Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

4 Answers 4

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 $\mathbb{R}P^3$.

share|improve this answer
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.

share|improve this answer
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.

share|improve this answer

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 $x(\mathbb{N}) \cap x'(\mathbb{N}) = \emptyset$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.