Every orientable 3-manifold can be obtained from the 3-sphere by doing surgery along a framed link. Kirby's theorem says that the surgery along two framed links gives homeomorphic manifolds if and only if the links can be related by a sequence of Kirby moves and isotopies. This is pretty similar to Reidemeister's theorem, which says that two link diagrams correspond to isotopic links if and only if they can be related by a sequence of plane isotopies and Reidemeister moves.

Note however that Kirby moves, as opposed to the Reidemeister moves, are not local: the second Kirby move involves changing the diagram in the neighborhood of a whole component of the link. In "On Kirby's calculus", Topology 18, 1-15, 1979 Fenn and Rourke gave an alternative version of Kirby's calculus. In their approach there is a countable family of allowed transformations, each of which looks as follows: replace a $\pm 1$ framed circle around $n\geq 0$ parallel strands with the twisted strands (clockwise or counterclockwise, depending on the framing of the circle) and no circle. Note that this time the parts of the diagrams that one is allowed to change look very similar (it's only the number of strands that varies), but still there are countably many of them.

I would like to ask if this is the best one can do. In other words, can there be a finite set of local moves for the Kirby calculus? To be more precise, is there a finite collection $A_1,\ldots A_N,B_1,\ldots B_N$ of framed tangle diagrams in the 2-disk such that any two framed link diagrams that give homeomorphic manifolds are related by a sequence of isotopies and moves of the form "if the intersection of the diagram with a disk is isotopic to $A_i$, then replace it with $B_i$"?

I vaguely remember having heard that the answer to this question is no, but I do not remember the details.

  • 1
    $\begingroup$ This is a great question! You actually learn a lot from reading it. And the level of difficulty looks ideal for an MO question. $\endgroup$
    – Gil Kalai
    Commented Jul 8, 2010 at 18:59
  • $\begingroup$ Gil -- thanks. Glad you liked the question. $\endgroup$
    – algori
    Commented Jul 8, 2010 at 20:44
  • 3
    $\begingroup$ In a sense Kirby moves are local, just not local in the sense of diagrams. They're local for surgery presentations. If you think of surgery presentations as describing handle attachments on a $4$-ball, handle attachments come from Morse functions on the total manifold (after attachments) and moving from one surgery presentation to another amounts to moving from one Morse function to a neighbouring Morse function. I don't know the answer to your actual question although I think several people have thought about this. $\endgroup$ Commented Jul 15, 2010 at 10:23
  • $\begingroup$ Ryan -- yes, indeed. By the way, did anyone describe an analog of Kirby's calculus for general manifolds in a given oriented cobordism class? $\endgroup$
    – algori
    Commented Jul 15, 2010 at 16:15
  • 1
    $\begingroup$ A full analogy would be a little rough to pull off -- 3-manifolds have the advantage that the surgery presentations can be made to only have 2-handles. In high dimensions you tend to have combinations of handle dimensions so you'd have more of a "surgery sequence of diagrams" than a "surgery diagram". So there'd be relations between surgery sequences corresponding to handle cancellations, and relations coming from "changing the cobordism". But I imagine you could say something. I don't know what may have been done in this direction. $\endgroup$ Commented Jul 16, 2010 at 12:30

3 Answers 3


There is a finite set of local moves. For instance, these:

Finite set of moves
(source: unipi.it)

In the second row, the number of encircled vertical strands is $n\leqslant 3$ on the left and $n\leqslant 2$ on the right move. So they are indeed finite. The bottom-right move is just the Fenn-Rourke move (with $\leqslant 2$ strands). The box is a full counterclockwise twist. We also add all the corresponding moves with $-1$ instead of $+1$.

We can prove that these moves generate the Fenn-Rourke moves as follows. Consider as an example the Fenn-Rourke move with 3 strands:

(source: unipi.it)

To generate this move, we first construct a chain of 0-framed unknots as follows:

(source: unipi.it)

Then we slide the vertical strands along the chain:

(source: unipi.it)

Now it is sufficient to use the Fenn-Rourke move with 2 strands, slide, and use another Fenn-Rourke with one strand:

small FR
(source: unipi.it)

Finally, iterate this procedure 3 times:

(source: unipi.it)

and you are done. The same algorithm works for the general Fenn-Rourke move with $n$ strands.

EDIT I have slightly expanded the proof and posted it in the arXiv as https://arxiv.org/abs/1102.1288

  • 2
    $\begingroup$ @Bruno: That is because your images are all .pdf, which are not universally supported as embedded web images. You need to use .jpg, .gif, .png for everyone to see your images. $\endgroup$ Commented Jan 13, 2011 at 13:28
  • 1
    $\begingroup$ Thanks a lot, I converted them into png. I would suggest to write this suggestion also in the FAQs, that could be helpful (but maybe it's already there and I didn't look at the appropriate page) $\endgroup$ Commented Jan 13, 2011 at 13:57
  • 1
    $\begingroup$ A correction: n=5 is enough, I don't know about n=4 or smaller. $\endgroup$ Commented Feb 8, 2011 at 10:13
  • 2
    $\begingroup$ Very nice! I think your set of moves can be shrunk, by allowing only stabilization (Kirby 1) with $+1$ framing, and only a $-1$ Fenn-Rourke move. See Figure 13 of Ning Lu's paper, cited in my answer, where he credits this observation to Lickorish. $\endgroup$ Commented Feb 8, 2011 at 14:09
  • 2
    $\begingroup$ I am glad that you like it. Thank you for the reference, I agree that we can eliminate from the list all Fenn-Rourke moves with sign $+1$ except the stabilization. I don't know if we can do more. $\endgroup$ Commented Feb 9, 2011 at 13:40

If you didn't like my first answer, here's a different one; again, slightly changing the question. This answer is better suited to the way the Kirby theorem is used in quantum topology. Consider the space KTG of framed oriented knotted trivalent graphs, modulo four operations:

  1. Switching the orientation of an edge
  2. Edge deletion.
  3. The unzip operation- see here for example.
  4. Connect sum.
Dylan Thurston proved that KTG is finitely generated by two elements- the tetrahedron with its two possible vertex-orientations.
You can realize a band-slide (Kirby II) by unzipping a KTG in two different ways. The unzip is an honest local move. One application is that band-slides become a well-defined move, both on the topological level (for links the band-slide isn't well-defined because it depends how you bring together the arc to slide, and the arc it slides over), and indeed even on the level of Jacobi diagrams.
The full story is work-in-progress by Bar-Natan and Dancso. In keeping with Dror's habit of mathematical open-ness, they posted a rough draft version (Wayback Machine). See Page 13 for how to realize a band-slide with unzips.
Thus, my answer this time is:
Yes, you can realize Kirby moves locally if you extend to KTG. And quantum topologically, extending to KTG is probably conceptually the right thing to do.
I don't know the corresponding 4-dimensional picture, although I have my fantasies.
  • 1
    $\begingroup$ Dear Daniel -- it's not that I didn't like your answer(s). Far from it. Thanks for the information and the references. They are very interesting indeed; it's just that I don't think they completely settle the question as it is stated, which is why I'm keeping it open for the moment. $\endgroup$
    – algori
    Commented Jan 2, 2011 at 17:42
  • 1
    $\begingroup$ I completely agree. I also want to know the answer to this question! The two answers I gave are close thoughts I hoped could be useful- but they don't answer the question, only various closely related questions. $\endgroup$ Commented Jan 2, 2011 at 19:44

I think that the answer to your question is yes EDIT: if you allow local to mean "local within a thickened surface" or allow local moves between tangles whose strands may not be part of the surgery link. So my answer is "yes to a slightly modified version of your question".
One idea is that the Kirby theorem follows from your favourite finite presentation of the mapping class group (say one of Wajnryb's), and that you can translate there and back between surgery presentations and Heegaard splittings of $3$-manifolds.
In one direction, a mapping class on a Heegaard surface $H\subset S^3$ is generated by Dehn twists, each of which can be realized by surgery along the curve along which you are twisting, or rather the inclusion of that curve into $S^3$. Thicken $H$ to $H\times I$, and push each curve off to a different height $H\times \{t_i\}$. Include the curves in $S^3$, and there's your surgery link. In the other direction, project your surgery link along a surface $H$ so that each component becomes a simple closed curve (the boundary of a thickened Seifert surface of the link provides one good surface), and you have surgery along those curve realized as a bunch of Dehn twists, hence a mapping class of $H$.
By the Reidemeister-Singer theorem, any two Heegaard splittings of a 3-manifold are stably equivalent, so adding a $\pm 1$-framed unknot away from the surgery presentation is your first local move. Then, you have all the moves that are induced by the relations in your favourite finite presentation of the mapping class group. Write the left hand side of the relation as one framed link, the right-hand side as another, set them equal, and voila. It isn't pretty though. EDIT: As Ian Agol commented, these latter moves are local within a thickened surface, but not necessarily in a ball.
You can find pictures of the local moves in Ning Lu's paper or in Matveev and Polyak's paper. In one direction, as proved in both papers, the Kirby moves generate these moves. In the other direction, the fact that they come from a finite presentation of the mapping class group tells you that they generate the Kirby moves.
EDIT: If you want local moves in a ball, Matveev-Polyak gives a tangle presentation of the mapping class group, and Section 5 tells you how to translate there and back between this and a surgery presentation. Roughly, you remove regular neighbourhoods of strands whose endpoints are at the bottom, and the plat closure of what you are left with is the surgery link. A complete set of local moves between such tangles is Figures 12-19 of that paper. Very similar constructions appear for example in papers of Habiro. Anyway, there is a complete set of local moves in a ball between tangles, and a clear easy algorithm to translate there and back from such tangles to surgery presentations.

  • 4
    $\begingroup$ Hmmm, I'm not sure these are local in the sense of the question. The Hatcher-Thurston relations lie in subsurfaces which are a surface of bounded complexity. So the relation occurs in a thickening of this surface inside S^3. However, this is not inside of a ball (so not a tangle of bounded complexity), since there may be other strands of the surgery link coming through. $\endgroup$
    – Ian Agol
    Commented Sep 3, 2010 at 3:53
  • $\begingroup$ Fair enough. The moves I just described are local in the thickened surface, just not in a ball. However, you can write them as local moves between tangles inside a ball- see Figures 12-19 of Matveev-Polyak, where they do just that. The tangles are not themselves part of the surgery link (at least, not all strands are), but they seem a good substitute, and there are clear algorithms to translate both ways, as in Matveev-Polyak. I'll edit this into my answer later. So I suppose my answer is "if you slightly modify the question, then yes". $\endgroup$ Commented Sep 3, 2010 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.