A recent MathOverflow post asked for a proof that the roots of certain polynomials were located on the unit circle. A comment by Richard Stanley pointed to a beautiful theorem of T. D. Lee and C. N. Yang. By the way, these two authors were physicists, got the Nobel prize in physics in 1957, and were the first Chinese scientists to be honored by the Nobel prize.
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 be a Hermitian matrix ( for all ). Define a polynomial
Theorem 1 (Lee-Yang). — If for all , then all roots of have absolute value .
This theorem follows from a multivariate result. Let us define
Say that a polynomial is good if it has no root such that for all .
Proposition 2 (Lee-Yang). — If for all , then is good.
For every pair , set if belongs to , but not , and set otherwise. Consequently,
In other words, if we define polynomials
then is the “coefficientwise product” of the polynomials .
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 are good, then so is their product.
Lemma 2. — If is good, then is good for every such that .
This is obvious if ; 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 , then is good.
This is obvious.
Lemma 4. — If , then is good.
If , then is good, as the product of two good polynomials.
Now assume that . Let be a root of such that . One has
Since the Möbius transformation defines a bijection from the unit open disk to itself, one has .
Lemma 5. — If is good, then is good.
Assume otherwise, so that . By symmetry, we assume . We write .
Choose such that and have the same argument; if, moreover is close enough to and satisfies , then
Consequently, the polynomial is not good; a contradiction.
Lemma 6. — If 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 and , so that their coefficientwise product is given by . By assumption and ;
similarly, and . Consequently, and , which shows that is good.
We prove the result by induction on . For every subset of , let and be the coefficients of and of in ; define similarly and with . The coefficientwise product of and is equal to
Let be such that , so that
is good, by lemma 2. Similarly, for such that , $Q(T_1,\dots,T_{n-1},w)
=\sum _S (c_S+d_S w) \prod_{i\in S} T_i$ is good. By induction, their coefficientwise product, given by
is good as well.
We now fix complex numbers of absolute value . 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 . According to lemma 4, the polynomial
is good. This proves that is good.
Proof of theorem 2. — We have already observed that the polynomial is the coefficientwise product of polynomials , each of them has degree at most one in each variable. On the other hand, one has
a product of good polynomials, so that is good. This proves that 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 is defined using the transpose matrix of . Consequently, has no root with for every .
Proof of theorem 1. — Let be a root of . Since the polynomial is good, so is the one-variable polynomial . In particular, implies . But the polynomial has a symmetry property, inherited by that of , namely , where is defined using the transpose matrix of . Consequently, and . We thus have shown that .
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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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