Decribe the $S^2$ fibration over $S^2$ that gives $\mathbf{CP}^2\#\overline{\mathbf{CP}}^2$

Question

According to this MO post, there is two possible $S^2$ fibration over $S^2$. One is obviously $S^2\times S^2$, another one is the connected sum of two copies of $\mathbf {CP}^2$ with different orientations. Can someone explicitly describe the $S^2$ fibration over $S^2$ that gives $\mathbf{CP}^2\#\overline{\mathbf{CP}}^2$?

Answer 1

$\mathbb{C}\mathbb{P}^{2} \# \bar{\mathbb{C}\mathbb{P}}^{2}$ is the blow-up of $\mathbb{C}\mathbb{P}^{2}$ at a point. Let $A$ be an affine chart of $ \mathbb{C}\mathbb{P}^{2}$ and $L_{\infty} = \mathbb{C}\mathbb{P}^{2} \setminus A$ be the line at infinity.

Let $p$ be the origin in $A$. Consider the set of all (projective) lines through $p \in \mathbb{C}\mathbb{P}^{2}$, then they cover $\mathbb{C}\mathbb{P}^{2}$ and only intersect pair-wise at $p$. Blowing up $p$ "seperates" all of these lines and creates a $\mathbb{C}\mathbb{P}^{1}$-bundle structure; the fibres being strict transforms of lines through $p$. The base can also be identified with $\mathbb{C}\mathbb{P}^{1}$ since each of the lines containing $p$ intersects the line $L_{\infty}$ at a unique point.

Answer 2

Cheeger described in "Some examples of manifolds with nonnegative curvature" the connected sum $\mathbb C\mathbb P^n\#\overline{\mathbb C\mathbb P^n}$ as a biquotient: Take $(S^{2n-1}\times S^2)/S^1$, where $S^1$ is acting freely on $S^{2n-1}$ and by rotations on $S^2$. This is the associated bundle to the principal fibration $S^1\to S^{2n-1}\to\mathbb C\mathbb P^n$ with fiber $S^2$. See Example 3 in Cheeger's article, why the above quotient is $\mathbb C\mathbb P^n\#\overline{\mathbb C\mathbb P^n}$.