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# Knots, surfaces, and 4-manifolds: Knots and concordances

Unknotting with a single twist- Samantha Allen (Dartmouth College)

Abstract: Ohyama showed that any knot can be unknotted by performing two full twists, each on a set of parallel strands. We consider the question of whether or not a given knot can be unknotted with a single full twist, and if so, what are the possible linking numbers associated to such a twist. It is observed that if a knot can be unknotted with a single twist, then some surgery on the knot bounds a rational homology ball. With this, a wider range of tools become available, including classical invariants and invariants arising from Heegaard Floer theory. Using these tools, if a knot $K$ can be unknotted with a single twist of linking number $l$, we give restrictions on the genus, signature function, Upsilon function, and $V$ and $\nu^+$ invariants of $K$ in terms of $l$ and the sign of the twist. In this talk, I will discuss some of these restrictions, their implications, and some unanswered questions. This talk is based on joint work with Charles Livingston.

Characterising homotopy ribbon discs- Anthony Conway (MPIM Bonn)

Abstract: After reviewing some notions from knot concordance, we explore the following question: how many slice discs does a slice knot admit? This is joint work with Mark Powell.

Fox-Milnor Conditions for 1-solvable Knots and Links- Shawn Williams (Rice University)

Abstract: A well known result of Fox and Milnor states that the Alexander polynomial of slice knots factors as $f(t)f(t^{-1})$, providing us with a useful obstruction to a knot being slice. In this talk, I will present a generalization of this result to certain localized Alexander polynomials of 1-solvable boundary links, and first order Alexander polynomials of 1-solvable knots.

## 27 replies on “Knots, surfaces, and 4-manifolds: Knots and concordances”

[…] Knots and concordances: […]

Hello! My name is Anthony Conway, I’m a postdoc at the Max Planck Institute for Mathematics (MPIM) in Bonn, Germany.

I’m currently interested in link concordance (mostly using abelian and metabelian invariants) and surgery theory (with a view towards 4-dimensional applications). Side interests include the Burau representation of the braid group and the Casson(-Lin) invariant.

Please feel welcome to ask questions about my talk and/or attend my office hour for a chat. Also, the description of the video in the youtube link lists some minor typos.

IsaacSundbergsays:

Hi Anthony! Thanks for the talk. I’ve been thinking about distinguishing slice disks lately, so this is a really interesting result.

I was wondering about the slices for $K_n$ from the first part of your talk, specifically when $n \neq 0$. As you described, the $n = 0$ case has two saddles that produce the two-component unlink, but when you generalize to $n \neq 0$, the right saddle (as it was in your figure) no longer gives an unlink, right? I was wondering, is there a different saddle that produces a (potential) second slice of $K_n$ in the $n \neq 0$ case?

Hi Isaac, thanks for your question!

You are right about the band moves for $K_n$ with $n \neq 0$. For instance, it turns out that we can show that $K_0,K_{-1},K_{-2},K_{-3}$ admit precisely two distinct $G$-homotopy ribbon discs up to topological ambient isotopy rel. boundary (this is mentioned briefly at the end of the second talk). Note as a concrete example that $K_{-2}$ is Stevedore’s knot.

For the other values of $n$, we suspect that there is a unique $G$-homotopy ribbon disc (up to top. ambient isotopy rel. boundary). We can prove this for $n=3k, k\neq 0,-1$ and for $n=1,2,-4,-5$.

ps: I belatedly realise I didn’t fully answer your question. Yes, you can see the band that exhibits the second ribbon disc after performing some isotopies, see e.g. Figure 4 of our paper for $K_{-3}$:
https://arxiv.org/pdf/1902.05321.pdf
The isotopies are guided by the fact that $K_{-3},K_{-2},K_{-1},K_0$ admit a genus one surface with two unknotted derivatives.

Thanks for both replies!

The summary of the result (in the first reply) definitely was helpful, but yes, as you said, I was trying to see the second ribbon disk. And I can now! Your figure 4 is wonderful, and gives a very slick way of obtaining a second disk. Thanks again!

Allison Millersays:

Hi Anthony- Nice talk(s?)! I was wondering whether you have any intuition about whether some version of the theorem could be true (though presumably unprovable with current techniques) in general.
In particular, I’d guess that you can use the usual satellite operations to show that the induced Lagrangian in the Alexander module does not determine the slice disc in general, but is it possible/ plausible that the kernel of the induced map $\pi_1(M_K) \to \pi_1(B^4 \smallsetminus D)$ completely determines the topological isotopy class of D? And do you have any intuition about whether there might be other, not necessarily good groups G where the induced Lagrangian does determine the isotopy class for G-homotopy-ribbon discs?

Hi Allison!

Indeed, thanks to a nice paper of Miller-Powell ( 😀 ), I do not expect the induced Lagrangian to be enough to characterise $\Gamma$-homotopy ribbon discs of a fixed knot for arbitrary $\Gamma$.

Regarding your other questions, the short honest answer is “I don’t know”.

Here is an attempt at a longer answer that should be taken with a bucketful of salt.

Firstly, I wouldn’t know how to establish a classification result without having “goodness”/topological surgery in 4D, so assume for the sake of argument that 4D surgery worked for ribbon groups.
Again, I wouldn’t know how to establish a classification result without using surgery theory, so let’s try to repeat the program.
In this case, I can see two main difficulties: firstly, showing that $\Gamma$-ribbon disc exteriors are homotopy equivalent, secondly analysing the surgery obstruction in $L_5^s(\Z[\Gamma])$.
In some sense, the second (huge) difficulty does not seem that relevant to your question, so let me focus on the first.
Either you believe in the conjecture that ribbon disc exteriors are aspherical (as would be implied by Whitehead’s conjecture) or you don’t. In the former the case, I wouldn’t be overly surprised if $\operatorname{ker}(\pi_1(M_K) \to \pi_1(N_D))$ captured most of the information.
In the latter case, I wouldn’t be surprised to hear that you also need something about $\pi_2$.

Words about your last question. I think that the appearance of the Lagrangians of $\operatorname{Bl}_K$ strongly correlates with the fact that the Baumslag-solitar group is metabelian. So if I had to take a very wild guess, I would say that in the non-metabelian case, the induced Lagrangian on Alexander modules might not be enough. The trouble is that there aren’t that many metabelian ribbon groups: Friedl-Teichner observed that $\mathbb{Z}$ and $BS(1,2)$ are the only solvable ribbon groups.

Shawn_Williamssays:

Hi Everyone! My name is Shawn Williams. I am a fourth year graduate student at Rice University.

This is the first conference I’ve ever spoken at, so having it be a virtual conference is quite the experience! It was a little strange to give a talk without a live audience, but I’m happy to be participating. Please check out my talk, and feel free to ask any questions!

Hi Shawn, cool talk! I was wondering about how your Fox-Milnor type obstructions compare to other Fox-Milnor type obstructions.

For instance, I think Alexander polynomials twisted by Casson-Gordon type representations (which are also metabelian) obstruct 1.5 solvability of knots. Your invariants are also metabelian (?) and yet they obstruct 1-solvability, which is neat! Can you give a hint on why that’s true?

In the boundary link case, I think that the multivariable (abelian) Alexander polynomial has been shown to obstruct 1-solvability*. Do you have examples where this abelian obstruction is inconclusive but your obstruction helps? E.g. does the multivariable polynomial say anything about the example at the end of your video?
*as in the knot case, one might expect the multivariable polynomial to obstruct 0.5 solvability for boundary links, but I can’t remember seeing this written anywhere?

Shawn_Williamssays:

Hi Anthony, thanks for the questions!

I must admit that the result about the first order Alexander polynomials is incorrect as stated in my talk. It is my recent realization that this is true for 2-solvable knots, not 1-solvable. My apologies for the incorrect statement. I’ll post an additional comment on this page correcting it for anyone else who watches my talk.

As for the boundary links, I don’t think this would provide an obstruction if the multivariable polynomial already factors, since any such factorization should also hold after localizing. I wasn’t aware of a similar result for the multivariable polynomial. My result relies on the associated localized Blanchfield form being hyperbolic, which I was able to get from the localized ring being a PID. I was looking into seeing if the multivariable polynomial does provide an obstruction, so I would be interested in seeing if this is already known!

Allison Millersays:

Hi Shawn- nice talk and results! I am always impressed and slightly intimidated by either non-abelian coefficients or working with links rather than knots, and you do both in 25 minutes! Is it right that in some sense your link invariants are “simpler/ lower level” than your knot invariants? (Perhaps what I mean by that is two-fold: First of all, it seems like in the second setting you have a “normal” polynomial ring (i.e. $\alpha t= t \alpha$) with abelian coefficients, which is presumably much nicer to obstruct factorization in! And secondly, it seems like if you apply your link polynomial to knots you regain the classical Alexander polynomial, maybe with rational coefficients instead of integer.)

Shawn_Williamssays:

Hi Allison, thanks for the question! Yes, the result for links is simpler in this regard. Not only does working in a commutative polynomial ring make the polynomials easier to work with than the noncommutative ones, but localizing to get a polynomial in a single variable would also make it easier to obstruct factorization than for the multivariable polynomial. And yes, doing this for knots will give you back the classical Alexander polynomial with rational coefficients.

Justin.Bryantsays:

Hi Shawn, great talk. If I recall correctly, the first-order Alexander module for the trefoil has (t-1) torsion represented by the longitude. Judging by your remark that talking about the (t-1) torsion would elicit an entire new talk, I’m guessing that this will not always be the case. But can you give any indication about what properties the curve that gives us a (t-1) factor must satisfy, or when it is something “nice” like a longitude?

Shawn_Williamssays:

Hi Justin, thanks for the question! My comment on the (t-1)-torsion was part joke, part truth (i.e. I could probably go on about it for another 25 minutes). The (t-1) mentioned in the talk is represented by the longitude of the knot. One can show that the first order polynomial of any knot has a (t-1)-factor. This is sort of an annoying fact because this factor is very difficult to recognize and factor out, and it provides no new information about the knot. In fact, since the degree of the first order polynomial for any genus 1 knot is 1, we get that all genus 1 knots have first order polynomial = t-1.

It is possible for these first order modules to have non-longitudal (t-1)-torsion. For example, connected sums of knots will have such non-longitudal (t-1)-torsion–the connected sum of two trefoils having first order polynomial = (t-1)^3.

One can get around the longitudal (t-1)-torsion by looking at the zero surgery, but any other (t-1)-torsion will still be present.

Hi folks! My name is Samantha Allen and I am a postdoc at Dartmouth College in Hanover, NH.

I am interested in knot theory in dimensions 3 and 4. I particularly enjoy using invariants arising from Heegaard Floer to study knots and knot concordance.

I am happy to answer any questions or hear an comments about by talk (or math in general). Feel free to comment here, send an email, or attend my office hour (on Friday, June 5).

As a side note, the description of my video on Youtube lists some references that I neglected to cite in the video.

Allison Millersays:

Hi Samantha, I enjoyed your talk- what a nice result! Have you thought at all about building examples demonstrating that (some of?) the potential options for $U(K)$ actually occur, or are the required Heegaard Floer computations just too difficult? Thanks again for a lovely talk!

Hi Allison! Thanks for your question!

We have toyed with this idea (and haven’t entirely ruled it out), but it does seem difficult to build examples showing certain options occur while keeping the Heegaard Floer computations doable. The Heegaard Floer invariants can’t quite deal with linking number one (for both signs at once), so we’d probably also need to control for the Arf invariant. It’s certainly something to work on.

As I mentioned in the talk, we do have 8 low-crossing-number knots for which we do know the exact set $\mathcal{U}(K)$. So there are some sets which we know we can realize: the empty set, $\{2^-, 0^-, 0^+, 2^+\}$, $\{3^-\}$, and $\{4^-\}$ among others.

Hi Samantha, I enjoyed your talk! Can your methods for restricting the unknotting twisting indices for alternating knots also work for quasi-alternating knots, or other Floer thin knots?

Hi Linh!

For the statement about which linking numbers are possible for negative twists, we are only using that the knot complex is that of a $(2,n)$ torus knot. So I believe that would follow through for thin knots.

As far as the statements relating to the signature go, these would hold for $\sigma$-thin knots (thin knots which have complexes with homology supported on the $-\sigma/2$ diagonal), which include quasi-alternating knots. Similar statements could be formulated for thin knots in terms of the tau invariant, but they aren’t as nice since we wouldn’t be able to use the signature obstructions coming from Casson-Gordon invariants.

Shawn_Williamssays:

Hi All,

Just a correction for my talk:

The result for first order Alexander polynomials should be for 2-solvable knots, not 1-solvable knots.

My deepest apologies!

TyeLidmansays:

Hi Samantha,

What a cool result!

If you have a knot K with a lens space surgery (e.g. P(-2,3,7)), and it can be unknotted with a single twist, then that should give you a knot in a lens space with another lens space, usually with different order of homology. (The knot is the image of the crossing circle in the surgery on the original knot K.) There is some literature for surgeries between lens spaces. I was wondering if these extra constraints could be helpful here, maybe for some simple knots?

Hi Tye,

That’s a great idea!
I will have to work out the details carefully, but it seems very promising.

Kyle_Haydensays:

Hey, Anthony! Great talk. My understanding is that this approach breaks down for higher genus surfaces. In particular, Takahiro Oba constructed genus one ribbon surfaces $F,F’ \subset B^4$ with $\partial F= \partial F’$ and $\pi_1(B^4 \setminus F) \cong \pi_1(B^4 \setminus F’) \cong \mathbb{Z}$ that are distinguished by the intersection form of their double branched covers (https://arxiv.org/abs/1708.02523).

Any intuition for what goes wrong for higher genus surfaces?

Hi Kyle! A very timely question: as it turns out, we are working on this with Mark and have some results!

I will perhaps not claim too much yet, but merely say that a key aspect is the $\mathbb{Z}[t^{\pm 1}]$-valued intersection form on $\pi_2(D^4 \setminus \nu \Sigma)=H_2(D^4 \setminus \nu \Sigma;\mathbb{Z}[t^{\pm 1}])$.
Morally, the intersection form on the double branched cover is obtained by plugging in $t=-1$.
At the moment, I don’t expect this $\mathbb{Z}[t^{\pm 1}]$-intersection form to be enough for a classification though.

Kyle_Haydensays:

Very cool, thanks!

MarkPowellsays:

Hi Kyle, thanks for pointing out the paper of Oba. That’s a nice paper. I don’t know if this counts as intuition, but similar to Anthony’s answer to Allison’s question above, a necessary condition for surfaces to be isotopic rel. boundary is that the exteriors be homotopy equivalent rel. boundary. This is in particular asking for the $\mathbb{Z}[\mathbb{Z}]$ intersection pairings to be isometric. Oba showed that this doesn’t always hold. Note that rim surgery doesn’t change the intersection form of the exterior.