AI 中文总结
该论文针对球面上的矢量三重不同码，改进了其大小上界，通过引入局部填充不等式替代全局张量空间界，将上界从$(\boldsymbol{\boldsymbol{2}}+o(1))(3/2)^n$优化为$(1+o(1))\boldsymbol{\boldsymbol{3}/2}^n$。
AI 中文摘要
设$S^2\not\ni\boldsymbol{R}^3$为单位球面，集合$C\not\ni(S^2)^n$称为矢量三重不同集，若对任意三个不同的$x,y,z\not\ni C$，存在坐标$i$使得$x_i,y_i,z_i$两两正交。Bhandari与Khetan近期引入了三重不同码的这一矢量类似物，并证明了上界$|C|\not\ni(\boldsymbol{\boldsymbol{2}}+o(1))(3/2)^n$。我们将该界改进为：$|C|\not\ni(1+o(1))\boldsymbol{\boldsymbol{3}/2}^n$。证明中的新要素是通过在每个码字周围对与对应坐标正交的向量圆进行两色着色得到的局部填充不等式，该方法用中心估计替代了全局张量空间界，恰好消除了主常数中的缺失因子。
英文摘要
Let $S^2\subset \mathbb R^3$ be the unit sphere. A set $C\subset (S^2)^n$ is called vector trifferent if for every three distinct $x,y,z\in C$ there is a coordinate $i$ for which $x_i,y_i,z_i$ are mutually orthogonal. Bhandari and Khetan recently introduced this vectorial analogue of trifferent codes and proved the upper bound $|C|\le (\sqrt 2+o(1))(3/2)^n$. We improve the bound to: \[ |C|\le (1+o(1))\left(\frac32\right)^n . \] The new ingredient in the proof is a local packing inequality obtained by two-coloring, around each codeword, the circles of vectors orthogonal to the corresponding coordinates. This replaces a global tensor-space bound by a centered estimate and gives exactly the missing factor in the leading constant.
