4

Let $X$ be a topological space. A subspace $A\subset X$ is said to be a strong neighbourhood-deformation retract of $X$, if there is an open neighbourhood $U$ of $A$ and a continuous map $h:U\times [0,1]\to U$ such that

$h(u,0)=u$, $\;h(u,1)\in A$ and $h(a,t)=a$ for all $u\in U$ and $(a,t)\in A\times [0,1]$. (Note that this is not exactly the same definition of $A$ strong deformation rectract in $X$ as you can see here https://en.wikipedia.org/wiki/Retract .. However, $A$ is a strong deformation retract in $U$).

Consider the complex projective space $\mathbb{C}P^n:=\mathbb{C}^{n+1}\setminus \{0\} /\sim$, where $x,y\in \mathbb{C}^{n+1}\setminus \{0\}$ satisfy $x\sim y$ $:\iff \exists \lambda\in \mathbb{C}\setminus \{0\}$ such that $x=\lambda y$. I know that $\mathbb{C}P^n\cong S^{2n+1}/x\sim \lambda x,\;\lambda\in S^1$.

My question: Why is $\mathbb{C}P^n$ a strong neighbourhood-deformation retraction of $\mathbb{C}P^{n+1}$? I need this fact for other calculations, but I don't know why this is true.

I appreciate your help.

Edit: Can I use that $S^{n-1}$ is a strong neighbourhood deformation retract of $D^n$ (with $U=\mathbb{R}^n\setminus \{0\}$ ) ?

Ryan
  • 337
  • If $(X,A)$ is a CW-pair, then $A$ is a SNDR of $X$, and you can obtain $\mathbb CP^{n+1}$ from $\mathbb CP^n$ by attaching a single disk. This gives a rather concrete image of the SNDR, and you needn't appeal to the general claim I cited. – Pedro Jan 05 '17 at 18:54
  • Thanks. I still don't know what a CW-pair is (This will be discussed later in lecture). And how to see that $\mathbb{C}P^{n+1}$ can be obtained from $\mathbb{C}P^n$ by attaching a disc? – Ryan Jan 05 '17 at 18:58
  • Adapt the ideas here. Note then that puncturing the attached disk gives a concrete NDR by collapsing the punctured disk to the boundary which lies in the smaller projective space. – Pedro Jan 05 '17 at 19:06
  • ok I understand now how to obtain $\mathbb{C}P^{n+1}$ from $\mathbb{C}P^{n}$ by attaching a disc (Considered in a pushout -diagram). But for me it's not clear why $\mathbb{C}P^{n}$ is a SNDR of $\mathbb{C}P^{n+1}$ considering pushouts (by attaching a disk). Can you elaborate it ? – Ryan Jan 05 '17 at 19:53

2 Answers2

3

Since projective space is a manifold it suffices to use the fact that any closed submanifold $S\subset M$ is a SNDR (in the ambient manifold $M$). More specifically, a neighborhood of $S$ in $M$ will be diffeomorphic to a vector bundle and using such an identification you can then use the linear structure to construct a deformation retraction explicitly.

Mikhail Katz
  • 47,573
  • thank you for the answer (I don't have the necessary knowledge to understand this. I will try to understand it, though) – Ryan Jan 05 '17 at 19:38
3

$\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$Let $x = [x_{0}: x_{1} : \dots : x_{n}]$ be homogeneous coordinates on $\Cpx\Proj^{n}$, identify $\Cpx\Proj^{n-1}$ with the locus $\{x_{0} = 0\}$, and let $U$ be the complement of the point $O = [1 : 0 : \dots : 0]$.

The mapping $$ h(x, t) = [(1 - t)x_{0} : x_{1} : \cdots : x_{n}] $$ satisfies the desired conditions.

To see geometrically what this does, look in $\Cpx^{n+1}$, with $\Cpx^{n}$ represented by the hyperplane $\{x_{0} = 0\}$ and $U$ the complement of the $x_{0}$-axis. The map $h$ continuously (in fact, smoothly) joins the identity map (at $t = 0$) to the coordinate projection from $\Cpx^{n+1}$ to $\Cpx^{n}$ (at $t = 1$), a.k.a. projection away from a point (namely $O$ in $\Cpx\Proj^{n}$). The points on the $x_{0}$-axis are precisely those whose image is the origin (and so does not define a point of $\Cpx\Proj^{n-1}$).

Andrew D. Hwang
  • 82,053