Non-trivial $S^2$-bundle over $S^2$

Question

So every $S^2$-bundle over $S^2$ is either trivial or $\mathbb C P^2\#-\mathbb C P^2$ i.e. $\mathbb CP^2$ blown-up at a point (see this question)

My question; Given the group $G$ with Lie algebra $\mathfrak g=\mathfrak {sl}(2,\mathbb C)\oplus \mathfrak {su}(2)$.

and the bundle $$F\hookrightarrow E \to B$$ where $F:=\mathrm{SL}(2,\mathbb C)\Big/\begin{pmatrix} *&*\\ 0&*\end{pmatrix}\; \cong S^2$ and $B:=\mathrm{SU}(2)\Big/\begin{pmatrix} *&0\\ 0&*\end{pmatrix}\; \cong S^2$. Can we construct a subgroup $H$ of $G$ such that $G/H\cong \mathbb C P^2\#-\mathbb C P^2$?

Answer 1

In fact, the classification of compact simply connected homogeneous $4$-manifolds is quite easy to state: there are only three up to diffeomorphism and they are $S^4, \mathbb{C}P^2$, and $S^2\times S^2$.

The proof is not too bad, compared to, say, the $5$-dimensional classification. (But even that proof is not so bad...)

So, suppose a Lie group $G$ acts transitively on a simply connected closed $4$-manifold $M$. We will assume throughout that the ineffective kernel $K = \{g\in G: gp = p \text{ for all }p\in M\}$ of the action is at most finite. This will allow us to pass to and from covers without worrying.

Montgomery proved the following:

Suppose $G$ is a Lie group acting transitively on a simply connected closed manifold $M$. Then the identity component of $G$ also acts transitively. Further, the maximal compact subgroup of the identity component acts transitively. Even more so, if we find a cover of $G$ which splits as $G = T^k\times G_0$ with $G_0$ simply connected, then $G_0$ acts transitively.

Thus, we may restrict attention to the case where $G$ is a compact simply connected Lie group. Compactness allows us to assume the action is isometric by averaging an arbitrary Riemannian metric. Let $p\in M$ and set $H = \{g\in G: gp = p\}$, the isotropy group of the $G$ action on $M$ at $p$.

The map $H\rightarrow O(T_p M)$ given by $h\mapsto d_p h$ has finite kernel, so up to finite cover, $H$ embeds into $O(T_p M) = O(4)$.

If we follow the proof I gave here, it now follows that the rank of $H$ is at most $2$ and that some cover of $H$ has the form $T^{b_2(M)}\times H_0$ with $H_0$ a simply connected Lie group. This already shows $b_2(M)\leq 2$, and now the rest of the classification breaks down by cases depending on $b_2(M)$.

Since you care about $\mathbb{C}P^2 \sharp -\mathbb{C}P^2$, let me restrict attention to the case $b_2(M) = 2$ (which happens to be the easiest of the three cases).

Since we know the rank of $H$ is at most $2$ and that, up to cover, $H = T^2 \times H_0$, it follows that $H = T^2$ on the nose.

Because $G/H = M$ is a $4$-manifold, $\dim G = 6$. Up to cover, there are just not very many $6$-dimensional closed Lie groups, and only one of them is simply connected: $G$ must be $SU(2)\times SU(2)$. So, $G$ has rank $2$, so $H$ is a maximal torus. By the maximal torus theorem, any two maximal tori are conjugate (and it's easy to check that $G/H$ is naturally diffeomorphic to $G/(gHg^{-1})$ for any $g\in G$), so we can replace $H$ by our favorite maximal torus in $G$.

My favorite maximal torus is the product of the usual $S^1\subseteq SU(2)$. But then $$M = G/H = (SU(2)\times SU(2))/(S^1\times S^1) = (SU(2)/S^1)\times (SU(2)/S^1) = S^2\times S^2.$$