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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$-action ? 4 added 1406 characters in body; added 131 characters in body Edit (Due to Misha's comment) : I am fine if you reply into two ways: Way I : They have different intersection forms. I want to know the calculating way. On$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$-action ? 3 edited tags; added 20 characters in body I want to distinguish$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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