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Suppose we are given an embedding of $S^2$ in $\mathbb{CP}^2$ with self-intersection 1. Is there a diffeomorphism of $\mathbb{CP}^2$ which takes the given sphere to a complex line?

Note: I suspect that either it is known that there is such a diffeomorphism, or the problem is open. This is because if there was an embedding for which no such diffemorphism existed, you could use it to produce an exotic 4-sphere. To see this, reverse the orientation on $\mathbb{CP}^2$ then blow down the sphere.

EDIT: for a counter-example, it is tempting to look for the connect-sum of a line and a knotted $S^2$. The problem is to prove that the result cannot be taken to a complex line. For example, the fundamental group of the complement $C$ is no help, since it must be simply connected. This is because the boundary of a small neighbourhood $N$ of the sphere is $S^3$ and so $\mathbb{CP}^2$ is the sum of $N$ and $C$ across $S^3$ and so in particuar $C$ must be simply-connected.

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What is meant by an embedding with self-intersection one? Does it mean the Euler class of the normal bundle is a generator? Thanks. –  Mark Grant Oct 25 '10 at 21:49
@Mark Grant, that's one way of putting it. The point is that the homology class of the sphere should generate the second homology group of the projective plane. I agree that to combine the term "embedding" with the term "self-intersection" like that was perhaps a little confusing! –  Joel Fine Oct 26 '10 at 12:30
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3 Answers

up vote 13 down vote accepted

The conjecture that every $S^2 \subseteq \mathbb{C}P^2$ is standard if it is homologous to flat is implied by the smooth Poincaré conjecture in 4 dimensions. It also implies a special of smooth Poincaré that is accepted as an open problem, the case of Gluck surgery in $S^4$. I can't prove or disprove the question of course, but since the question is sandwiched between two open problems, I can "prove" that it is an open problem.

It is easier to consider $\mathbb{C}P^2$ minus a tubular neighborhood of the $S^2$, rather than to "blow it down". The condition on the homology class is equivalent to the condition that the boundary of this tube is $S^3$; the projection to the core is a Hopf fibration. The blowdown consists of attaching a 4-ball to this 3-sphere; let's skip this step. As Joel had in mind, the complement of the $S^2$ is simply connected. In fact, it is a homotopy 4-ball with boundary $S^3$. Thus, Freedman's theorem implies that it is homeomorphic to a 4-ball and smooth Poincaré would imply that it is diffeomorphic to a 4-ball. When it is, this 4-ball is still standard with its Hopf-fibered boundary (the Hopf fibration is unique up to orientation), so the $S^2$ is unknotted.

In the other direction, the $S^2$ could be the direct sum of a standard complex line in $\mathbb{C}P^2$ with a 2-knot $K$ in $S^4$. I argue that in this case, the blowdown is equivalent to the Gluck surgery along $K$. What is a Gluck surgery? It looks like Dehn surgery in 3 dimensions, except with peculiar behavior. The official definition is that you remove a neighborhood of $K$ (which here is $D^2 \times S^2$, not the twisted bundle in Joel's construction), then glue it back after applying the non-trivial diffeomorphism of $S^1 \times S^2$. That diffeomorphism comes from the non-trivial element in $\pi_1(\text{SO}(3)) = \mathbb{Z}/2$. One thing that is peculiar is that the Gluck surgery does not change the homotopy type of its 4-manifold, which is why it produces many candidate counterexamples to smooth Poincaré.

Again, it is easier to think about the closed complement to Joel's $S^2$ than the blowdown. The corresponding version of Gluck surgery is to remove all of $D^2 \times K$, but only glue back a thickened $D^2$ (a 2-handle) along an attaching circle, and not glue back in the remaining 4-ball along the rest of $K$. What is peculiar here is that the attaching circle does not change; it is still a vertical circle in $S^1 \times S^2$. What changes instead is that the framing of the attachment is twisted by 1. (Or it can be twisted by some other odd number, since $\pi_1(\text{SO}(3)) = \mathbb{Z}/2$ and not $\mathbb{Z}$. More prosaically, the "belt trick" lets you change the twisting by an even number.) Anyway, if Joel's sphere is $K$ connect summed with a complex line $L$, then you can represent this crucial 2-handle with another complex line $J$ in $\mathbb{C}P^2$. The question is whether the framing of its attachment to $L$ is odd or even. The fact that a perturbation $J'$ of $J$ intersects $J$ once tells me that the framing is odd. So the result is Gluck surgery.

The old version of this answer was less developed (and at first I made the $\pi_1$ mistake that is corrected in the comments and the edit to the question). But it is still worth noting that there are many open special cases of smooth Poincaré that consist of just one homotopy 4-sphere. Some topologists interpret this as strong evidence that smooth Poincaré is false. Others suppose that we just might not be very good at finding diffeomorphisms with $S^4$. A few examples, including some Gluck surgeries, were shown to be standard only after many years, for instance in this paper by Selman Akbulut.

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Regarding your first paragraph: don't some (at least) of those knotted spheres unknot after the connected sum? –  Mariano Suárez-Alvarez Dec 12 '09 at 16:15
I was thinking that they can't by van Kampen's theorem. But did I reason through this correclty? Now am I worried; I'll think about it some more. –  Greg Kuperberg Dec 12 '09 at 16:21
No, my bad, the calculation is that $\pi_1$ is exactly trivial again. –  Greg Kuperberg Dec 12 '09 at 16:25
Thanks for the answer Greg. Yes, t's true that the complement must be simply connected. For a calculation-free argument see my comment to Tim Perutz's answer. I agree that these seemingly knotted spheres are candidates, but it seems to me that the only real way to prove that there is no diffeomorphism taking them to the line is to prove the blow-down is an exotic 4-sphere. –  Joel Fine Dec 12 '09 at 16:38
The remaining question is whether you have found a variation of Gluck surgery, or whether in the case of a 2-knot in R^4 it is a no-op. –  Greg Kuperberg Dec 12 '09 at 16:45
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Take a self-intersection one $S^2$ in $\mathbb CP^2$ -- it appears to me that several people have made the observation in this thread that the complement is simply-connected. If you don't see this right away then a nice way to see this is that the unit normal bundle is $S^3$ and the Hopf fibration $S^3 \to S^2$ then gives a relator that kills the $S^2$-linking elements of $\pi_1$ but these generate $\pi_1 (\mathbb CP^2 \setminus S^2)$ (generalized Wirthinger presentation). So Poincare/Alexander duality tells you $\mathbb CP^2 \setminus S^2$ is contractible. So "blowing-down" on this embedded $S^2$ basically amounts to a twisted sphere construction -- gluing two discs $D^4$ together along their common boundary. There's the issue of whether or not one of the discs has an exotic smooth structure or not, but that's the only issue.

So I guess there's a question lurking in this -- is there a "natural" way to unknot a self-intersection one embedded $S^2$ in $\mathbb CP^2$? If there were, you couldn't produce any interesting $S^4$'s via this construction.

I've seen this kind of "unnatural unknotting" phenomena with other constructions -- deform-spinning / twist-spinning knots is another example.

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Hey Joel, long time etc. It looks to me like blowing down your knotted $S^2$ will only produce a homology 4-sphere. And one could presumably produce examples by taking some known 2-knot in $S^4$ and connect-summing it with the line in $\mathbb{CP}^2$, distinguishing the resulting 2-knots in $\mathbb{CP}^2$ from the line via $\pi_1$ of their complements.

[EDIT: I fell into Joel's heffalump trap. Still, at least there's company down here...]

You could rephrase the question (with a bit of help from Gromov) as asking whether a 2-knot in $\mathbb{CP}^2$ with self-intersection $1$ and simply connected complement is isotopic to a symplectic sphere. You could invoke Taubes too, and see that, to produce a diffeo with the line, it's enough to extend a symplectic form on the image of $S^2$ to one on $\mathbb{CP}^2$. Well, the complement of a neighbourhood of $S^2$ is then a homotopy 4-ball, bounding $S^3$ with its usual contact structure, and the goal is to build a symplectic form which is a convex filling of the contact boundary... Yep, that's probably an open problem.

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Hi Tim. Good to hear from you. I had a feeling you'd try your hand at this question! First, I think blowing down should give a homotopy 4-sphere. It should be simply connected since blowing back up gives CP<sup>2</sup> which is simply connected. Secondly, I seem to get that the complement of this sort of sphere must be simply connected, for essentially the same reason: if the sphere has self-intersection +1, then we can write CP<sup>2</sup> as the connected sum of a small neighbourhood of the sphere and the complement. Since CP^<sup>2</sup> is simply connected, the two parts must be also. –  Joel Fine Dec 12 '09 at 16:34
Quite right - as Greg just noticed too. –  Tim Perutz Dec 12 '09 at 16:37
Re convex fillings, it seems to me that if you could prove any homotopy 4-ball with bdry $S^3$ has a symplectic form giving a convex filling of the standard contact structure on $S^3$, you'd have proved the Poincaré conjecture. (So yep, definitely an open problem!). This is because it would follow from Gromov that all homotopy 4-balls would be standard. But any homotopy 4-sphere can be written as the sum of a 4-ball and a homotopy 4-ball across their common $S^3$ bdry. So now the only chance of a counter example would be a so-called "twisted sphere" and these are known to be standard in dim 4. –  Joel Fine Dec 12 '09 at 23:16
It wasn't a trap! At least it wasn't meant to be... In the interests of brevity, I left out the proof that the blow-down is a homotopy 4-sphere. If, in the interests of clarity, I'd included it then I might not have mistakenly caught any heffalumps! –  Joel Fine Dec 13 '09 at 21:03
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