Rank-two binomial positivity conjecture for Schur coefficients

Proof for every Schur index

Define

$$C_{k,d}(u) =e_k\bigl(i+(d-i)u:0\le i\le d\bigr), \qquad A_{k,j}(d)=[u^j](1-u)C_{k,d}(u),$$

and write

$$A_{k,j}(d)=\sum_{r\ge0}B_{k,j,r}\binom dr.$$

We prove, simultaneously for every $k$ and $0\le j\le\lfloor k/2\rfloor$,

$$B_{k,j,r}\ge0,$$

together with the stronger support bounds

$$B_{k,0,r}=0\quad(r<k), \qquad B_{k,j,r}=0\quad(r<k-1,\ j\ge1).$$

First, if a polynomial $F(x)$ has a nonnegative binomial expansion supported on indices at least $I$, then

$$(x-c)\binom xr =(r-c)\binom xr+(r+1)\binom x{r+1} \tag{1}$$

shows that multiplication by $x-c$ preserves binomial nonnegativity whenever $c\le I$. If $c=I$, the minimum support increases by one.

For $h\ge0$ and $t\ge1$, set

$$R_{h,t}(x) =\binom{x-h+1}{t} =\frac1{t!}\prod_{v=0}^{t-1}(x-(h-1+v)).$$

Successive applications of (1) show that if $F$ has minimum support at least $h-1$, then $R_{h,t}F$ is binomially nonnegative with minimum support at least $h+t-1$. The same conclusion holds when its initial support is at least $h$.

For the zeroth index,

$$A_{k,0}(d)=e_k(0,1,\ldots,d) =\genfrac{[}{]}{0pt}{}{d+1}{d+1-k}.$$

Grouping permutations by the size $N$ of their nonfixed support gives

$$A_{k,0}(d) =\sum_N D(N,k) \left(\binom dN+\binom d{N-1}\right),$$

where $D(N,k)\ge0$. A permutation of cycle defect $k\ge1$ has nonfixed support $N\ge k+1$, proving the claimed zeroth-index positivity and support bound.

Now use the exact factorization

$$E_d(T,u) :=\prod_{i=0}^d(T+i+(d-i)u) =(T+d)E_{d-1}(T+u,u).$$

Comparing coefficients, multiplying by $1-u$, extracting $[u^j]$, and writing $x=d-1$ gives the all-index recurrence

$$\begin{aligned} \Delta A_{k,j}(x) &= \sum_{t=1}^{j} \binom{x-k+t+1}{t}A_{k-t,j-t}(x)\\ &\quad+ (x+1)\sum_{t=0}^{j} \binom{x-k+t+2}{t}A_{k-1-t,j-t}(x), \end{aligned} \tag{2}$$

where $\Delta F(x)=F(x+1)-F(x)$.

There is only one apparent out-of-range term: when $k=2j$, the $t=0$ term in the second sum involves $A_{2j-1,j}$. Since $C_{h,d}(u)$ is palindromic of degree $h$,

$$A_{h,t}(d)=-A_{h,h+1-t}(d),$$

and therefore $A_{2j-1,j}=0$. All remaining terms in (2) involve legitimate lower-degree indices.

Induct on $k$. In the first sum, put $h=k-t$; the prefactor is $R_{h,t}(x)$, so the support lemma shows that each summand is binomially nonnegative with minimum support at least

$$h+t-1=k-1.$$

In the second sum, for $t\ge1$, put $h=k-1-t$; each summand has minimum support at least

$$h+t-1=k-2.$$

Multiplication by $x+1$ preserves positivity. The $t=0$ term either vanishes by the palindromic boundary above or already has support at least $k-2$.

Hence $\Delta A_{k,j}(x)$ has a nonnegative binomial expansion supported on indices at least $k-2$. But

$$\Delta A_{k,j}(x) =\sum_{r\ge0}B_{k,j,r+1}\binom xr,$$

and $B_{k,j,0}=A_{k,j}(0)=0$ for every positive $k$. Therefore

$$B_{k,j,r}\ge0 \quad\text{for all }k,j,r, \qquad B_{k,j,r}=0\quad(r<k-1,\ j\ge1).$$

This proves Conjecture 3.2 for every Schur index, extending the previously established cases $j=0,1,2$.