Let $\mathcal{H}(n,q^2)$ be a non–degenerate Hermitian variety of $\mathrm{PG}(n,q^2)$, $n\ge 2$. Let $\mathrm{NU}(n+1,q^2)$ be the graph whose vertices are the points of $\mathrm{PG}(n,q^2)\setminus \mathcal{H}(n,q^2)$ and two vertices $P_1,P_2$ are adjacent if the line joining $P_1$ and $P_2$ is tangent to $\mathcal{H}(n,q^2)$. Then $\mathrm{NU}(n+1,q^2)$ is a strongly regular graph. In this paper we show that $\mathrm{NU}(n+1,q^2)$, $n\ne 3$, is not determined by its spectrum.

中文翻译：

---

NU(n + 1,q2), n ≠ 3 的共谱图

让 $\mathcal{H}(n,q^2)$ 是非退化厄米变种 $\mathrm{PG}(n,q^2)$, $n\ge 2$. 让$\mathrm{NU}(n+1,q^2)$ 是顶点为点的图 $\mathrm{PG}(n,q^2)\setminus \mathcal{H}(n,q^2)$ 和两个顶点 ${磷}_1,{磷}_2$ 如果连接线是相邻的 ${磷}_1$ 和 ${磷}_2$ 与 $\mathcal{H}(n,q^2)$. 然后$\mathrm{NU}(n+1,q^2)$是一个强正则图。在本文中，我们表明$\mathrm{NU}(n+1,q^2)$, $n\ne 3$, 不是由它的频谱决定的。