Double disk bundle: A smooth, closed manifold $M \cong DB^{-} \cup_L DB^{+}$ where · $B^{±}, L$ smooth, closed manifolds · $D^{l± +1} → DB^{±} → B^{±}$ smooth disk bundles such that $S^{l±} → L \cong ∂DB^{−} \cong ∂DB^{+} → B^{±}$ is sphere bundle. In the above Double disk bundle, I do not understand what is base space and fibre of a double disk bundle are. Please give some ideas to understand.
Thanks
The basic idea is that you have a manifold $L$ which fibers in two different ways as a sphere bundle, over two different base spaces $B_\pm$. So you have two locally trivial fibrations $$ S_\pm\to L\xrightarrow{p_\pm} B_\pm. $$ Associated to each of these fibrations is its mapping cylinder $M_{p_\pm}$, which is a disk bundle (the cone over a a sphere is a disk), $$ D_\pm\to M_{p_\pm}\xrightarrow{r_\pm} B_\pm. $$ The disk bundle maps $r_\pm$ are deformation retractions (since they are obtained from the mapping cylinders of $p_\pm$) and we have inclusion maps $i_\pm:L\to M_{p_\pm}$. Note that $M_{p_\pm}$ are manifolds with boundary $L$. We may therefore glue $M_{p_+}$ and $M_{p_-}$ along $L$. The resulting manifold $M=M_{p_-}\cup_L M_{p_+}$ is called the double mapping cylinder. It may or may not be a sphere, depending on $p_+$ and $p_-$
A good example is the case where $L$ is the flag manifold of $\mathbb C^3$, which is (real) 6-dimensional and fibers in the obvious way over $\mathbb CP^2$ (by mapping a flag $(E\subseteq F)$ to the one-dimensional subspace $E$), and which fibers in a different way over $\mathbb CP^2$ by mapping $(E\subseteq F)$ to the one-dimensional subspace $F^\perp$. In this case one can show that $M\cong \mathbb S^7$, for example by representing flags as traceless hermitian $3\times3$-matrices of norm $1$.
