Embeddings of $S^2$ in $\mathbb{CP}^2$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:03:20Z http://mathoverflow.net/feeds/question/8665 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8665/embeddings-of-s2-in-mathbbcp2 Embeddings of $S^2$ in $\mathbb{CP}^2$ Joel Fine 2009-12-12T11:27:50Z 2009-12-13T20:14:05Z <p>Suppose we are given an embedding of $S^2$ in $\mathbb{CP}^2$ with self-intersection 1. Is there a diffeomorphism of <code>$\mathbb{CP}^2$</code> which takes the given sphere to a complex line?</p> <p>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.</p> <p>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 <i>prove</i> 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. </p> http://mathoverflow.net/questions/8665/embeddings-of-s2-in-mathbbcp2/8683#8683 Answer by Greg Kuperberg for Embeddings of $S^2$ in $\mathbb{CP}^2$ Greg Kuperberg 2009-12-12T16:12:59Z 2009-12-13T20:00:47Z <p>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 <a href="http://atlas-conferences.com/c/a/b/o/02.htm" rel="nofollow">Gluck surgery</a> 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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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 <em>just one</em> 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 <a href="http://www.math.msu.edu/~akbulut/papers/cs.pdf" rel="nofollow">this paper</a> by Selman Akbulut.</p> http://mathoverflow.net/questions/8665/embeddings-of-s2-in-mathbbcp2/8685#8685 Answer by Tim Perutz for Embeddings of $S^2$ in $\mathbb{CP}^2$ Tim Perutz 2009-12-12T16:19:57Z 2009-12-13T20:14:05Z <p>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. </p> <p>[EDIT: I fell into Joel's heffalump trap. Still, at least there's company down here...]</p> <p>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. </p> http://mathoverflow.net/questions/8665/embeddings-of-s2-in-mathbbcp2/8709#8709 Answer by Ryan Budney for Embeddings of $S^2$ in $\mathbb{CP}^2$ Ryan Budney 2009-12-12T20:02:38Z 2009-12-12T20:34:38Z <p>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. </p> <p>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.</p> <p>I've seen this kind of "unnatural unknotting" phenomena with other constructions -- deform-spinning / twist-spinning knots is another example. </p>