9
$\begingroup$

Let $u,a,b,n$ be nonnegative integers such that $n\le a+b$. Define the quantity $$ L(u,a,b,n):= (u+a+b-n)!\times\sum_{i,k,\ell}\ \frac{(-1)^k\ \ (u+a+b-i)!\ (k+\ell)!\ (a+b-k-\ell)!\ (u+a+b-k-\ell)!}{i!\ (n-i)!\ k!\ (a-k)!\ \ell!\ (b-\ell)!\ (a+b-k-\ell-i)!\ (k+\ell-n+i)!\ (u+a+b-k-\ell-i)!} $$ where the range of summation is $i,k,\ell\in\mathbb{Z}$ such that the arguments of all the factorials are nonnegative. Namely, this means the inequalities $$ 0\le i\le n\ ,\ 0\le k\le a\ , $$ and $$ \max(0,n-i-k)\le \ell\le\min(b,a+b-i-k)\ . $$

My question is: how to show that $L(u,a,b,n)$ is always nonnegative?

As per Timothy's comment, let me add some context.

Let $\mathbb{C}^{N}$ be equipped with the Hermitian inner product (with the physics convention of linearity on the right, and antilinearity on the left) $$ \langle v,w\rangle:=\sum_{i=1}^{N}\overline{v_i} w_i\ . $$ Let $v,w,\tilde{v},\tilde{w}$ be four independent uniform random vectors on the unit sphere of $\mathbb{C}^{N}$. Define the two independent random variables (or $U(N)$ invariant observables) $$ X:=|\langle v,w\rangle|^2\ ,\ Y:=|\langle \tilde{v},\tilde{w}\rangle|^2 $$ and consider the expectation $$ G(u,a,b):=\mathbb{E}\left[ X^u Y^u(X-Y)^a (X+Y)^b \right]\ . $$ With a bit of work, one can show that $$ G(u,a,b)=a! b!\ (N)_{u+a+b}^{-2}\sum_{n=0}^{a+b} L(u,a,b,n)\ (N-1)_{n} $$ where $(x)_n=x(x+1)\cdots(x+n-1)$ is the Pochhammer symbol.

If $a$ is odd then $G(u,a,b)=0$. If $a$ is even, then the integrand is nonnegative, and therefore we always have $G(u,a,b)\ge 0$. The case $u=0$, is the Ginibre inequality for the $\mathbb{C}\mathbb{P}^{N-1}$ model of statistical mechanics, with only two lattice sites. The consideration of the case $u>0$ was introduced in my paper "Non-Abelian correlation inequalities and stable determinantal polynomials".

If my conjecture on the positivity of the $L$'s is true, this would establish the positivity of $G$, even when $N\ge 1$ is not an integer, somewhat in the spirit of the work of Deligne on representations of the symmetric group of non-integer order, or the article "Deligne categories in lattice models and quantum field theory, or making sense of O(N) symmetry with non-integer N" by Binder and Rychkov.

Update: In the extreme case $n=a+b$, with $a$ even, I was able to prove $$ L(u,a,b,a+b)=u!^2 \binom{2u+a+b+1}{b} \binom{u+\frac{a}{2}}{\frac{a}{2}}\ . $$ This is in line with the predictions in the comments by Peter Taylor about the degree as a polynomial in $u$.

Improve this question
$\endgroup$
10
  • 4
    $\begingroup$ Where does this sum come from and/or why do think it is always nonnegative? $\endgroup$ Commented Jul 24 at 22:05
  • 3
    $\begingroup$ @TimothyChow: The inequality comes from the study of correlation inequalities for the $\mathbb{C}\mathbb{P}^{N-1}$ model, as in the paper arxiv.org/abs/2207.07603 It is a strengthening of an inequality which is easy to prove. I also did quite a few numerical checks on Mathematica to have a reasonable level of confidence that $L$ should be nonnegative. $\endgroup$ Commented Jul 25 at 13:48
  • $\begingroup$ @AbdelmalekAbdesselam The symbol $u$ is used for two different things. $\endgroup$ Commented Jul 28 at 12:34
  • 3
    $\begingroup$ We can rearrange to $$L(u,a,b,n):= \frac{(u+a+b-n)!^2}{(a+b-n)!} \sum_{i,k,\ell} (-1)^k \binom{k+\ell}{k} \binom{a+b-k-\ell}{a-k} \binom{a+b-n}{i+k+\ell-n} \binom{u+a+b-i}{n-i} \binom{u+a+b-k-\ell}{i}$$ You want the sum (which is a polynomial in $u$ of degree at most $n$) to be non-negative for $u \ge 0$, but a stronger conjecture appears to hold: that all of its coefficients are non-negative. (NB empirically the sum appears to be $0$ for odd $a$ and of degree $n-\frac a2$ for even $a$). $\endgroup$ Commented Jul 29 at 15:27
  • 3
    $\begingroup$ (Corrected) Your comment on Pochhammer symbols prompted me to look at the inner sum in the binomial basis rather than the monomial basis, and empirically some strong structure emerges, leading to the further conjecture that for even $a$ $$L(u,a,b,n):= \frac{(u+a+b-n)!^2}{(a+b-n)!} \sum_{m=0}^n \binom{u}{m} \sum_{d=\lceil b/2 \rceil}^b (-1)^{b-d} 2^{2d-b} \binom{d}{2d-b} \binom{a/2+d}{d} \binom{a/2+d}{m+a+b-n}$$ I'm unsure how to prove that it is equivalent, nor will I have much time immediately to work on it, but perhaps you can see how to derive this from the original setup. $\endgroup$ Commented Aug 3 at 7:05

1 Answer 1

Reset to default
3
+500
$\begingroup$

This is not a full answer, and is based on the comments above, mainly on the inspiring comments of @PeterTaylor, as well as on own thoughts.

Denote the expression in the op's sum, including the prefactor, but without the sign $(-1)^k$, as $A(\cdot)>0$. The partial sums over $\{i,\ell\}$ result in (anti-) symmetric terms under $k\mapsto a-k$, i.e., \begin{align}\tag{1a}\label{eq:1a} L(u,a,b,n;k) &= (-1)^k \sum_{i,\ell} A(\cdot) = (-1)^a L(u,a,b,n;a-k), \end{align} and therefore $L$ vanishes for odd $a$, where there is an even number ($a{+}1$) of terms.

On the other hand, the partial sums over $\{i,k\}$ under $\ell\mapsto b-\ell$, \begin{align}\tag{1b}\label{eq:1b} L(u,a,b,n;\ell) &= \sum_{i,k} (-1)^k A(\cdot) = L(u,a,b,n;b-\ell) \geq 0 \end{align} are all nonnegative and symmetric.

As Peter already noted in the comments, the inner sum over $d$ in his conjectured expression \begin{align} L(u,a,b,n) &= \frac{(u+a+b-n)!^2}{(a+b-n)!} \sum_{m=0}^n \binom{u}{m} \times{}\\ \tag{2}\label{eq:2} &\quad\times\sum_{d=\lceil b/2 \rceil}^b (-1)^{b-d} 2^{2d-b} \binom{d}{2d-b} \binom{a/2+d}{d} \binom{a/2+d}{m+a+b-n} \end{align} is always nonnegative. It can be simplified as follows:

Let $a$ even, $c=\frac a2+b\in\mathbb N_0$, and $\nu=a+b-n \in\{0,\ldots,c\}$. Then $L$ can be written as \begin{align}\tag{3}\label{eq:3} \tilde L(u,c,b,\nu)= 2^b\frac{(u+\nu)!^2}{\nu!}\binom{c}{b} \sum_{\mu=0}^c \binom{u}{\mu}\binom{c}{\mu+\nu} \, {}_{3}F_{2}\big(\tfrac{1-b}2,-\tfrac b2,\mu+\nu-c;-c,-c;1\big). \end{align}

I guess one can show that the hypergeometrics are always positive for integer $b,\mu,\nu \in \{0,\ldots,c\}$.

Finally note that $\binom{u}{\mu} = \frac{u^{\underline\mu}}{\mu!}$ can be written as (falling) factorial power, and that therefore $L(u,a,b,n) =\tilde L(u,\frac a2+b,b,a+b-n)$ is a Newton series of degree $c$ in $u$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.