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Let $\Delta\subset S^2\times S^2$ be the diagonal. Then $S^2\times S^2\setminus \Delta$ is an open four dimensional manifold. By a compactification of it, I mean a closed four dimensional manifold $X$ with an embedding $S^2\times S^2\setminus \Delta\to X$ on to a dense open subset.

Of course $X=S^2\times S^2$ is an obvious candidate. My question is that do we have other compactifications which is not (diffeo)homeomorphic to $S^2\times S^2$, with $h^2(X)=rank(H^2(X))=2$?

And more generally, in four-dimensional, replace $S^2$ by a compact Riemann surface and the same question.

Even more generally, let $S^k\Sigma$ be the $k$-th symmetric product of a compact Riemann surface, which is a $2k$-dimensional manifold. Consider $S^k\Sigma\times S^l\Sigma\setminus \Delta$, where $\Delta= \{ (x, y)\in S^k\Sigma\times S^l\Sigma| x\cap y\neq \emptyset \}$. What kind of compatification of $S^k\Sigma\times S^l\Sigma\setminus \Delta$ can we have? (How to describe the minimum condition in this case?)

p.s. Many blow-up constructions give candidate compactifications, but they will increase the second betti number $h^2$. So I restrict this question to "minimal" compactifications.

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Any compactification of $S^2\times S^2\setminus\Delta$ which is simply connected and has the same homology as $S^2\times S^2$ will be at least homeomorphic to $S^2\times S^2$ (in particular it can't be the nontrivial $S^2$-bundle over $S^2$). Here's a sketch of an argument.

For later convenience let's work instead with the complement of the antidiagonal $\bar{\Delta}$ in $S^2\times S^2$, i.e. $\bar{\Delta}$ consists of points $(p,Ap)$ where $A:S^2\to S^2$ is the antipodal map. Of course this is equivalent to $S^2\times S^2\setminus \Delta$ by an orientation reversing diffeomorphism.

Note then that $S^2\times S^2\setminus \bar{\Delta}$ is diffeomorphic to the open unit disc bundle of the tangent bundle to $S^2$: send a tangent vector $v$ based at $p$ to the pair $(p,exp_p(v))$, where the metric is a standard round metric normalized so that lines of longitude have length exactly one. So if $U(\bar{\Delta})$ is a small tubular neighborhood of $\bar{\Delta}$ the closed set $S^2\times S^2\setminus U(\bar{\Delta})$ is homeomorphic as a manifold with boundary to the closed unit disc bundle which I'll write as $DS^2$. It's a standard fact (and an amusing exercise) to show that $\partial DS^2=\mathbb{R}P^3$.

Consequently any hypothetical compactification of $S^2\times S^2\setminus \bar{\Delta}$ which is a manifold can be written as a union $X=DS^2\cup_{\mathbb{R}P^3} N$ where $N$ is a manifold with boundary $\mathbb{R}P^3$. (Ordinarily one should say that its oriented boundary is $\mathbb{R}P^3$ with reversed orientation, but since $\mathbb{R}P^3$ admits an orientation-reversing diffeomorphism this is immaterial.) Now there are various constraints on manifolds with boundary $\mathbb{R}P^3$; see for instance Lemma 2.1 in math.GT/0308073 for some relevant ones. From these and from toying with the Mayer-Vietoris sequence one infers various things [what follows is EDITED from the original version, which contained some misstatements that don't affect the conclusion], for instance that the map $H_2(N;\mathbb{Z})\oplus H_2(DS^2;\mathbb{Z})\to H_2(X;\mathbb{Z})$ is injective and has image of index $2$. Thus if $b_2(X)=2$, we have $b_2(N)=1$; further the above-cited result then shows that a generator of the $H_2(N)/torsion$ will have self-intersection $-2$. Consequently $H_2(X)/torsion$ has an index 2 subgroup which is generated by a surface $A$ in $N$ with self-intersection $-2$ together with the diagonal $\Delta$ in $S^2\times S^2\setminus \bar{\Delta}\cong DS^2$, which has self intersection $2$. These generators for the index 2 subgroup don't intersect each other. You can then obtain that the whole group $H_2(X;\mathbb{Z})/torsion$ is generated by $\frac{1}{2}(\Delta+A)$ and $\frac{1}{2}(\Delta-A)$. These generators form a standard hyperbolic pair (i.e. they have intersection number one with each other and zero with themselves), and so the intersection form of $X$ is the same as the intersection form of $S^2\times S^2$. If $X$ is simply-connected then Freedman's results show that $X$ is homeomorphic to $S^2\times S^2$.

To conclude that it's necessarily diffeomorphic to $S^2\times S^2$ one would need to know more than I do about manifolds with boundary $\mathbb{R}P^3$.

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Thanks Mike, I have to take some time read your answer. – Guangbo Xu Oct 12 '10 at 16:15
There's one other possibility that my answer above doesn't take into account: the surface that I call A in N could have self-intersection +2 instead of -2. In this case one finds that X has the same intersection form as the connect sum of CP2 with itself, so again Freedman's results show that if X is simply connected it's homeomorphic to CP2#CP2. (This shouldn't be confused with the nontrivial sphere bundle, which is CP2#(CP2-bar)). So bearing in mind that in the second paragraph of the answer the orientation was switched, the other possible compactification is (CP2-bar)#(CP2-bar). – Mike Usher Oct 13 '10 at 12:54

The answer to your first and second questions is "yes". You can blow up a product of complex projective lines (or Riemann surfaces) at any point on the diagonal, and you can identify the preimage of your open 4-fold as an open dense subset of the resulting manifold. This blowing-up process can be iterated, so you have a lot of possible compactifications. The product is in some sense "minimal" as a complex variety containing your 4-fold, since the diagonal has non-negative self-intersection.

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Then the rank of $H^2$ increase by 1 after one blow-up, right? But can we get like "the" nontrivial $S^2$-bundle over $S^2$? – Guangbo Xu Oct 11 '10 at 13:52
I doubt it, but I'm really the wrong person to ask. – S. Carnahan Oct 11 '10 at 16:51

Using the nice answer of Scott Carnahan, you can also prove that there are find infinitely many smooth compactifications of $S^2 \times S^2 \setminus \Delta$ which are pairwise homeomorphic but not diffeomorphic.

In fact, the complex quadric $\mathbb{CP}^1 \times \mathbb{CP}^1$ blown-up in one point is isomorphic to $\mathbb{CP}^2$ blown-up in two points; therefore $\mathbb{CP}^1 \times \mathbb{CP}^1$ blown-up in four points is isomorphic to $\mathbb{CP^2}$ blown-up in five points.

But the topological 4-manifold $\mathbb{CP}^2 \sharp 5 \overline{\mathbb{CP}^2}$ supports infinitely many different smooth structures, see [Park-Stipsics-Szabo, Exotic smooth structures on $\mathbb{CP}^2 \sharp 5 \overline{\mathbb{CP}^2}$, Math. Research Letters 12 (2005)], and this proves the assertion.

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EDIT: Keeping it here because I don't like deleting stuff, but you really should disregard this as it A) fails to notice the 1-point compactification (I don't know what came over me...), and B) doesn't answer the problem of keeping a manifold structure.

This is a completely elementary approach that seems to work:

If $(x,x)$ is a point in the diagonal and $(x_n)$ and $(x_n')$ are "essentially different" sequences(*) that converge to $x$, then clearly any compactification of $S_2 \times S_2 \setminus \Delta$ has to contain $(x,x)$. And since $S_2 \times S_2$ is a compactification of $S_2 \times S_2 \setminus \Delta$ and it contains all the points it absolutely has to to be a compactification, it must be the smallest one.

As to whether there are larger ones... probably, but it's fairly obvious that $S_2 \times S_2$ is the Stone–Čech compactification of $S_2 \times S_2 \setminus \Delta$, so that puts some pretty clear limits on what other interesting compactifications you can have.

(*): More formally we could write $x(\mathbb{N}) \cap x'(\mathbb{N}) = \emptyset$.

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Clearly the one point compactification is the "smallest" one. However it is not a manifold. So I don't see why $S^2\times S^2$ should be the smallest manifold compactification. – HenrikRüping Oct 11 '10 at 14:12
You are correct. I should obviously not be writing when I'm sleepy. – K. Henriksen Oct 11 '10 at 14:28

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