Minimally Knotted Spatial Graphs are Totally Knotted

Abstract

Applying Jaco's Handle Addition Lemma, we give a condition for a 3-manifold to have an incompressible boundary. As an application, we show that the boundary of the exterior of a minimally knotted planar graph is incompressible.

Bridge position and the representativity of spatial graphs

“…Then the innermost disk D in S bounded by an innermost loop of F ∩S is a compressing disk for F since S is an essential tangle decomposing sphere. It follows that r(F, Γ) ≤ |∂D ∩ Γ| ≤ |Γ ∩ S| < n. Figure 5 shows a pair (F, Γ) of a genus two Heegaard surface F and a minimally knotted (hence totally knotted by [15]) handcuff graph Γ which is in a Morse bridge position. To show r(F, Γ) ≥ 2, let D i 1 and D i 2 be meridian disks in a handlebody V i as in Figure 5 and then, following Lemma 2.3, we have a 4-punctured sphere F i 1 .…”

Tunnel number and bridge number of composite genus 2 spatial graphs

“…On the other hand, it turns out that a genus 2 graph with a bridge sphere having four or fewer punctures, or a vp-bridge torus with two or fewer punctures cannot be Brunnian. We'll use the following theorem of Ozawa and Tsutsumi [24], though not in full generality. The version for genus 2 graphs is not difficult to prove directly.…”

Newly Found Forbidden Graphs for Trivializability

“…A spatial embedding of a planar graph G is said to be minimally knotted if it is non-trivial but f | H is trivial for any proper subgraph H of G. A spatial embedding f : G → S 3 is said to be totally knotted if the natural homomorphism i * : π 1 (∂E(f (G))) → π 1 (E(f (G))) is injective, where E(f (G)) denotes the spatial graph-exterior. The second and the last authors showed the following in [1]. Proof.…”

“…18. In the eye graph there exist exactly three pairs (γ (1,4,5), γ(6, 9, 12)), (γ (2,3,4), γ(7, 9, 11)), (γ(1, 2, 6, 7), γ(8, 10)) of disjoint cycles, where γ denotes the cycle uniquely determined by the specified labels of edges. It is easy to check that each of the six cycles appeared in these pairs forms a trivial knot, and each corresponding 2-component link is split.…”