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Figure 2: A connected sum κ of a trefoil knot and a figure 8 knot. The dashed ellipse represents a sphere (the boundary of the 3-balls in the construction described above) in S3 separating the two summands.

Our first proof is probably the simplest possible, though it provides slightly less information about the Kinoshita graph than the second. The we’ll use for this proof is the bridge number of a knot. Bridge number, like connected sum, is defined using a certain way of gluing two 3-balls together to obtain S3. Consider two 3-balls, each containing the same number n of strands. Unlike in the definition of connected sum, we require that in each 3-ball the strands can be simultaneously isotoped into the boundary of the 3-ball, as in Figure 3, where n = 3. These 3-balls, together with the strands they contain, are called trivial n-tangles. Trivial n-tangles do have diagrams with no crossings, as on the top left and bottom left of Figure 3; however, they also have diagrams with lots of crossings as on the top right and bottom right of Figure 3. We then choose a homeomorphism φ between the boundaries of the 3-balls, taking the endpoints of the strands in one 3-ball to the endpoints of the strands in the other 3-ball. Gluing the 3-balls together along their boundary using φ produces the 3-sphere S3 and the union of the strands is a knot or in S3. Every knot or link in S3 can be obtained this way (for some choice of n and φ). For a given knot or link K, the bridge number b(K) is smallest value of n such that there exists a homeomorphism φ such that the resulting knot or link is equivalent to K. The knot K is the unknot if and only if b(K)= 1. Schubert [19] proved the following marvelous theorem. (See [20] for a different proof.) 3 Theorem (Schubert). Suppose that K1 and K2 are knots in S and that K1#K2 is any connected sum of them. Then b(K1#K2)= b(K1)+ b(K2) − 1.

2 Figure 3: On the left is a schematic depiction of gluing two trivial 3-tangles together to produce a knot or link K with b(K) ≤ 3. On the right, we have a knot K in S3 with b(K) ≤ 3. The thick circle denotes a sphere separating the knot into two trivial 3-tangles.

One consequence of Schubert’s theorem is that if K is a knot with b(K)= 2, then K is prime. Suppose now, for a contradiction, that the Kinoshita graph K is trivial. Then there is an ambient isotopy of K to the trivial θ-graph T. If e is an edge of K, then this isotopy takes e to an edge e′ of T. It also takes an open regular neighborhood of e to an open regular neighborhood of e′. The complement of the open regular neighborhood of e is a closed 3-ball B. The isotopy takes B to the complement B′ of the open regular neighborhood of e′. The graph K intersects B in two strands. Similarly, the graph T intersects B′ in two strands. The isotopy induces a homeomorphism of pairs taking (B,K ∩ B) to (B′,T ∩ B′). For each edge of T, it is easy to verify that the complement of an open regular neighborhood of that edge is a trivial 2-tangle. Thus, (B,K ∩ B) is a trivial 2-tangle. We will use bridge number to show that this is impossible. Inside the ball N(e) complementary to B replace K∩N(e) with a trivial 2-tangle, as in the first step of Figure 4. We arrive at a knot K, which must be a two-bridge knot since it was created by gluing together two trivial 2-tangles. (One tangle is obviously trivial, the other we have shown must be trivial if K is trivial.) However, as shown in the second step of Figure 4, the knot K is the composite knot κ! This contradicts Schubert’s theorem, and so K is nontrivial. Remark. A somewhat more involved argument lets us bypass the use of Schubert’s theorem, at the expense of more work. By keeping track of which trivial tangle we place in the ball N(e) and how it is affected by the hypothetical isotopy from K to T, it is possible to show that the knot K we constructed must be both a torus knot and the connected sum of a trefoil and a figure 8. This is impossible, as no torus knot is composite. (See [3] for the idea of how to prove this.)

We now embark on our second proof, during which we will learn an important fact about the exterior of the Kinoshita graph: it’s not a handlebody. A handlebody is the result of attaching the ends of solid tubes to 3-balls so as to arrive at a connected, orientable 3-manifold with boundary. More precisely, it is a closed regular neighborhood N(G) of a finite spatial graph G ⊂ S3, called a spine of the handlebody. The genus of the handlebody is, by definition, the genus of the bounding surface. Thus, every spatial θ-graph is a spine for a genus 2 handlebody. The trivial θ-graph T has the property that its exterior X(T) (the closure of S3 \ N(G)) is also a handlebody. Indeed, every spatial graph contained in a tame 2-sphere has this property. The handlebody X(T) can be constructed using two 3-balls and attaching solid tubes that run through the disk faces. The two handlebodies N(T) and X(T) form what is called a Heegaard splitting of S3. But there are many other spatial graphs with this property, for instance, the one appearing in Figure 5.

3 Figure 4: On the left the dashed ellipse encloses a 3-ball N(e) which is a regular neighborhood of the edge e of K. The first arrow denotes the action of replacing (N(e),K ∩ N(e)) with a certain trivial tangle. The second arrow depicts an isotopy taking the resulting knot to the knot κ that is clearly composite.

An ambient isotopy of a spatial graph G to a spatial graph G′ can be extended to an ambient isotopy of N(G) to N(G′); however, an ambient isotopy from one handlebody to another need not restrict to an ambient isotopy between two given spines. As no ambient isotopy of a handlebody changes the homeomorphism type of its exterior, this allows us to construct many potentially distinct spatial graphs having homeomorphic exteriors. In particular, every spatial graph G such that N(G) is ambient isotopic to N(T) has handlebody exterior. Figure 6 shows an ambient isotopy of a regular neighborhood of the graph from Figure 5 to N(T). On the other hand, we will show that K does not have a handlebody exterior, in which case there can be no ambient isotopy taking N(K) to N(T), much less one taking K to T. This also will show that K is not equivalent to any graph having handlebody exterior.

Figure 5: An example of a nontrivial spatial θ-graph having handlebody exterior. To see that the graph is nontrivial, note that it contains a knotted cycle (a trefoil knot). Figure 6 below shows that the θ-graph has handlebody exterior, and hence that the trefoil knot has tunnel number one. Remark. Kinoshita’s original proof [6] that K is knotted also proceeds by showing that K does not have handlebody exterior, but uses new algebraic techniques rather than classical knot invariants.

A nontrivial knot K (such as the trefoil knot in Figure 5) has tunnel number one if it has the property that we can attach an arc in such a way as to create a spatial θ-graph with handlebody exterior. In general, a knot K has tunnel number t(K)= n if n is the smallest integer such that we may attach n arcs to K to arrive at a spatial graph with handlebody exterior [2]. Unlike bridge number, tunnel number need not be additive under connected sum (see [7, 11], for example); however, Norwood did prove the following inequality [13]. See [17, 23] for other proofs and [4, 12, 18] for generalizations. 3 Theorem (Norwood). If K1 and K2 are nontrivial knots in S , then t(K1#K2) ≥ 2.

As a consequence of Norwood’s theorem, we see that knots of tunnel number one are prime. Equivalently,

4 Figure 6: Some frames from a movie showing how to perform an ambient isotopy of a regular neighborhood of the graph from Figure 5 to a regular neighborhood of the trivial θ-graph T. the connected sum of two nontrivial knots will never be a cycle in a spatial θ-graph with handlebody exterior. In particular, the composite knot κ appearing in Figure 2 and in the middle and right of Figure 4 cannot be a cycle in a spine of spatial θ-graph having handlebody exterior. However, Figure 7 shows an ambient isotopy of a regular neighborhood N(K) of the Kinoshita graph to a handlebody that is a regular neighborhood of a certain spatial θ-graph G. The right side of Figure 7 shows G. We see that G contains the knot κ as a cycle. Since κ is composite, N(G), and hence N(K), does not have handlebody exterior. This concludes our second proof of Theorem 1. We end with a challenge.

Figure 7: The first arrow shows an isotopy of N(K). The second arrow focuses attention on a spine (different from K) for the handlebody.

Challenge. Modify the preceding proofs to show the nontriviality of the Kinoshita–Wolcott θn-graphs [26]. (These graphs generalize Suzuki’s [22] generalization of the Kinoshita graph.)

References

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