This theorem appears in an appendix to their paper, Statistical Theory of Equations of State and Phase Transitions.IL Lattice Gas and Ising Model, published in Phys. Review in 1952, and devoted to properties of the partition function of some lattice gases. Here is a discussion of this theorem, following both the initial paper and notes by Shayan Oveis Gharan.

Let $A=(a_{i,j})_{1\leq i,j\leq n}$ be a Hermitian matrix ($a_{i,j}=\overline{a_{j,i}}$ for all $i,j$). Define a polynomial

\[ F(T) = \sum_{S\subset\{1,\dots,n\}} \prod_{\substack{i\in S \\ j\not\in S}} a_{i,j} T^{\# S}. \]

**Theorem 1 (Lee-Yang).**—

*If $\lvert a_{i,j}\rvert\leq 1$ for all $i,j$, then all roots of $F$ have absolute value $1$.*

This theorem follows from a multivariate result. Let us define

\[ P(T_1,\dots,T_n) = \sum_{S\subset\{1,\dots,n\}} \prod_{\substack{i\in S \\ j\not\in S}} a_{i,j} \prod_{i\in S} T_i .\]

Say that a polynomial $F\in\mathbf C[T_1,\dots,T_n]$ is good if it has no root $(z_1,\dots,z_n)\in\mathbf C^n$ such that $\lvert z_i\rvert <1$ for all $i$.

**Proposition 2 (Lee-Yang).**—

*If $\lvert a_{i,j}\rvert\leq 1$ for all $i,j$, then $P$ is good.*

For every pair $(i,j)$, set $a^S_{i,j}=a_{i,j}$ if $i$ belongs to $S$, but not $j$, and set $a^S_{i,j}=1$ otherwise. Consequently,

\[ P(T_1,\dots,T_n) = \sum_{S\subset\{1,\dots,n\}}\prod_{i,j} a^S_{i,j} \prod_{k\in S} T_k. \]

In other words, if we define polynomials

\[ P_{i,j} (T_1,\dots,T_n) = \sum_{S\subset\{1,\dots,n\}} a^S_{i,j} \prod_{k\in S} T_k, \]

then $P$ is the “coefficientwise product” of the polynomials $P_{i,j}$.

We also note that these polynomials have degree at most one with respect to every variable. These observations may motivate the following lemmas concerning good polynomials.

*Lemma 1. — If $P,Q$ are good, then so is their product.*

*Lemma 2. — If $P(T_1,\dots,T_n)$ is good, then $P(a,T_2,\dots,T_n)$ is good for every $a\in\mathbf C$ such that $\lvert a\rvert \leq 1$.*

This is obvious if $\lvert a\rvert <1$; in the general case, this follows from the Rouché theorem — the set polynomials (of bounded degree) whose roots belong to some closed subset is closed.

*Lemma 3. — If $\lvert a\rvert \leq 1$, then $1+aT$ is good.*

This is obvious.

*Lemma 4. — If $\lvert a\rvert\leq 1$, then $P=1+aT_1+\bar a T_2+ T_1T_2$ is good.*

If $\lvert a\rvert =1$, then $P=(1+aT_1)(1+\bar aT_2)$ is good, as the product of two good polynomials.

Now assume that $\lvert a\rvert <1$. Let $(z_1,z_2)$ be a root of $P$ such that $\lvert z_1\rvert<1$. One has

\[ z_2 = - \frac{1+az_1}{z_1+\bar a}. \]

Since the Möbius transformation $z\mapsto (z+\bar a)/(1+a z)$ defines a bijection from the unit open disk to itself, one has $\lvert z_2\rvert >1$.

*Lemma 5. — If $P=a+bT_1+cT_2+dT_1T_2$ is good, then $Q=a+dT$ is good.*

Assume otherwise, so that $\lvert a\rvert <\lvert d\rvert$. By symmetry, we assume $\lvert b\rvert\geq \lvert c\rvert$. We write $P (T_1,T_2) = (a+cT_2) + (b+dT_2) T_1$.

Choose $z_2\in\mathbf C$ such that $dz_2$ and $b$ have the same argument; if, moreover $z_2$ is close enough to $1$ and satisfies $\lvert z_2\rvert <1$, then

\[ \lvert b+dz_2\rvert=\lvert b\rvert +\lvert dz_2\rvert > \lvert a\rvert+\lvert c\rvert>\lvert a+cz_2\rvert. \]

Consequently, the polynomial $P(T_1,z_2)$ is not good; a contradiction.

*Lemma 6. — If $P,Q$ are good polynomials of degree at most one in each variable, then so is their coefficientwise product.*

We first treat the case of one variable: then $P=a+bT$ and $Q=a'+b'T$, so that their coefficientwise product is given by $R=aa'+bb'T$. By assumption $\lvert a\rvert \geq \lvert b\rvert$ and $a\neq 0$;

similarly, $\lvert a'\rvert \geq \lvert b'\rvert $ and $a'\neq 0$. Consequently, $aa'\neq0$ and $\lvert aa'\rvert \geq \lvert bb'\rvert$, which shows that $R$ is good.

We prove the result by induction on $n$. For every subset $S$ of $\{1,\dots,n-1\}$, let $a_S$ and $b_S$ be the coefficients of $\prod_{i\in S}T_i$ and of $\prod_{i\in S} T_i \cdot T_n$ in $P$; define similarly $c_S$ and $d_S$with $Q$. The coefficientwise product of $P$ and $Q$ is equal to

\[ R= \sum_S (a_S c_S +b_S d_S T_n ) \prod_{i\in S} T_i . \]

Let $z\in\mathbf C$ be such that $\lvert z\rvert \leq 1$, so that

\[ P(T_1,\dots,T_{n-1},z)= \sum _S (a_S+b_S z) \prod_{i\in S} T_i \] is good, by lemma 2. Similarly, for $w\in\mathbf C$ such that $\lvert w\rvert\leq 1$, $Q(T_1,\dots,T_{n-1},w)

=\sum _S (c_S+d_S w) \prod_{i\in S} T_i$ is good. By induction, their coefficientwise product, given by

\[ R_{z,w} = \sum_S (a_S+b_S z)(c_S+d_S w) \prod_{i\in S} T_i \]

is good as well.

We now fix complex numbers $z_1,\dots,z_{n-1}$ of absolute value $<1$. By what precedes, the polynomial

\[ S(T,U) = (\sum_S a_S c_S z_S) + (\sum_S b_Sc_S z_S) T + (\sum_S a_S d_S z_S) U

+ (\sum_S b_S d_S z_S) TU \]

is good, where $z_S=\prod_{i\in S}z_i$. According to lemma 4, the polynomial

\[ R(z_1,\dots,z_{n-1},T) = (\sum_S a_S c_S z_S) + (\sum_S b_S d_S z_S) T \]

is good. This proves that $R$ is good.

*Proof of theorem 2. —*We have already observed that the polynomial $P$ is the coefficientwise product of polynomials $P_{i,j}$, each of them has degree at most one in each variable. On the other hand, one has

\[ P_{i,j} = (1+a_{i,j} T_i + a_{j,i} T_j + T_i T_j) \prod_{k\neq i,j} (1+T_k), \]

a product of good polynomials, so that $P_{i,j}$ is good. This proves that $P$ is good.

In fact, more is true. Indeed, one has

\[

\begin{align*} T_1\dots T_n P(1/T_1,\dots,1/T_n)

& = \sum_ S \prod_{\substack{i\in S \\ j\notin S}} a_{i,j} \prod_{i\notin S} T_i \\

& =P^*(T_1,\dots,T_n)

\end{align*}

\]

where $P^*$ is defined using the transpose matrix of $A$. Consequently, $P$ has no root $(z_1,\dots,z_n)$ with $\lvert z_i\rvert >1$ for every $i$.

*Proof of theorem 1. —*Let $z$ be a root of $P$. Since the polynomial $P$ is good, so is the one-variable polynomial $F(T)=P(T,\dots,T)$. In particular, $F(z)=0$ implies $\lvert z\rvert \geq 1$. But the polynomial has a symmetry property, inherited by that of $P$, namely $ T^n F(1/T)=F^*(T)$, where $F^*$ is defined using the transpose matrix of $A$. Consequently, $F^*(1/z)=0$ and $\lvert 1/z\rvert \geq 1$. We thus have shown that $\lvert z\rvert=1$.

Nice! Thanks Antoine for this post

ReplyDeleteC'était un de mes oraux pour l'entrée à l'ENS ! Malheureusement j'avais été complètement incapable de le résoudre.

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