CONSTRUCTIONS OF SMOOTHLY SLICE KNOTS
TETSUYA ABE

1. Abstract
A slice-ribbon conjecture is a long standing conjecture. In this note, we
explain constructions of smoothly slice knots which might be non-ribbon anddiscuss related topics.
The idea of constructions is the following: Let HD be a non-trivial handle
decomposition of the standard 4-ball B 4 and h2 a 2-handle of HD. A slice
knot K is obtained as the belt-sphere of h2 . The cocore disk of h2 is a slice
disk for K. There is no apparent reason for K to be ribbon. Typical examples
are explained in Section 11.
If HD has, at least, two 2-handles, then we can construct more complicatedslice knots. Let h21 and h22 be 2-handles of HD. Let Ki (i = 1, 2) be thebelt-sphere of h2i . Then Ki is a slice knot. Furthermore any band sum of K1
and K2 , denoted by K1 ♯b K2 , is a slice knot1. There is no apparent reason forK1 ♯b K2 to be ribbon. Gompf, Scharlemann and Thompson [GST] gave such aslice knot, which will be explained in Section 10.
1This is because the link consists of K

1 and K2 bounds disjoint smooth disks in B

1

4

.

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The problem is how to obtain a “good” handle decomposition of B 4 . If
we consider a complicated handle decomposition of B 4 , then we will obtaincomplicated slice knots. However it may be too diﬃcult to check whether
these slice knots are ribbon or not. One of the purposes of this note is to give
simple and enough complicated handle decompositions of B 4 explicitly.
2. Notations and organization
Throughout this note, we only consider the smooth category unless otherwise stated. The symbol A ≈ B means that A and B are diﬀeomorphic. We
prefer to use the term“handle calculus, handle diagram” rather than “Kirby
calculus, Kirby diagram”. We sometimes identify a given handle diagram withthe corresponding handle decomposition or the 4-manifold itself represented
by the handle decomposition.
Section 3–Section 12 are based on the talks given by the author in Mini-workshop on knot concordance, Sep. 17-20, 2013 at Tokyo Institute of Technology.
In Section 13, we explain how to obtain ribbon presentations via handle calculus. The rest of this note, we give a brief overview of related topics.
3. 2-handles
We recall some deﬁnitions for the reader who is not familiar with handle
theory.
A (4-dimensional) 2-handle h2 is a copy of D2 ×D2 , attached to the boundaryof a 4-manifold X along ∂D2 × D2 by an embedding ϕ : ∂D2 × D2 → ∂X.
We call ∂D2 × D2 the attaching region of h2 , ∂D2 × {0} the attachingsphere of h2 , D2 × {0} the core of h2 , {0} × D2 the cocore of h2 , {0} × ∂D2
the belt-sphere of h2 . The belt-sphere of h2 is a knot in ∂(X ∪ϕ h2 ) whichbounds a smooth disk(=cocore) in X ∪ϕ h2 . A schematic picture may help us
understanding, see the left picture in Figure 1.

Figure 1. A 2-handle and related terminologies.

CONSTRUCTIONS OF SMOOTHLY SLICE KNOTS

3

Note that the belt sphere of h2 is isotopic to the meridian of the attachingsphere of h2 , see the right side picture in Figure 1. This fact is used to draw
a slice knot in a handle diagram in Section 10.
4. The slice-ribbon conjecture
Let S 3 be the 3-sphere and B 4 the standard 4-ball such that ∂B 4 = S 3 .
A knot K in S 3 is called smoothly slice if it bounds a properly embedded
smooth disk D2 in B 4 . We call D2 a slice disk for K, see Figure 2.

Figure 2. A schematic picture of a slice knot K and a slice disk D2 .
The class of slice knots are conjectured to be that of knots which are deﬁned
3-dimensionally. A knot K in S 3 is called ribbon if it bounds an immersed
disk D in S 3 with only ribbon singularities.
Lemma 4.1. A ribbon knot is slice.
Proof. Let K be a ribbon knot. By deﬁnition, it bounds an immersed disk D2
in S 3 with only ribbon singularities. By pushing intD2 toward the interior of
B 4 , we obtain a slice disk for K.
Here we give a more precise proof. Let N be the color neighborhood of B 4 .
Then N is diﬀeomorphic to S 3 × [0, 1] and S 3 × {0} ≈ S 3 = ∂B 4 . If we deform
D2 as Figure 4, then we obtain a slice disk for K.
□The slice-ribbon conjecture
All slice knots are ribbon.

Figure 3. A ribbon singularity and an example of a ribbon knot.




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S 3 × {1}1
S3 × { }2S 3 × {0} ≈ S 3

Figure 4. A slice disk obtained from an immersed disk.
This conjecture is due to Fox [F2]. I can not guess whether this conjectureis true or not.
Positive direction.
There are some results which suggest that the slice-ribbon conjecture is true.
In 2007, Lisca [Li] proved that the slice-ribbon conjecture is true for twobridge knots by a gauge theoretic method. Note that this result is based on
the work of [CG2]. Further development was done by Greene and Jabuka [GJ]
and Lecuona [Le1], [Le2].
Open problem. Is the slice-ribbon conjecture true for three-bridge knots ?
Negative direction.
There are some results which suggest that the slice-ribbon conjecture is nottrue. Indeed, there exist slice knots which may not be ribbon. Such slice knots
are obtained a byproduct of the study of the smooth Poincaré conjecture indimension four. See also Section 14.
5. The smooth Poincaré conjecture in dimension four
1980’s, Freedman classiﬁed simply-connected closed 4-manifolds. As a corollary, Freedman solve the topological Poincaré conjecture in dimension four.
Theorem 5.1 (The topological Poincaré conjecture in dimension four). Let X
be a topological 4-manifold. If X is homotopic to S 4 , then X is homeomorphic
to S 4 .

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On the other hand, the smooth Poincaré conjecture in dimension four (SPC4)
is not yet solved. The statement is the following.
 The smooth Poincaré conjecture in dimension four (SPC4)

If a smooth 4-manifold X is homeomorphic to S 4 , then X isdiﬀeomorphic to S 4 .
This conjecture is one of the biggest unsolved problems in low dimensional
topology. Note that Donaldson proved that there exists an exotic R4 = S 4 \{pt}, which suggests that there might exist an exotic S 4 . Furthermore it iswell known that there exist uncountably many exotic R4 ’s due to Taubes.
6. Cappell-Shaneson (homotopy) 4-spheres
For the matrix




0 1
0
1  ∈ SL(3, Z),
An =  0 1
1 0 n+1

Cappell and Shaneson2 associated a homotopy 4-sphere Σn . The homotopy
4-sphere Σn is obtained as follows: First, consider the mapping torus of thepunctured 3-torus T03 with the diﬀeomorphism induced by An . Then by gluinga S 2 × D2 to it with the non-trivial diﬀeomorphism of S 2 × S 1 , we obtain Σn .
In 1991, Gompf [G1] proved that
Σ0 ≈ S 4 .
For a long time, many people thought that Σ1 might be an exotic S 4 . In 2010,
Akbulut [A1] proved thatΣn ≈ S 4 .
7. Another story: an idea to disprove SPC4This is a story before Akbulut’s work [A1] and this section can be skipped.
Let W be a homotopy 4-ball with ∂W ≈ S 3 . If there exists a knot K in S 3such that• K bounds a smooth disk in W , and• K does not bound a smooth disk in B 4 ,2They constructed more homotopy 4-spheres, see [G2]. In this note, we only consider Σ .
n

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then W is an exotic 4-ball. In particular, the 4-manifold W ∪id B 4 is an exotic
4-sphere 3. In [FGMW], Freedman, Gompf, Morrison and Walker constructeda candidate of such a knot.
The construction is the following: Consider Σ1 and its handle decompositionh0 ∪ h11 ∪ h12 ∪ h21 ∪ h22 ∪ h4
given by Gompf in [G1], where h0 is a 0-handle, h1i (i = 1, 2) is a 1-handle, h2j(j = 1, 2) is a 2-handle, h4 is a 4-handle. Note thath0 ∪ h11 ∪ h12 ∪ h21 ∪ h22
is a homotopy 4-ball whose boundary is diﬀeomorphic to S 3 and we denote it
by W1 . Let Ki (i = 1, 2) be the belt-sphere of h2i . Then Ki bounds a smoothdisk in W1 . Freedman, Gompf, Morrison and Walker consider a band sum of
K1 and K2 , denoted by K1 ♯b K2 , where b is a certain band in S 3 , see [FGMW].
The knot K1 ♯b K2 has the following properties:• K1 ♯b K2 bounds a smooth disk in W1 , and• K1 ♯b K2 may not bound a smooth disk in B 4 .
The problem is how to prove that K1 ♯b K2 does not bound a smooth disk in
B . In this century, two strong obstructions for sliceness are introduced. Oneof them is the τ -invariant which is derived from the knot Floer homology. The
other is the s-invariant which is derived from the Khovanov homology. The
properties of these invariants are the following.
4

Theorem 7.1. Let K be a smoothly slice knot. Thenτ (K) = s(K) = 0.
Furthermore, if K bounds a smooth disk in a homotopy 4-ball, thenτ (K) = 0.
The point is that, in 2010, it was not known whether s(K) is zero or not
for a knot K which bounds a smooth disk in a homotopy 4-ball. Therefore
there was a hope that s(K1 ♯b K2 ) ̸= 0 which implies that the smooth Poincaré
conjecture in dimension four is false.
Freedman, Gompf, Morrison and Walker calculated the s-invariant of K1 ♯b K2
using a supercomputer, turning out to be s(K1 ♯b K2 ) = 0. Soon after theirwork, as described before, Akbulut [A1] proved that W1 (and Σ1 ) is standard.
3A proof of this statement is the following. Suppose that W ∪

id B

4

is standard, that is,which is diﬀeomorphic to the standard 4-sphere B ∪id B . The embedding of B 4 into a
connected 4-manifold is unique up to isotopy. This implies that W ≈ B 4 , which contradicts
the assumption. Therefore W ∪id B 4 is exotic.
4

4

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Figure 5. A 2/3-cancelling pair.
This implies that K1 ♯b K2 is a slice knot4 which implies that s(K1 ♯b K2 ) = 0.
In 2013, Kronheimer and Mrowka [KM] proved the following.
Theorem 7.2. Let K a knot in S 3 which bounds a smooth disk in a homotopy
4-ball. Thens(K) = 0.
Therefore the s-invariant can not be used to detect counterexamples to thesmooth Poincaré conjecture in dimension four.
8. Remark on attaching a 3-handle
A (4-dimensional) 3-handle h3 is a copy of D3 ×D1 , attached to the boundaryof a 4-manifold X along ∂D3 × D1 by an embedding ϕ : ∂D3 × D1 → ∂X. The
attaching sphere of h3 is ∂D3 × {0}. A typical Kirby diagram of a canceling
2/3-handle pair is depicted in Figure 5.
The aim of this section is to prove the way of attaching of a 3-handle isunique in some sense (Corollary 8.3). Here we recall a well known Laudenbach
and Poénaru’s theorem.
Theorem 8.1 ([LP]). Any self diﬀeomorphism of #n S 1 × S 2 extends to thatof #n S 1 × D3 .
As an immediate corollary, we obtain the following.
Corollary 8.2. Let X be a 4-manifold with ∂X ≈ #n S 1 ×S 2 . Let fi (i = 1, 2)
be a self diﬀeomorphism of #n S 1 × S 2 . Then
X ∪f1 #n S 1 × D3 ≈ X ∪f2 #n S 1 × D3 ,X ∪fi #n S 1 × D3 is the 4-manifold obtained by gluing X and #n S 1 × D3 withfi .
The following is the main result in this section.
Corollary 8.3. Let X be a 4-manifold represented by a handle decomposition
which consists of a 0-handle, 1-handles and 2-handles. Suppose that h3i (i =1, 2) is a 3-handle attached to ∂X such that ∂(X ∪ h3i ) ≈ S 3 . Then
X ∪ h31 ≈ X ∪ h32 .
4This result is not a good news for the person who try to disprove the smooth Poincaré

conjecture in dimension four. On the other hand, it is a good news for the person who try
to disprove the slice-ribbon conjecture since K1 ♯b K2 might be non-ribbon.

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Figure 6. The deﬁnition of Hn,k .
Proof. We consider X ∪ h3i ∪ h4 , where h4 is a 4-handle. Then ∂(h3i ∪ h4 ) is
diﬀeomorphic to S 1 × S 2 and h3i ∪ h4 is diﬀeomorphic to S 1 × D3 . By (a special
case of) Corollary 8.2,
X ∪ h31 ∪ h4 ≈ X ∪ h32 ∪ h4 .
Since an embedding of B 4 is unique,X ∪ h31 ≈ X ∪ h32 .
□This means that, for any 3-handle attached to ∂X satisfying the condition
in Corollary 8.3, resulting manifolds are always diﬀeomorphic.
9. Gompf’s work
Gompf’s work in [G1] is important to construct slice knots. Here we recall
his work with detail.
In 1991, Gompf [G1] gave a homotopy 4-ball Hn,k (n, k ∈ Z) such that
∂Hn,k ≈ S 3 by Figure 6. He proved the following.
Theorem 9.1 ([G1]). The homotopy 4-ball Hn,k is diﬀeomorphic to B 4 .
As a corollary, he proved the following.
Corollary 9.2 ([G1]). The homotopy 4-sphere Σ0 is standard.
Proof. Gompf showed that H4,1 ∪id B 4 ≈ Σ0 . Then Theorem 9.1 implies that
H4,1 ∪id B 4 ≈ S 4 .
This completed the proof.

□

CONSTRUCTIONS OF SMOOTHLY SLICE KNOTS

9Figure 7. Two handle diagrams represent diﬀeomorphic manifolds.

Figure 8. The two handle diagrams are related by a sequenceof handle slides and isotopies.
To prove Theorem 9.1, we need the following three lemmas. Note that eachbox in Figures means a full twist.
Lemma 9.3 ([G1]). The two handle diagrams in Figure 7 represent diﬀeomorphic manifolds.
Lemma 9.4 ([G1]). The two handle diagrams in Figure 8 are related by asequence of handle slides and isotopies.
Lemma 9.5 ([G1]). We have the following.
H0,k ≈ B 4 .
Proof of Theorem 9.1. First, we add a 2/3 canceling pair to Hn,k . Then we
obtain Hn,k ∪ h2 ∪ h31 , where h2 is a 2-handle and h31 is a 3-handle. Here we

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Figure 9. The two slice knots K1 and K2 .
remove h31 . The resulting manifold Hn,k ∪ h2 is represented by the ﬁrst picturein Figure 7. By Lemmas 9.3, 9.4 and 9.5 5, it is not diﬃcult to see that it isdiﬀeomorphic to h0 ∪ h2 , where h0 is a 0-handle. We attach a 3-handle h32 sothat cancels h2 . Then by Corollary 8.3,
Hn,k ∪ h2 ∪ h31 ≈ h0 ∪ h2 ∪ h32 .
This implies that Hn,k is diﬀeomorphic to h0 (= B 4 ).

□We sometimes call the way of the above proof the canceling 2/3 handle
pair technique. This technique is also essential in [A1] and [AT1].
10. The construction of smoothly slice knots 1
In this section, we recall Gompf, Scharlemann and Thompson’s slice knot
which might be non-ribbon. As we will see, this slice knot is obtained from
H3,1 .
Let h21 be the 0-framed 2-handle and h22 the -1-framed 2-handle of H3,1 . Let
Ki (i = 1, 2) be the belt sphere of h2i , see Figure 9. There is no apparent
reason for Ki to be ribbon. However, by handle calculus, it is not diﬃcult
to see that K1 is the connected sum of the right handed trefoil and the left
handed trefoil and K2 is the connected sum of T (3, 4)#T (3, 4), where T (3, 4)
is the (3, 4)-torus knot and T (3, 4) is its mirror image. Therefore Ki is a ribbon
knot.
In 2010, Gompf, Scharlemann and Thompson [GST] explicitly wrote down
the picture of the 2-component link which consists of K1 and K2 as in Figure
10. Then they considered the band sum of K1 and K2 , denoted by K1 #b K2 ,where b is attached along the dashed line. By the construction, K1 #b K2 is a
5Note that we use Lemma 9.3 twice.

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Figure 10. The 2-component link which consists of K1 and K2 .
slice knot. However, there is no apparent reason for K1 #b K2 to be ribbon.
Open problem. Is K1 #b K2 a ribbon knot ?
Note. We can construct slice knots from Hn,1 with the same manner. The
handle decompositions H1,1 and H2,1 are too simple to produce interesting slice
knots and it is known that all slice knots obtained from H1,1 and H2,1 in thisway are ribbon. Slice knots obtained from Hn,1 (n ≥ 4) are more complicated,see [GST].
11. The construction of smoothly slice knots 2
In this section, we give another construction of slice knots which might be
non-ribbon.
For a knot K in S 3 , we denote by MK the 0-surgery manifold along K.
Here we recall a folklore result, which was mentioned by Akbulut in Kirby’s
problem list [Ki].
Lemma 11.1. Let K and K ′ be knots such that MK (0) ≈ MK ′ (0). If K is
slice, then K ′ bounds a smooth disk in W , where W is a homotopy 4-ball with∂W ≈ S 3 .
If we try to construct slice knots by using Lemma 11.1, we will encountertwo problems. The ﬁrst problem is the following.
Problem 1. How to construct a pair of knots such that MK (0) ≈ MK ′ (0) ?

12TETSUYA ABE

Osoinach [O] invented the annulus twist construction which enables us to
produce inﬁnitely many knots whose 0-surgery manifolds are diﬀeomorphic.
Here we brieﬂy recall the deﬁnitions of an annulus presentation and an annulus
twist. For the details, see [AT1].
Annulus presentation. A knot admits an annulus presentation if it is obtained from the Hopf link by a single band-surgery. For example, the knot K
in the left side picture in Figure 11 is clearly obtained from the Hopf link by asingle band-surgery. Therefore K admits an annulus presentation. Note that
K is also represented as in the middle picture in Figure 11. We call this pic-

Figure 11. A knot admitting an annulus presentation.

Figure 12. The knot Kn obtained from K by the n-fold annulus twist.
ture an annulus presentation of K 6 since we can easily ﬁnd an unknotted
annulus (which intersects with K at two points) as in the right side picture in
Figure 11.
6In [AJOT], they call it a band presentation.

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Annulus twist. Instead of giving the deﬁnition of an annulus twist, we justexplain one example. Let K be the knot in the left side picture in Figure 11.
Then the knot obtained from K by the n-fold annulus twist, denoted by
Kn , is the knot represented by the left side picture in Figure 12. The explicit
picture of K1 is depicted in Figure 12.
One of the answers to Problem 1 is the following.
Lemma 11.2 ([AJOT]). Let K be a knot with an annulus presentation and
Kn (n ∈ Z) the knot obtained from K by the n-fold annulus twist. Then
MK (0) ≈ MKn (0).
Let K be a slice knot with an annulus presentation and Kn (n ∈ Z) the
knot obtained from K by the n-fold annulus twist. Then, by lemmas 11.1 and
11.2, Kn bounds a smooth disk in Wn , where Wn is a certain homotopy 4-ballwith ∂Wn ≈ S 3 . The second problem is the following.
Problem 2. How to prove Wn ≈ B 4 ?
To solve Problem 2, Abe and Tange used the canceling 2/3 handle pairtechnique and obtain the following.
Theorem 11.3 ([AT1]). Let K be a slice knot with an annulus presentation
and Kn (n ∈ Z) the knot obtained from K by the n-fold annulus twist. Then
Kn is a slice knot.
It is important to know which slice knots admit annulus presentations. Abe,
Jong, Omae and Takeuchi [AJOT] proved that an unknotting number one knot
admits an annulus presentation. Therefore, for a given unknotting number one
ribbon knot K, we obtain inﬁnitely many slice knots Kn by Theorem 11.3.
There is no reason for Kn to be a ribbon knot.
Here we consider a concrete example. Let K be the knot in the left side
picture in Figure 11 again.

Rolfsen’s presentation

ribbon presentation

annulus presentation

Figure 13. Three presentations of 820 .

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We can check that K is 820 which is known to be ribbon, see Figure 13.
As the author knows, this is the simplest ribbon knot admitting an annulus
presentation. Let Kn be the knot obtained from K by the n-fold annulus twist.
Then Kn is a slice knot by Theorem 11.3. There is no apparent reason for Kn
to be ribbon.
12. The construction of smoothly slice knots 2 revisited
In the previous section, we gave a construction of slice knots which might be
non-ribbon. In this section, we consider this construction again with emphasison handle decomposition of B 4 .
The construction of W in Lemma 11.1 is important. Here we recall the
proof of Lemma 11.1.
Sketch of Proof of 11.1. Let D2 be a slice disk for K and X the 4-manifold
obtained from B 4 by removing an open tubular neighborhood of D2 in B 4 , see
FigureFigure 14. Schematic pictures.
Note that ∂X is diﬀeomorphic to MK (0). By the assumption, MK (0) ≈MK ′ (0). Therefore we can identify ∂X with MK ′ (0). The 4-manifold W is
obtained from X by attaching a 2-handle along the meridian µ′ of K ′ withframing 0. We can check the following.
• W is a homotopy 4-ball.
• The belt-sphere is isotopic to K ′ .
Therefore K ′ bounds a smooth disk (= the cocore disk of the 2-handle) in a
homotopy 4-ball W .
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Remark. For a movie presentation of D2 , then we can associate a handle
decomposition of X. Therefore we also obtain a handle decomposition of W .
In particular, if D2 is a ribbon disk, it naturally has a movie presentation andthe corresponding handle decomposition of W consists of a 0-handle, 1-handles
and 2-handles.
Summary. For a slice knot K with an annulus presentation, we gave handle
decompositions HDn of B 4 such that the belt-sphere of a certain 2-handle of
HDn is isotopic to Kn , where Kn is the knot obtained from K by the n-foldannulus twist (This is a version of Theorem 11.3).
Note. It may be natural to ask the relation between HD0 and HDn . In a
special case, Abe and Tange [AT1] proved that HDn is obtained from HD0 by
a log transformation along the torus (=elliptic ﬁber) of a ﬁshtail neighborhood
embedded into HDn .
13. Ribbon presentations via handle calculus
So far, we constructed slice knots which might be non-ribbon via handle
calculus. In this section, we explain how to obtain ribbon presentations via
handle calculus.
As an example, let HD be a handle diagram as in Figure 15. Handle calculus

Figure 15. A handle diagram (of B 4 ).
in Figure 16 tells us that HD represents B 4 . Let K be the belt-sphere of a2-handle as in Figure 17. Then K is a slice knot. Furthermore, K is a ribbon
knot. To see this, we recall the following well known fact.
Fact 13.1. A knot K is ribbon if and only if K is changed into the (n + 1)component unlink by n band-surgeries. In particular, if K is changed into
the 2-component unlink by a single band-surgery, then K is a ribbon
knot.
The knot K is changed into the 2-component unlink by a single band-surgery
as follows: By a handle slide, we obtain the ﬁrst handle diagram in Figure 18.

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Figure 16. The handle diagram HD represents B 4 .

Figure 17. The deﬁnition of the knot K.
Then we attach a band to K along the dashed arc. After isotopy, we obtainthe second handle diagram in Figure 18. By annihilating canceling 1/2 handle
pair, we obtain the 2-component unlink. This result is generalized as follows.
Lemma 13.2 ([AT1]). Let HD be a handle diagram of B 4 . Suppose that HD
is changed into ∅ by handle moves without adding canceling 2/3-handle pairs,
where ∅ is the empty handle diagram of B 4 . Then the belt sphere of any 2handle of HD is a ribbon knot.

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Figure 18. The knot K is changed into the 2-component unlink by a single band-surgery.
Remark. To draw a ribbon presentation of K explicitly, we need a little more
work. See ﬁgure 19.
In Section 11, we gave a way to produce slice knots which might be nonribbon. Here we consider the simplest case. That is, when K is the left sideknot in Figure 11 with the annulus presentation. Let Kn (n ∈ Z) the knot
obtained from K by the n-fold annulus twist and HDn the handle diagram of
the corresponding handle decomposition of Wn ≈ B 4 .
Theorem 13.3 ([AT1]). The handle diagram HDn is changed into ∅ by asequence of handle moves without adding canceling 2/3-handle pairs. In particular, Kn is a ribbon knot.
Remark. In the proof of Theorem 11.3, Abe and Tange proved that HDn ischanged into ∅ by a sequence of handle moves including adding a canceling
2/3-handle pair. In the proof of Theorem 13.3, They proved that HDn ischanged into ∅ by a sequence of handle moves without adding canceling 2/3handle pairs. The later needs a rather long handle calculus.

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Figure 19. A ribbon presentation of K.
Open problem. Does there exist a sequence of handle moves without adding
any canceling 2/3-handle pairs which changes H3,1 to ∅ ?
If there exists a such sequence, we can prove that the slice knot K1 #b K2 in
Section 10 is ribbon. However, it is expected that there does not exist a such

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19

sequence by a group theoretical reason (the Andrews-Curtis conjecture), see[GST].
14. Non-Ribbon knots which might be slice
In this section, we recall another type of potential counterexamples of theslice-ribbon conjecture, non-ribbon knots which might be slice.
Casson and Gordon gave a ribbon obstruction for ﬁbered knots as follows.
Theorem 14.1 ([CG1]). Let K be a ﬁbered knot in S 3 . If K is a ribbon knot,
then the closed monodromy of K extends over a handlebody.
Miyazaki proved the following.
Theorem 14.2 ([M]). The (2,1)-cable of the ﬁgure eight knot is not ribbon.
Outline of proof. Let Φ be the closed monodromy the (2,1)-cable of the ﬁgure
eight knot. Φ does not extend over a handlebody. By Theorem 14.1, the(2,1)-cable of the ﬁgure eight knot is not ribbon.
□
Open problem. Is the (2,1)-cable of the ﬁgure eight knot is slice ?
Remark. Livingston and Melvin [LM] and Kawauchi [Ka1] proved that it isalgebraically slice. Furthermore Cha [C] and Kawauchi [Ka2] showed that it
is rationally slice.
For related work, see [AS], [GM].
Remark. Another ribbon obstruction was given by Friedl [Fr].
15. Differential geometry and the slice-ribbon conjecture
Hass formulated the slice-ribbon conjecture in terms of diﬀerential geometryin [H]. See also Appendix B in [HedKL].
16. Seifert surfaces of slice knots
Cochran and Davis [CD] formulated the slice-ribbon conjecture in terms ofSeifert surfaces. See also [JMP].
17. Fox’s characterization of slice knots
In this section, we recall a characterization of slice knots which is essentially
due to Fox [F1].
Theorem 17.1. A knot K is slice if and only if K#R is ribbon for someribbon knot R.
In particular, we obtain the following.

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Corollary 17.2. The slice-ribbon conjecture is true if we can chose R to be
the trivial knot in Theorem 17.1.
For a given slice knot K, it is diﬃcult to ﬁnd a ribbon knot R such thatK#R is ribbon. As the author knows, the ﬁrst non-trivial example was given
by Herald, Kirk and Livingston [HerKL]. They proved that the knot 12a990 #R
is ribbon where R is the connected sum of the right-handed trefoil and lefthanded trefoil. Abe and Tange [AT2] proved that 12a990 is a ribbon knot7 andgeneralize this result.
Acknowledgments. The author thanks In Dae Jong for comments to thedraft, Brendan Owens for his question to Lemma 13.2 which makes Section
13 readable, Ryan Budney for telling his the papers [H] and [HedKL], Kengo
Kawamura for providing us the picture in the ﬁrst page. He also thanks theinvited speakers, Hye Jin Jang, Min Hoon Kim and Min kyoung Song who
are students of Jae Choon Cha. He learned much from them the ﬁltration
theory on the topological knot concordance group. He was supported by JSPS
KAKENHI Grant Number 23840021, 255998.
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knot.

CONSTRUCTIONS OF SMOOTHLY SLICE KNOTS

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Department of Mathematics, Tokyo Institute of Technology, 2-12-1 Ookayama,
Meguro-ku, Tokyo 152-8551, Japan
E-mail address: abe.t.av@m.titech.ac.jp