1
$\begingroup$

A Bott tower of height $n$ is a sequence of $\mathbb CP^1$ bundles $\require{AMScd}$ \begin{CD} B_n @>{\pi_n}>> B_{n-1} @>{\pi_{n-1}}>> \cdots @>{\pi_2}>>B_1@>{\pi_1}>> B_0 \end{CD} Where $B_0$ is a point, $B_k=\mathbb P(1\oplus L_{k-1})$, $1$ is the trivial complex line bundle over $B_{k-1}$ and $L_{k-1}$ is a line complex bundle over $B_{k-1}$. Here $\mathbb P(-)$ is the complex projectivization.

Question - Given the zero section of the line bundle $p_{k-1}:L_{k-1}\to B_{k-1}$ can we define a section of $\pi_k:B_k\to B_{k-1}$?

My attempt :

for $k=1$ I tried the following - The only $\mathbb C$ - bundle over a point is the trivial bundle. Let $0:B_0\to \mathbb C\times B_0\cong \mathbb C$ be the zero section. Now $B_1\cong\mathbb CP^1$. Consider the map $\mathbb C\to\mathbb CP^1$ given by $z\mapsto [1:z]$. Then the composition of these maps gives a section of $\pi_1:B_1\to B_0$.

How do I go about it for the general case? Is there a natural map from $L_{k-1}\to\mathbb P(1\oplus L_{k-1})=B_k$?

Thank you.

$\endgroup$

1 Answer 1

2
$\begingroup$

Given two one dimensional vector spaces $L_1,L_2$ over $\mathbb{C}$ we can form the projectivized space $P(L_1\oplus L_2)$. This is of course a sphere. Note that there are two special points: the points $[z,0]$ and $[0,w]$, with $z\in L_1$ and $w\in L_2$ non-zero. Note that in the projectivization these points are well defined, i.e. do not depend on the choice of $z$ and $w$. Now for a bundle $L_1\oplus L_2\rightarrow B$ we can write down two sections of $P(L_1\oplus L_2)$: Namely the section that sends $x\mapsto [z,0]$ or the section that sends $x\mapsto [0,w]$ for some choices of non-zero $z\in (L_1)_x$ and $w\in (L_2)_x$. These sections do not depend on the choice of non-zero $z$ and $w$ hence are well defined sections of the bundle $P(L_1\oplus L_2)$.

$\endgroup$
1
  • $\begingroup$ I am a little confused by the definition of the two sections you mentioned. I have asked a separate question to get this clarified. Please have a look at it if you can. Thank you. $\endgroup$
    – R_D
    Jul 15, 2016 at 5:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .