论文标题
关于广义折线的均等的cobordismism的注释
A note on the equivariant cobordism of generalized Dold manifolds
论文作者
论文摘要
令$（x，j）$为一个几乎复杂的流形，具有（平滑）涉及$σ：x \ to x $，使$ fix（σ）\ neq \ emptyset $。假设$σ$是一种复杂的共轭，即$ j $ $ j $的$σ$ antial差异。空间$ p（m，x）：= \ m athbb {s}^m \ times x/\！\ sim $其中$（v，x）\ sim（-v，σ（x）$假设一个组$ g \ cong \ mathbb z_2^s $在$ x $上表现顺利，以便$ g \circtσ=σ\ circ g $ for G $中的所有$ g \。使用对角线亚组的动作$ d = o（1）^{m+1} \ subset o（m+1）$在球体上$ \ mathbb s^{m} $，其中只有许多对$ d $的反po点cabl $ \ n $ \ nime $ \ n $ \ n $ \ n $ \ n $ \ s o \ y $ \ n $下降到$ p（m，x）$上的$ \ mathcal g $的（平滑）动作。当固定点在$ x $上设置$ g $ action的$ x^g $是有限的，对于$ p（m，x）$上的$ \ nathcal g $ Action也是如此。本说明的主要结果是，当且仅当$ [x，g] $消失时，Equivariant Clobordism类$ [P（M，X），\ Mathcal G] $消失。我们在$ x $是复杂的标志歧管的情况下说明了这一结果，$σ$是天然复杂的共轭，而$ g \ cong（\ mathbb z_2）^n $包含在$ u（n）$的对角线亚组中。
Let $(X,J) $ be an almost complex manifold with a (smooth) involution $σ:X\to X$ such that $Fix(σ)\neq \emptyset$. Assume that $σ$ is a complex conjugation, i.e, the differential of $σ$ anti-commutes with $J$. The space $P(m,X):=\mathbb{S}^m\times X/\!\sim$ where $(v,x)\sim (-v,σ(x))$ is known as a generalized Dold manifold. Suppose that a group $G\cong \mathbb Z_2^s$ acts smoothly on $X$ such that $g\circ σ=σ\circ g$ for all $g\in G$. Using the action of the diagonal subgroup $D=O(1)^{m+1}\subset O(m+1)$ on the sphere $\mathbb S^{m}$ for which there are only finitely many pairs of antipodal points that are stablized by $D$, we obtain an action of $\mathcal G=D\times G$ on $\mathbb S^m\times X$, which descends to a (smooth) action of $\mathcal G$ on $P(m,X)$. When the stationary point set $X^G$ for the $G$ action on $X$ is finite, the same also holds for the $\mathcal G$ action on $P(m,X)$. The main result of this note is that the equivariant cobordism class $[P(m,X),\mathcal G]$ vanishes if and only if $[X,G]$ vanishes. We illustrate this result in the case when $X$ is the complex flag manifold, $σ$ is the natural complex conjugation and $G\cong (\mathbb Z_2)^n$ is contained in the diagonal subgroup of $U(n)$.
