发表机构
Fudan University; AI for Scientific Simulation and Discovery Lab, Westlake University; Westlake University; University of Glasgow; ShanghaiTech University（复旦大学; 西湖大学科学模拟与发现人工智能实验室; 西湖大学; 格拉斯哥大学; 上海科技大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文证明复Banach等距猜想：若赋范空间所有固定维数（大于1）子空间线性等距，则必为Hilbert空间，通过丛-度方法刻画复超平面截口为Hermite椭球。
AI 中文摘要
Banach提出了一个问题：如果一个赋范空间在某个大于1的固定维数下，所有该维数的子空间都线性等距，那么这个空间是否必须是Hilbert空间。我们证明了复情形。通过余维数一约化，本质的有限维问题是在复$(n+1)$维空间中刻画一个平衡凸体，其所有复超平面截口都复线性等价。我们证明这样的凸体是Hermite椭球。证明改编了Lu和Yang针对实问题提出的近期丛-度方法，并具有复设定特有的两个特征。在协方差归一化后，记$G=\operatorname{Aut}_{\mathbb C}(S)$为模型截口$S$的复线性对称群。精确截口映射构成$S^{2n+1}$上的主$G$-丛；约化到$G^{\circ}$将其障碍置于$\pi_{2n}(G^{\circ})$中，该群是有限的。通过适当正度自映射拉回使该丛平凡化，并产生精确复线性截口映射的全局Lipschitz族。Brouwer度和带符号度公式随后蕴含$p_S^{2n+2}$和$p_S^{2n+4}$是实齐次多项式，其中$p_S$是模型截口的范数。唯一分解迫使$p_S^2$为二次的，相位不变性使所得二次型为Hermite型。平行四边形恒等式随后得出一般复Banach空间的结论。
英文摘要
Banach asked whether a normed space must be Hilbert if, for one fixed dimension greater than one, all subspaces of that dimension are linearly isometric. We prove the complex case. By the codimension-one reduction, the essential finite-dimensional problem is to characterize a balanced convex body in a complex $(n+1)$-space whose complex hyperplane sections are all complex-linearly equivalent. We show that such a body is a Hermitian ellipsoid. The proof adapts the recent bundle--degree method of Lu and Yang for the real problem, with two features specific to the complex setting. After covariance normalization, write $G=\operatorname{Aut}{\mathbb C}(S)$ for the complex-linear symmetry group of a model section $S$. The exact section maps form a principal $G$-bundle over $S^{2n+1}$; reduction to $G^{\circ}$ places its obstruction in $π{2n}(G^{\circ})$, which is finite. Pulling back by a suitable positive-degree self-map trivializes this bundle and yields a global Lipschitz family of exact complex-linear section maps. Brouwer degree and a signed degree formula then imply that $p_S^{2n+2}$ and $p_S^{2n+4}$ are real homogeneous polynomials, where $p_S$ is the norm of a model section. Unique factorization forces $p_S^2$ to be quadratic, and phase invariance makes the resulting quadratic form Hermitian. The parallelogram identity then yields the general complex Banach-space statement.
Comments20 pages
