发表机构
Department of Mathematics Education, Gyeongsang National University（庆尚国立大学数学教育系）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
该研究给出共形乘积结构中法拉第形式的Weyl曲率表达式，推导相关曲率恒等式，在四维情形得到ambi-Hermitian结构的局部闭性结果，并证明紧致四维爱因斯坦流形的适配Weyl联络等于黎曼联络，其通用覆叠为两常曲率曲面乘积。
AI 中文摘要
设$(M^n,g)$带有共形乘积结构，其适配Weyl联络$D$保持秩$p,q\ge2$的正交分布。我们将法拉第形式$dθ$直接用$g$的Weyl曲率表示。若$S$是由该分裂确定的正交对合，且对任意局部标准正交标架有$\mathcal K_W(X)=\sum_i W_{X,e_i}(Se_i)$，则$(dθ)^\sharp =\frac{n-2}{4(p-1)(q-1)}[\mathcal K_W,S]$，无需假设Ricci曲率。同一曲率计算给出$\nablaθ$对称部分的配套公式：$SΘ^s-Θ^sS+θ^\sharp\wedge Sθ^\sharp =\frac{p-q}{n-2}(dθ)^\sharp +\frac1{n-2}[S,\Ric^\sharp]$，其中$Θ=\nablaθ$。当Ricci张量关于乘积分裂为块对角时，这些公式导出Lee形式两个分量的显式恒等式。在四维情形，秩$(2,2)$分裂确定一个ambi-Hermitian（双埃尔米特）对。经典的ambi-Hermitian曲率恒等式在Ricci张量在两个复结构下不变时蕴含局部闭性，我们也通过上述混合Weyl迹重新得到该结论。最后，若$(M^4,g)$是紧致爱因斯坦流形，则无需特殊假设：$D=\nabla^g$，因此其通用覆叠是两个具有相同常高斯曲率的单连通曲面的乘积。
英文摘要
Let $(M^n,g)$ carry a conformal product structure whose adapted Weyl connection $D$ preserves orthogonal distributions of ranks $p,q\ge2$. We express the Faraday form $dθ$ directly in terms of the Weyl curvature of $g$. If $S$ is the orthogonal involution determined by the splitting and \[ \mathcal K_W(X)=\sum_i W_{X,e_i}(Se_i) \] for any local orthonormal frame, then \[ (dθ)^\sharp = \frac{n-2}{4(p-1)(q-1)}[\mathcal K_W,S]. \] No Ricci-curvature assumption is required. The same curvature calculation gives a companion formula for the symmetric part of $\nablaθ$, \[ SΘ^s-Θ^sS+θ^\sharp\wedge Sθ^\sharp = \frac{p-q}{n-2}(dθ)^\sharp +\frac1{n-2}[S,\Ric^\sharp], \qquad Θ=\nablaθ. \] When the Ricci tensor is block diagonal with respect to the product splitting, these formulas lead to explicit identities for the two components of the Lee form. In dimension four, the rank-$(2,2)$ splitting determines an ambi-Hermitian pair. The classical ambi-Hermitian curvature identities imply local closedness when the Ricci tensor is invariant under both complex structures; we also recover this conclusion from the mixed Weyl trace above. Finally, if $(M^4,g)$ is compact and Einstein, no specialness assumption is needed: $D=\nabla^g$. Hence the universal cover is the product of two simply connected surfaces with the same constant Gaussian curvature.
