On the degeneration of tunnel numbers under connected sum
Abstract.
We show that, for any integer , there is a prime knot such that (1) is not meridionally primitive, and (2) for every bridge knot with , the tunnel numbers satisfy . This gives counterexamples to a conjecture of Morimoto and Moriah on tunnel number under connected sum and meridionally primitive knots.
1. Introduction
Let be a compact 3–manifold. If there is a closed surface that cuts into two compression bodies and such that , then we say that is a Heegaard surface of , and the decomposition, denoted by , is called a Heegaard splitting of . If is minimal among all Heegaard surfaces of , then is called a minimal Heegaard splitting, and the Heegaard genus is defined to be . Any orientable compact 3–manifold admits a Heegaard splitting.
Let be a knot in the 3–sphere and let , where is an open tubular neighborhood of . There is always a collection of disjoint and embedded arcs in , such that for each and is a handlebody. This means that is a Heegaard surface of . These arcs are called unknotting tunnels of and we say that they form a tunnel system for . The tunnel number of , denoted by , is the minimal number of arcs in a tunnel system for . Let be the Heegaard genus of the knot exterior . Clearly .
For two knots and , we denote the connected sum of and by . Given any tunnel systems for and , one can obtain a tunnel system for by putting together the tunnel systems for and plus an extra tunnel lying in the decomposing 2–sphere. This means that the tunnel numbers of these knots satisfy the following inequality:
An interesting question in knot theory and 3–manifold topology is to study the relation between the tunnel number of a composite knot and the tunnel numbers of its factors. The following is a list of some results in this direction:
(1) Norwood proved that tunnel number one knots are prime [15], Scharlemann and Schultens proved that the tunnel number of the connected sum of nontrivial knots is at least and of the sum of the tunnel numbers of the factors. See [18] and [19].
(2) For two knots and , the equality is called the super additivity of tunnel number under connected sum. Morimoto [12] first showed that the super additivity does not always hold. Kobayashi [4] gave examples of composite knots with for any integer . Nogueira [14] found the first prime knots , with . Schultens [23] proved that if both and are small knots, then , and Morimoto [13] proved that if both and are small knots, then the super additivity holds if and only if none of and is meridionally primitive (see section 2 for definition). As an extension of Morimoto’s result, Gao, Guo and Qiu [2] proved that if a minimal Heegaard splitting of has high distance while a minimal Heegaard splitting of has distance at least 3, then .
Note that if a knot is meridionally primitive, then for any knot (see section 2 for details). Based on the result in [13], Morimoto conjectured that if and only if none of and is meridionally primitive (this was also conjectured by Moriah [10]). Kobayashi and Rieck [5] gave a counterexample to this conjecture, but none of the two factors and in their example is prime. The following is a modified version of Morimoto’s conjecture (this is also conjectured by Moriah, see [11, Conjecture 7.14]).
Conjecture 1.1.
Suppose that both and are prime. Then if and only if neither nor is meridionally primitive.
The main result of this paper is the following:
Theorem 1.2.
For any integer , there is a prime knot such that

is not meridionally primitive, and

for every bridge knot with , the tunnel numbers satisfy , in other words, the Heegaard genera of the knot exteriors satisfy .
By choosing the knot in Theorem 1.2 to be prime and not meridionally primitive, we have a counterexample to Conjecture 1.1. Furthermore, we have the following immediate corollaries.
Corollary 1.3.
For any collection of knots , there is a prime knot such that is not meridionally primitive and for all .
Corollary 1.4.
For any integer , there exist knots and that are prime and not meridionally primitive, such that .
In the proof of the main theorem, we only give an upper bound on . However, it is conceivable that the tunnel number of the connected sum should be at least as large as the tunnel number of any of its factors. This is an interesting question in its own right.
Question 1.5.
Are there two knots and in with ?
If there is a degreeone map between two closed orientable 3–manifolds and , a difficult question in 3–manifold topology is whether or not always holds. Note that there is a degreeone map from to . Thus a positive answer to Question 1.5 also gives an example of 3–manifolds and such that there exists a degreeone map but .
Acknowledgments.
The first author would like to thank Yoav Moriah for many helpful conversations.
2. Annulus sum and meridionally primitive knots
Notation.
Throughout this paper, for any subspace in a manifold, denotes an open regular neighborhood of , denotes the interior of and denotes the closure of . For any knot in , we use to denote the knot exterior . For any compact 3–manifold , we use to denote its Heegaard genus.
Definition 2.1.
Let be a knot exterior and a Heegaard splitting of with . So is a handlebody. Let be a slope in the boundary torus . If there are an essential disk in and a properly embedded annulus in with such that

is a curve of slope , and

and is a single point,
then we say that the Heegaard splitting is primitive. Note that, since is a single point, is a primitive curve in the handlebody .
Definition 2.2.
Given two 3–manifolds and with nonempty boundary, an annulus sum of and is a 3–manifold obtained by identifying an annulus and an annulus . Note that, for any two knots and in , is an annulus sum of and , identifying a pair of meridional annuli in and (in this paper, an annulus is meridional if its core curve has meridional slope in the boundary torus).
Let be an annulus sum of two compact 3–manifolds and . One can obtain a Heegaard surface of by connecting Heegaard surfaces of and using a tube through the gluing annulus . Thus . Suppose and a minimal Heegaard splitting of is primitive, where is the slope of the core curve of the annulus . Then it is easy to see that the Heegaard splitting of constructed this way (using the primitive Heegaard splitting of ) is stabilized and hence . Thus a meridionally primitive Heegaard splitting of provides a clear picture how the Heegaard genus of the connected sum of two knots can degenerate. We summarize this wellknown fact as the following lemma, see [13] for a proof.
Lemma 2.3.
Let be a nontrivial knot in and any compact irreducible 3–manifold with boundary. Let be an annulus in and let be an annulus sum identifying to a meridional annulus in . If is primitive, then .
The following is an immediate corollary of Lemma 2.3, and we will use it later to prove that a knot is not primitive.
Corollary 2.4.
Let and be as in Lemma 2.3. Suppose . If , then is not primitive.
Before we proceed, we would like to clarify some terminology that we will use in constructing surfaces: Given a surface and an embedded arc with , we can construct a new surface by first removing two disk neighborhoods of the two endpoints of in and then connecting the resulting boundary circles by an annulus (or tube) along . We say that the surface is obtained by adding a tube to along the arc .
3. The construction
Let be any integer. The knot in Theorem 1.2 is constructed by gluing together two string tangles. We require that each tangle exterior has a highdistance Heegaard splitting. The following is a construction of such tangles.
Let be the integer above. We fix a large number ( is assumed to be sufficiently large relative to ). By [9], for any number , there is a knot in with tunnel number such that has a Heegaard splitting of distance at least . Let be the unknotting tunnels of . We may assume that are disjoint and view as a trivalent graph in . Since , we may find a tree in the graph that contains exactly tunnels and is disjoint from the remaining tunnels. Let be a small open neighborhood of . We require that each tunnel is either totally in or totally outside . Since is a tree, is an open 3–ball. Let . So is a 3–ball and is an string tangle in .
Let be a neighborhood of in that contains . Let and . By our construction of , is a handlebody and is a Heegaard surface of . Moreover, the distance of the Heegaard splitting along is at least , where is a number that can be assumed to be arbitrarily large. Since , . Let be the tangle exterior, where is a small open neighborhood of in . By our construction, can be obtained from the compression body by removing a small neighborhood of part of its spine. Thus is also a compression body and is a Heegaard surface of the tangle exterior . Moreover, the disk complex of the compression body is a nontrivial subcomplex of the disk complex of the compression body . Thus the distance of the Heegaard splitting of along is at least . In particular, by assuming , we may assume that the Heegaard splitting of (along ) is strongly irreducible, and hence is incompressible in [1].
We take two copies of , denoted by and . Then we glue to along the boundary sphere in such a way that the union of the two tangles is a knot. This is our knot . We will show next that is a prime knot and it is not primitive. These follow from that the tangle exterior has a highdistance Heegaard splitting.
Let be the 2–sphere, , and (). So is basically the exterior of the tangle in . Let and let , where is a small tubular neighborhood of in . Thus , and the knot exterior can be obtained by gluing and to along and respectively. So we may view and view , and as submanifolds of . By our earlier conclusion, is incompressible in . Moreover, each tangle exterior has a Heegaard surface such that the distance of the Heegaard splitting of along is at least , where is a number that we can arbitrarily choose.
Next we consider . Note that has 3 components: , and the torus . By our construction, can be obtained by gluing a product to a product , where , is a hole sphere and is a torus, and the gluing map identifies each annulus in to an annulus in . Thus , and each can be obtained from a component of by adding tubes/annuli (from ) connecting its boundary circles. We view and as submanifolds of .
Let be the components of . So divides into two submanifolds and . It is clear from our construction that each is an incompressible and incompressible annulus properly embedded in . This also implies that and are incompressible in . Thus and are incompressible in .
Lemma 3.1.
If , the knot is prime.
Proof.
Suppose is not prime. Then contains an essential annulus with meridional slope. Since and are incompressible in , after isotopy, we may assume that if , is a collection of essential annuli in . Since the Heegaard distance of the splitting in is at least , by [17, Theorem 3.1], contains no essential annulus. Thus for both and hence . The following claim is well known.
Claim.
Let be an incompressible surface in with , then either is parallel to a subannulus of a component of , or is in the form , where .
Let be as above. Since each is an essential annulus in , after isotopy, each component of is an incompressible annulus in . Since , by the claim, each component of must be an annulus parallel to a subannulus of for some . This means that after isotopy, with . Clearly contains no such essential annulus, a contradiction. ∎
Next we give two constructions of a genus Heegaard surface for . These constructions imply that the Heegaard genus . The first construction gives a Heegaard surface of that separates from . In the second construction, and lie on the same side of the Heegaard surface.
Construction 1: We start with a peripheral surface in parallel to , which has genus . We describe a neighborhood of in using a (2dimensional) schematic picture Figure 3.1(a). Note that in Figure 3.1(a) is a picture of an annulus and is our neighborhood of in , Figure 3.1 is a picture with and the shaded regions in Figure 3.1 denote parts of that are neighborhoods of the annuli ’s. Now we add tubes to along unknotted arcs which are parallel to arcs in and circularly connect the annuli of , see the dashed arcs in Figure 3.1(b) for a picture of these arcs (this is slightly misleading since all other arcs in Figure 3.1 denote surfaces but the dashed arcs denote arcs connecting the surfaces). Since , the resulting surface has genus . Moreover, if one maximally compresses outside the added tubes, one ends up with a surface parallel to and a torus parallel to . Thus is a Heegaard surface for of genus .
Construction 2: Let () be a peripheral surface in parallel to . We may assume is of the form (). There is a subarc of an fiber of connecting to , and we add a tube along such a vertical arc to . Since , the resulting surface has genus . If we compress this surface along the meridional curve of the tube, we get , and if we maximally compress it on the other side, we get a peripheral torus parallel to . This means that the resulting surface is also a Heegaard surface for of genus .
Let be a meridional annulus in . Let be the annulus sum of and a product , identifying to an essential annulus in . We view as an annulus properly embedded in , dividing into and . Thus has 4 components: , , a torus which is the union of and an annulus in , and a torus component from . Since , by section 2, .
Our next goal is to compute the Heegaard genera of and . As , we may view and as submanifolds of both and .
Lemma 3.2.

Either or contains a minimal genus Heegaard surface such that is a connected surface in the form ().

Either or contains a minimal genus Heegaard surface such that is a connected surface in the form ().
Proof.
We prove the two parts of the lemma at the same time, since and are similar. Let be a minimal genus Heegaard surface in our 3–manifold ( or ). We will show that we can either isotope or change to a possibly different Heegaard surface satisfying the conditions in the lemma. We have two cases.
Case (a). is strongly irreducible.
Since consists of essential annuli in our 3–manifold, by [7, Lemma 3.7] (also see [6]), after isotopy, we may assume that at most one component of is strongly irreducible and all other components are incompressible in (recall that a surface is strongly irreducible if it is compressible on both sides and every compressing disk on one side meets every compressing disk on the other side). Moreover, after pushing components of out of , we may assume that no component of lies in a collar neighborhood of the annuli . So by the claim in the proof of Lemma 3.1, each incompressible component of is of the form ().
Claim.
There is a minimal genus Heegaard surface of the 3–manifold ( or ) such that each component of is of the form ().
Proof of the Claim.
By our assumption above, either satisfies the claim or exactly one component of is strongly irreducible. Suppose is a strongly irreducible component of . Next we are basically repeating an argument in [7, Lemma 3.9 and Lemma 3.10]. We call the two sides of plus and minus sides. Let and be the surfaces obtained by maximally compressing in on the plus and minus sides respectively and deleting all the 2–sphere components. We may assume that . There is a connected submanifold of between and (similar to a compression body), which we denote by , and similarly, a connected submanifold between and , denoted by . Furthermore, . Hence there is a connected region between and which contains . Clearly . Since is strongly irreducible, by [1] (also see [17, Lemma 5.5]), is incompressible in . Hence each component of is either a parallel annulus or a surface of the form .
Let be a component of or . We define the shadow of , denoted by , as follows. If is a parallel annulus, then is the solid torus bounded by and a subannulus of . By our assumption that is not in a collar neighborhood of , must lie outside . If is in the form , then divides into two intervalbundles, one of which contains , and we define to be the intervalbundle that does not contain . Thus, in either case, lies outside . Moreover, since , and lie on the two sides of and .
Let and be two components of . We say that and are nonnested if , and we claim that and are always nonnested. By our definition of , if , then either or . Suppose the claim is false and . Since and since , this means that , which contradicts that lies outside .
Since for each component of , and since , the union of and all the shadows ’s is the whole of . Thus is isotopic to and the isotopy can be realized by pushing each to in . This implies that has exactly two components in the form and all other components are nonnested parallel annuli.
Without loss of generality, we may assume that has at least one component in the form , and has components. As shown in Figure 3.2(a, b), we can connect the components of by adding tubes to along unknotted arcs which can be isotoped into , and the resulting connected surface, which we denote by , is a connected sum of all the components of . and have some similar properties: If one compresses along the meridional curves of the tubes, one gets back ; if one maximally compresses on the other side, one gets a collection of nonnested parallel surfaces in , which means that the resulting surface is . Moreover, by our construction, can be obtained by adding 1–handles (or tubes) to on the same side of . Since has components and is connected, to obtain , we need to add at least tubes to . This implies that . Now we can replace by . The union of and all the other components of and is a closed surface, which we denote by . Since is a Heegaard surface, the properties of and above imply that is also a Heegaard surface (see the proof of Lemma 3.9 in [7]). Since , we have and . Since is a minimal genus Heegaard surface, must also be a minimal genus Heegaard surface and .
As shown in Figure 3.2(c), we can isotope by pushing the parallel annuli in and all the added tubes across and out of . After this isotopy, each component of is of the form . ∎
Next we consider the number of components in . If has at least 3 components, then and . Since , we have .
Recall that our 3–manifold is either or , and by our construction prior to the lemma, we have and .
We now separately discuss and . Suppose our 3–manifold is . Since is a minimal genus Heegaard surface and since , the conclusion above implies that has at most 2 components. If has exactly two components, then the two components of lie on the same side of the Heegaard surface , and hence and are boundary components of the same compression body in the Heegaard splitting of along . As , this implies that . As is a minimal genus Heegaard splitting and since , this means that . Thus either or has a single component. Hence part (1) of the lemma holds in Case (a).
Suppose our 3–manifold is . Since is a minimal genus Heegaard surface, we have . Let be the Heegaard splitting along . We may assume is strongly irreducible, otherwise we are in Case (b) below. By the conclusion above, if has at least 3 components, then , hence and part (2) of the lemma holds. To prove part (2), it remains to consider the possibility that has 2 components. Suppose this is the case, then similar to the argument above, both and lie in the same compression body, say , in the Heegaard splitting along . As contains both and , we have .
Note that two components of are tori. If also contains a torus component of , then , which implies that , and hence part (2) of the lemma holds. Thus it remains to consider the case that both tori in lie in the compression body . Suppose the lemma is false in this case and . By our conclusion earlier that , we have and hence the compression body can be obtained by adding a single 2–handle to (a product neighborhood of) along a separating curve in . Thus can be obtained by adding a single 2–handle to . Since the core curve of the 2–handle is a separating (and nullhomologous) curve in , we have the identity on the first homology.
The compression body can be obtained by adding 1–handles to a product neighborhood of . Since contains both boundary tori and of , we may view the homology groups and as direct summands of . By our conclusion above, and are direct summands of . Recall that is an annulus sum of and the product . Thus there is a vertical annulus that is also a properly embedded annulus in connecting to . This means that an essential closed curve in is homologous (in ) to a curve in , which contradicts our conclusion that and are direct summands of .
Case (b). is weakly reducible.
We consider the untelescoping of the Heegaard splitting, see [21]. Since our 3–manifold is a submanifold of , every closed surface in our 3–manifold is separating. So by [21], there are a collection of separating closed incompressible surfaces that divide our 3–manifold ( or ) into submanifolds . There is a strongly irreducible Heegaard surface in each such that is an amalgamation of these ’s along the incompressible surfaces ’s. Since is assumed to be weakly reducible, and for each . Since is a minimal Heegaard surface, is not parallel in the 3–manifold ( or ), and is not parallel to if .
Recall that is viewed as a submanifold and is a collection of essential annuli in our 3–manifold ( or ). By the claim in the proof of Lemma 3.1, after isotopy, we may assume that each component of is of the form . Thus each component of is a product (). Since each is strongly irreducible, similar to Case (a), after isotopy, we may assume that

each component of is of the form ,

at most one component of is strongly irreducible and all other components are incompressible in ,

no component of lies in a collar neighborhood of the annuli .
Let () be a component of for some . We may assume that each component of is an essential annulus in . Now we apply the argument in the proof of the Claim in Case (a) on and . As in the Claim in Case (a), after isotopy and replacing by a possibly different Heegaard surface , we may assume that each component of is of the form . Note that since is a minimal genus Heegaard surface, and must be minimal genus Heegaard surfaces of .
If has more than two components, then . Similar to Case (a), since , this means that . However, since , we have . This contradicts that and . Thus has at most two components.
Suppose has two components. Then since is a Heegaard surface of , both components of lie on the same side of . Since is not parallel to a component of , we have . This means that . Since , we have . If our 3–manifold is , then and this is a contradiction. If our 3–manifold is , then . So we have and part (2) of the lemma holds. Thus, to finish the proof of the lemma, we may assume that has exactly one component for each component of and for any .
Note that, by amalgamating these Heegaard surfaces along the incompressible surfaces , we can obtain a minimal genus Heegaard surface for our 3–manifold ( or ). Since intersects each component of in a connected surface of the form , after the amalgamation, is a connected surface of the form (see [22, Figure 3] for a picture of amalgamation). We can also see this through the untelescoping, which is a rearrangement of 1– and 2– handles, see [21]. Since is a connected surface of the form for every component of and for each , if we maximally compress on either side, we can choose the compressions to be disjoint from . Thus we may view that all the 1– and 2–handles (in the untelescoping for ) are outside . Hence is a connected surface of the form and the lemma holds in Case (b). ∎
Lemma 3.3.
and .
Proof.
The idea of the proof is to first replace by , which changes the 3–manifold ( or ) into a product of a planar surface and . Then we apply the classification of Heegaard surfaces in such products [22] to get a lower bound on the Heegaard genus.
Similar to the proof of Lemma 3.2, we consider the two cases and at the same time. Let be a minimal genus Heegaard surface in our 3–manifold ( or ).
By Lemma 3.2, we may assume that is a connected surface of the form . This implies that

and lie on different sides of , and

if we maximally compress on either side, all the compressions can be chosen to be disjoint from .
Recall that is the union of a component of and annuli which we denote by . Let be the annuli in . Without loss of generality, we may suppose connects to , . Now we construct a new 3–manifold by replacing with copies of ( is an annulus) so that each copy of connects a pair of annuli (i.e. ). This construction is for both and , and we use and to denote the new 3–manifolds constructed from and respectively.
Recall that and , where is an annulus and is a pair of pants, see Figure 3.3(a, b) for pictures of and (with ). Figure 3.3(a) can be compared with Figure 3.1(a). Note that these annuli ’s are in . Moreover, since is a boundary component of , after renaming these annuli if necessary, we may assume that lie in in a cyclic order, see Figure 3.3(a,b). Thus, as shown in Figure 3.3(c, d), the new 3–manifolds and , where is a planar surface with boundary circles and