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• CommentRowNumber1.
• CommentAuthorRichard Williamson
• CommentTimeJan 17th 2016
• (edited Jan 18th 2016)

Dear all,

I would like to announce that in this thread I will begin, starting today, another ’mathematical serial’ of the kind I have tried a couple of times before here at the nForum. Though I did not finish the serial on either of the previous occasions (I plan to return to them one day!), I rather enjoyed the opportunity for discussion that arose as I posted the mathematics.

In this thread, I will be discussing an argument towards giving a new, rather short and simple, proof of the Poincaré conjecture. A version of this argument in outline form was distributed a few months ago, and a paper with most of the details has been distributed to some people since early December. A number of people have very kindly been working through this paper. I am especially grateful to Tobias Barthel and Emily Riehl, who have spent a great deal of time on the paper; their sharp eyes have led to a number of significant corrections and other improvements. I am also very grateful to Louis Kauffman, who has spent considerable time studying the paper. The open-mindedness and intellectual integrity of Tobias, Emily, and Louis is most heart-warming.

I would like to acknowledge the efforts of several other people in studying the paper: Richard Garner, who made some very useful observations both on the earliest, outline version of the argument, and on a recent pdf version; Markus Szymik, with whom I discussed the argument over email as it came to me; and Bruno Martelli, who asked some helpful questions about one section of the pdf version of the argument.

Finally, I would like to acknowledge the work of my students, Therese Mardal Hagland and Marte Lovise Nilsen, on diagrammatic knot theory and higher knot theory. Although this work was not directly related to the argument that will be presented in this thread, the ideas and points for exploration which arose through Therese’s and Marte’s creativity and tenacity played an important role in shaping and sharpening my overall vision of, and way of approaching, knot theory and higher knot theory. In addition, I would like to thank Reidun Persdatter Ødegaard for much discussion of her master thesis, out of which the idea to approach the Poincaré conjecture in the way I will present came.

The pdf version of the argument is quite long, and is formulated in an explicitly category theoretic way. It has proven difficult so far for low-dimensional topologists to appreciate the underlying ideas. However, I realised recently that a ’self-contained’ argument, which does not rely on the category theoretic framework that I develop in the paper, can be extraced from the paper. It is this ’self-contained’ argument that I will present here.

The plan is to present the argument, with expositional remarks, in this thread. Alongside this, I will create an nLab page which will contain just the mathematics (I will update it as I proceed with the ’serial’ here in the nForum), without any gloss. I hope that this combination will give a good balance between offering the possibility for discussion (here) and having a document that is easy to refer to for the mathematics (the nLab page).

The nLab page, and the argument here, will not be explicitly category theoretic. Moreover, the nLab page will probably be somewhat divorced from the rest of the nLab, at least for a while. However, I certainly think in a ’category theoretic way’, and this will, I would have thought, be implicitly evident in the mathematics, so I hope that it will be considered acceptable that I am presenting the argument here at the nForum. If the nLab page is not considered a good fit with the rest of the nLab, I am happy to remove it.

I would like to welcome all discussion. Ask any questions at all, either in this thread, or over email, to richard (at) rwilliamson-mathematics.info.

In the next message in the thread, I will make a few remarks about the overall approach (more remarks of this nature will no doubt come as I proceed). Then I will begin with the mathematics.

• CommentRowNumber2.
• CommentAuthorRichard Williamson
• CommentTimeJan 17th 2016
• (edited Jan 21st 2016)

The approach that I will take to the Poincaré conjecture is entirely diagrammatic. Indeed, three key theorems establish that closed, connected, orientable 3-manifolds can be thought of as link diagrams up to the framed Reidemeister moves and certain ’local Kirby moves’.

Thes first of these theorems, usually attributed to Lickorish and Wallace, establishes that any closed, connected, orientable 3-manifold can be obtained by integral Dehn surgery on a framed link in $S^{3}$.

The second, due to Kirby, establishes that a pair of closed, connected, orientable 3-manifolds $M_{1}$ and $M_{2}$, obtained by integral Dehn surgery on framed links $L_{1}$ and $L_{2}$ respectively in $S^{3}$, are isomorphic if and only if the diagram of $L_{1}$ can be obtained from the diagram of $L_{2}$ by a finite sequence of the framed Reidemeister moves and certain additional moves, known as Kirby moves. I like to think of these two theorems together as the analogue for 3-manifolds of the (surgery proof of the) classification of surfaces.

The third theorem, due to Fenn and Rourke, refines Kirby’s theorem, demonstrating that $M_{1}$ and $M_{2}$ are isomorphic if and only if the diagram of $L_{1}$ can be obtained from the diagram of $L_{2}$ by a finite sequence of the framed Reidemeister moves and certain additional local moves, usually referred to as Fenn-Rourke moves. Recently, Martelli, in this paper, refined Kirby’s theorem still further, demonstrating that a finite collection of local moves suffice in addition to the framed Reidemeister moves: for instance, two local handle slide moves and three Fenn-Rourke moves (see Figure 3 in Martelli’s paper). It was principally Martelli’s work that originally inspired the author to consider seriously working with 3-manifolds in the way I will do here.

The Poincaré conjecture, when viewing 3-manifolds diagrammatically, says the following. Let $L$ be a link diagram such that the group $\pi_{1}(L) / \langle l_1, ..., l_n \rangle$, namely the fundamental group of $L$ modulo the longitudes $l_1$, …, $l_n$ of the components of $L$, is trivial. Then there must be a sequence of framed Reidemeister moves and local Kirby moves which take $L$ to the empty link diagram.

The first reaction of the reader who is an expert on low-dimensional topology, upon taking a glance at the argument that I will give in this thread, may well be to dismiss it as not possibly being able to give a proof of the Poincaré conjecture. This is because there is a deep-rooted expectation that an approach to the Poincaré conjecture which involves the Kirby calculus must entail a highly delicate consideration of Tietze transformations of a presentation of the fundamental group of a link modulo the longitudes of its components, showing that these transformations correspond to certain Kirby moves, in such a way that if we have a sequence of Tietze transformations taking a presentation to a trivial presentation, then there is a corresponding sequence of Kirby moves taking the link to the empty link.

This is not the approach taken in the present paper. Indeed, the argument which I will give does not exhibit an explicit sequence of Kirby moves taking a link diagram $L$ such that $\pi_{1}(L) / \langle l_1, ..., l_n \rangle$ is trivial to the empty link diagram; the argument works in the same way in every example. Yet, our approach is still constructive: it is just different, of a more ’structural’ nature.

Thus I ask the reader to set aside their expectations, and to try to understand the structure of the argument on its own merits.

The way in which I make use of the Kirby calculus to approach the Poincaré conjecture is new. Roughly speaking, we ‘wrap up’ all link diagrams equivalent under the framed Reidemeister moves and local Kirby moves to a given one into a single equivalence class. To prove the Poincaré conjecture, it suffices to show that the empty link diagram is a representative of the equivalence class of a link diagram $L$ for which $\pi_{1}(L) / \langle l_1, ..., l_n \rangle$ is trivial.

We proceed in the opposite direction from the expected approach. Rather than trying to show that a link diagram belongs to the same equivalence class as the empty link diagram by means of an explicit sequence of Kirby moves, we begin with the equivalence class of the link diagram, and seek to understand this equivalence class, in a structural way, sufficiently well to be able to show that the the empty link diagram is a representative of it.

We achieve this by means of a certain group which we associate to the equivalence class of a link diagram, its operator group. We show that the longitude of any component of a link diagram $L$ is trivial in this operator group. In addition, we show that this operator group is a quotient of $\pi_{1}(L) / \langle l_1, .., l_n \rangle$. Combining these facts, we find that triviality of the latter group implies triviality of the operator group.

Triviality of the operator group of the equivalence class of a link diagram is a very strong structural property of this equivalence class. We show that it is so strong that it allows us to construct a representative of this equivalence class which is equivalent to the empty link diagram, as required.

This outline does not mention one fundamental ingredient in our approach, which I regard as the most important technical innovation of it: we do not work only with link diagrams in the usual/classical sense, we work with link diagrams in a more general sense; and, in forming the equivalence class of a link diagram in the classical sense, we identify it with certain non-classical link diagrams. The latter identifications are indispensable to our approach.

I will begin, in the next message to the thread, by giving the details of the previous paragraph.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJan 17th 2016

It sounds very exciting, although I have severe doubts I’m going to be able to follow this. Two questions, however: (1) where is the pdf for this? And (2) why will you not be giving the category-theoretic framework here, at the nLab?

1. Thanks for the questions, Todd! Don’t be put off by what I have written so far, I am rather sure that you will be able to follow it once I get down to the actual mathematics.

I do not wish to distribute the pdf publically, because one or two things need to be tweaked in the last available version. Several people have received copies of some version of the paper at some point, but all such pdfs should now be regarded as ’bootlegs’ if you happen to get hold of one! Things have no doubt been corrected or tweaked since that version was distributed (even if the fundamental ideas have remained the same throughout).

The things that need tweaking are relatively minor, but my present circumstances mean that a ’mathematical serial’ fits the time that I have to work on mathematics well (short but regular intervals), so, since I would like to write a ’self-contained’ version of the argument in any case, I have decided to write it here, where there is the opportunity for discussion along the way; the necessary tweaks will be made here.

I am fond of the category theoretic framework, and certainly plan to make it available eventually. But I think it could be enlightening for me to give the argument without it first, as doing this will, I think, allow the essence of the approach to the Poincaré conjecture itself to be seen more clearly, even for a category theorist. I am willing to distribute the pdf privately to anybody interested, but would ask people to wait for a little while, until I have posted a little more in the thread here, so that it can be seen how to make the necessary tweaks.

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJan 17th 2016

Please carry on then! I’m grateful for your confidence that I’ll be able to follow after all, and let me repeat that it sounds very exciting.

2. Thank you!

In this message, I will begin with the mathematics.

A link diagram (in the usual, or as I will from now on write, classical, sense) is, formally, a 4-valent planar graph, each vertex of which is decorated with one of two symbols. One can be a little more formal, but this is already much more formal than the intuitive conception of a link diagram which one makes use of when working with them! The vertices are to be thought of as crossings, and the decoration indicates which arc of the crossing is an ’over’ arc.

The first key idea of the approach that I will take to the Poincaré conjecture is that I will make use of ’generalised link diagrams’, which I will take to be arbitrary graphs, each 4-valent vertex of which may be (but need not be) decorated with one of two symbols. I will refer to such ’generalised link diagrams’ simply as link diagrams, and refer to link diagrams in the classical sense as classical link diagrams. I will denote the set of (generalised) link diagrams by $\mathsf{Link}$ $\mathsf{Diagrams}$, and denote the set of classical link diagrams by $\mathsf{Classical}$ $\mathsf{Link}$ $\mathsf{Diagrams}$.

In other words, a link diagram in the sense I will use the term still consists of arcs and crossings; but arcs may be glued together at their endpoints in arbitrary ways (that is to say, a vertex may have arbitrary valency); and we drop the requirement of planarity. A vertex may be 4-valent without being a crossing (i.e. it may have no decoration).

A fully formal definition can now be found at the nLab page which I have begun to accompany this thread.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeJan 17th 2016

It’s a very good idea to have created an accompanying nLab page.

• CommentRowNumber8.
• CommentAuthorDavidRoberts
• CommentTimeJan 17th 2016
• (edited Jan 17th 2016)

Richard, this is crazy bold, and I’m going to thoroughly enjoy this. If this works…. !!!

3. Richard,

I just have a couple questions about your definition of link diagram. What exactly do you mean by “graph”? (Are you using one of the definitions found at the graph article?) And perhaps you could compare your definition of link diagram to the definition in virtual knot theory?

• CommentRowNumber10.
• CommentAuthorRichard Williamson
• CommentTimeJan 18th 2016
• (edited Jan 18th 2016)

Thank you for the interest, David!

Thank you too for your interest, Noam, and for the questions. Your first question allows me to discuss something that I had intended to mention, namely that the definition of a link diagram that I give on the accompanying nLab page is far more general than I will actually need. That is to say, the non-classical link diagrams that I will make use of arise in quite a specific way from classical link diagrams, as we shall see shortly. But it is convenient to have a definition that is simple to state and easy to understand, and which clearly encompasses the non-classical link diagrams that we shall need.

It will now not be surprising that my answer to your question is: it doesn’t really matter! I think that any of the variants would be fine for my purposes. I myself am thinking of what is referred to as a ’multigraph’ at the graph page, which could be defined formally as consisting of a set $V$ of vertices, a set $E$ of edges, and a map from $E$ to the set of subsets of $V$ with two elements. But it would not do any harm to allow loops, and I don’t think that I ever actually need multiple edges between a given pair of vertices.

Let me now turn to your second question. I am very happy to have the opportunity to say a few words about virtual knot theory with regard to the argument that I will give.

To answer your actual question, one of the ideas that led to virtual knot theory is the beautiful observation, due to Kauffman, that we still obtain a very nice diagrammatic theory if we drop the requirement that a classical link diagram be planar. Of course there is much more to it than that, as with any good concept, but the relaxing of planarity is a significant aspect. The notion of a link diagram that I define can be thought of taking this idea further, arguing there are useful generalisations of link diagrams that not only are not planar, but in which arcs may be glued together in arbitrary ways.

However, one should be careful with taking this analogy with virtual knot theory too far. In virtual knot theory, we are actually working with a different kind of diagram (we need to allow a third kind of decoration, corresponding to virtual crossings), and, crucially, we extend the Reidemeister moves by ’virtual Reidemeister moves’. In the work that I will present, I do not introduce any new moves in a way analogous to this, and I would not say that I am introducing a new kind of knot theory in the way that virtual knot theory is a new kind of knot theory. Rather, I will be using certain non-classical link diagrams to prove something about classical knot theory, or more precisely, classical diagrammatic 3-manifold theory.

A very good question, which Louis Kauffman asked me, is: why does the argument that I will give, being diagrammatic in nature, not give a proof of the Poincaré conjecture for virtual 3-manifolds, which is known to be false (a counter-example is given in this paper of Dye and Kauffman)? My answer is: I think that my argument, assuming that is correct in the classical case, will actually give a proof of a form of the Poincaré conjecture for virtual 3-manifolds. Indeed, I think that the argument will be able to be adapted to prove that any virtual 3-manifold with trivial fundamental group will be $S^{3}$ ’up to virtual crossings’. The latter would need to be defined carefully, but I understand intuitively what this should mean: the counter-example in the paper of Dye and Kauffman is of this form, for instance (if one ’replaces’ the virtual crossing simply by a parallel pair of arcs, one has a Hopf link with 0 framing on one component, and +1 or -1 (depending on one’s sign conventions) framing on the other component; this framed Hopf link is equivalent under the Kirby moves to the empty link diagram). Later, if there is an interest, I can explain how this ’up to virtual crossings’ condition arises naturally from the argument that I will give.

• CommentRowNumber11.
• CommentAuthorRichard Williamson
• CommentTimeJan 18th 2016
• (edited Jan 18th 2016)

Back to the argument itself (continuing from #4).

In classical knot theory, we are interested not in the set $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams}$ of classical link diagrams itself, but in the quotient $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$ of $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams}$ by the equivalence relation generated by the relation identifying a pair of classical link diagrams $L_{1}$ and $L_{2}$ if the one can be obtained from the other by applying one of a collection of local moves. In knot theory itself, for instance, our collection of local moves consists of the Reidemeister moves $\mathsf{R1}$, $\mathsf{R2}$, and $\mathsf{R3}$; and it is the ’fundamental theorem of diagrammatic knot theory’, due of course to Reidemeister, that a pair of links are isotopic if and only if any given diagram of one is identified under the afore-mentioned equivalence relation (with respect to the moves $\mathsf{R1}$, $\mathsf{R2}$, and $\mathsf{R3}$) with any given diagram of the other.

If we are interested not in links but in framed links, then (viewing framed links as defined by ordinary link diagrams via the blackboard framing) our collection of local moves consists of the framed Reidemeister moves: the $\mathsf{R2}$ and $\mathsf{R3}$ moves as before, and a variant of the $\mathsf{R1}$ move which does not change framings. Again, we have the theorem (a simple corollary of Reidemeister’s theorem) that a pair of framed links are isotopic if and only if any given diagram of one is identified under the afore-mentioned equivalence relation (with respect to the moves $\mathsf{Framed} \mathsf{R1}$, $\mathsf{R2}$, and $\mathsf{R3}$) with any given diagram of the other.

If we are interested in 3-manifolds, then (making use of the theorems that I mentioned in #2), our collection of moves consists of the framed Reidemeister moves together with a collection of local Kirby moves, which I will take to be those of Figure 3 in this paper of Martelli. From now on, when working with $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$, I shall always be doing so with respect to this collection of moves.

We now come to what, as I have already mentioned, I feel is the most important technical innovation of the argument that I will give. Instead of working with $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$, I will work with a quotient $\mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$ of the set $\mathsf{Link} \mathsf{Diagrams}$ of (generalised) link diagrams as I define them. This quotient will be defined by an equivalence relation generated by the relation identifying not only, as above, a pair of classical link diagrams $L_{1}$ and $L_{2}$ if the one can be obtained from the other by applying one of a collection of local moves; but also identifying certain classical link diagrams with certain non-classical link diagrams.

Crucially, I will demonstrate that a pair of classical link diagrams are identified in passing from $\mathsf{Link} \mathsf{Diagrams}$ to $\mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$ if and only if they are identified in passing from $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams}$ to $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$.

To elaborate slightly, it is a priori possible that the identification of classical link diagrams with non-classical link diagrams could lead, in the equivalence relation generated by these identifications (along with the classical ones), to a pair of classical link diagrams being identified which are not identified classically. But I will demonstrate that this is not the case.

If one accepts this, one might already begin to believe that this could be quite a powerful observation: we have ’strengthened’ our diagrammatic theory, without losing anything. The equivalence classes of $\mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$ are much ’richer’ than those of $\mathsf{Classical} \mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$, and it is this additional richness, i.e. the non-classical representatives of the equivalence class, that will play a crucial role in allowing us to develop the structural theory of these equivalence classes, that I previewed in #2.

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeJan 18th 2016

Sorry for this diversion, but I’m not convinced the description of John Baez given at virtual knot theory is quite right. Shouldn’t it be more like the study of the walking braided monoidal category with duals equipped with a Yang-Baxter operator $R$ on an object $X$? (Even that might not be sufficiently refined – one has to think carefully about the first Reidemeister move – but the point of this comment is to replace his ’symmetric’ with ’braided’.)

• CommentRowNumber13.
• CommentAuthorRichard Williamson
• CommentTimeJan 18th 2016
• (edited Jan 24th 2016)

I will now outline the details of the identifications that I make in defining $\mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$. Next, not tonight but probably tomorrow evening, I will give the fully formal details at the accompanying nLab page. After that, I will give the proof of the crucial fact stated in the third paragraph before the end of #11.

Let us define $\sim$ to be the equivalence relation on $\mathsf{Link} \mathsf{Diagrams}$ generated by the relation which makes the following three kinds of identifications.

1) We identify classical link diagrams $L_{1}$ and $L_{2}$ if the one can be obtained from the other by applying one of the local moves with respect to which I am working (framed Reidemeister moves and the local Kirby moves in Martelli’s paper).

2) Let $M_{1}$ be one half of either the $\mathsf{R2}$ or the $\mathsf{R3}$ move. This is a fragment of a classical link diagram, but a perfectly valid link diagram in the sense I have defined. Let us denote the other half of the $\mathsf{R2}$ or $\mathsf{R3}$ move respectively by $M_{2}$.

Let us then denote by $M^{\mathsf{R2}}$ or $M^{\mathsf{R3}}$ respectively the link diagram obtained by glueing together $M_{1}$ and $M_{2}$ at their endpoints. The link diagram $M^{\mathsf{R2}}$ or $M^{\mathsf{R3}}$ is not classical.

Let $L$ be a classical link diagram whose underlying graph is connected. We identify $L$ with $M^{\mathsf{R2}} \sqcup L$ and $M^{\mathsf{R3}} \sqcup L$.

This file contains a few hand-drawn figures to illustrate this kind of identification in the case of an $\mathsf{R2}$ move.

3) Let $M_{1}$ be a link diagram which looks as in the figure at the top of this file, for some number of crossings. Let $M_{2}$ be a link diagram as in the second figure in this file, with the same number of crossings. Let $M^{l}$ be a link diagram obtained by glueing $M_{2}$ to $M_{1}$ at certain of the endpoints of $M_{1}$. See the third figure in this file. The link diagram $M^{l}$ is not classical.

We identify $L$ with $M^{l} \sqcup L$, for all possible $M^{l}$ (i.e. any number of crossings).

I define $\mathsf{Link} \mathsf{Diagrams} / \mathsf{Moves}$ to be the quotient of $\mathsf{Link} \mathsf{Diagrams}$ by $\sim$.

The role of the new kind of identifications 2) and 3) will become clear as we proceed. It will be easier to motivate them later: for now, I recommend just to focus on understanding the definitions of them, and on following the proof that I will give of the ’crucial fact’.

[Edit: I have made certain tweaks to these identifications since the initiial post, the most significant of which are a simplification of both 2) and 3), and the removal of a fourth kind of identification.]

• CommentRowNumber14.
• CommentAuthorRichard Williamson
• CommentTimeJan 18th 2016
• (edited Jan 18th 2016)

Regarding #12: when I looked at the page virtual knot theory, I also reacted to that statement, Todd. If you take a look at the comment of John Baez at the n-café linked to at that page, you will see that there is supposed to be both a braided structure and a symmetric structure, with the braided structure coming from the classical part of virtual knot theory, and the symmetric structure coming from the virtual part. [Edit: perhaps I misread your comment, Todd, and you are suggesting that both structures should be braided?]

Whilst I haven’t thought about it in any depth, my first reaction was that I am not convinced that the statement is correct. At the very least, there are many things which need making precise, and need justification. I think that Baez makes it clear at the n-café that he is just sharing an (interesting!) idea; the quote at the nLab has maybe been taken a little out of context.

• CommentRowNumber15.
• CommentAuthorTodd_Trimble
• CommentTimeJan 18th 2016
• (edited Jan 18th 2016)

Right, it seemed to me that we should have two ’braidings’: the classical one coming from tangles in 3-space, and the virtual one coming from the Yang-Baxter $R$. (It looks like you said it the other way around in #14 from what Baez said; maybe he should have been said your way.) Tangles in a sufficiently high dimensional space (4 would be enough for 1-dimensional tangles) would have a symmetric classical braiding (where there is no intrinsic difference between over- and under-crossing), but not for virtual knots in 3-space, which is what I thought we were discussing based on the diagrams (where under- vs over- does make a difference). And insofar as $R$ need not be involutive, the virtual braiding isn’t required to be symmetric either. Should we be considering involutive $R$?

The quote at the nLab reads like a bald, almost apodictic assertion; it should be tempered perhaps by marking it as a rough speculation. I agree that Baez’s comment at the Café had a less assertive-sounding tone to it.

4. Todd: I inserted that quote from John Baez into the article because it made sense to me at the time, but I grant that it looks more authoritative than is probably warranted and should be put more in context. However, I do think that Baez got it essentially right. The idea is that the braiding of the ambient symmetric monoidal category $C$ provides virtual crossings “for free”, while the Yang-Baxter operator $R : X \otimes X \to X \otimes X$ on the given object $X \in C$ represents the “real” (a.k.a. classical) over- and under-crossings of the virtual knot/link. Keep in mind that every virtual knot can be realized as an embedding of a circle into a thickened, compact oriented 2-manifold, and that virtual crossings only arise from projecting a higher genus surface onto the plane.

Rather than talking explicitly about virtual crossings, another way to define a virtual link diagram that is closer to Richard’s definition (which is what made me want to ask about the relationship) is as a pair of

1. an embedding of a 4-valent graph in a compact oriented surface (which can be fixed up to isomorphism by placing a cyclic ordering on the half-edges incident to each vertex, in other words as a combinatorial map or ribbon graph), and
2. a (binary) labelling of each vertex as an over-/under-crossing
• CommentRowNumber17.
• CommentAuthorTodd_Trimble
• CommentTimeJan 19th 2016

@Noam, @Richard Very sorry; I misread. So it’s the $R$ that’s supposed to correspond to classical crossings in that set-up. Got it now.

• CommentRowNumber18.
• CommentAuthorRichard Williamson
• CommentTimeJan 19th 2016
• (edited Jan 19th 2016)

Regarding the first part of #16: I agree that this is the idea, and can see what Baez was getting at. It is more the details that worry me: with the caveat, again, that I have hardly thought about it, I do not see how the virtual Reidemeister moves are captured (the one which is analogous to the R3 move, but which has two virtual crossings and a single classical crossing, for instance). In particular, it seems to me that the symmetric and braided structures must be intertwined somehow, and I do not see how this intertwining is present in the category Baez describes. But it is definitely an interesting idea, which it would be nice to see further investigated.

Regarding the last part of #16: thanks, I see what you had in mind, now. Link diagrams in my sense strictly generalise virtual link diagrams thought of in this way.

PS - I did not find the time to write any details in the nLab page tonight after all, but expect to find the time tomorrow.

5. @Richard #18: thanks for clarifying that your link diagrams strictly generalise virtual link diagrams in the style of #16. Something that specifically caught my attention when I read your definition 2 is that you ask that every +/- annotated vertex is equipped with a 4-tuple on edges, rather than a 4-cycle (i.e., 4-tuple quotiented by cyclic permutation) on half-edges. Maybe the edges vs half-edges has to do with allowing loops? But in any case, these are just low-level details and I’m looking forward to seeing how your link diagrams work in practice.

• CommentRowNumber20.
• CommentAuthorRichard Williamson
• CommentTimeJan 20th 2016
• (edited Jan 20th 2016)

The accompanying nLab page has now been updated. I set out to give the details of #13, but found time this evening only for the following.

1. Some clarification of the notion of a classical link diagram in the setting in which I’m working. I thank Noam for his questions regarding this, which gave me a spur to write down carefully what I had in mind.

2. The introduction of a little terminology: that of an ’arc’ of a link diagram. When working formally, one has to be a little careful to pin this down correctly, even in the classical setting, in such a way as to match the standard terminology. This notion of an arc will be used crucially later, so it is a good idea to reflect upon it, and ask here if anything is unclear.

6. In reply to #19: I think that there are several ways to define a classical link diagram formally. They all have an underlying 4-valent planar graph together with some additional data, including decoration of the vertices in one of two ways, but there is still, as you say, a little more that one has to say to be completely rigorous. As I just mentioned in #20, I have written up what I had in mind quite a bit more carefully.

I’m not quite sure what you mean by a ’half-edge’ in this context?

But yes, as you say, it probably does not matter all that much with regard to what I am going to: probably any correct definition of a classical link diagram would be able to be generalised to encompass the non-classical link diagrams that I shall need.

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeJan 20th 2016

Hi Richard, is it possible for you to indicate what it is, or will be, that allows you to give a “rather short and simple, proof of the Poincaré conjecture”? What do you claim is it that all the people who worked on the rather long existing proof missed?

• CommentRowNumber23.
• CommentAuthorRichard Williamson
• CommentTimeJan 20th 2016
• (edited Jan 24th 2016)

To keep things moving whilst I work on the accompanying nLab page, I will now outline a proof that a pair of classical link diagrams $L_{1}$ and $L_{2}$ are identified under the equivalence relation $\sim$ of #13 if and only if they are identified under the equivalence relation generated by the classical identifications of 1) of #13.

Firstly, we need a lemma. Let $L_{1}$, $L_{2}$, $L_{1}'$, and $L_{2}'$ be link diagrams. Suppose that $L_{1}$ is classical, that $L_{2}'$ is non-classical, and that the underlying graphs of $L_{1}$, $L_{2}$, and $L_{2}'$ are connected. If $L_{1} \sqcup L_{1}'$ is isomorphic to $L_{2} \sqcup L_{2}'$, then $L_{1}$ is isomorphic to $L_{2}$.

To prove this, note that since $L_{1} \sqcup L_{1}'$ is isomorphic to $L_{2} \sqcup L_{2}'$, and since all three of $L_{1}$, $L_{2}$, and $L_{2}'$ are connected, we must have that $L_{1}$ is isomorphic to $L_{2}$ or to $L_{2}'$. Since $L_{2}'$ is not classical, $L_{1}$ cannot be isomorphic to $L_{2}'$. Thus $L_{1}$ is isomorphic to $L_{2}$, as claimed.

Now, suppose that $L_{1}$ and $L_{2}$ are classical, and that $L_{1}$ is identified with $L_{2}$ with respect to $\sim$. This means that, for some integer $n \geq 1$, there is an $n$-tuple of pairs $\big( (L^{0}, L^{1}), (L^{1}, L^{2}), \ldots, (L^{n-1}, L^{n}) \big)$, where $L^{0}$ is $L_{1}$, where $L^{n}$ is $L_{2}$, and where, for every $1 \leq i \leq n$, we have that the unordered pair $\{ L^{i-1}, L^{i} \}$ arises either as $\{ L_{1}, L_{2} \}$ as in 1) of #13; as $\{ L, M^{\mathsf{R2}} \sqcup L \}$ or $\{ L, M^{\mathsf{R3}} \sqcup L \}$ as in 2) of #13; or as $\{ L, M^{l} \sqcup L \}$ as in 3) of #13.

Let $i$ be the first integer for which the unordered pair $\{ L^{i-1}, L^{i} \}$ does not arise as in 1) of #13. Then $\{ L^{i-1}, L^{i} \}$ arises as in 2) or 3) of #13. Since $L^{1}$ is classical, we must have, by definition of $i$, that $L^{i-1}$ is classical. Hence $L^{i}$ is of the form $L^{i-1} \sqcup L_{i-1}'$, where $L_{i-1}'$ is non-classical, and where $L^{i-1}$ is classical and has a connected underlying graph.

Since identifications as in 1) of #13 involve only classical link diagrams, and since $L^{i}$ is not classical, the unordered pair $\{ L^{i}, L^{i+1} \}$ must arise as in 2) or 3) of #13. Since $L^{i}$ is not connected, we deduce that $L^{i+1}$ is classical and has a connected underlying graph; and that $L^{i}$ is of the form $L^{i+1} \sqcup L_{i+1}'$, where $L_{i+1}'$ is non-classical.

We now have that $L^{i} = L^{i-1} \sqcup L_{i-1}'$ and that $L^{i} = L^{i+1} \sqcup L_{i+1}'$, and thus that $L^{i-1} \sqcup L_{i-1}' = L^{i+1} \sqcup L_{i+1}'$. By the lemma, we deduce that $L^{i-1} = L^{i+1}$.

Thus we can remove the pairs $(L^{i-1}, L^{i})$ and $(L^{i}, L^{i+1})$ from the $n$-tuple with which we began.

We have demonstrated that if $i$ is the first integer of our $n$-tuple for which the pair $(L^{i-1}, L^{i})$ arises by means of a non-classical identification, then there is an $(n-2)$-tuple which does not involve this pair; for which the first $i-1$ pairs arise by means of classical identifications; and which also exhibits that $L_{1}$ is identified with $L_{2}$ under $\sim$.

By induction, it follows that $L_{1}$ is identified with $L_{2}$ under $\sim$ using only classical identifications, i.e. those arising as in 1) of #13, as required.

This fact, as I have mentioned earlier in the thread, is indispensable for everything that we will do later. Thus you should scrutinise every last detail of the proof! I am indebted to Emily Riehl and Tobias Barthel for help (by pointing out oversights of mine!) with getting earlier versions of this argument watertight.

The requirement that $L$ be connected in 2) and 3) of #13, and the fact that we work with $M^{\mathsf{R2}} \sqcup L$ rather than simply $M^{\mathsf{R2}}$ (and similarly for the other two cases), are both technical choices, made exactly so as to ensure that the above proof goes through (without preventing the arguments we shall give later from going through).

• CommentRowNumber24.
• CommentAuthorRichard Williamson
• CommentTimeJan 20th 2016
• (edited Jan 20th 2016)

Hi Urs, I will reply to your question tomorrow. For now, let me just say that the argument that I am giving is completely independent of the proof via Ricci flow. Thus a more apt version of your question would be: how does my argument differ from previous attempts (the most well-known probably being that of Rourke and Rego in the 80s) to prove the Poincaré conjecture using the Kirby calculus, which have led to a deep-rooted feeling amongst low-dimensional topologists that such a proof is either impossible, or extraordinarily hard and intricate? I have already said a few words in this direction in #2, but can elaborate a little tomorrow.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeJan 20th 2016

Ah, thanks, I had missed the later paragraphs in #2, sorry. Interesting.

7. Richard #21: by “half-edge” I meant it as used for example in ribbon graph. One way to represent a connected 4-valent graph embedded in a compact oriented surface is as a pair of permutations $(v,e)$ acting transitively on a set $H$, where $v^4 = e^2 = 1$, and where no element of $H$ is fixed by $v$, $e$, or $v^2$. You can find examples in this paper, in particular check out Figures 6 (a planar 4-valent map) and 7 (a toric 4-valent map drawn with a virtual crossing).

• CommentRowNumber27.
• CommentAuthorRichard Williamson
• CommentTimeJan 21st 2016
• (edited Jan 25th 2016)

[This post became too long after an edit! It is now contained in #37 and #38 of this thread.]

• CommentRowNumber28.
• CommentAuthorRichard Williamson
• CommentTimeJan 21st 2016
• (edited Jan 24th 2016)

I will now prove two crucial facts concerning operator equivalence. For this, we will make use of the identifications of 2) in #13 (and indeed this is the entirety of their purpose). Not tonight, but in the next post of this serial, I will prove a third crucial fact concerning operator equivalence, for which the identifications of 3) in #13 will be used.

Throughout this post, I will work with a classical link diagram $L$ whose underlying graph is connected, and for which the choice of ordering of the edges at each crossing of $L$ has been determined with respect to an orientation of a link giving rise to $L$.

Firstly, let $a$ and $b$ be arcs of $L$. Then $a^{bb^{-1}} = a$ with respect to $L$ and $q(L)$, and $a^{b^{-1}b} = a$ with respect to $L$ and $q(L)$.

To prove this, note that the pair of arcs $a$ and $b$ define a sub-link diagram of $L$ which is, up to subdivison of edges, the half $M_{1}$ of the $\mathsf{R2}$ move of #13. Thus we have that $q(L) = q(M^{\mathsf{R2}} \sqcup L)$, where $M^{\mathsf{R2}}$ is as defined in #13. We can view $M_{1}$ as a sub-link diagram of $M^{\mathsf{R2}}$, and hence of $M^{\mathsf{R2}} \sqcup L$, in the obvious way. To show that $a^{bb^{-1}} = a$ with respect to $L$ and $q(L)$, it suffices to show that $a^{bb^{-1}} = a$ with respect to $M^{\mathsf{R2}}$, because we then have that $a^{bb^{-1}} = a$ with respect to $M^{\mathsf{R2}} \sqcup L$.

But it is clear that $a^{bb^{-1}} = a$ with respect to $M^{\mathsf{R2}}$. Indeed, within $M^{\mathsf{R2}}$ we have the sub-link diagram which is the first figure in this file. Suppose that we begin with the top right hand corner of the figure with the arc $a$. Travelling downwards and to the left, we encounter a crossing $v$ whose ’over arc’ is $b$; we can choose which edge of $v$ is the left and which is the right in such a way that the right edge belongs to $a$; continuing to travel along the other side of this crossing, we next encounter a crossing $v'$ whose over arc is also $b$; we can choose which edge of $v'$ is the left and which is the right in such a way that the left edge belongs to $a$.

Swapping the choice of right and left edges at both crossings, we similarly have that $a^{b^{-1}b} = a$ with respect to $M^{\mathsf{R2}}$, and hence with respect to $L$ and $q(L)$.

Secondly, let $a$, $b_{1}$, $b_{2}$, $c_{1}$, and $c_{2}$ be arcs of $L$. Suppose that $b_{1}^{b_{2}} = c_{1}$ with respect to $L$, and that $a^{b_{2}c_{1}} = c_{2}$ with respect to $L$. Then $a^{b_{1}b_{2}} = c_{2}$ with respect to $L$ and $q(L)$.

To prove this, note that the fact that $b_{1}^{b_{2}} = c_{1}$ with respect to $L$, and that $a^{b_{2}c_{2}} = c_{1}$ with respect to $L$, means that $L$ has a sub-link diagram which is the half $M_{1}$,of the $\mathsf{R3}$ move of #13, depicted in the second figure in the same file. We now use exactly the same technique as in the previous proof.

Namely, we have that $q(L) = q(M^{\mathsf{R3}} \sqcup L)$, where $M^{\mathsf{R3}}$ is as defined in #13. We can view $M_{1}$ as a sub-link diagram of $M^{\mathsf{R3}}$, and hence of $M^{\mathsf{R3}} \sqcup L$, in the obvious way. To show that $a^{b_{1}b_{2}} = c_{2}$ with respect to $L$ and $q(L)$, it suffices to show that $a^{b_{1}b_{2}} = c_{2}$ with respect to $M^{\mathsf{R3}}$.

But this is clear. Indeed, within $M^{\mathsf{R3}}$ we have the sub-link diagram which is the third figure in the same file.

• CommentRowNumber29.
• CommentAuthorRichard Williamson
• CommentTimeJan 21st 2016
• (edited Jan 21st 2016)

We now draw some consequences from the two results proven in #28. Let $F(L)$ be the free group on the arcs of $L$. The first result implies that operator equivalence is well-defined for elements of $F(L)$. Indeed, an element of $F(L)$ corresponds exactly to a word in the arcs of $L$, up to adding/deleting sub-words of the form $bb^{-1}$ or $b^{-1}b$.

The first result also implies that the set of elements of $F(L)$ which are operator equivalent to the identity with respect to $L$ and $q(L)$ is a normal subgroup of $F(L)$. That it is a subgroup is clear. To prove normality, let $w$ be a word in the arcs of $L$ which is operator equivalent to the identity with respect to $L$ and $q(L)$. Then, purely from the definition of operator equivalence with respect to $L$ and $q(L)$, we have that $b^{-1}wb$ is operator equivalent to $b^{-1}b$ with respect to $L$ and $q(L)$. By the first result, we have that $b^{-1}b$ is operator equivalent to the identity with respect to $L$ and $q(L)$. Putting these two observations together, we have that $b^{-1}wb$ is operator equivalent to the identity with respect to $L$ and $q(L)$, as required.

We define the operator group with respect to $L$ and $q(L)$ to be the quotient of $F(L)$ by this normal subgroup of elements which are operator equivalent to the identity with respect to $L$ and $q(L)$.

Now, $\pi_{1}(L)$, thought of in terms of the Wirtinger presentation, is the quotient of $F(L)$ by the normal closure in $F(L)$ of the set of words of the form $a_{3}a_{2}^{-1}a_{1}^{-1}a_{2}$, where $a_{1}^{a_{2}} = a_{3}$ with respect to $L$.

Making use of both the first and the second result proven in this post, we have that $a_{3}a_{2}^{-1}a_{1}^{-1}a_{2}$ is operator equivalent to the identity with respect to $L$ and $q(L)$. Indeed, we have the following identifications, taking $=$ to mean operator equivalence with respect to $L$ and $q(L)$.

$a_{3}a_{2}^{-1}a_{1}^{-1}a_{2}$

$= a_{2}^{-1}a_{2}a_{3}a_{2}^{-1}a_{1}^{-1}a_{2}$

$= a_{2}^{-1}a_{1}a_{2}a_{2}^{-1}a_{1}^{-1}a_{2}$

$= a_{2}^{-1}a_{1}a_{1}^{-1}a_{2}$

$= a_{2}^{-1}a_{2}$

$= \emptyset$.

The second equality is a consequence of the second result; the others are a consequence of the first.

We deduce that the operator group with respect to $L$ and $q(L)$ is a quotient of $\pi_{1}(L)$.

The third crucial fact about operator equivalence, that I will prove in the next installment of this serial by means of exactly the same technique, will allow us to strengthen this observation, to prove that the operator group with respect to $L$ and $q(L)$ is a quotient of $\pi_{1}(L)$ modulo the longitudes of its components.

• CommentRowNumber30.
• CommentAuthorRichard Williamson
• CommentTimeJan 21st 2016
• (edited Jan 21st 2016)

The terminology operator equivalence and operator group comes from the basic theory of racks, as outlined in the first few sections of this paper of Fenn and Rourke. Indeed, the notions for racks inspired the definitions that I have given. However, the details are significantly different, in several ways.

Roughly speaking, the identifications of 2) in #13 play the role of the rack axioms in allowing us to define the operator group, and to ensure that it is a quotient of the fundamental group. In this way, these non-classical identifications can be fruitfully thought of as equipping the equivalence class of a link diagram with ’extra structure’ which resembles the imposition of axioms in an algebraic setting.

• CommentRowNumber31.
• CommentAuthorRichard Williamson
• CommentTimeJan 22nd 2016
• (edited Jan 29th 2016)

I will now turn to the third crucial fact concerning operator equivalence. Let us begin by recalling the notion of the longitude of a component of an oriented classical link diagram $L$ (thought of as blackboard framed). This is defined by picking an arc of $L$ (any arc will do), and then applying the following procedure.

1) Begin with the empty word in the arcs of $L$.

2) Travel around this component (in $L$), following the orientation of $L$. For each crossing which we encounter which we travel ’under’, we replace the word $w$ in the arcs of $L$ which we have constructed up to this point by $wa$, where $a$ is the ’over arc’ of the crossing, if the configuration of orientations of $L$ at the crossing is as in the first figure of this file, and by $wa^{-1}$ if the configuration of orientations of $L$ at the crossing is as in the second figure of this file.

3) Stop when we return to the arc with which we began.

Our third fact concerning operator equivalence is that the longitude $l$ of any component of $L$ is strictly operator equivalent to the identity with respect to $L$ and $q(L)$.

Let us prove this. We shall make use of exactly the same technique as in #28, but this time using identifications as in 3) of #13.

Our first observation is that, by definition of $l = a_{1}^{\epsilon_{1}} \cdots a_{n}^{\epsilon_{n}}$, there is an arc $a$ of $L$ (the arc at which we started when defining $l$) and a sub-link diagram of $L$ which is as in the third figure in this file, where the two edges labelled $a$ both belong to $a$. Let $b$ be any arc of $L$ (which may be $a$ itself). Then we have a sub-link diagram of $L$ which is as in the fourth figure in this file, that is to say, which is a link diagram of the form $M_{1}$ of 3) in #13. There may (recalling our definition of a sub-link diagram in #27) be other arcs of $L$ which do not belong to the component of $L$ of which $l$ is the longitude, but which cross the horizontal arcs in this figure

Thus we have that $q(L) = q(M^{l} \sqcup L)$, where $M^{l}$ is as defined in #13, using an $M_{2}$ which is glued to the arcs $a_{1}$, \ldots, $a_{n}$ and $b$ to obtain the fifth figure in this file. We can view $M_{1}$ as a sub-link diagram of $M^{l}$ in the obvious way. We shall demonstrate that $b^{l} = b$ with respect to $L$ and $q(L)$ by demonstrating that $b^{l} = b$ with respect to $M^{l}$, and hence with respect to $M^{l} \sqcup L$.

But it is clear that $b^{l} = b$ with respect to $M^{l}$ (see again the fifth figure in this file).

Since $b$ was arbitrary, we conclude that $l$ is strictly operator equivalent to the identity with respect to $L$ and $q(L)$, as claimed.

Suppose now that $L$ has $n$ components. Let $l_1$, $\ldots$, $l_n$ be the longitudes of these components, and let $\langle l_1, \ldots, l_n \rangle$ be the normal closure of the set $\{ l_1, \ldots, l_n \}$ in $\pi_{1}(L)$. Since each of the $l_i$’s is, as just demonstrated, strictly operator equivalent to the identity with respect to $L$ and $q(L)$, it follows immediately from the definitions in #29 of $\pi_{1}(L)$ and the operator group with respect to $L$ and $q(L)$ that the latter is a quotient of the group $\pi_{1}(L) / \langle l_1, \ldots, l_n \rangle$.

• CommentRowNumber32.
• CommentAuthorRichard Williamson
• CommentTimeJan 23rd 2016
• (edited Jan 25th 2016)

I will now prove the final fact that we shall require. Namely, let $L$ be a classical link diagram such that the operator group with respect to $L$ and $q(L)$ is trivial, and such that no component of $L$ is a $\pm 1$ framed unknot. Then $L$ is the empty link diagram.

Let us prove this. Since the operator group with respect to $L$ and $q(L)$ is trivial, every word in the arcs of $L$ is operator equivalent to the identity with respect to $L$ and $q(L)$. Let $a$ be any arc of $L$. We then have that $a$ is operator equivalent to the identity with respect to $L$ and $q(L)$. Since $a$ is a word of length one, we have that $a$ is in fact strictly operator equivalent to the identity. Thus there is a link diagram $L'$ which has a sub-link diagram $M$ in common with $L$ to which parts of $a$ belong, for which $q(L') = q(L)$, and for which $a^{a} = a$ with respect to $L'$.

Since $a^{a} = a$ with respect to $L'$, there must be a crossing of $L'$, all three edges of which belong to an arc to which parts of $a$ belong, as in the first figure in this file. We deduce, by 1) in the definition (in #37)of the equallity $a^{a} = a$ with respect to $L$ and $q(L)$, that $M$ has a crossing of the same kind, namely for which all three edges are $a$. Since $M$ is a sub-link diagram of $L$, we deduce that $L$ has a crossing of the same kind, all three edges of which are $a$. Since $L$ is classical, we conclude that the component of $L$ containing $a$ is a $\pm 1$ framed unknot, as in the second figure in this file.

But, by assumption, $L$ has no components which are a $\pm 1$ framed unknot. We conclude that $L$ in fact has no arcs, that is to say, that it is the empty link diagram.

[This post was edited after the message of Emily Riehl in #36, which refers to the original version of the post.]

• CommentRowNumber33.
• CommentAuthorRichard Williamson
• CommentTimeJan 23rd 2016
• (edited Jan 24th 2016)

We now bring everything together, to complete the argument. Let $M$ be a closed, connected 3-manifold with trivial fundamental group. We then have that $M$ is orientable. Thus the theorem of Lickorish and Wallace implies that there is a framed link $L^{\mathsf{link}}$ such that $M$ is obtained by the integral Dehn surgery on $L^{\mathsf{link}}$ determined by its framing. Let $L$ be a classical link diagram determined (with the blackboard framing) by $L^{\mathsf{link}}$ equipped with an orientation (any orientation will do). Using $\mathsf{R2}$ moves, $\mathsf{R3}$ moves, and local Kirby moves, we can replace $L$ by a classical link diagram $L'$ which does not have any components which are $\pm 1$ framed unknots. I will explain this in #39. Moreover, using $\mathsf{R2}$ moves, we can replace $L'$ by a classical link diagram $L''$ which does not have any components which are $\pm 1$ framed unknots, and whose underlying graph is connected. Thus I will assume that $L$ has these two properties.

Suppose that $L$ has $n$ components, and let $l_{1}$, $\ldots$, $l_{n}$ be the longitudes of these components as defined in #31. It follows more or less immediately from the definition of Dehn surgery, the van Kampen theorem, and the fact the fundamental group of the complement of $L^{\mathsf{link}}$ is $\pi_{1}(L)$ as defined in #29, that the fundamental group of $M$ is isomorphic to $\pi_{1}(L) / \langle l_1, \ldots, l_n \rangle$. Since the fundamental group of $M$ is trivial, we have that the group $\pi_{1}(L) / \langle l_1, \ldots, l_n \rangle$ is trivial.

By #30 and #31, the operator group with respect to $L$ and $q(L)$ is a quotient of the group $\pi_{1}(L) / \langle l_1, \ldots, l_n \rangle$. Since the latter group is trivial, we deduce that the operator group with respect to $L$ and $q(L)$ is trivial. By #32, we conclude that $L$ is the empty link diagram. By the theorems of Kirby, Fenn and Rourke, and Martelli mentioned in #2, we deduce that $M$ is obtained by Dehn surgery on $\emptyset$, that is to say, that $M$ is $S^{3}$. That’s it!

8. I will give a summary of the argument later today, or tomorrow. For now, I just wish to mention that the same kind of argument may well be able to demonstrate that any 3-manifold with finite cyclic fundamental group must be a lens space or $S^{3}$ (i.e. obtained by integral Dehn surgery on an unknot with $\pm n$ framing for some $n \geq 1$).

Indeed, the crucial structural observation in the argument I have given is that if a 3-manifold has trivial fundamental group, then the operator group of a (blackboard framed) link diagram giving rise to this 3-manifold must also be trivial. If a 3-manifold has finite cyclic fundamental group, then the operator group with respect to $L$ and $q(L)$, where $L$ is a blackboard framed link diagram giving rise to this 3-manifold, must also be finite cyclic. This is also a very strong structural property of $L$ and $q(L)$, and I expect it to be strong enough, with an appropriate modification of 4) in #13, to demonstrate that $q(L)$ must have a representative which is an unknot with $\pm n$ framing for some $n \geq 1$.

I thank Marc Lackenby for a question, back in December, which prompted these reflections.

Now that the argument has been given, I would be delighted to hear from anybody who has worked through it and feels that they understand it. If you are trying to work through it but do not follow something, just let me know, either here or via email (richard (at) rwilliamson-mathematics.info). I will work gradually on a slightly more formal exposition, completing the accompanying nLab page, but I would already regard the details that I have given here, together with what is already present on the nLab page, as being sufficient to be regarded as a proof (assuming that there is no significant error, of course!).

• CommentRowNumber35.
• CommentAuthorRichard Williamson
• CommentTimeJan 23rd 2016
• (edited Jan 24th 2016)

[This message is in response to the mesage of Emily Riehl in #36.]

Thank you very much for your questions, Emily! You are correct that there were some slight gaps in #32 and the material which relates to it. These can be fixed/clarified without much change, so I will just edit the earlier posts to correct them. I will edit this message to indicate when I am done. [Edit: now done!]

• CommentRowNumber36.
• CommentAuthorEmily Riehl
• CommentTimeJan 23rd 2016

[Note: this is a self-plagiarized re-post to try and solve an nForum login problem.]

It’s possible that I don’t understand the new definitions but I’m skeptical about several of the claims made in post #32.

I agree that if $L$ is a classical link diagram with trivial operator group, then for any arc $a$ there is some abstract link diagram $L'$ (which can be constructed with the particular arc $a$ in mind) so that $q(L)=q(L')$, and so that there is injection of abstract link diagrams $L \hookrightarrow L'$ so that $L'$ contains a chain of arcs (including the image of $a$) arranged in a “figure eight” meaning it is a maximal chain that crosses over itself.

I. Now if this $L'$ is classical, this chain defines a +/- framed unknot. But I don’t see why $L'$ couldn’t have other components (i.e., other links):

(i) Firstly, is it clear that $L$ and $L'$ have the same operator group? Its operator group is generated by arcs that aren’t in the image of $L$ in addition to the arcs coming from $L$. Perhaps these lie off in another component somewhere and don’t satisfy the same sorts of relations.

(ii) To illustrate this sort of confusion, if $L$ is two disjoint unknots then an R2 move gives $L'$ with two overlapping circles and there is an injection of abstract link diagrams $L \hookrightarrow L'$. But I don’t think $L'$ injects into $L$ because it has two crossings that need to map somewhere.

(iii) Supposing $L$ and $L'$ do have the same operator group, just because arcs $b$ and $a$ lie in different components in $L'$ doesn’t prohibit $b^a =b$, I think. The point is that this relation would be “witnessed” in some larger $L''$ so that $q(L') = q(L'')$. Indeed, doesn’t your proof that $a^l= a$ implies $b^l=b$ for all $b$ force this relation to hold?

II. If $L'$ is not classical, its underlying graph might nonetheless be disconnected. The relationship $b^a = b$ relative $q$ does not force $b$ and $a$ to be in the same component of $L'$; they just have to lie in the same component of the “larger” $L''$ that witnesses this relation.

• CommentRowNumber37.
• CommentAuthorRichard Williamson
• CommentTimeJan 24th 2016
• (edited Jan 25th 2016)

[This post, along with #38, was originally #27, but became too long after an edit, so needed to be split into two.]

I thought that this evening I would continue from #23, with an outline of the next steps in the argument. I will first introduce the notion of operator equivalence to the identity of an arc of a link diagram.

Let $L$ be a link diagram (in my generalised sense, as usual). Let us define a word in the arcs of $L$ to be an $n$-tuple $\big( (a_{1}, \epsilon_{1}), \ldots, (a_{n}, \epsilon_{n}) \big)$ of pairs $(a_{i}, \epsilon_{i})$, where $a_{i}$ is an arc of $L$, and $\epsilon$ is either $+1$ or $-1$. We shall denote such an $n$-tuple by $a_{1}^{\epsilon_{1}} \cdots a_{n}^{\epsilon_{n}}$. This is, for now, just notation.

Here, and everywhere from now on, I am using arc in the sense defined precisely on the accompanying nLab page. In the case of a classical link diagram, an arc has its usual meaning.

Let us introduce one more piece of terminology. Let $a$ be an arc of $L$. If $v$ is a crossing (4-valent vertex decorated by $+$ or $-$) of $L$ for which $n(v)$ and $s(v)$ belong to $a$, then we refer to $w(v)$ as the ’left edge’ of $v$, and refer to $e(v)$ as the ’right edge’ of $v$. If $v$ is a crossing (4-valent vertex decorated by $+$ or $-$) of $L$ for which $w(v)$ and $e(v)$ belong to $a$, then we refer to $n(v)$ as the ’left edge’ of $v$, and refer to $s(v)$ as the ’right edge’ of $v$.

Let $a$ and $b$ be arcs of $L$, and let $w=a_{1}^{\epsilon_{1}} \cdots a_{n}^{\epsilon_{n}}$ be a word in the arcs of $L$. I will write that $a^{w} = b$ with respect to $L$ if the following hold.

1) There is a crossing of $L$ whose ’over arc’ is part of $a_{1}$, the right edge of which belongs to $a$ if $\epsilon_{1}$ is $+1$, and the left edge of which belongs to $a$ if $\epsilon_{1}$ is $-1$. Let us denote the fourth edge of this crossing by $e_{1}$.

2) There is a crossing of $L$ whose ’over arc’ is part of $a_{2}$, the right edge of which belongs to an arc to which $e_{1}$ belongs if $\epsilon_{2}$ is $+1$, and the left edge of which belongs to an arc to which $e_{1}$ belongs if $\epsilon_{2}$ is $-1$. Let us denote the fourth edge of this crossing by $e_{2}$.

3) There is a crossing of $L$ whose ’over arc’ is part of $a_{3}$, the right edge of which belongs to an arc to which $e_{2}$ belongs if $\epsilon_{3}$ is $+1$, and the left edge of which belongs to an arc to which $e_{2}$ belongs if $\epsilon_{3}$ is $-1$. Let us denote the fourth edge of this crossing by $e_{3}$.

n) There is a crossing of $L$ whose ’over arc’ is part of $a_{n}$, the right edge of which belongs to an arc to which $e_{n-1}$ belongs if $\epsilon_{n}$ is $+1$, and the left edge of which belongs to an arc to which $e_{n-1}$ belongs if $\epsilon_{n}$ is $-1$. We require that the fourth edge of this crossing belongs to the arc $b$.

It is important here to understand that $a^{w} = b$ means exactly what I have just written, nothing more and nothing less! This file contains some figures illustrating the definition.

We shall now make use of the notion of a sub-link diagram of a link diagram $L$. By this we shall mean a link diagram $L'$ such that there is, up to subdivision of edges of $L'$ or $L$ into a pair of edges joined at a 2-valent vertex, a monomorphism from the underlying graph of $L'$ to the underlying graph of $L$, which takes vertices decorated with $+$ or $-$ to vertices decorated with $+$ or $-$.

Note that this definition allows that an undecorated vertex (i.e. a vertex decorated with $\emptyset$) of $L'$ be mapped to a vertex of $L$ which is decorated with $+$ or $-$.

Now, let $q(L)$ be the equivalence class of $L$ with respect to the equivalence relation $\sim$ of #13. I will define $a^{w} = b$ with respect to $L$ and $q(L)$ if there is a link diagram $L'$ for which the following hold.

1) The link diagrams $L'$ and $L$ have a common sub-link diagram $M$ which, when viewed as a sub-link diagram of $L$, contains parts of the arcs $a$ and $b$, and parts of the arcs $a_{1}$, $\ldots$, $a_{n}$. We require that if there is one or more crossing of $L'$, all three edges of which are $a$, then at least one such crossing belongs to $M$ (and hence to $L$).

2) We have that $q(L') = q(L)$.

3) We have that $a^{w} = b$ with respect to $L'$, viewing $a$, $b$, and the $a_{i}$’s as arcs of $L'$, which we can do because of 1).

If different arcs of $M$ are parts of the same arc of $L$, say $c$, then we may think of $c$ as an arc of $L'$ in any of the possible ways in the course of establishing that $a^{w} = b$ with respect to $L'$ as in 3) (i.e. we may think of several of the arcs of $M$ as being $a$).

It is absolutely crucial to appreciate here that I am working with two different meanings of $a^{w} = b$. One is with respect to $L$. The other is with respect to $L$ and $q(L)$. The latter notion is weaker: we are allowing ourselves to replace $L$ by another link diagram sharing a common sub-link diagram with it.

• CommentRowNumber38.
• CommentAuthorRichard Williamson
• CommentTimeJan 24th 2016
• (edited Jan 24th 2016)

[Continuing from #37.]

Now, let $w_{1}$ and $w_{2}$ be words in the arcs of $L$, for a link diagram $L$ whose underlying graph is connected. Let us define $w_{1}$ and $w_{2}$ to be strictly operator equivalent with respect to $L$ and $q(L)$ if, for any arc $a$ of $L$, there is an arc $b_{a}$ of $L$ such that $a^{w_{1}} = b_{a}$ with respect to $L$ and $q(L)$, and such that $a^{w_{2}} = b_{a}$ with respect to $L$ and $q(L)$.

Note again (I cannot stress it strongly enough, it is vital!) that I am asking that $a^{w_{1}} = b_{a}$ and that $a^{w_{2}} = b_{a}$ with respect to $L$ and $q(L)$ here, not with respect to $L$.

We define a pair of words in the arcs of $L$ to be operator equivalent with respect to $L$ and $q(L)$ if they are equivalent under the equivalence relation on words in the arcs of $L$ generated as follows.

1) So long as neither is of length $1$, we identify words of the form $u \cdot v_{1} \cdot w$ and $u \cdot v_{2} \cdot w$ for which $v_{1}$ and $v_{2}$ are strictly operator equivalent with respect to $L$ and $q(L)$, where $\cdot$ denotes concatenation, and either or both of $u$ and $w$ may be $0$-tuples, i.e. empty. We also allow that $v_{1}$ or $v_{2}$ be $0$-tuples, setting $a^{\emptyset} = a$ with respect to $L$ and $q(L)$ for any arc $a$ of $L$, and extending the definition of strictly operator equivalent accordingly, to allow that a word be strictly operator equivalent to the empty word.

2) We identify an arc $a$ with the empty word if $a$ is strictly operator equivalent to the identity with respect to $L$ and $q(L)$.

We now define a word $w$ in the arcs of $L$, for a classical link diagram $L$ whose underlying graph is connected, to be operator equivalent to the identity with respect to $L$ and $q(L)$ if $w$ is operator equivalent to the empty word with respect to $L$ and $q(L)$. This definition is crucial for us. We will use it to define the ’operator group’ mentioned in #2.

• CommentRowNumber39.
• CommentAuthorRichard Williamson
• CommentTimeJan 24th 2016
• (edited Jan 24th 2016)

In this post, I will explain the fact appealed to in #33, that, using $\mathsf{R2}$ moves, $\mathsf{R3}$ moves, and local Kirby moves, we can replace any classical link diagram $L$ by one which does not have any components which are $\pm 1$ framed unknots.

Indeed, suppose that we have such a component $U$, as in the first figure of this file. If no other arcs of $L$ cross this component, we can apply a Fenn-Rourke move with zero strands (which simply adds or deletes a disjoint $\pm 1$ framed component) to remove it.

If any other arc of $L$, belonging to a component $K$, crosses this component $U$, then there must be a corresponding arc, also belonging to $K$, which also crosses $U$. That is to say, if $K$ and $U$ cross once, they must cross twice, etc. Thus, we have that $L$ looks locally as in the second figure of the same file. I have drawn the crossings only as 4-valent vertices; in each case, the arc of $U$ could be either the ’over arc’ of the crossing or the ’under arc’. The box can contain any fragment of a classical link diagram which can be joined to the arcs which enter and leave it.

Now, if the over arc of the crossing indicated 1 belongs to $U$, then we can slide this arc over the box in the manner indicated in the third figure of the same file. Formally, we make use of $\mathsf{R2}$ moves and $\mathsf{R3}$ moves to carry out this slide. Similarly, if the under arc of the crossing indicated 1 belongs to $U$, then we can slide this arc under the box in the manner indicated in the fourth figure of the same file.

Carrying out the same kind of manoeuvre for each of the crossings indicated 2, $\ldots$, n, we can replace $L$ by a link diagram which looks locally as in the fifth figure of the same file.

Now, suppose that the over arc of both of the crossings indicated $1'$ and $1$ does not belong to $U$. Then we can slide this arc so that it does not cross $U$, as in the sixth figure of the same file. Formally, again, we make use of $\mathsf{R2}$ and $\mathsf{R3}$ moves to carry out this slide.

Suppose that the under arc of both of the crossings indicated $1'$ and $1$ does not belongs to $U$. Then we can slide this arc so that it does not cross $U$, as in the seventh figure of the same file. Formally, again, we make use of $\mathsf{R2}$ and $\mathsf{R3}$ moves to carry out this slide.

Finally, suppose that none of the pairs of crossings $i$ and $i'$ can be removed by either of the preceding two procedures. Then, using $\mathsf{R2}$ and $\mathsf{R3}$ moves, we can re-arrange the crossings so that $L$ looks locally as in the eighth figure of the same file. Now, the Fenn-Rourke move with $n$ vertical strands precisely allows us to remove $U$ from $L$, replacing the vertical strands with a twist, depicted in the ninth figure of the same file in the case of three vertical strands.

We have demonstrated that we can remove a $\pm 1$ framed unknot component from $L$ without adding any new components. By induction, it follows that we can replace $L$ by a link diagram which has no components which are $\pm 1$ framed unknots, as claimed.

9. Thank you very much again for your questions, Emily, I am very grateful to your alerting me to the deficiencies of the original version of #32. I have sharpened a few things, notably the definition of the non-classical identifications in 2) and 3) of #13 (removing that of the fourth kind), with consequent tweaking here and there later in the thread. But, most significantly, I have made use of the fact discussed in #39.

If you have the opportunity to take another look, I would be very happy to hear if you feel that the dénouement now holds up to scrutiny.

With regard to your question of I. (i) in #36, though this fact is no longer needed, it is clear that if $L$ and $L'$ define a 3-manifold with trivial fundamental group, which is the case at hand, then they have the same operator group (it is trivial). In general, it is not at all clear to me.

As a general remark to all, there may be other ways to get from triviality of the operator group to the kind of consequence that I require.

• CommentRowNumber41.
• CommentAuthorRichard Williamson
• CommentTimeJan 28th 2016
• (edited Jan 30th 2016)

I realised a couple of days ago that it seems to be possible to give an argument along the lines of the one given earlier in the thread, but which does not make use of generalised link diagrams. Indeed, this argument is outrageously simple. I would be very happy to hear any thoughts about it.

Here goes. Everything below is independent of the material earlier in the thread, except in the one place indicated.

Let $L$ be a link diagram (in the usual sense). Let F(L) be the free group on the arcs of $L$. Let us define a subset $N$ of the underlying set of $F(L)$ by stipulating that a word w in the arcs of $L$ belongs to $N$ if it is possible to manipulate $w$ to become the empty word (i.e. the identity in $F(L)$) by the following rules.

1) A sub-word can be deleted if it is equal to the longitude of a component of $L$.

2) A sub-word can be deleted if it is of the form $a_3 a_2^-1 a_1^-1 a_2$, for a crossing of L as in this file.

Now, I claim that $N$ defines a normal subgroup of $F(L)$. That it defines a sub-group is obvious (if $v$ and $w$ can both be manipulated to become the empty word using the above rules, then so can $v \cdot w$). To see that it is normal, let $a$ be any arc of $L$, and suppose that $w$ belongs to $N$. It suffices to check that $a^-1 \cdot w \cdot a$ belongs to $N$. But this is obvious: since we can manipulate $w$ by rules 1) and 2) to the empty word, we can manipulate $a^-1 \cdot w \cdot a$ to become $a^-1 \cdot a$, and the latter is equal to the empty word in $F(L)$.

I also claim that $F(L) / N$ is a quotient of $\pi_1(L) / \langle l_1, ..., l_n \rangle$, where $l_1$, …, $l_n$ are the longitudes of the components of $L$. This is yet again obvious. Indeed, $\pi_1(L) / \langle l_1, ..., l_n \rangle$ is the quotient of $F(L)$ by the normal closure $M$ of the set consisting of the $l_i$’s and the set of words $a_3 a_2^-1 a_1^-1 a_2$ as in 2) above. This set is a subset of $N$, and since $N$ is normal, the normal closure of it is also a subset of $N$. Hence, by the third isomorphism theorem, we have that $F(L) / N$ is isomorphic to $\big( F(L) / M \big) / (N / M)$, that is to say, to $\big( \pi_1(L) / \langle l_1, ..., l_n \rangle \big) / (N / M)$.

Without loss of generality, we can assume that $L$ does not have any components that are $\pm 1$ framed unknots. This is demonstrated in #39 above, but I would assume that this is a well-known fact.

Suppose that $\pi_1(L) / \langle l_1, ..., l_n \rangle$ is trivial. Then, because $F(L) / N$ is a quotient of $\pi_1(L) / \langle l_1, ..., l_n \rangle$, we must have that $F(L) / N$ is trivial. This means that $N$ is all of $F(L)$. In particular, for any given arc $a$ of $L$, we must have that both $a$ and $a^{-1}$ can be manipulated according to 1) and 2) to obtain the empty word. Rule 2) can never be applied to a word of length 1, so we must in fact be able to apply 1). That is to say, we must have that $a$ is the longitude of a component of $L$, and the same for $a^{-1}$.

There are only two ways in which this is possible.

1) This component is a $\pm 1$ framed unknot (i.e. a figure of eight, where the single crossing may be either of the two possibilities). But $L$ does not, by assumption, have any components that are $\pm 1$ framed unknots.

2) This component is a $0$ framed unknot, and is not the same component as that to which $a$ belongs. Let us denote the single arc of this $0$ framed unknot by $b$. But $b$ is also the longitude of a component of $L$,. The only possibility is that $b$ is the longitude of the component to which $a$ belongs. Thus $a$ must be the single arc of a $0$ framed unknot. We thus have that the components of $L$ to which $a$ and $b$ define a Hopf link in which both components are $0$ framed. But then $a^{-1}$ is not the longitude of any component of $L$, which it must be.

Since $a$ was arbitrary, we conclude that $L$ is the empty link diagram, as required.

[Edit: in the original post of yesterday evening, case 2) was omitted.]

• CommentRowNumber42.
• CommentAuthorRichard Williamson
• CommentTimeJan 29th 2016
• (edited Jan 30th 2016)

As an additional remark, the argument of #41 also, in the case of a link which is in fact a knot, seems to give a proof of the Gordon-Luecke theorem that no integral Dehn surgery on a knot can give $S^{3}$. Indeed, the argument of #41 yields that the only framed knot which can give $S^{3}$ is a $\pm 1$ framed unknot. (It is well-known that no other framing of an unknot can give $S^{3}$, so that this statement is equivalent to the Gordon-Luecke theorem.)

• CommentRowNumber43.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 7th 2016

I feel a little bad that there has been little discussion about Richard’s work in this thread, besides what Richard has written (with a brief interlude from Emily Riehl).

While I find the prospect of trying to read through this intimidating (which is not meant to be a reflection on Richard – not at all! – only on my own limitations), I’m thinking about giving it a try anyway, just to see what happens. I can’t and won’t make any promises how far I’ll get, of course.

If Richard doesn’t mind, I may be adding to the nLab bits and pieces I pick up from his comments as I go through them. Certainly that should be okay in cases of results of others like Lickorish & Wallace, etc. Mainly I think I’d want to go about this in a modest way (at least in the beginning), mostly at the level of definitions and linking up with what we already have here in the nLab. The progress and pace on this may be leisurely.

• CommentRowNumber44.
• CommentAuthorRichard Williamson
• CommentTimeFeb 8th 2016
• (edited Feb 8th 2016)

Thank you very much for your kind interest, Todd. Your plan sounds great, and I will be happy to try to help out with the nLab entries.

I feel that knot theory (and higher and generalised versions) is rather intimately and deeply intertwined with higher category theory (and not only in the ’obvious’ cases of braided monoidal categories, etc), and thus that any material relating to it is a good fit for the nLab in the long run.

With regard to the arguments that I have given in the thread: there is a somewhat subtle but significant oversight to the argument of #41 that seems to put it to the sword for practical purposes. However, I am actually rather happy about this, because it illustrates that the use of operator equivalence in the longer argument has some special aspects, which take care of the tricky point in #41!

But the longer argument as presented in the thread is also not quite right in one particular (this is not the matter that Emily raised, but something earlier). I am in the process of writing a letter to Lou Kauffman, with whom I have been discussing these matters, which will tidy up the argument; it involves operator equivalence, but I am able to avoid the use of generalised link diagrams, so that it also bears some resemblance to the proposed shorter argument of #41.

I have been quite busy and have not had much time to work on this in the last week or so, but hope to be finished with the letter sometime this week. I will then post a copy of it here.

I apologise for any confusion resulting from these to-ings and fro-ings; I am trying to find a way to present the essence of the ideas behind my approach in a way that can readily be understood. I thought for a short time, because of the proposed argument in #41, that some things that I thought were essential were not; and now, through understanding why #41 does not go through, have a deeper understanding of what working with the notion of operator equivalence brings us.

Just for lack of time, I will clarify what the error in #41 is when I post the letter to Lou Kauffman, rather than now.

All this is to say that I think that adding the background material concerning viewing 3-manifolds diagrammatically to the nLab, and perhaps, if you so wish, relevant notions from the theory of racks (operator equivalence as defined in this setting, for instance), sounds great, and could not be anything other than a useful addition to the nLab in any case. When I have slightly more time, probably from next week, I will try to help out. You are also entirely welcome to add any ideas that you find interesting from my work; but, to avoid spending time on things that may change, you may wish to wait until seeing the letter to Kauffman before looking closely at the original ideas (though of course I will be happy to discuss anything that you do take a look at). Thanks again!

• CommentRowNumber45.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 8th 2016

Obviously Lou Kauffman is far better placed to evaluate the argument than someone like me.

But I’m happy for the time being just to try to bring myself up to speed on some of the classical background material like Kirby calculus and the Lickorish-Wallace theorem, before wading into the arguments here. I expect there will be some time before the dust settles completely; I can certainly wait.

• CommentRowNumber46.
• CommentAuthorEmily Riehl
• CommentTimeFeb 9th 2016

Richard, I’m sorry to be so slow to comment. I haven’t had a chance to think about this for the past few weeks.

I’ve just looked at the argument in #41.

One thing that occurs to me is that representations of an element of $F(L)$ as a word in arcs and their inverses are not unique. For instance $a = a b^{-1} c^{-1} b b^{-1} c b$ so in the presence of a crossing as displayed in the figure we get a relation $a \sim b^{-1}cb$. There is a well-defined notion of “minimum word length” but this demonstrates that the relations of type (2) have implications for single arcs as well.

Is this the “subtle but significant oversight” you were referring to, or is it something else?

• CommentRowNumber47.
• CommentAuthorRichard Williamson
• CommentTimeFeb 9th 2016
• (edited Feb 9th 2016)

No problem at all! Thank you very much for taking a look now.

Yes, this is exactly the principal oversight (there is also a small one regarding $N$ having inverses, but that one can be repaired). Well spotted! The fact that $a$ could be equal to something much longer seems to put the argument to the sword for all practical purposes, because both types of ’deletions’ can then be applied in all kinds of complicated ways.

For a simple but illustrative example, which Lou Kauffman drew my attention to, consider a Hopf link in which one component is zero framed, and the other is $\pm 1$ framed. Then a way in which $a$, the arc of the zero framed component, can be ’deleted to become the empty word’ is by first replacing it by $ab^{-1}b$ (or $abb^{-1}$, depending on whether has a $+1$ or $-1$ framing), where $b$ is the arc of the other component which crosses $a$, and then using that $ab^{-1}$ is the longitude of the component to which $b$ belongs, and that $b$ is the longitude of the component to which $a$ belongs.

In any particular example that I have tried, it is possible to find an argument to show that we have must have a link diagram which is equivalent to the link diagram. But the problem of course is to be able to find a systematic way to do it, and I do not see any way to approach this; one of the principal aspects of my approach, as we have discussed, is precisely that it aims to avoid a ’case-by-case argument’ of this nature, which is what 3-manifold topologists expect a proof of the Poincaré conjecture using the Kirby calculus to involve.

One of the virtues of operator equivalence is that whether or not a word is operator equivalent to the identity is independent of its representative in the free group on the arcs of $L$. I did not formulate things quite correctly in this regard, in the passage from ’strict operator equivalence’ to ’operator equivalence’, earlier in the thread; but it can be done. This is one the principal things which the letter I am writing to Lou Kauffman takes care to treat carefully.

In #41, one could try to replace a word by a reduced one before deleting according to (1) and (2). But then, as far as I see, there is no reason for $N$ to be a subgroup (or normal), so that the argument falls through in a different way.

• CommentRowNumber48.
• CommentAuthorEmily Riehl
• CommentTimeFeb 10th 2016

It seems to me that for $F(L)/N$ to be a quotient of $\pi_1(L)$ modulo the longitudes you need $N$ to be the normal subgroup generated by the elements (1) and (2). And in fact, aren’t these quotients are isomorphic? (Confession: I’m not 100% certain I know how $\pi_1(L)$ is defined by I thought it was supposed to be the free group on the arcs modulo the relations (2) for each crossing.)

• CommentRowNumber49.
• CommentAuthorRichard Williamson
• CommentTimeFeb 10th 2016
• (edited Feb 10th 2016)

For $F(L) / N$ to be a quotient of $\pi_{1}$ modulo the longitudes of its components, one certainly needs $N$ to contain these longitudes and to contain words of the form $a_{3} a_{2}^{-1} a_{1}^{-1} a_{2}$, where $a_{1}$, $a_{2}$, and $a_{3}$ come from a crossing as depicted in #41. But other than that one is free to define $N$ as one wishes.

The idea is to find a definition of $N$ which allows us to draw ’controllable’ geometric consequences from $N$ being all of $F(L)$; in particular, from words of length $1$ belonging to $N$. The point being that $N$ as used to define $\pi_{1}(L)$ modulo the longitudes does not allow us to do this; but we could consider an $N$ that in general could be larger, or an $N$ that turns out to be the same, but which has a more useful description. In the case at hand, where $\pi_{1}(L)$ modulo the longitudes is trivial, all possible $N$ must of course be the same (namely all of $F(L)$), but one description could be much more useful than another.

For a short time, I thought that perhaps something that I had thought was a significant aspect of the usefulness of the definition using operator equivalence was not essential; and that one could obtain a useful definition just by defining $N$ as in #41. But the oversight indicates that the $N$ of #41 is no more useful than the one used to define $\pi_{1}(L)$ modulo the longitudes.

Yes, your description of $\pi_{1}(L)$ is correct, and I think that $N$ as defined in #41 will indeed always (i.e. not just when the quotient is trivial) give $\pi_{1}(L)$ modulo the longitudes.

• CommentRowNumber50.
• CommentAuthorRichard Williamson
• CommentTimeFeb 13th 2016
• (edited Feb 13th 2016)

As promised, here is a link to the letter that I wrote to Lou Kauffman.

It is incomplete: (Claim 2) and (Claim 3) are not demonstrated. But I have been quite busy, and decided to send what I had, as it may be that it contains enough of the ideas for a fruitful discussion to be able to be had.

I also outlined the proof of (Claim 2) and (Claim 3) over email. I have copied this below. It is formatted in the same way as an email, in order that the ascii figures display correctly!

Any questions, comments, and feedback are very welcome. I am happy to elaborate on any points. Over time, I would like to use the nLab page a diagrammatic proof of the Poincaré conjecture to complete the details of this (what is there at the moment could be removed; it will still be archived in the revision history). If anybody wishes to help out with, or make a start on, that, for example in the course of trying to follow the argument, they are very welcome.

(Claim 2). We need four facts.

i) A concatenation of words which are operator equivalent to the
identity is operator equivalent to the identity.

ii) The inverse of a word which is operator equivalent to the
identity is operator equivalent to the identity.

iii) The empty word is operator equivalent to the identity.

iv) If w is op. equiv to the identity, then so is a^-1 w a for
any a.

I think that i) is quite clear. One of the principal aspects of
the definition of operator equivalence, allowing a 'box' to the
left of the arc a, is to ensure that i) holds.

For ii), switch the orientation of the component to which the arc
a belongs (note that I allow op. equiv. to the identity to be
exhibited by any choice of orientation), and appeal to the fact
that all occurrences of a or a^-1 are removed, so that none of
the vertical arcs can be a (and hence cannot belong to the same
component as a, because this component consists exactly of a and
the horizontal arcs between the vertical ones). One needs to be
careful about the boxes, but there is no problem.

We have that iii) holds by prescription, namely (A).

For iv), use essentially the same argument as is used to prove
Claim 1.

(Claim 3). We need two facts.

i) Given a crossing

↑
|
←------ | -------
a_3  |   a_1
|
|
a_2

we have that a_3 a_2^-1 a_1^-1 a_2 is op. equiv. to the identity.

ii) The longitude of any component of L is op. equiv. to the
identity.

For i), just slide any arc a under the crossing, to look as
follows, with an appropriate orientation of a (applying a framed
R1 move first if needed).

a  ↑
--- | ---
|    |
a      |    |
----    |    |
|   |    |
|   |    |
←------ | -------
|   |    |
|   |    |
--- | ---
|

For ii), we give essentially the argument that we have discussed
before, involving a handle slide. The first step is to observe
that if a longitude of a component, calculated by starting at an
arc a, contains occurrences of a or a^-1, we can manipulate L so
that the longitude of this component does not contain occurrences
of a or a^-1 except possibly for a twist at the ends.  For any
other arc b, we can then apply a handle slide (after first
sliding it using R2 and R3 moves if needed) to this figure.  Note
again that we remove all occurrences of a and a^-1 in the
definition of operator equivalence in case (B), so that the
figure after applying the handle slide move has a local piece
that is of the required form.  Again, we have to be careful about
the boxes, but there is no problem.

There is only one exceptional case to consider here in ii),
namely when the longitude is exactly a or a^-1 (after applying R2
moves). In that case, (B) cannot hold for any b, because of the
condition that n ≥ 1. But the only way in which a or a^-1 can
occur is in a component which is a ±1 framed unknot, which we can
delete, so that we can show that our link diagram is equivalent
to one with strictly fewer components; and thus every word is
operator equivalent to the identity (in particular, all
longitudes).

• CommentRowNumber51.
• CommentAuthorRichard Williamson
• CommentTimeJun 21st 2016
• (edited Jun 21st 2016)

After the last comment in this thread, things lay in limbo for several months. Lou Kauffman was busy with teaching duties, and did not find the time I believe to look into the letter. I did not myself find the motivation to return to the argument until recently. I have now been going through it with my former student, Reidun Persdatter Ødegaard. Prompted by these discussions, I have found a way to simplify one aspect of the letter. In addition, I have tweaked various other things, and am optimistic that the argument has reached a watertight form.

I have begun writing down this latest version of the argument at the page a diagrammatic proof of the Poincaré conjecture that I began back in January. I should be able to complete this over the next few days. As it states at the page, a plain text version of the argument is available by email in the meantime, should anybody be impatient.

• CommentRowNumber52.
• CommentAuthorUrs
• CommentTimeJun 22nd 2016
• (edited Jun 22nd 2016)

While this is under construction, would it make sense to modify the first sentence in the entry from

We present a short and simple … proof of the Poincaré conjecture

to something more cautious? Maybe: “The following is meant to develop a short and simple … proof of the Poincaré conjecture”

?

• CommentRowNumber53.
• CommentAuthorDavidRoberts
• CommentTimeJun 22nd 2016

I wouldn’t object to a sneak peek at the raw text version.

10. You can now view it here. There is one possibility (or, better, family of possibilities) missing from the case $m=0$ of showing that longitudes belong to $N$. It is straightforward to treat this possibility as well; one needs a handle slide move. If you make it through the rest of the argument and I have not yet added it to the nLab page, I will add it to the plain text version :-).

11. Hmm, it does not display too well in my browser, at least; it does not display the unicode correctly. It should work if you download it and open it in some text editor which can cope with unicode (I use vim, but I think almost anything should do). Let me know if you don’t have any luck with this, and I’ll send it directly to you over email.

12. Thanks for the suggestion, Urs. But for now I think I’ll leave it, since there is a message near the top that the article is under construction. The proof seems to be holding up, so I do wish to claim to have a proof; should it prove to be erroneous, I will of course retract the claim.

• CommentRowNumber57.
• CommentAuthorDavidRoberts
• CommentTimeJun 22nd 2016

@Richard, got it. :-)

13. It turns out that there is a serious problem with the argument; in fact, as is often the case, the actual error is in one of the simplest points, but the problem is more fundamental. I’ve now renamed the nLab page to Poincaré conjecture - diagrammatic formulation, and left only things that are known there, so that hopefully something useful for the nLab comes out of this. The page could of course benefit from some links, and some of the content could probably be better placed on other pages.

I still think that there are good ideas in the approach that I outlined, but at present I do not see how to get from them to a proof of the Poincaré conjecture, even in a special case (such as for knots).

• CommentRowNumber59.
• CommentAuthorDavidRoberts
• CommentTimeJun 26th 2016

Oh well – fun while it lasted :-/

14. Indeed!