Dimension conjecture for Schur squares of HRS codes

Complete proof of Conjecture 3.5 for all stipulated parameters.

In fact, the conclusion holds over every field $K$ of characteristic zero or characteristic $p\ge t$. Suppose that $\alpha_1,\ldots,\alpha_r\in K$ are pairwise distinct and that

$$t\le r,\qquad t\le 2s,\qquad r\ge 2t-2s+1.$$

For $0\le i<t$, let $g_i$ denote the HRS evaluation vector of $X^i$. Its coordinate in derivative block $j$ at the evaluation point $\alpha_v$ is

$$g_i(j,v)=\binom{i}{j}\alpha_v^{\,i-j}.$$

Consequently the Schur products $R_{i,k}=g_i\star g_k$, indexed by $0\le i\le k<t$, have coordinates

$$R_{i,k}(j,v) =\binom{i}{j}\binom{k}{j}\alpha_v^{\,i+k-2j}. \tag{1}$$

Interpolation functionals. Set

$$\Phi(X)=\prod_{v=1}^{r}(X-\alpha_v).$$

For $0\le j<s$ and $0\le e<r$, let $L_{j,e}$ interpolate the $r$ entries in derivative block $j$ by their unique polynomial of degree less than $r$, then extract its $X^e$-coefficient. Thus, for every polynomial $P$,

$$L_{j,e}\!\left((P(\alpha_v))_{v=1}^{r}\right) =[X^e](P\bmod\Phi). \tag{2}$$

For each total degree $h\in\{0,\ldots,2t-2\}$, define

$$\begin{aligned} I_h&=\max\{0,h-t+1\},\\ c_h&=\left\lfloor\frac h2\right\rfloor-I_h+1,\\ a_h&=\max\left\{0,\left\lceil\frac{h-r+1}{2}\right\rceil\right\}. \end{aligned} \tag{3}$$

Exactly $c_h$ product vectors have total degree $h$, namely

$$(i,k)=(I_h+\ell,h-I_h-\ell), \qquad 0\le\ell<c_h. \tag{4}$$

Pair these vectors with the $c_h$ functionals

$$L_{j,h-2j}, \qquad a_h\le j<a_h+c_h. \tag{5}$$

All these indices are valid. Indeed, $a_h\le I_h$ and $0\le h-2j<r$. If $a_h=0$, then

$$c_h\le\left\lceil\frac t2\right\rceil\le s.$$

If $a_h>0$, the hypothesis $r\ge 2t-2s+1$ gives

$$a_h\le s-t+\left\lceil\frac h2\right\rceil, \qquad c_h=t-\left\lceil\frac h2\right\rceil,$$

and hence $a_h+c_h\le s$.

Block-triangularity. A product vector of smaller total degree $h'<h$ has, in derivative block $j$, either the zero polynomial or a polynomial of degree

$$h'-2j<h-2j<r.$$

The corresponding functional $L_{j,h-2j}$ therefore annihilates it. Ordering both product vectors and functionals by total degree makes their square pairing matrix block triangular. Its diagonal block of degree $h$ is

$$D_h= \left[ \binom{I_h+\ell}{a_h+u} \binom{h-I_h-\ell}{a_h+u} \right]_{0\le\ell,u<c_h}. \tag{6}$$

For $h=0$, this is the invertible block $[1]$. For $h>0$, put $I=I_h$, $a=a_h$, and $c=c_h$. The previously established determinant formula in Lemma 2.5 of Gu, Zhu, and Zhang, applied to $M(a,a+c-1,I,h-I)$, gives

$$\det D_h = \prod_{\ell=0}^{c-1} \frac{ \binom{I+\ell}{a}\binom{h-I-\ell}{a} }{ \binom{a+\ell}{\ell}^{2} } \prod_{w=0}^{c-2} \frac{ \displaystyle\prod_{u=1}^{c-1-w}(h-2I+1-w-2u) }{ (c-1-w)! }. \tag{7}$$

Every upper binomial argument and every factorial argument is less than $t$. Every remaining linear factor belongs to $\{1,\ldots,t-1\}$: if $h<t$, its minimum is $1$ or $2$ and its maximum is $h-1<t$; if $h\ge t$, its minimum is

$$2\left\lceil\frac h2\right\rceil-h+1\in\{1,2\},$$

and its maximum is $2t-h-3<t$. Thus $\det D_h\ne0$ in characteristic zero and in every characteristic $p\ge t$.

Every diagonal block is therefore invertible. Since

$$\sum_{h=0}^{2t-2}c_h =\#\{(i,k):0\le i\le k<t\} =\binom{t+1}{2},$$

all Schur products are linearly independent. Consequently,

$$\boxed{ \dim_K\!\left( \operatorname{HRS}_t(\boldsymbol\alpha,s)^{\star 2} \right) =\binom{t+1}{2} =\frac{t(t+1)}2. }$$

This proves the full conjecture for every permitted finite field and every choice of distinct evaluation points, with no generic-position assumption.