发表机构
University of Minnesota; Purdue University（明尼苏达大学; 普渡大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
研究Cartwright--Steger曲面实商的光滑拓扑，通过Albanese纤维、分支曲面及正球面性质，揭示其与$\mathbb{CP}^2$及$S^1\times S^3$的关联。
AI 中文摘要
设$X$为Cartwright--Steger曲面，具有其实对合$c$和Albanese映射$\alpha:X\to E$，并令$Y=X/\langle c\rangle$。每个光滑实Albanese纤维有两个卵形且不分割。分支曲面$B\cong\\#_3\mathbb{RP}^2$满足$j_*\pi_1(B)=4\mathbb Z\subset\pi_1(Y)\cong\mathbb Z$且模二零调。每个表示$H_2(Y;\mathbb Z)$正生成元的光滑嵌入$+1$球面与$B$相交至少六点。其提升具有平方二，与$K_X$配对平凡，且无亏格零或一的连通嵌入表示。对Albanese生成元进行圆环手术，对于任一正规框架，同胚于$\mathbb{CP}^2$。若存在正球面，则吹下后得到$\mathbb Z[\mathbb Z]$-同调$S^1\times S^3$，且同伦等价于$S^1\times S^3$。
英文摘要
Let $X$ be the Cartwright--Steger surface with its real involution $c$ and Albanese map $α:X\to E$, and put $Y=X/\langle c\rangle$. Every smooth real Albanese fiber has two ovals and is nondividing. The branch surface $B\cong\#_3\mathbb{RP}^2$ satisfies $j_*π_1(B)=4\mathbb Z\subsetπ_1(Y)\cong\mathbb Z$ and is nullhomologous modulo two. Every smoothly embedded $+1$-sphere representing the positive generator of $H_2(Y;\mathbb Z)$ meets $B$ in at least six points. Its lift has square two, pairs trivially with $K_X$, and has no connected embedded representative of genus zero or one. Circle surgery on an Albanese generator is homeomorphic to $\mathbb{CP}^2$ for either normal framing. Blowing down a positive sphere, if one exists, gives a $\mathbb Z[\mathbb Z]$-homology $S^1\times S^3$ homotopy equivalent to $S^1\times S^3$.
