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I want to distinguish $S^2$-bundles over $S^2$ from $CP^2\sharp CP^2$.

As you know, a $S^2$-bundle over $S^2$ is $S^2\times S^2$ or $M= S^3\times_{S^1}S^2$ where $M$ is diffeomorphic to a cohomogeneity one manifold, i.e, $M/G=[0,1]$, whose group diagram is $G=S^3 \supset K_- = S^1, K_+ = S^1 \supset H={ 1}$.

In detail, $G=S^3$ acts on $M^4$ isometrically whose principal isotropy group is $H={1}$. If $\pi : M \rightarrow M/G=[0,1]$ is a quotient map, then $\pi^{-1}(0)$ and $\pi^{-1}(1)$ are only two singular orbits diffeomorphic to $G/K_- = G/K_+ = S^2$

That is to say $M$ is the union of two two dimensional disk bundles $D_i$ over $S^2$ whose intersection is $S^3$. Here $\partial D_i = S^3 \rightarrow S^2$ is a Hopf fibration.

But I can not describe $CP^2\sharp CP^2$. By definition $CP^2\sharp CP^2 = (CP^2-D^4) \cup_{\partial D^4=S^3} (CP^2-D^4)$. But I can not draw the manifold in my head.


Edit (Due to Misha's comment) :

I am fine if you reply into two ways:

Way I : They have different intersection forms. I want to know the calculating way.

On $M=S^2\times S^2$, $H_2(S^2\times S^2; Z) = Z^2$. The corresponding surfaces to the generators of $H_2(S^2\times S^2; Z)$ are $S_1= S^2\times { pt}$ and $S_2={ pt } \times S^2$. We can push $S_i$ onto $S_i'$ in $M$ so that $S_i\cap S_i'=\emptyset$. But even though we push $S_i$ in any direction, $S_1'\cap S_2' \neq \emptyset$ This implies that the intersection form is $ \left( \left( \begin{array}{cc} 0 & 1 \ 1 & 0 \ \end{array} \right)$. right) $ But I do not understand why the intersection form of $CP^2\sharp −CP^2$ is $ \left( \begin{array}{cc} 1 & 1 \ 1 & 0\ \end{array} \right)$

Way II : Assume that the above three manifolds is nonnegatively curved. I want to know the geometry when $T^2$ acts on them.

Searle and Yang (See [SY]) have shown that if a closed simply connected nonnegatively curved $M^4$ admits an isometric $S^1$-action, then $M$ is homeomorphic to $S^4$, $CP^2$, $S^2\times S^2$ or $CP^2\sharp \pm CP^2$.

[SY] C. Searle and D. Yang, On the topology of nonnegatively curved simply connected 4-manifolds with continuous symmetry, Duke Math. J. Volume 74, Number 2 (1994), 547-556.

Here in the above theorem, is there the possibility of no $S^1$-action on $CP^2\sharp CP^2$ ? Or is there exact example of $S^1$-action ?

show/hide this revision's text 4 added 1406 characters in body; added 131 characters in body

Edit (Due to Misha's comment) :

I am fine if you reply into two ways:

Way I : They have different intersection forms. I want to know the calculating way.

On $M=S^2\times S^2$, $H_2(S^2\times S^2; Z) = Z^2$. The corresponding surfaces to the generators of $H_2(S^2\times S^2; Z)$ are $S_1= S^2\times { pt}$ and $S_2={ pt } \times S^2$. We can push $S_i$ onto $S_i'$ in $M$ so that $S_i\cap S_i'=\emptyset$. But even though we push $S_i$ in any direction, $S_1'\cap S_2' \neq \emptyset$ This implies that the intersection form is $\left( \begin{array}{cc} 0 & 1 \ 1 & 0 \ \end{array} \right)$.

But I do not understand why the intersection form of $CP^2\sharp CP^2$ is $ \left( \begin{array}{cc} 1 & 1 \ 1 & 0 \ \end{array} \right)$

Way II : Assume that the above three manifolds is nonnegatively curved. I want to know the geometry when $T^2$ acts on them.

Searle and Yang (See [SY]) have shown that if a closed simply connected nonnegatively curved $M^4$ admits an isometric $S^1$-action, then $M$ is homeomorphic to $S^4$, $CP^2$, $S^2\times S^2$ or $CP^2\sharp \pm CP^2$.

[SY] C. Searle and D. Yang, On the topology of nonnegatively curved simply connected 4-manifolds with continuous symmetry, Duke Math. J. Volume 74, Number 2 (1994), 547-556.

Here in the above theorem, is there the possibility of no $S^1$-action on $CP^2\sharp CP^2$ ? Or is there exact example of $S^1$-action ?

show/hide this revision's text 3 edited tags; added 20 characters in body

I want to distinguish $S^2$-bundles over $S^2$ from $CP^2\sharp CP^2$.

As you know, a $S^2$-bundle over $S^2$ is $S^2\times S^2$ or $M$ M= S^3\times_{S^1}S^2$ where $M$ is diffeomorphic to a cohomogeneity one manifold, i.e, $M/G=[0,1]$, whose group diagram is $G=S^3 \supset K_- = S^1, K_+ = S^1 \supset H={ 1}$.

In detail, $G=S^3$ acts on $M^4$ isometrically whose principal isotropy group is $H={1}$. If $\pi : M \rightarrow M/G=[0,1]$ is a quotient map, then $\pi^{-1}(0)$ and $\pi^{-1}(1)$ are only two singular orbits diffeomorphic to $G/K_- = G/K_+ = S^2$

That is to say $M$ is the union of two two dimensional disk bundles $D_i$ over $S^2$ whose intersection is $S^3$. Here $\partial D_i = S^3 \rightarrow S^2$ is a Hopf fibration.

But I can not describe $CP^2\sharp CP^2$. By definition $CP^2\sharp CP^2 = (CP^2-D^4) \cup_{\partial D^4=S^3} (CP^2-D^4)$. But I can not draw the manifold in my head.

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