Fibered knots and Property 2R
Abstract.
It is shown, using sutured manifold theory, that if there are any component counterexamples to the Generalized Property R Conjecture, then any knot of least genus among components of such counterexamples is not a fibered knot.
The general question of what fibered knots might appear as a component of such a counterexample is further considered; much can be said about the monodromy of the fiber, particularly in the case in which the fiber is of genus two.
1. Introductory remarks
Recall the famous Property R theorem, proven in a somewhat stronger form by David Gabai [Ga2]:
Theorem 1.1 (Property R).
If framed surgery on a knot yields then is the unknot.
There is a natural way of trying to generalize Theorem 1.1 to links in . In fact, there are several ways in which it can be generalized, but in the discussion here we restrict to the least complex one, known as the Generalized Property R Conjecture (see Conjecture 1 below). Other options are described in [GSch] and briefly below. The interest in this conjecture, as in the case of the original Property R Conjecture, is motivated in part by manifold questions. The viewpoint taken here derives almost entirely from 3manifold theory, in particular new insights that can be brought to the question by developments in sutured manifold theory beyond those used by Gabai in his original proof of Property R.
Unless explicitly stated otherwise, all manifolds throughout the paper will be compact and orientable.
2. Handle slides and Generalized Property R
To make sense of how Property R might be generalized, recall a small bit of manifold handlebody theory [GS]. Suppose is a link in a manifold and each component of is assigned a framing, that is a preferred choice of cross section to the normal bundle of the component in . For example, if , a framing on a knot is determined by a single integer, the algebraic intersection of the preferred crosssection with the longitude of the knot. (In an arbitrary manifold a knot may not have a naturally defined longitude.) Surgery on the link via the framing is standard Dehn surgery (though restricted to integral coefficients): a regular neighborhood of each component is removed and then reattached so that the meridian is identified with the crosssection given by the framing. Associated to this process is a certain manifold: attach dimensional handles to along , using the given framing of the link components. The result is a dimensional cobordism, called the trace of the surgery, between and the manifold obtained by surgery on . The collection of belt spheres of the handles constitute a link called the dual link; the trace of the surgery on can also be viewed as the trace of a surgery on .
The manifold trace of the surgery on is unchanged if one handle is slid over another handle. Such a handle slide is one of several moves allowed in the Kirby calculus [Ki1]. When the handle corresponding to the framed component of is slid over the framed component of the effect on the link is to replace by the band sum of with a certain copy of , namely the copy given by the preferred crosssection given by the framing of .
If is there is a simple formula for the induced framing on the new component . Suppose give the framing of the original components and respectively, and is the algebraic linking number of the components and in . Then the framing of the new component that replaces is given by the formula [GS, p.142]:
Any statement about obtaining manifolds by surgery on a link will have to take account of this move, which we continue to call a handleslide, in deference to its role in dimensional handle theory.
Suppose is obtained from components and by the handleslide of over as described above. Let and be the dual knots to and . It will be useful to note this counterintuitive but elementary lemma:
Lemma 2.1.
The link in that is dual to is , where is obtained by a handleslide of over .
Proof.
It suffices to check this for the simple case in which the manifold is a genus handlebody, namely a regular neighborhood of , , and the arc between them along which the bandsum is done. A sketch of this is shown in Figure 1. The dual knots , and are displayed as boundaries of meridian disks for regular neighborhoods of , and respectively.
Alternatively, a dimensional schematic of the dimensional process is shown in Figure 2. The handle corresponding to is shown halfway slid across the handle corresponding to . Each disk in the figure is given the same label as its boundary knot in or as appropriate. ∎
Let denote the connected sum of copies of . The Generalized Property R conjecture (see [Ki2, Problem 1.82]) says this:
Conjecture 1 (Generalized Property R).
Suppose is an integrally framed link of components in , and surgery on via the specified framing yields . Then there is a sequence of handle slides on that converts into a framed unlink.
In the case no slides are possible, so Conjecture 1 does indeed directly generalize Theorem 1.1. On the other hand, for it is certainly necessary to include the possibility of handle slides. Figure 3 shows an example of a more complicated link on which framed surgery creates . To see this, note that the Kirby move shown, bandsumming the square knot component to a copy of the unknotted component, changes the original link to the unlink of two components, for which we know surgery yields . Even more complicated links with this property can be obtained, simply by using Kirby moves that complicate the link rather than simplify it. See Figure 4; the free ends of the band shown can be connected in an arbitrarily linked or knotted way.
The conjecture can be clarified somewhat by observing that the only framed links that are relevant are those in which all framings and linking numbers are trivial. There is a straightforward dimensional proof, using the intersection pairing on the trace of the surgery. Here is an equally elementary dimensional proof:
Proposition 2.2.
Suppose is a framed link of components in , and surgery on via the specified framing yields . Then the components of are algebraically unlinked and the framing on each component is the framing.
Proof.
It follows immediately from Alexander duality that In particular, filling in the solid tori via whatever framing we are given yields an epimorphism, hence an isomorphism . For each torus component of , the filling will kill some generator of , so the homomorphism is not injective. It follows that the homomorphism cannot be injective and, moreover, must contain the framing curve. But must be contained in the subgroup generated by the standard longitude, since this is the only subgroup that is trivial when we just replace all the other components of . It follows that the framing at each component is that of the standard longitude, namely the framing. Since the longitude of each is nullhomologous in it follows that all linking numbers are trivial. ∎
There is also an immediate topological restriction on the link itself, which carries over to a restriction on the knots can appear as individual components of such a link:
Proposition 2.3.
Suppose is a framed link of components in , and surgery on via the specified framing yields . Then bounds a collection of smooth  disks in a dimensional homotopy ball bounded by .
An equivalent way of stating the conclusion, following Freedman’s proof of the dimensional topological Poincare Conjecture [Fr], is that (and so each component of ) is topologically slice in .
Proof.
Consider the manifold trace of the surgery on . has one end diffeomorphic to and the other end, call it , diffeomorphic to . has the homotopy type of a oncepunctured . Attach to via the natural identification . The result is a homotopy ball, and the cores of the original handles that are attached to are the required disks. ∎
Somewhat playfully, we can turn the Generalized Property R Conjecture, which is a conjecture about links, into a conjecture about knots, and also stratify it by the number of components, via the following definition and conjecture.
Definition 2.4.
A knot has Property nR if it does not appear among the components of any component counterexamples to the Generalized Property R conjecture.
Conjecture 2 (Property nR Conjecture).
All knots have Property nR.
Thus the Generalized Property R conjecture for all component links is equivalent to the Property nR Conjecture for all knots. Following Proposition 2.3 any nonslice knot has Property nR for all . The first thing that we will show (Theorem 3.3) is that if there are any counterexamples to Property 2R, a least genus such example cannot be fibered. We already know that both of the genus one fibered knots (the figure 8 knot and the trefoil) cannot be counterexamples, since they are not slice. So these simplest of fibered knots do have Property 2R. On the other hand, for reasons connected to difficulties proving the AndrewsCurtis Conjecture, there is strong evidence (see [GSch]) that Property 2R may fail for as simple a knot as the square knot. Since the square knot is fibered, it would then follow from Theorem 3.3 that there is a counterexample to Property 2R among genus one knots.
3. Special results for Property 2R
Almost nothing is known about Generalized Property R, beyond the elementary facts noted in Propositions 2.2 and 2.3 that the framing and linking of the components of the link are all trivial and the links themselves are topologically slice. A bit more is known about Property 2R. The first was shown to us by Alan Reid:
Proposition 3.1 (A. Reid).
Suppose is a component link with tunnel number . If surgery on gives then is the unlink of two components.
Proof.
The assumption that is tunnel number means that there is a properly embedded arc so that is a genus handlebody . Let . There is an obvious epimorphism (fill in a meridian disk of ) and an obvious epimorphism (fill in solid tori via the given framing). But any epimorphism is an isomorphism, since free groups are Hopfian, so in fact . It is then a classical result that must be the unlink on two components. ∎
This first step towards the Property 2R conjecture is a bit disappointing, however, since handleslides (the new and necessary ingredient for Generalized Property R) do not arise. In contrast, Figure 3 shows that handle slides are needed in the proof of the following:
Proposition 3.2.
The unknot has Property 2R.
Proof.
Suppose is the union of two components, the unknot and another knot , and suppose some surgery on gives . Following Proposition 2.2 the surgery is via the framing on each and, since is the unknot, framed surgery on alone creates . Moreover, the curve that is dual to is simply for some point .
A first possibility is that is a satellite knot in , so lies in a solid torus in such a way that the torus is essential in . Since there is no essential torus in , compresses after the surgery on . Since contains no summand with finite nontrivial homology, it follows from the main theorem of [Ga3] that is a braid in and that surgery on has the same effect on as some surgery on . Proposition 2.2 shows that the surgery on must be along a longitude of , but that would imply that has winding number in . The only braid in a solid torus with winding number is the core of the solid torus, so in fact is merely a core of and no satellite. So we conclude that cannot be a satellite knot.
Consider the manifold . If is irreducible, then it is a taut sutured manifold (see, for example, [Ga1]) and two different fillings (trivial vs. framed) along yield reducible, hence nontaut sutured manifolds. This contradicts [Ga1]. We conclude that is reducible. It follows that is isotopic in to a knot lying in a ball in and that surgery on creates a summand of the form . By Property R, we know that is the unknot in . Hence is the unlink of two components.
The proof, though, is not yet complete, because the isotopy of to in may pass through . But passing through can be viewed as bandsumming to the boundary of a meridian disk of in . So the effect in is to replace with the band sum of with a longitude of . In other words, the knot , when viewed back in , is obtained by from by a series of handle slides over , a move that is permitted under Generalized Property R. ∎
In a similar spirit, the first goal of the present paper is to prove a modest generalization of Proposition 3.2. A pleasant feature is that, since the square knot is fibered, Figure 4 shows that the proof will require handle slides of both components of the link.
Theorem 3.3.
No smallest genus counterexample to Property 2R is fibered.
Proof.
Echoing the notation of Proposition 3.2, suppose there is a component counterexample to Generalized Property R consisting of a fibered knot and another knot . Let be the manifold obtained by framed surgery on alone. Since is a fibered knot, fibers over the circle with fiber , a closed orientable surface of the same genus as . The dual to in is a knot that passes through each fiber exactly once.
The hypothesis is that framed surgery on creates . Following [ST, Corollary 4.2], either the knot lies in a ball, or is cabled with the surgery slope that of the cabling annulus, or can be isotoped in to lie in a fiber, with surgery slope that of the fiber. If were cabled, then the surgery on would create a Lens space summand, which is clearly impossible in . If can be isotoped into a ball or into a fiber, then, as argued in the proof of Proposition 3.2, the isotopy in is realized in by handleslides of over , so we may as well regard as lying either in a ball that is disjoint from or in a fiber . The former case, in a ball disjoint from would, as in Proposition 3.2, imply that the link is the unlink. So we can assume that .
The surgery on that changes to has this local effect near : is cut open along , creating two copies , a handle is attached to the copy of in each of , compressing the copies of the fiber to surfaces . The surfaces are then glued back together by the obvious identification to give a surface . (See the Surgery Principle Lemma 4.1 below for more detail.) This surface has two important features: each component of (there are two components if and only if is separating in ) has lower genus than ; and intersects in a single point.
Let be the dual knot to and let be the component of that intersects . intersects in some collection of points (in fact, two points, but that is not important for the argument). Each point in can be removed by a handleslide of over along an arc in . Let be the final result of these handleslides. Then is an orientable surface that has lower genus than , is disjoint from and intersects in a single point.
Following Lemma 2.1 the handleslides of over in correspond in to handleslides of over . Call the knot in that results from all these handleslides . Since is disjoint from , and intersects in a single meridian, is a surface in whose boundary is a longitude of . In other words, the knot , still a counterexample to Property 2R, has
as required. ∎
4. Fibered manifolds and Heegaard splittings
We have just seen that a fibered counterexample to Property 2R would not be a least genus counterexample. We now explore other properties of potential fibered counterexamples. In this section we consider what can be said about the monodromy of a fibered knot in , and the placement of a second component with respect to the fibering, so that surgery on the component link yields . Perhaps surprisingly, the theory of Heegaard splittings is useful in answering these questions. Much of this section in fact considers the more general question of when can be created by surgery on a knot in a manifold that fibers over a circle. The application to Property 2R comes from the special case in which the manifold is obtained from framed surgery on a fibered knot in .
Suppose is a surface in a manifold and is an essential simple closed curve in . A tubular neighborhood intersects in an annulus; the boundary of the annulus in defines a slope on . Let denote the manifold obtained from by surgery on with this slope and let be the surface obtained from by compressing along .
Lemma 4.1 (Surgery Principle).
can be obtained from by the following step process:

Cut open along , creating two new surfaces in the boundary, each homeomorphic to .

Attach a handle to each of along the copy of it contains. This changes each of the new boundary surfaces to a copy of . Denote these two surfaces .

Glue to via the natural identification.
Proof.
The surgery itself is a step process: Remove a neighborhood of , then glue back a copy of so that is the given slope. The first step is equivalent to cutting along an annulus neighborhood of in , creating a torus boundary component as the boundary union of the two copies of . Thinking of as the boundary union of two intervals, the second step can itself be viewed as a twostep process: attach a copy of to each annulus along (call the attached copies ), then identify the boundary disks with in the natural way. This creates a threestage process which is exactly that defined in the lemma, except that in the lemma is first cut apart and then reglued by the identity. ∎
The case in which fibers over a circle with fiber is particularly relevant. We will stay in that case throughout the remainder of this section (as always, restricting to the case that and are orientable) and use the following notation:

is the monodromy homeomorphism of .

is an essential simple closed curve in .

is the surface obtained by compressing along

is the manifold obtained by doing surgery on along using the framing given by .
Note that may be disconnected, even if is connected.
Proposition 4.2.
Suppose is isotopic to in

If is nonseparating in , or if is separating and the isotopy from to reverses orientation of , then , where fibers over the circle with fiber .

If separates so , and the isotopy from to preserves orientation of , then , where each fibers over the circle with fiber .
Proof.
We may as well assume that and consider first the case where is orientation preserving. In this case, the mapping cylinder of in is a torus containing . The stage process of Lemma 4.1 then becomes:

is cut along to give a manifold with two torus boundary components. fibers over the circle with fiber a twicepunctured . ( is connected if and only if is nonseparating.)

A handle is attached to each torus boundary component , turning the boundary into two spheres.

The two spheres are identified.
The second and third stage together are equivalent to filling in a solid torus along each , giving an fibered manifold , then removing a ball from each solid torus and identifying the resulting spheres. Depending on whether is connected or not, this is equivalent to either adding to or adding the two components of together.
The case in which is orientation reversing is only slightly more complicated. Since is orientable, the mapping cylinder of is a sided Klein bottle , so is a single torus . The argument of Lemma 4.1 still mostly applies, since has an annulus neighborhood in , and shows that the surgery can be viewed as attaching two handles to along parallel curves, converting the boundary into two spheres, then identifying the spheres. This is again equivalent to filling in a solid torus at (which doublecovers ) and then adding . But filling in a solid torus at changes the fiber from to . (Note that if separates , so , then since is orientation preserving on but orientation reversing on , must exchange the . So also fibers over the circle with fiber .) ∎
Corollary 4.3.
If and is isotopic to in , then is a torus.
Proof.
According to Proposition 4.2, the hypotheses imply that fibers over the circle with fiber (a component of) . But this forces and so . ∎
Surgery on fibered manifolds also provides a natural connection between the surgery principle and Heegaard theory:
Definition 4.4.
Suppose , are two copies of a compression body and is a given homeomorphism. Then the union of , along their boundaries, via the identity on and via , is called the Heegaard double of (via ).
Lemma 4.1 gives this important example:
Example 4.5.
For as above, let be the compression body obtained by attaching a handle to along . Then is the Heegaard double of via .
Note that the closed complement of in any Heegaard double is a manifold with Heegaard splitting . Here is a sample application, using Heegaard theory:
Proposition 4.6.
For as above, suppose some surgery on gives a reducible manifold. Then the surgery slope is that of and either

can be isotoped in so that it is disjoint from or

is nonseparating and , where

fibers over the circle with fiber and

is either or a Lens space.

Note in particular that possibility (2) is not consistent with .
Proof.
Choose distinct fibers in , with . Via [ST, Corollary 4.2] and the proof of Theorem 3.3 we know that the surgery on must use the framing given by the fiber , so the result of surgery is . Example 4.5 shows that is a Heegaard double via , so the complement of a regular neighborhood of has a Heegaard splitting . That is, .
If , so , then . Since , the Heegaard splitting of is of genus , so is either , a Lens space, or . But the last happens only if the same curve in compresses in both and ; in our context, that implies and are isotopic in , and so can be isotoped to be disjoint.
If , choose a reducing sphere with a minimal number of intersection curves with . If the reducing sphere is disjoint from , then is reducible. If the reducing sphere intersects , then at least one copy of in must be compressible in . We conclude that in either case the Heegaard splitting of is weakly reducible (and possibly reducible), see [CG]. That is, there are essential disjoint simple closed curves in which compress respectively in and .
Case 1: The curve is separating.
In this case, since the compression bodies each have only the handle with boundary attached, any curve in that compresses in is isotopic to . In particular, fixing the identification , must represent in and represents . Hence and are disjoint.
Case 2: The curve is nonseparating, and so is at least one of the curves .
If both curves are nonseparating then, as in Case 1, and , when viewed in the handlebodies , must each be isotopic to and the case concludes as Case 1 did.
If is nonseparating, and is separating, then is isotopic to whereas bounds a punctured torus on which lies. If is disjoint from , then and are disjoint, as required. If lies in then also bounds a disk in . The union of the disks in and bounded by is a sphere that decomposes into . This implies that , where fibers over with fiber and is Heegaard split by into two solid tori, with meridian disks bounded by and respectively. If then is a Lens space. If then . If then is disjoint from .
Case 3: The curve is nonseparating, but both are separating.
In this case, much as in Case 2, each cuts off a torus from , with and . Since the are disjoint, the two tori either also are disjoint (and the proof is complete) or the two tori coincide. If the two tori coincide, the argument concludes as in Case 2. ∎
5. Could there be fibered counterexamples of genus ?
In applying Proposition 4.6 to the manifold obtained from framed surgery on a fibered knot , note that the isotopy in the Proposition takes place in a fiber of , the closed manifold obtained by framed surgery on , not in the fiber of the knot itself. The importance of the distinction is illustrated by the following Proposition which, without the distinction, would (following Propositions 3.2 and 4.6) seem to guarantee that all genus fibered knots have Property 2R.
Proposition 5.1.
Suppose is a fibered knot, with fiber the punctured surface and monodromy . Suppose a knot has the property that framed surgery on the link gives and can be isotoped to be disjoint from in . Then either is the unknot or .
Proof.
Case 1: bounds a disk in or is parallel in to .
In this case, framed surgery on would be , where is the result of framed surgery on . Our hypothesis is that which, by classical Property R [Ga2], implies that is the unknot. Hence .
Case 2: is essential in .
If is a punctured torus, then the fact that is essential and can be isotoped off of imply that is isotopic to , and we may as well assume that . The mapping torus of is then a nonseparating torus in , which is absurd.
Suppose is a punctured genus surface, and let denote the closed surface obtained by capping off the puncture. We may as well assume that , and, following Corollary 4.3, is not isotopic to in . In particular, must be nonseparating. Since and are nonseparating and disjoint in , but not isotopic in , if is compressed along both and simultaneously, becomes a disk. Apply the Surgery Principle Lemma 4.1 to and conclude that bounds a disk after framed surgery on . In particular, if is the manifold obtained by framed surgery on alone, then surgery on would give . For this to be would require hence, again by classical Property R, would be the unknot. ∎
Return to the general case of fibered manifolds and surgery on a curve in the fiber, and consider the case in which the fiber has genus two. According to Corollary 4.3, if the result of surgery on is , then is not isotopic to in . The following Proposition is a sort of weak converse.
Proposition 5.2.
For as above, suppose has genus and can be isotoped off of in . If is not isotopic to in then , where is , , or a Lens space.
Proof.
We may as well assume that is disjoint from but not isotopic to in . Since is genus two, this immediately implies that is nonseparating.
Take the Heegaard viewpoint of Example 4.5. The complement of a regular neighborhood of in has a Heegaard splitting , with the splitting surface a fiber not containing . Since can be isotoped off of in , the Heegaard splitting is a weakly reducible splitting, with bounding a disk in and bounding a disk in .
Now do a weak reduction of this splitting. That is, consider the handles with boundary and with boundary in in . Since and are disjoint, can also be regarded as the union of compression bodies and . Each can be regarded as obtained from by attaching a single handle. Moreover it is that is identified with to get . A genus count shows that this new surface is a sphere. Put another way, the manifold is Heegaard split by the torus , so is the connected sum of with a manifold that has a genus one Heegaard splitting. ∎
Corollary 5.3.
For as above, suppose has genus and is reducible.
If is not isotopic to in then , where is , , or a Lens space and is either or a torus bundle over the circle.
Proof.
Via [ST, Corollary 4.2] and the proof of Theorem 3.3 we know that the surgery on must use the framing given by the fiber in which it lies. Apply Proposition 4.6. If the first conclusion holds, and can be isotoped off of in , then Proposition 5.2 can be applied and that suffices. If the second conclusion holds then is nonseparating, so is a torus, as required. ∎
Corollary 5.4.
Suppose is a genus two fibered knot and is a disjoint knot. Then framed surgery on gives if and only if after possible handleslides of over ,

lies in a fiber of ;

in the closed fiber of the manifold obtained by framed surgery on , can be isotoped to be disjoint from ;

is not isotopic to in ; and

the framing of given by is the framing of in .
.
Proof.
Suppose first that framed surgery on gives . Apply [ST, Corollary 4.2] as in the proof of Theorem 3.3 to handleslide over until it lies in the fiber of in a way that the framing on is the framing given by the fiber in which it lies. Proposition 4.6 shows that satisfies the second condition and Corollary 4.3 gives the third: is not isotopic in to .
For the other direction, suppose lies in a fiber of and the four conditions are satisfied. The last condition says that the surgery on is via the slope of the fiber. By Proposition 5.2, the surgery gives , for either , a Lens space, or . But and are unlinked in (push off of ), so framed surgery on must give a homology . This forces to be a homology , hence precisely. ∎
6. Connected sums of fibered knots
There is a potentially useful addendum to Proposition 4.6 in the case that has a separating curve that is invariant under the monodromy. Suppose, as usual, is an orientable closed manifold that fibers over the circle, with fiber and monodromy . Suppose further that there is a separating simple closed curve , with complementary components in , so that and . Let .
Proposition 6.1.
Suppose is a simple closed curve so that framed surgery on in creates and has been isotoped in to minimize . For any element represented by an arc component of , the algebraic intersection satisfies
Proof.
Recall the following standard fact about curves in surfaces: Suppose are simple closed curves in a surface so that neither or can be reduced by an isotopy of or . Then there is an isotopy of in that minimizes via an isotopy that never changes or .
Apply this fact to the curves in . Since , the second conclusion of Proposition 4.6 does not hold, so the first conclusion must: can be isotoped to be disjoint from in . It follows that for any arc component of , there is a proper isotopy of in so that after the isotopy and are disjoint. Connect the ends of in by a subarc of to get a simple closed curve representing . Similarly connect the ends of to get a representative of . Depending on whether the pairs of ends are interleaved in , the resulting circles can be isotoped either to be disjoint or to intersect in a single point. ∎
Proposition 6.1 might give useful information about the monodromy of a connected sum of fibered knots. Suppose the knot is the connected sum of two fibered knots and . Then the knot is also fibered; its fiber is the connected sum of the fibers for and . This structure carries over to the manifold obtained by framed surgery on . If each is the manifold obtained by framed surgery on , with monodromy , then the fiber of is and the monodromy is the connected sum of the along an invariant simple closed curve . The closed curve intersects the fiber of in an invariant arc . The arc can be viewed as the intersection of a decomposing sphere of with the fiber .
With this as background, Proposition 6.1 immediately yields:
Corollary 6.2.
Suppose is a fibered composite knot, and framed surgery on a link creates . Following [ST], handleslide over and isotope so that lies in a fiber of and, among all such positionings, choose one that minimizes in . For any element represented by a component of ,
If the summand is a genus one knot, this puts severe restrictions on the set of possible curves of . For example, suppose is a figureeight knot. Then there is a symplectic basis on with respect to which the monodromy is given by the matrix
For a class to have the property
requires
An elementary descent argument shows that solutions are pairs such that and are successive Fibonacci numbers or and are successive Fibonacci numbers. As many as three of these curves may be present simultaneously: if are successive Fibonacci numbers, then a similar calculation shows that the three successive pairs
in may be represented in the punctured torus by disjoint arcs.
Similarly, for the trefoil knot, there is a symplectic basis on with respect to which the monodromy is given by the matrix
For a class to have the property
requires
This allows only three possible curves:
These three can be represented in the punctured torus by disjoint arcs.
Afterword: It will be shown in [GSch] that a similar analysis gives a precise catalog of all possible curves in the complement of a square knot (up to bandsum with ) so that surgery on gives . It will also be shown that the central example of [Go] gives rise to a likely counterexample to Property 2R in which one of the link components is the square knot. The other component can be precisely described, but it remains a puzzle how, even after bandsums with the square knot, it fits into this catalog. So it also remains mysterious how, via 3.3 it then gives rise to a probable genus one nonfibered counterexample to Property 2R.
References
 [CG] A. Casson and C. McA. Gordon, Reducing Heegaard splittings, Topology and its applications, 27 (1987) 275–283.
 [Fr] M. Freedman, The topology of fourdimensional manifolds. J. Differential Geom. , 17 (1982) 357–453.
 [Ga1] D. Gabai, Foliations and the topology of manifolds. II. J. Differential Geom. 26 (1987) 461–478.
 [Ga2] D. Gabai, Foliations and the topology of manifolds. III, J. Differential Geom. 26 (1987) 479–536.
 [Ga3] D. Gabai, Surgery on knots in solid tori Topology 28 (1989) 16.
 [GSch] R. Gompf and M. Scharlemann, Does the square knot have Property 2R? to appear.
 [GS] R. Gompf and A. Stipsicz, Manifolds and Kirby Calculus, Graduate Studies in Mathematics 20 American Mathematical Society, Providence, RI, 1999.
 [Go] R. Gompf, Killing the AkbulutKirby sphere, with relevance to the AndrewsCurtis and Schoenflies problems Topology 30 (1991), 97–115.
 [Ki1] R Kirby, A calculus for framed links in Invent. Math 45 (1978) 35–56.
 [Ki2] R Kirby, , Problems in LowDimensional Topology, in Geometric Topology, Edited by H. Kazez, AMS/IP Vol. 2, International Press, 1997.
 [ST] M. Scharlemann, A. Thompson, Surgery on knots in surface I, ArXiv 0807.0405