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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( \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 ?

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Try computing the intersection form on $H^2$ (and use spell-checker). – Misha May 5 '12 at 12:01
It is not. Concerning circle actions: All the manifolds you listed have smooth circle actions, I do not know about isometric actions preserving a metric of nonnegative curvature. If the latter is your real question, you should either completely rewrite the current question or post a new one with an appropriate title. – Misha May 5 '12 at 21:43
@HKL If you change basis from $a,b$ to $a, a-b$, the form $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ changes to $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, which is the intersection form of $CP^2 \# -CP^2$. – Paul May 6 '12 at 21:46
up vote 3 down vote accepted

Summary about the manifolds $S^2\times S^2$, $S^2\tilde\times S^2$ and $\mathbb CP^2\#\mathbb CP^2$:

  1. There are only two $S^2$-bundles over $S^2$: the trivial bundle $S^2 \times S^2$ or the non-trivial bundle usually denoted $S^2\tilde\times S^2$, which is diffeomorphic to $\mathbb C P^2\#-\mathbb C P^2$, i.e., $\mathbb C P^2$ blown-up at a point. (Recall $-\mathbb C P^2$ denotes $\mathbb C P^2$ with the opposite orientation). Both of these manifolds admit cohomogeneity one actions, $SO(3)$ acts diagonally on $S^2\times S^2$ and $SU(2)$ acts on $S^2\tilde\times S^2$ when viewed as the quotient $S^3\times_{S^1} S^2$, see respectively p. 5 and 12 of this paper for details.

  2. The connected sum $\mathbb C P^2\#\mathbb C P^2$ does not admit a cohomogeneity one action, even though it can be decomposed in two disk bundles that do (however cannot be glued respecting the action). For a description in terms of a cohomogeneity two torus action, see below.

Maybe a good geometric way to visualize $\mathbb C P^2\# \mathbb C P^2$ that can help you is the following. The $2$-torus $T^2$ acts on $\mathbb C P^2$ with cohomogeneity two. Recall the construction of this action can be seen as having $T^{n+1}$ act on $\mathbb C^{n+1}$ by multiplication coordinate-wise, then restrict the action to the unit sphere $S^{2n+1}\subset \mathbb C^{n+1}$ and take a quotient of $T^{n+1}$ acting on $S^{2n+1}$ by the circle in $T^{n+1}$ whose action on $S^{2n+1}$ is the Hopf action that gives $\mathbb C P^n=S^{2n+1}/S^1$. In this way we get $T^n$ acting on $\mathbb C P^n$ (and the example above is the case $n=2$). The orbit space of this $T^2$ action on $\mathbb CP^2$ is a spherical triangle with the three angles equal to $\pi/2$. Each of the vertices corresponds to a torus fixed point in the manifold (and the sides have singular isotropy a circle).

Now, take two copies of this spherical triangle (corresponding to two copies of $\mathbb CP^2$ - and if we want to be careful I think they have to have opposite orientations), remove a neighborhood of one vertex in each triangle (corresponding to removing a ball in each copy of $\mathbb CP^2$, since the orbits of these vertices are points) and glue them together along this deleted neighborhoods. The resulting Alexandrov space is a rectangle (since the two remaining vertices on each of the triangles glued together had angle $\pi/2$); and this is exactly the orbit space of the cohomogeneity $2$ torus action on $\mathbb C P^2\#\mathbb CP^2$. [Since you were asking, in particular, you get lots of circle sub actions, but with larger orbit spaces.]

This picture is similar to the one you had of the cohomogeneity one manifold $S^3\times_{S^1} S^2$ in the following way. Imagine we are foliating the rectangle (orbit space of $T^2$-action on $\mathbb C P^2\#\mathbb CP^2$) by vertical segments, from side to side. Each of the vertical (actually, any) sides of the rectangle is a cohomogeneity one manifold with singular orbits equal to points and principal isotropy group a circle. Indeed, they are equivariantly diffeomorphic to $S^2$ with the $S^1$ rotation action. In this way, you get two $S^2$'s on opposite sides, and in the "middle", we're left with vertical segments that correspond to cohomogeneity one manifolds with singular orbits equal to circles and trivial principal isotropy groups. They are in fact spheres $S^3$ with the standard $T^2$-action. So we have a similar picture: two $S^2$'s on opposite sides and $S^3$'s in the middle. But this is different from the cohomogeneity one manifold $S^3\times_{S^1} S^2$, where you have an $S^3$ action, even though it has a similar topological structure of two $S^2$'s on opposite sides and $S^3$'s filling the "middle". Sorry for the rather informal description, I hope it helps...

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I just noticed I should maybe clarify a detail here: the nontrivial $S^2$-bundle over $S^2$, $S^3\times_{S^1} S^2$, is in fact diffeomorphic to $\mathbb CP^2\# -\mathbb CP^2$. The "difference" I mean in the above description is regarding the actions considered (cohom 1 $S^3$-action on $S^3\times_{S^1} S^2$ vs. cohom 2 $T^2$-action on $\mathbb CP^2\# -\mathbb CP^2$). If you look carefully, one action is actually a restriction of the other. In other words, they are diffeomorphic manifolds; but we described two different isometric actions on them (one is a restriction of the other). – Renato G. Bettiol Oct 1 '12 at 15:42
Perhaps another quick comment that may be useful for future reference is that $\mathbb C P^2\#\mathbb C P^2$ does not admit a cohom 1 action, only $\mathbb C P^2\#-\mathbb C P^2$ does (the difference is that the latter has the orientation opposite from the one given by the complex structure in one of the summands). The manifold $\mathbb C P^2\#\mathbb C P^2$ can be written as a biquotient $S^3\times S^3//T^2$, and is not diffeomorphic to the nontrivial $S^2$-bundle over $S^2$, which coincides with $\mathbb C P^2\#-\mathbb C P^2$. – Renato G. Bettiol Feb 23 '13 at 3:29
For more, see: – Renato G. Bettiol Feb 23 '13 at 3:32
Thank you for your description. So I concluded as follows : Two manifolds is obtained from $T^2$-action on $S^3\times S^3 $. In case of ${\bf CP}^2\sharp - {\bf CP}^2$ there exists $S^1$-action whose fixed point component has dimension 3. And note that $ {\bf CP}^2\sharp {\bf CP}^2$ is obtained from $(a,b)\cdot (Z_1,Z_2)\times (Z_3, Z_4) = (aZ_1, bZ_2)\times ( aZ_3, bZ_4)$ where $ (a,b)\in T^2$ and $(Z_1,Z_2), (Z_3,Z_4) \in S^3$. Note that this action does not have $S^1$-action whose fixed point component has dimension 3. – Hee Kwon Lee May 1 '13 at 7:29
The above comment is not correct. In the first one, $T^2$-action is free. But latter case has circle orbits. – Hee Kwon Lee Feb 6 '14 at 1:36

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