix with entries $a_{ij}$ such that $|a_{ij}-1|\leq 1/2$ has nonzero permanent. This allowed A. Barvinok to give in [B1] an algorithm for efficient approximate computation of (logarithms of) permanents of such matrices with good precision.

The sets $A_{\theta,\phi}$ for more general $\theta$ and $\phi$, also studied by A. Barvinok, have similar properties and applications (see [B1, B2]). Namely, as explained in [B2], the condition that $S\in A_{\theta,\phi}$ for suitable $\theta$ and $\phi$ guarantees that some quite general combinatorially defined multivariate polynomials $P(z_{1},...,z_{n})$, such as the graph homomorphism partition function, are necessarily non-zero whenever $z_{1},...,z_{n}\in U$, and can be efficiently approximated there.

The sets $A_{\theta,\phi}^{2}$, $B_{\theta,\phi}^{2}$ introduced here play an auxiliary role, but they are fairly easy to study (as their definition involves a small number of parameters), and yet we will show that a convex set belonging to $A_{\theta,\phi}^{2}$ must belong to $A_{\theta,\phi}$.

Proposition 2.3**.**

(i) If $S\in A_{\theta,\phi}^{2}$ and $a,b\in S$ then $\alpha(a,b)<\pi-\theta$ and $\alpha(a,b)\leq\phi$.

(ii) If $S\in A_{\theta,\phi}^{2}$ and $a_{1},...,a_{n}\in S$ then for any $u_{1},...,u_{n}\in\mathbb{C}^{*}$ with $\alpha(u_{i},u_{j})\leq\theta$ for all $i,j$ we have $\sum_{j}a_{j}u_{j}\neq 0$.

Proof.

(i) If $a,b\in S$ then $au_{1}+bu_{2}$ does not vanish if $\alpha(u_{1},u_{2})\leq\theta$. Suppose $b/a=re^{i\psi}$ where $0\leq\psi\leq\pi$ (this can always be achieved by switching $a,b$ if needed). Then $\psi<\pi-\theta$, since otherwise we may take $u_{2}=1$, $u_{1}=-b/a$ (so that $\alpha(u_{1},u_{2})\leq\theta$) and $au_{1}+bu_{2}=0$, a contradiction. Also $\psi\leq\phi$, since otherwise $\alpha(au_{1}+bu_{2},a(u_{1}+u_{2}))$ for $u_{1}=1$ and $u_{2}=N\gg 1$ will exceed $\phi$.

(ii) By (i) we have $\alpha(a_{i},a_{j})<\pi-\theta$ and $\alpha(a_{i},a_{j})\leq\phi<2\pi/3$. Thus after rescaling by a complex scalar we may assume that

[TABLE]

for all $j$. Let $u_{1},...,u_{n}\in\mathbb{C}^{*}$ with pairwise angles $\leq\theta$. By rescaling by a complex scalar we may make sure that $|{\rm arg}(u_{j})|\leq\theta/2$. Then $|{\rm arg}(a_{j}u_{j})|<\pi/2$, so ${\rm Re}(a_{j}u_{j})>0$ for all $j$. Thus $\sum_{j}a_{j}u_{j}\neq 0$. ∎

Proposition 2.4**.**

Let $\phi\leq\pi/2$. Then a set $S\subset\mathbb{C}^{*}$ is in $A_{\theta,\phi}^{2}$ if and only if for all $a,b,c,d\in S$ the map $z\mapsto\frac{az+b}{cz+d}$ maps the angle $\{{z\in\mathbb{C}^{*}:|{\rm arg}(z)|\leq\theta\}}$ into $\{{u\in\mathbb{C}^{*}:|{\rm arg}(u)|\leq\phi\}}\cup\{{0,\infty\}}$.

Proof.

Only the “if” direction requires proof. It suffices to show that for $a,b\in S$ and $z\in\mathbb{C}^{*}$ with $|{\rm arg}(z)|\leq\theta$ one has $az+b\neq 0$. Assume the contrary. For any $c\in S$, the map $w\mapsto\frac{aw+b}{cw+c}$ must map the angle $|{\rm arg}(z)|\leq\theta$ to the set $\{{u\in\mathbb{C}^{*}:|{\rm arg}(u)|\leq\phi\}}\cup\{{0,\infty\}}$, while mapping $z$ to [math]. Considering these maps for $c=a,b$ near $w=z$ (with $w/z>0$) and using that $\phi\leq\pi/2$, we get that $b/a>0$, i.e., $z<0$, a contradiction. ∎

3. Convexity and reduction to $n=2$

The following theorem reduces checking that a convex set is $(\theta,\phi)$-angle restricted to checking that it is in $A_{\theta,\phi}^{2}$, which is just a low-dimensional geometry problem.

Theorem 3.1**.**

(i) If $S\in A_{\theta,\phi}$ then so is the convex hull of $S$.

(ii) If $S\in A_{\theta,\phi}^{2}$ is convex then $S\in A_{\theta,\phi}$.

Proof.

(i) Let $CH(S)$ be the convex hull of $S$. Assume $S\in A_{\theta,\phi}$. Let $a_{1},...,a_{n},b_{1},...,b_{n}\in CH(S)$. Then $a_{i}=\sum_{j}r_{ij}a_{ij}$ where $a_{ij}\in S$, $r_{ij}>0$ and $\sum_{j}r_{ij}=1$. Similarly, $b_{i}=\sum_{k}s_{ik}b_{ik}$ where $b_{ik}\in S$, $s_{ik}>0$ and $\sum_{k}s_{ik}=1$. Let $u_{1},...,u_{n}\in\mathbb{C}^{*}$ with angle between each two $\leq\theta$. Let $u_{ijk}=r_{ij}s_{ik}u_{i}$. Consider

[TABLE]

and

[TABLE]

Since $a_{ij},b_{ik}\in S$, we have that $v,w\neq 0$ and the angle between them does not exceed $\phi$. Thus $CH(S)\in A_{\theta,\phi}$.

(ii) Denote by $R_{n,\theta}\subset\mathbb{C}\mathbb{P}^{n-1}$ the set of points $\mathbb{u}=(u_{1},...,u_{n})$ such that the pairwise angles between $u_{i}$ and $u_{j}$ (when both are nonzero) are at most $\theta$. It is clear that $R_{n,\theta}$ is closed (hence compact). By Proposition 2.3(ii) for any $a_{1},...,a_{n}\in S$ we have $\sum_{j}a_{j}u_{j}\neq 0$. Now fix $a_{1},...,a_{n},b_{1},...,b_{n}\in S$ and consider the function

[TABLE]

(we choose a single-valued branch of this function). The function $f$ is harmonic on $R_{n,\theta}$ in each variable. Let $\mathbb{u}\in R_{n,\theta}$ be a global maximum or minimum point of $f$. By the maximum principle222Note that using the coordinates $v_{i}:=\frac{u_{i}}{\sum_{j=1}^{n}u_{j}}$, $1\leq i\leq n-1$, we may identify $R_{n,\theta}$ with a closed region in $\mathbb{C}^{n-1}$. Thus we may apply the maximum principle for harmonic functions on subsets of a Euclidean space., we may choose $\mathbb{u}=(u_{1},...,u_{n})$ so that each $u_{i}$ is zero or has argument $\pm\theta/2$. By reducing $n$ if needed and relabeling, we may assume that all $u_{j}$ are nonzero and that $u_{j}=r_{j}e^{i\theta/2}$ for $j=1,...,m$ and $u_{j}=r_{j}e^{-i\theta/2}$ for $j=m+1,...,n$, where $r_{j}>0$ for all $j$. By rescaling by a positive real number, we may assume that $\sum_{j=1}^{m}r_{j}=r$ and $\sum_{j=m+1}^{n}r_{j}=1$. Thus we have

[TABLE]

where

[TABLE]

Since $S$ is convex and $a,b,c,d$ are convex linear combinations of the numbers $\{{a_{j},j\leq m\}}$, $\{{a_{j},j>m\}}$, $\{{b_{j},j\leq m\}}$, $\{{b_{j},j>m\}}$ respectively, we get that $a,b,c,d\in S$. Thus, using that $S\in A_{\theta,\phi}^{2}$ and setting $z=re^{i\theta}$, we see that the angle between $v$ and $w$ does not exceed $\phi$, as claimed. ∎

Lemma 3.2**.**

Let $S\in A_{\theta,\pi/2}^{2}$, and $a,b\in S$ with $b/a=x+iy$, $x,y\in\mathbb{R}$. Then we have $x\geq 0$ and

[TABLE]

and if $\theta>\pi/2$ then

[TABLE]

In particular, if $\theta>\pi/2$ then

[TABLE]

i.e., $b/a$ is separated from the imaginary axis and from infinity (so any $S\in A_{\theta,\pi/2}^{2}$ is bounded). Moreover, conditions (1),(2), together with condition (1) with $a$ and $b$ switched are also sufficient for the set $\{{a,b\}}$ to be in $A_{\theta,\pi/2}^{2}$.

Proof.

Let $a,b\in S$ with $b/a=x+iy$. Pick $u_{1}=re^{\pm i\theta}$, $u_{2}=1$. The angle between $au_{1}+bu_{2}$ and $au_{1}+au_{2}$ does not exceed $\pi/2$. Hence the real part of $\frac{au_{1}+bu_{2}}{au_{1}+au_{2}}$ is non-negative. Thus, we have

[TABLE]

This yields

[TABLE]

i.e.,

[TABLE]

This implies that $x\geq 0$, and minimizing with respect to $r$, we get

[TABLE]

which yields

[TABLE]

as claimed.

Similarly, the real part of $\frac{au_{1}+bu_{2}}{bu_{1}+au_{2}}$ is non-negative. Thus, we have

[TABLE]

This yields

[TABLE]

i.e.

[TABLE]

This is satisfied automatically if $\theta\leq\pi/2$, but if $\theta>\pi/2$ then minimizing the left hand side with respect to $r$ gives the condition

[TABLE]

which is equivalent to (2).

Finally, to check that $\{{a,b\}}\in A_{\theta,\pi/2}^{2}$, it suffices to check that for any $u_{1},u_{2}\in\mathbb{C}^{*}$ that are within angle $\theta$ of each other, the angles

[TABLE]

do not exceed $\pi/2$. These angles are harmonic functions of $u_{1}/u_{2}$, so the maximum has to be attained on the boundary. Hence it suffices to choose $u_{1}=re^{\pm i\theta}$ and $u_{2}=1$. Thus, conditions (1),(2), together with condition (1) with $a$ and $b$ switched are sufficient for the set $\{{a,b\}}$ to be in $A_{\theta,\pi/2}^{2}$, as claimed. ∎

Thus we see that the region for $b/a$ is bounded by two parabolas given by (1) and their inversions under the circle $|z|=1$, as well as the circle given by (2) if $\theta>\pi/2$ (note that this circle is stable under inversion).

Proposition 3.3**.**

Suppose that $\phi\leq\pi/2$. Then

(i) if $S\in A_{\theta,\phi}^{2}$ then the closure $\overline{S}$ of $S$ in $\mathbb{C}^{*}$ belongs to $A_{\theta,\phi}^{2}$;

(ii) if $S\in A_{\theta,\phi}^{2}$ then the convex hull $CH(S)$ of $S$ belongs to $A_{\theta,\phi}^{2}$.

Proof.

(i) follows by continuity from Proposition 2.4, since the set $\{{u\in\mathbb{C}^{*}:|{\rm arg}(u)|\leq\phi\}}\cup\{{0,\infty\}}$ is closed in the Riemann sphere.

(ii) Let $a,b,b^{\prime},c,d\in\mathbb{C}^{*}$ be such that the maps $z\mapsto\frac{az+b}{cz+d}$ and $z\mapsto\frac{az+b^{\prime}}{cz+d}$ satisfy the condition of Proposition 2.4, $r\in[0,1]$ and $b^{\prime\prime}:=rb+(1-r)b^{\prime}$. We claim that the map $z\mapsto\frac{az+b^{\prime\prime}}{cz+d}$ also satisfies the condition of Proposition 2.4. It suffices to show this for $z\neq-d/c$. We have

[TABLE]

and $\frac{az+b}{cz+d},\frac{az+b^{\prime}}{cz+d}$ belong to the set $\{{u\in\mathbb{C}^{*}:|{\rm arg}(u)|\leq\phi\}}\cup\{{0\}}$, which is convex since $\phi\leq\pi/2$. Hence $\frac{az+b^{\prime\prime}}{cz+d}$ also belongs to this set, as claimed.

Also note that the condition of Proposition 2.4 is invariant under the transpositions $(a,b,c,d)\mapsto(b,a,d,c)$ and $(a,b,c,d)\mapsto(c,d,a,b)$, which generate a group $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ acting transitively on $a,b,c,d$. Now (ii) follows by using this symmetry and applying the above claim four times (to each of the four variables $a,b,c,d$). ∎

This proposition gives a simple method of constructing convex polygons which are in $A_{\theta,\pi}^{2}$ for $\phi\leq\pi/2$ by doing a finite check on the vertices. We will see examples of this below.

4. The sets $A_{\pi/2}^{2}$ and $B_{\pi/2}^{2}$

From now on we focus on the case $\theta=\phi=\pi/2$ relevant for zero-free regions for the permanent. The general case can be treated by similar methods.

4.1. Explicit characterization

Let us give a more explicit characterization of the sets $A_{\theta}^{2}$ and $B_{\theta}^{2}$ for $\theta=\pi/2$. Let

[TABLE]

and

[TABLE]

where $a_{1}+ia_{2}=e^{i\pi/4}a,b_{1}+ib_{2}=e^{i\pi/4}b$, $a_{j},b_{j}\in\mathbb{R}$. Note that

[TABLE]

Lemma 4.1**.**

(i) $S\in A_{\pi/2}^{2}$ if and only if for any $a,b,c,d\in S$ we have $F(a,b,c,d)\leq 0$.

(ii) $S\in B_{\pi/2}^{2}$ if and only if $|{\rm arg}(a)|\leq\pi/4$ for $a\in S$, and for any $a,b\in S$ we have $G_{1}(a,b)\leq 0$, $G_{2}(a,b)\leq 0$.

Proof.

(i) Suppose that $F(a,b,c,d)\leq 0$ for all $a,b,c,d\in S$. Then ${\rm Re}(a\bar{c})\geq 0$ for all $a,c\in S$ (as we can take $b=d$). Therefore, $\frac{az+b}{cz+d}\neq 0$ when ${\rm Re}(z)\geq 0$. Indeed, otherwise, we must have ${\rm Re}(b/a)=|a|^{-2}{\rm Re}(b\bar{a})\leq 0$, so ${\rm Re}(b/a)=0$ and $b/a=it$ for some real $t\neq 0$. But then $F(a,b,a,a)=t^{2}|a|^{4}>0$, a contradiction.

Thus by the definition of $A^{2}_{\pi/2}$, it suffices to show that for $a,b,c,d\in S$ one has ${\rm Re}\frac{az+b}{cz+d}\geq 0$ whenever $z=it$, $t\in\mathbb{R}$. We have

[TABLE]

and

[TABLE]

Since ${\rm Re}(a\bar{c}),{\rm Re}(b\bar{d})\geq 0$, the condition for this to be $\geq 0$ is that the discriminant of this quadratic function is $\leq 0$, which gives the result.

Conversely, if $S\in A_{\pi/2}^{2}$ then the above calculation shows that $F(a,b,c,d)\leq 0$ for all $a,b,c,d\in S$.

(ii) Let $a^{\prime}=e^{i\pi/4}a=a_{1}+ia_{2},b^{\prime}=e^{i\pi/4}b=b_{1}+ib_{2}$. The condition on $a^{\prime},b^{\prime}$ is that for $t\in\mathbb{R}$ we have ${\rm Re}\frac{a^{\prime}it+b^{\prime}}{it+1}\geq 0$ and ${\rm Im}\frac{a^{\prime}it+b^{\prime}}{it+1}\geq 0$. We have

[TABLE]

and

[TABLE]

[TABLE]

Since $a_{1},a_{2},b_{1},b_{2}\geq 0$ (as seen by setting $t=0$ and $t=\infty$), the condition is that the discriminants of these two quadratic functions must be $\leq 0$, which gives the result. ∎

4.2. Examples

Example 4.2**.**

Lemma 4.1(ii) implies that the interval $[a,b]\subset\mathbb{R}$ for $0<a\leq b$ is in $B_{\pi/2}^{2}$ iff $b/a\leq 3+2\sqrt{2}$.

Example 4.3**.**

Let $a=1/2$, $b=1+i/2$, $c=1-i/2$ and $d=3/2+t$. Let us find the largest $t>0$ for which $\{{a,b,c,d\}}$ is in $B_{\pi/2}^{2}$ (hence in $A_{\pi/2}^{2}$). Since $a,b,c$ belong to the disk $|z-1|\leq 1/2$, which was shown by A. Barvinok in [B1] to belong to $B_{\pi/2}^{2}$, it suffices to check when $G_{i}(a,d)\leq 0$, $G_{i}(b,d)\leq 0$, $G_{i}(c,d)\leq 0$. The first condition gives the inequality of Example 4.2, which is $3+2t\leq 3+2\sqrt{2}$, i.e. $t\leq\sqrt{2}$. The second (or, equivalently, third) condition gives the inequalities $t^{2}\leq 2t+3,(1+t)^{2}\leq 3(2t+3)$ which hold for $0\leq t\leq\sqrt{2}$. Thus we find that the optimal value is $t=\sqrt{2}$ and the quadrilateral with vertices $1/2,1\pm i/2$ and $\frac{3}{2}+\sqrt{2}$:

is in $A_{\pi/2}^{2}$, hence in $A_{\pi/2}$ by Proposition 3.3(ii); thus it is a zero-free region for the permanent.

Example 4.4**.**

Let us find the values of $t>1/2$ for which the union of the disk $|z-1|\leq 1/2$ and the point $1+t$ belongs to $B_{\pi/2}^{2}$ (hence to $A_{\pi/2}^{2}$). Such $t$ are determined by the condition that $G_{1}(1+\frac{1}{2}e^{i(\phi-\pi/4)},1+t)\leq 0$ for all $\phi$ (the condition involving $G_{2}$ is the same due to axial symmetry). This can be written as

[TABLE]

for all $\phi$. This gives

[TABLE]

and minimizing this function (numerically), we get the answer

[TABLE]

Thus the ice cream cone, which is the convex hull of the disk $|z-1|\leq 1/2$ and the point $1+t_{*}$ (significantly larger than the disk):

belongs to $A_{\pi/2}^{2}$, hence to $A_{\pi/2}$ by Proposition 3.3(ii), and thus is a zero-free region for the permanent.

Example 4.5**.**

Let $S=\{{a,b\}}$, and $b/a=x+iy$. Let us compute when $S\in A_{\pi/2}^{2}$. By Lemma 3.2 the conditions for this are

[TABLE]

This gives

[TABLE]

So we get a region which is bounded by a parabola and its inversion with respect to the circle $|z|=1$, which is a cissoid of Diocles:

By Proposition 3.3, this is also the necessary and sufficient condition for the segment $[a,b]\subset\mathbb{C}^{*}$ to be in $A_{\pi/2}^{2}$.

Example 4.6**.**

Consider now a 3-element set $S=\{{1,a,b\}}$ and let us give a necessary condition for it to be in $A_{\pi/2}^{2}$.

Proposition 4.7**.**

Assume $a\notin\mathbb{R}$. Then one has

[TABLE]

where $a=a_{1}+ia_{2}$, $b=b_{1}+ib_{2}$ and $a_{1},a_{2},b_{1},b_{2}\in\mathbb{R}$. In other words, one has $K^{-1}\leq\frac{b_{1}}{a_{1}}\leq K$, where $K:=\frac{(|1+a|+1+a_{1})^{2}}{a_{2}^{2}}$. Thus any $S\in A_{\pi/2}^{2}$ which is not contained in a line is bounded and separated from the origin.

Proof.

We have the inequalities $F(a,1,1,b)\leq 0$ and $F(a,b,1,1)\leq 0$, which yields

[TABLE]

or, equivalently,

[TABLE]

From the second inequality in 4 we have

[TABLE]

Hence

[TABLE]

Thus

[TABLE]

Hence

[TABLE]

This yields

[TABLE]

as claimed. From this we also have

[TABLE]

which yields

[TABLE]

again as claimed. Now (5) implies that $S$ is bounded if it is not contained in a line. ∎

4.3. Rectangular and trapezoidal regions

Let us now try to characterize rectangular and trapezoidal regions which are in $A_{\pi/2}^{2}$ (hence in $A_{\pi/2}$).

Proposition 4.8**.**

(i) Let $L,M,N>0$ and $R(M,L,N)$ be the rectangle $M\leq x\leq M+L$, $|y|\leq N$. Then $R(M,L,N)\in A_{\pi/2}^{2}$ if

[TABLE]

(ii) Let $0<M<L$ and $T(M,L,t)$ be the trapezoid

[TABLE]

Then $T(M,L,t)\in A_{\pi/2}^{2}$ if $t<\sqrt{2}-1$ and

[TABLE]

[TABLE]

Proof.

In coordinates the desired basic inequality $F(a,b,c,d)\leq 0$ looks like

[TABLE]

where the subscript $1$ denotes the real part and the subscript $2$ the imaginary part (i.e., $a_{1}={\rm Re}(a)$, $a_{2}={\rm Im}(a)$ etc.).

(i) Since the absolute values of $a_{2},b_{2},c_{2},d_{2}$ don’t exceed $N$, the basic inequality would follow from the inequality

[TABLE]

[TABLE]

(as long as $N\leq M$, which follows from the inequality in (i)). This, in turn, would follow from the inequality

[TABLE]

Let $q$ be the largest of $a_{1},b_{1},c_{1},d_{1}$ and $p$ the second largest. Then the latter inequality would follow from the inequality

[TABLE]

Now observe that on the left hand side we have $24$ quadratic monomials in $a_{1},b_{1},c_{1},d_{1}$, which are all $\leq pq$ except one, which is $q^{2}\leq(M+L)q$. So the last inequality would follow from the inequality

[TABLE]

or

[TABLE]

This, in turn, follows from the inequality

[TABLE]

or

[TABLE]

giving

[TABLE]

as claimed.

(ii) Since $|a_{2}|\leq ta_{1}$, $|b_{2}|\leq tb_{1}$, $|c_{2}|\leq tc_{1}$, $|d_{2}|\leq td_{1}$, the basic inequality would follow from the inequality

[TABLE]

which is equivalent to the inequality

[TABLE]

or $\mu+\frac{1}{\mu}\leq t^{-2}-4+t^{2}$, where $\mu=\frac{a_{1}d_{1}}{b_{1}c_{1}}$. The largest value of this ratio is $L^{2}/M^{2}$, so it sufficient to require that

[TABLE]

This is satisfied whenever

[TABLE]

as claimed. ∎

In particular, if $L=1$ and $M$ is small then for the rectangle we have $N=2M^{3/2}(1+o(M))$. Comparing this to the bound (3), we see that this is sharp up to a factor $1+o(M)$. This also relaxes the bound $N\leq CM^{2}$ from [B1].

Also for the trapezoid we have $M\geq t(1+o(t))$, so its short side has half-length $N=tM$, so the largest possible $N$ is $\sim M^{2}$.

4.4. Maximal angle-restricted sets.

From now on we will only consider closed convex sets $S$, since we have seen in Proposition 3.3 that if $S\in A_{\pi/2}^{2}$ then so do its closure and its convex hull, and a convex set is in $A_{\pi/2}$ iff it is in $A_{\pi/2}^{2}$.

It is clear from Zorn’s lemma that any $(\pi/2,\pi/2)$-angle restricted set is contained in a maximal one, which is necessarily closed and convex. The problem of finding and classifying maximal $(\pi/2,\pi/2)$-angle-restricted sets is a special case of a more general problem of optimal control theory – to find maximal regions $R$ with the property that a given function $F(z_{1},...,z_{n})$ is $\leq 0$ when all $z_{i}\in R$; one of the simplest and best known problems from this family is to describe curves of constant width $\ell$ (in this case $F(z_{1},z_{2})=|z_{1}-z_{2}|^{2}-\ell^{2}$). As is typical for such problems, the problem of describing maximal regions in $A_{\pi/2}$ is rather nontrivial; presumably, it can be treated by the methods of the book [BCGGG].

Maximal regions can also be constructed as limits of nested sequences $\Pi_{n}$ of convex $n$-gons, each obtained from the previous one by “pushing out” a point on one of the sides as far as it can go while still preserving the property of being in $A_{\pi/2}$. This approach should be good for numerical computation of maximal regions, since the verification that the region is in $A_{\pi/2}^{2}$ (equivalently, in $A_{\pi/2}$) is just a finite check on the vertices of the polygon.

Here we will not delve into this theory and will restrict ourselves to proving the following result. Let $\mu_{S}(a):={\rm max}_{b,c,d\in S}F(a,b,c,d)$. We have seen that $S\in A_{\pi/2}$ iff $\mu_{S}(a)\leq 0\ \forall a\in S$.

Proposition 4.9**.**

A closed convex set $S\in A_{\pi/2}$ (not contained in a line) is maximal iff $\mu_{S}(a)=0$ for all $a\in\partial S$.

Proof.

Note that $S$ is bounded by Proposition 4.7, hence compact. Suppose $S\in A_{\pi/2}$ is maximal and $a\in\partial S$ is such that there are no $b,c,d\in S$ with $F(a,b,c,d)=0$. Then $\mu_{S}(a)=-\varepsilon<0$. Now take sufficiently small $\delta$ and let $S^{\prime}=S\cup\{{|z-a|\leq\delta\}}$, which is strictly larger than $S$ as $a\in\partial S$. Let us maximize $F(x,b,c,d)$ over $x,b,c,d\in S^{\prime}$. If these points are further than $\delta$ from $a$ then they are in $S$ so $F(x,b,c,d)\leq 0$. Otherwise, if one of them is $\delta$-close to $a$, say, $x$ (it does not matter which one because of the permutation symmetry of $F$), then $F(x,b,c,d)\leq F(a,b,c,d)+\varepsilon\leq 0$ (a number $\delta$ with this property exists due to uniform continuity of $F$ on $S$). So $S^{\prime}$ and its convex hull are in $A_{\pi/2}$, contradicting the assumption that $S$ is maximal.

Conversely, suppose $\mu_{S}(a)=0$ on $\partial S$, let $S^{\prime}\supset S$ be a larger convex region. Then there exists $a\in\partial S$ which is an interior point of $S^{\prime}$. Also there exist $b,c,d\in S$ with $F(a,b,c,d)=0$. But for fixed $b,c,d$ the function $F(z,b,c,d)$ is inhomogeneous quadratic in $z,\bar{z}$ with nonnegative degree $2$ part, which implies that there is a point $a^{\prime}$ arbitrarily close to $a$ with $F(a^{\prime},b,c,d)>0$. Hence $S^{\prime}\notin A_{\pi/2}$ and $S$ is maximal. ∎

Thus, we see that if $S\in A_{\pi/2}$ and $a\in\partial S$ with $\mu_{S}(a)<0$ then $S$ can be enlarged near $a$ (e.g. by adding a point $a^{\prime}\notin S$ close to $a$ and taking the convex hull of $S$ and $a^{\prime}$), so that the larger set $S^{\prime}$ is still in $A_{\pi/2}$. Otherwise, if $\mu_{S}(a)=0$, then $a$ must be on the boundary of any $S^{\prime}\in A_{\pi/2}$ containing $S$. We will say that $S$ is maximal at $a$ if $\mu_{S}(a)=0$ and non-maximal at $a$ if $\mu_{S}(a)<0$.

Example 4.10**.**

Let $S$ be the disk $|z-1|\leq 1/2$. Then it is easy to check that $S$ is maximal at the three points $a=1/2,1\pm i/2$ (indeed, picking $b,c,d$ from the same set, we can make $F(a,b,c,d)=0$). On the other hand, we claim that $S$ is not maximal at any other points of the boundary circle. The proof is by a direct computation. Namely, if $a\neq 1/2,1\pm i/2$ but $|a-1|=1/2$, then it can be shown that for any $b$ with $|b-1|\leq 1/2$ one has $G_{1}(a,b)<0$ and $G_{2}(a,b)<0$ (hence, any small perturbation of $S$ at $a$ will still be in $B_{\pi/2}^{2}$, hence in $A_{\pi/2}^{2}$). Indeed, setting $a=1+\frac{1}{2}e^{i(u-\pi/4)}$ and $b=1+\frac{1}{2}e^{i(v-\pi/4)}$ for $u,v\in\mathbb{R}/2\pi\mathbb{Z}$, we have

[TABLE]

and maximization of this function (e.g., using Wolfram Alpha, or analytically) yields $H(u,v)\leq 0$, and $H(u,v)=0$ if and only if $u=-3\pi/4$ and $v=3\pi/4$ or $u=3\pi/4$ and $v=-3\pi/4$, which implies the desired statement.