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Knots, surfaces, and 4-manifolds: Knotted surfaces

Exotic slice disks and symplectic surfaces- Kyle Hayden (Columbia University)

Abstract: One approach to understanding the smooth topology of a 4-dimensional manifold is to study the embedded surfaces it contains. I'll construct new examples of "exotically knotted" surfaces in 4-manifolds, i.e. surfaces that are isotopic through ambient homeomorphisms but not through diffeomorphisms. We'll begin with simple examples of exotic disks in the 4-ball. Then we'll turn to the symplectic setting and exhibit new types of exotic phenomena among symplectic, holomorphic, and Lagrangian surfaces.

Doubly slice pretzels- Clayton McDonald (Boston College)

Abstract: A knot K in S^3 is slice if it is the cross section of an embedded sphere in S^4, and it is doubly slice if the sphere is unknotted. Although slice knots are very well studied, doubly slice knots have been given comparatively less attention. We prove that an odd pretzel knot is doubly slice if it has 2n+1 twist parameters consisting of n+1 copies of a and n copies of -a for some odd integer a. Combined with the work of Issa and McCoy, it follows that these are the only doubly slice odd pretzel knots. Time permitting, we might go over some preliminary results involving links as well.

Topologically embedding spheres in knot traces- Patrick Orson (Boston College)

Abstract: Knot traces are smooth 4-manifolds with boundary, that are homotopic to the 2-sphere, and obtained by attaching a 2-handle to the 4-ball along a framed knot in the 3-sphere. I will give a complete characterisation for when the generator of the second homotopy group of a knot trace can be represented by a locally flat embedded 2-sphere with abelian exterior fundamental group. The answer is in terms of classical and computable invariants of the knot. This result is directly analogous to the the result of Freedman and Quinn that says a knot with Alexander polynomial 1 is topologically slice, and can be used to exhibit new exotic 4-dimensional phenomena. This is a joint project with Feller, Miller, Nagel, Powell, and Ray.

Regular homotopy and Gluck twists- Hannah Schwartz (MPIM Bonn)

Abstract: Any 2-sphere K smoothly embedded in the 4-sphere is related to the unknotted sphere through a finite sequence of locally supported homotopies called finger moves and Whitney moves. We call the minimum number of finger moves (or equivalently Whitney moves) in any such a homotopy the "Casson-Whitney number" of the sphere K. In this talk, I will discuss joint work with Joseph, Klug, and Ruppik showing that if the Casson-Whitney number of K is equal to one, then the unknotted torus can be obtained by attaching a single 1-handle to K. I will also present an application of this result, from joint work with Naylor, that the Gluck twist of any sphere with Casson-Whitney number equal to one is diffeomorphic to the standard 4-sphere, and give some well-known families of 2-spheres for which this is the case.

18 replies on “Knots, surfaces, and 4-manifolds: Knotted surfaces”

[…] Knotted Surfaces: […]

rubermansays:

Hi Hannah–Nice talk (and fun setup!). Are there simple examples of unknotted spheres that require an arbitrarily large number of stabilizations (and/or Whitney moves)? Are there examples where the minimal number of stabilizations and minimal number of Whitney moves are different?

Hannahsays:

Hi Danny!!!
Are there examples where the minimal number of stabilizations and minimal number of Whitney moves are different?Yes! In the paper referenced in the talk that is joint with Jason Joseph, Mike Klug, and Ben Ruppik (to appear probably in a few weeks on the arxiv?), we give examples of such spheres in S^4. Although we have no examples where they differ by an arbitrarily large amount, we show that at least two Whitney moves are needed in any regular homotopy from K to the unknot, when K is the spin of the connected sum of torus knots T(2,p) # T(2,q) for any p and q. However, Miyazaki in 1986 showed that one stabilization suffices, for an infinite but carefully chosen family of integers p and q.

Are there simple examples of unknotted spheres that require an arbitrarily large number of stabilizations (and/or Whitney moves)?
I’m unsure how exactly to interpret this question… you probably don’t mean “unknotted” here? So, I’ll give two remarks. First, in S^4, the minimum number of generators of the Alexander module of the 2-sphere bounds the minimum number of stabilizations and Whitney moves from below. So, for instance, the connected sum of n copies of a 2-knot K with non-trivial Alexander module, such as the spin of any 1-knot with non-trivial Alexander polynomial, needs at least n stabilizations and Whitney moves to get to a sphere with cyclic fundamental group — and hence in particular to a smoothly unknotted sphere. (Thank you Jason Joseph for helping me with this part of the answer!)  Also, although we don’t deal with this in our paper, it is worthwhile to mention that in the setting of non-simply connected manifolds, the Freedman-Quinn invariant can be used to cook up examples of 2-spheres with fundamental group isomorphic to that of the ambient manifold (unlike the situation in the 4-sphere) requiring arbitrarily many Whitney moves in any regular homotopy taking one to the other.

MarkPowellsays:

Hi Hannah and Danny,

Arbitrarily many stabilisations is Theorem A from this paper with Allison Miller.
https://arxiv.org/abs/1908.06701

Cool video.

rubermansays:

Hi Mark and Hannah,

Yes, I meant “knotted”, not “unknotted”. (Very confusing without the quotation marks.) I think I also had in the back of my mind the question of whether there’s a difference between smooth and topological in looking for stabilizations.

Jason_Josephsays:

Hi Danny, Mark, and Hannah:
Just to clarify, we do not claim to be the first to prove the result about stabilizations. Actually, this was first proved in Miyazaki ’86 “On the relationship among unknotting number, knotting genus and Alexander invariant for 2-knots”, the same paper mentioned before which had examples of nonadditivity.

MarkPowellsays:

Thanks, I didn’t know about that paper.

Hey Patrick,

Could you say something about what motivated your (+FMNPR)’s lemma that ks(X)=arf(K), for X the n-trace of K? I just watched your video and thought that was interesting.

Patrick_Orsonsays:

Hey Maggie! Thanks for the question.

Something like this statement was already known for $n=1$; this is one way to construct the Chern manifold $\ast\mathbb{C}P^2$. More specifically, take the trace of $+1$ framed surgery on an Arf invariant $1$ knot. You get an integer homology sphere on the boundary, which can be capped off by a contractible topological manifold – this guy is the Chern manifold, and it is homotopy equivalent to the normal $\mathbb{C}P^2$, with the same intersection form, but not homeomorphic as it has different $ks$ invariant, which we can see in a moment. The mod $2$ Rochlin invariant of this surgery manifold is equal to the Arf invariant of the knot (this is from “Dehn’s construction on knots” by Gonzalez-Acuna). It is also equal to the $ks$ invariant of the contractible manifold (Freedman-Quinn textbook p.$165$). So we get $ks$ equal to Arf here. (Also, that $\ast\mathbb{C}P^2$ is not smooth by Novikov additivity of $ks$.)

This was the motivating idea, and what we did was to extend the result to different $n\neq 1$ using the $\tau$ invariant.

Patrick_Orsonsays:

Ah, I have just seen something in your question. The honest $n$-trace is smooth so it always has $ks(X_n(K))=0$. What we did is construct this putative sphere exterior $V$, a homology bordism from $S_n^3(K)$ to $L(n,1)$, then cap on $L(n,1)$ with the euler number $n$ disc bundle over the $2$-sphere to obtain $X$. We hope $X$ is the honest knot trace. When $n$ is odd, this $X$ is the manifold we prove has $ks$ equal to Arf of the knot.

Oh okay, I totally missed that point! Thanks for explaining.

Hey Kyle,

Could you explain more about why there is no hope of your smooth isotopy obstruction working in closed symplectic manifolds, if there is anything easy to say about that?

Kyle_Haydensays:

Hi Maggie! I’m happy to elaborate. It’s a bit subtle, so I apologize in advance for the long answer.

The key obstruction in my argument is an adjunction inequality, which gives a lower bound on the genus $g(F)$ of an embedded surface in a symplectic 4-manifold $(Z,\omega)$. However, the precise bound depends on the symplectic structure by pairing a certain characteristic cohomology class $c_1(\omega)$ against the homology class $[F]$.

My argument uses an adjunction inequality to help distinguish the branched double covers $Z,Z’$ of the closed symplectic surfaces $\Sigma,\Sigma’ \subset X$ (by distinguishing the types of embedded surfaces they contain). But here’s the dirty little secret: While the branched covering induces natural symplectic structures $\omega,\omega’$ on $Z,Z’$, the corresponding Chern classes $c_1(\omega),c_1(\omega’)$ are morally equivalent and don’t help us distinguish $Z,Z’$. Instead, we find some mysterious alternative symplectic structure on $Z$, and its corresponding adjunction inequality provides the bound needed to distinguish the smooth structures on $Z$ and $Z’$.

Now, what about closed symplectic 4-manifolds? It’s no problem to symplectically embed the original 4-manifold $X$ containing $\Sigma,\Sigma’$ into a closed symplectic 4-manifold $\hat{X}$. Now look at the branched double covers of $\Sigma,\Sigma’ \subset \hat{X}$: They again have symplectic structures that provide identical are closed symplectic 4-manifolds, but those symplectic structures will provide identical restrictions on embedded surfaces. So now we want to try to find some alternative symplectic structure on $\hat{X}$ (or at least some alternative Seiberg-Witten basic class) that provides a sharper restriction. This seems harder to construct than in the original setting, and now there are more constraints on the possible symplectic structures and Seiberg-Witten basic classes (due to Taubes, especially).

However, there’s no obvious reason that the surfaces should always become isotopic in every possible $\hat{X}$. And I think some creativity and elbow grease should give a version of the argument that does extend. Finally, I think it should be easy to embed $X$ in \emph{some} closed, symplectic 4-manifold $\hat{X}$ so that $\Sigma,\Sigma’$ remain exotic, but it’s hard to make that embedding $X \hookrightarrow \hat{X}$ symplectic at the same time.

Kyle_Haydensays:

Oops, *”They again have symplectic structures that provide identical restrictions on embedded surfaces.”

Thanks for explaining! So if I’m parsing your answer right, it seems like you’re saying that a similar argument to yours might work for closed symplectic manifolds, but it’s not clear how to arrange all the conditions simultaneously and get the correct example?

Kyle_Haydensays:

That’s exactly right!

Anubhavsays:

Hi Kyle,

This is a great talk.

Can you please explain me the step3 of the proof of your main theorem (18th min of the video) how you are using the adjunction inequality to show that such knot $J$ cannot bound a disk?

Kyle_Haydensays:

Thanks, Anubhav! Following the notation in the talk, we look at the branched cover $\Sigma_2(B^4,D)$. An explicit Stein handle diagram for $\Sigma_2(B^4,D)$ is constructed in Figure 15 (pg 19-20) of my paper. In this diagram, you can realize the lift $\tilde J$ as a Legendrian knot with Thurston-Bennequin number $tb(\tilde J)=0$. Therefore, if we attach a 2-handle along $\tilde J$ with framing $tb(\tilde J)-1=-1$, we get another Stein domain.

Now suppose $J$ bounds a disk in $B^4 \setminus D$. Then $\tilde J$ bounds a disk in $\Sigma_2(B^4,D)$. Capping off this disk with the core of the new 2-handle yields an embedded 2-sphere with self-intersection $-1$. This can’t exist in a Stein domain, yielding a contradiction.