Toroidal embeddings of abstractly planar graphs are knotted or linked

We give explicit deformations of embeddings of abstractly planar graphs that lie on the standard torus $T^2 \subset \mathbb{R}^3$ and that contain neither a nontrivial knot nor a nonsplit link into the plane. It follows that ravels do not embed on the torus. Our results provide general insight into properties of molecules that are synthesized on a torus.

The main result (Theorem 1) proves this conjecture without assuming 3-connectivity or simpleness using topological graph theory. The argument gives an explicit deformation from embeddings of abstractly planar graphs on the torus that contain neither a nontrivial knot nor a nonsplit link into the plane. A much shorter but less intuitive proof that relies on deep theorems of topological graph theory is given in [25]. The argument on hand not only presents a selfsufficient argument but will hopefully also give the reader a better feeling for the nature of graphs that are embedded on the torus.
Theorem 1 (Existence of knots and links). Let G be an abstractly planar graph and f : G → R 3 be an embedding of G with image G. If G lies in the torus T 2 and contains no subgraph that is a nontrivial knot or a nonsplit link, then the embedding f is planar.
Corollary (Ravels do not embed on the torus). Every nontrivial embedding of a θ n -graph on the torus contains a nontrivial knot.
The topological structure of the surface on which the spatial graph is embedded is crucial for the theorem. For all closed orientable surfaces of genus g > 1, there exist examples of abstractly planar spatial graphs that are neither knotted or linked nor embeddable on a closed orientable surface with genus less than g. A famous example for the closed oriented genus two surface is Kinoshita's theta curve [27] (Figure 1). A big class of examples is given by ravels: A ravel corresponds to an abstractly planar spatial graph and does not embed on the torus by Theorem 1 as a ravel is nonplanar but contains no nontrivial knot or nonsplit links by definition. As every spatial graph embeds on a compact closed surface of some genus, it follows that ravels correspond to spatial graphs that are neither planar embedded nor knotted or linked but which are realisations of abstractly planar graphs on higher genus surfaces. Again, one need arbitrarily high genera to accommodate all ravels which can be shown using a Borromean construction as given by Suzuki [28]. Castle, Evans and Hyde [1] proved that polyhedral toroidal molecules which contain a nontrivial knot are chiral. The chirality of polyhedral toroidal molecules which contain a nonsplit link is shown in [29]. Note that topological chirality implies chemical chirality.
Theorem 2 (Chirality [1], [29]). Let G be a simple 3-connected abstractly planar graph and f : G → T 2 ⊂ R 3 be an embedding of G with image G on the torus T 2 . If G ⊂ T 2 is nonplanar embedded, then G is topologically chiral in R 3 .
Acknowledgements. The first author thanks Tom Coates, Erica Flapan, Youngsik Huh, Stephen Hyde, Danielle O'Donnol, Makoto Ozawa, and Kouki Taniyama for helpful comments and discussions. Matt Rathbun is particularly thanked for long discussions and the outline of the second step in the proof. The research was financially supported by the Roth studentship of Imperial College London mathematics department, the DAAD, the Evangelisches Studienwerk, the Doris Chen award, and by a JSPS grant awarded to Kouki Taniyama.
2. Proof of the theorem by giving an explicit isotopy 2.1. Outline of the proof of Theorem 1 and preparations. The idea of the proof of Theorem 1 is the following: Let G be an abstractly planar graph and G be the spatial graph that is the image of the embedding f : G → T 2 . Assume that G contains neither a nontrivial knot nor a nonsplit link. Find a general construction for an explicit ambient isotopy in R 3 from the spatial graph G to a spatial graph G which is embedded in the plane R 2 ⊂ R 3 . This demonstrates that any embedding of an abstractly planar graph that is embedded on the torus and contains neither a nontrivial knot nor a nonsplit link is planar. The ambient isotopy is illustrated in Figure 2.
To construct the ambient isotopy in R 3 from the embedding G ⊂ T 2 of an abstractly planar graph G that contains neither a nontrivial knot nor a nonsplit link to a spatial graph G which lies in the plane R 2 ⊂ R 3 , we first note that it is sufficient to consider connected graphs by Lemma 1. Furthermore, we will see in step (1) of the proof that it is sufficient to restrict to spatial graphs that have a subgraph of the form T (1, n), n > 0 as otherwise, the graphs would be abstractly nonplanar or would be already planar embedded. Then, the desired ambient isotopy of the graph consists of three deformations. The first one is a twist around the core of the torus which transforms T (1, n) into the longitude l. This twist is described in step (3).
Step (2) of the proof is a technicality that ensures the existence of the twist given in step (3) by arguing that there exists a meridian of T 2 that intersects G in only one point. As G is abstractly planar by assumption, it follows from Theorem 3 that the conflict graph of G with respect to l is bipartite. The bipartiteness of the conflict graph is used together with the property of the spatial graph being embedded on the torus to show in step (4) and step (5) of the proof that a second ambient isotopy can be performed. This ambient isotopy is given in step (6). It rotates the spatial fragments in space around the longitude l, so that G is ambient isotoped to a spatial graph that is embedded on the surface 8 × S 1 , so that conflicting spatial fragments lie in different components of (8 × S 1 ) \ (P × S 1 ). We use 8 to denote a loop with one point P of self-intersection. Step (7) shows that each spatial fragment can individually ambient isotoped to a planar embedded spatial fragment in space, independently from all other spatial fragments. The combination of the ambient isotopies given in step (3), step (6) and step (7) gives the desired deformation of G into the plane which proves Theorem 1. To work on the torus, define the following: A meridian of a solid torus T is an essential simple closed curve in ∂T that bounds a disc in T . (An essential simple closed curve in ∂T does not bound a disc in ∂T .) The preferred longitude is a simple closed curve in ∂T that intersects the meridian once and has linking number zero with the core of the torus T . For the standard torus T 2 in R 3 define the meridian and preferred longitude analogously by taking the interior of the solid torus to be the bounded component of R 3 \ T 2 . Whenever we use the term longitude in this paper, we will refer to the preferred longitude. To describe knots and links on the torus, we define a torus knot or torus link to be a knot or link that is embedded on the standard torus T 2 following the longitude of the standard torus p times and the meridian q times. Those knots or links are denoted by T(p,q) with p, q integers. Therefore, a meridian is denoted by m = T (0, 1) and a longitude by l = T (1, 0). An unknot that bounds a disc in T 2 is denoted by T (0, 0). The concept of the 'conflict graph' defined below will be needed (compare Figure 3), to ensure abstract planarity of the considered graphs during the proof: A cycle is a simple closed path in a graph or in a spatial graph. Let C be a cycle in a graph G. The connected components f i of the graph G \ C are called fragments of G with respect to C and two fragments f i and f j conflict if and only if at least one of the following conditions is satisfied: • There exist three points on C to which both components f i and f j are attached to.
• There exist four interlaced points v 1 , v 2 , v 3 , v 4 on C in cyclic order so that f i is attached to C at v 1 and v 3 and f j is attached to C at v 2 and v 4 . Two sets {p 1 . . . p n } and {q 1 . . . q m } of points on C are nested, if all points of one set lie in between two points of the other set. The elements of two nested sets do not conflict by definition. Let f i be a fragment of G with respect to C. Call the set of points in which f i is attached to C the endpointset v( f i ) of the fragment f i . Note that vertices of C can be elements of different endpointsets v( f i ) and v( f j ) where f i and f j might or might not conflict. Two fragments f a and f b are nested if their endpointsets on C are nested. If a basepoint on C is given and a pair of nested fragments, call the fragment outer fragment whose endpoints are first reached form the basepoint. Its endpointset is called outer points. Call the other fragment inner fragment and its endpointset inner points. The restriction of a spatial graph to a fragment is a spatial fragment. The conflict graph of a cycle C in a graph G is constructed by introducing a vertex u i for every fragment f i of G with respect to C and adding an edge between the vertices u i and u j if and only if the fragments f i and f j conflict. Fix a point p on C, take an orientation of C and a parametrisation f : that respects the orientation. Then for two points a and b on C, we say that During the following proof, we will use the abstract planarity of G in form of the following statement: Theorem 3 (Tutte's Theorem [30]). A graph G is abstractly planar if and only if the conflict graph of every cycle in G is bipartite.
A graph is bipartite if its vertices can be divided into two disjoint sets S 1 and S 2 such that every edge of the graph has one endpoint in S 1 and the other endpoint in S 2 . ( The sets v a1 , . . . , v ar and v b1 , . . . , v br in the second case above are nested; the inner points are the points v b1 , . . . , v br . Lemma 1 (Connectivity Lemma [25]). The image G of an embedding f : G → T 2 ⊂ R 3 of a graph G with n > 1 connected components on the torus T 2 contains either a nonsplit link, or contains no nonsplit link and decomposes into n disjoint components of which at least n − 1 components are planar embedded in R 3 .

Proof. (Theorem 1)
(1) Reducing the types of spatial graphs: We show that it is sufficient to consider connected abstractly planar spatial graphs that are embedded on the torus T and do neither contain a nontrivial knot nor a nonsplit link, but do contain a torus unknot T (1, n), n > 0. We can assume that the graph G is connected by Lemma 1. Furthermore, we claim that it is sufficient to consider spatial graphs G that contain a torus unknot of the form T (1, n), n > 0 (respectively T (n, 1)) since if the only knot types contained in G are T (0, 0), T (0, 1) and T (1, 0), then G is planar. We see this below by a case study where we restrict the knot types that occur in the spatial graph G. Let #K denote the number of disjoint copies of the knot K. If the only knot type contained in the spatial graph G is T (0, 0), there exists a meridian and a longitude of the torus that do not intersect G. Therefore, G is planar.
There exist either a meridian or a longitude of the torus that does not intersect G. Therefore, G is planar.  Note that the case where G is the union of T (1, n) and a longitude does not fulfil the assumptions since there exists a meridian that intersect G in one point only (there are n such meridian, namely one for each intersections of T (1, n) with the longitude). To construct a contradiction, assume that every meridian of T 2 intersects G in at least two points. Cut T 2 along a meridian that intersects the graph with minimal number to get a cylinder [0, 1] × S 1 . Define the projection function f : Let S be the set of all pairs of pairwise different paths on the cylinder where one path forms a knot of type T (1, n) and the other path has an endpoint in f −1 (0) (Figure 5a). The set S is non-empty because there exists at least one cycle T (1, n) by step (1) and a second path π with endpoint in f −1 (0) by the assumption that every meridian of T 2 intersects G in at least two points. Note that π either intersects T (1, n) or has the second endpoint on f 1 (1) otherwise since the number of intersections between f −1 (0) ∩ G is minimal by assumption. Now take a pair of paths in S which maximises the value f (t n ) where t n is the intersection point of the two paths in the pair.
If T (1, n) and π do not intersect, then there exist two disjoint paths on the cylinder. There exists a nontrivial knot or a nonsplit link since the graph is connected as we show now: If π is a cycle, a nonsplit link is formed by T (1, n) and π (Figure 5b). So let π be a path with endpoints (0, α) ∈ f −1 (0) and (1, β) ∈ f −1 (1), α β. As the graph intersects f −1 (0) and f −1 (1) in a minimal number, wlog we can assume that there exist paths π 0 from (0, β) and π 1 from (1, α) to T (1, n) or to π (since the graph is connected). (a) If π 0 intersects π in a point v before intersecting T (1, n) (Figure 5c), denote the segment of π between v and (1, β) by π β . Such a graph contains a nonsplit link where one component is T (1, n) and the other consists of π 0 and π β . If π 1 intersects π in a point v before intersecting T (1, n) (Figure 5d), denote the segment of π between (0, α) and v by π α . Again, such a graph contains a nonsplit link where one component is T (1, n) and the other consists of π α and π 1 . Up to this point the arguments applies to k = T (1, 1) as well. (b) If both paths π 0 and π 1 intersect T (1, n) before they intersect π (Figure 5e), denote the segment (possibly a point) of T (1, n) that lies between the endpoints of π 0 and pi 1 by T i . The cycle that runs through π, π 0 , T i and π 1 is a nontrivial knot for n > 1.
If T (1, n) = T (1, 1), there are possibilities to connect (0, β) and (1, α) to T (1, 1) without introducing a nontrivial knot or a nonsplit link. But considering T (1, 1), we could exchange the meridian and the longitude in the argument. This gives the extra condition that not only each meridian but also each longitude intersects the graph in at least two points. An elementary investigation shows directly that Theorem 1 is valid in those cases. If T (1, n) and π do intersect, there exists a point of maximal intersection t max , 0 < f (t max ) < 1. Now, consider the set of all paths that are different from T (1, n) and that have one endpoint on f −1 (1). Take one path of this set which minimises the value of f (t min ) where t min is the point of intersection between the path and T (1, n) and call that path π min . Denote the component of G − T (1, n) that contains π min (π max ) by c min (c max ). If f (t min ) < f (t max ), the argument is very similar to the case above.
So let us finally consider the case where f (t max ) ≤ f (t min ) ( Figure 5f). As before, since the graph intersects every meridian at least twice by assumption, there exists a point x π in the graph so that f (x) = f (t max ). No path disjoint from T (1, n) containing x can connect to c min as this would contradict the maximality of t max . Similarly, no path disjoint from T (1, n) containing x can connect to c max as this would contradict the minimality of t min . Therefore, every path through x connects to T (1, n) before and after t max (and similarly before and after t min ) and is disjoint from c max and c min . Take such a path and denote it by σ. Replacing the segment T i of T (1, n) that runs between the endpoints of σ with σ gives us a torus knot T (1, n). Since T (1, n) is disjoint from π max , there exists a path π c consisting of π and T i disjoint from T (1, n) until some time after t max . This contradicts the maximality of the pair we selected from S .
(3) Deforming T (1, n) to a longitude and finding a diagram D R of G: Recall that the spatial graph G contains no nontrivial knots or links by assumption. By putting G into a general position, there exists a meridian m of the torus that intersects G in exactly one point P. The existence of P is given by the previous step (2). The point P lies on T (1, n) since every meridian intersects T (1, n). Now perform the following twist: By cutting the torus T along the meridian m, then twisting it n-times around the core of T and identifying the same points again afterwards, an ambient isotopy i : G → G of the spatial graph is induced that maps T (1, n) onto the longitude l = T (1, 0) of a new torus T (not isotopic to T ). We denote the image i(G) on T by G . Restricted to the meridian m of T , the isotopy is the identity by construction. Therefore, G and m intersect in l only. Define Z := T \ (m \ P). Let us furthermore consider the diagram D R of G that we obtain the following way: We project Z onto a rectangle R = (0, 1) so that l is the bottom line of R and take a generic position so that the top line of R does not intersect the graph in vertices. As usual, we indicate the over and under crossings of G ( Figure 6). Wlog we can assume that the diagram is regular, i.e. the diagram has only finitely many multiple points which all are transversal double points and no vertex is mapped onto a double point. Furthermore, let D R be a reduced diagram, i.e. a diagram with the minimal number of crossings that can be achieved from projecting Z onto a rectangle R as described. (4) Showing that pairs of spatial fragments in a reduced diagram D R have no crossings if they are nonconflicting and only one type of crossings if they are conflicting: By Tutte's Theorem 3, any cycle of an abstractly planar graph has a bipartite conflict graph. As G is abstractly planar by assumption, it follows that the conflict graph of l in G is bipartite. As the graph G is connected, all fragments of G with respect to l fall into two sets S 1 and S 2 so that fragments which are elements of the same set do not conflict. Choose an orientation of l. Starting at the point P, enumerate along the orientation all vertices v 1 , . . . , v k of l that are endpoints of fragments of G with respect to l. (P might or might not be the element of a fragment's endpointset.) Denote the spatial fragments of G by f 1 , . . . , f n respecting the orientation of l and so that v 1 ∈ v( f 1 ) . For each fragment f i of G with respect to l, assign its endpointset v( f i ) = v l i ≤ · · · ≤ v r i ⊂ {v 1 , . . . , v k }. We show that each pair of spatial fragments f i and f j has either no crossings or wlog f i over crosses f j at every crossing of the diagram D R : (a) This is clear if i = j since a fragment overcrosses (as well as undercrosses) itself at every self-crossing. (b) If i j and f i and f j do not conflict, they have no crossings in a reduced diagram D R since one of the cases in Remark 1 holds: It is clear that there are no crossings between f i and f j in the first case of Remark 1. In the second case, let wlog f j be the inner fragment. Then, f j lies entirely inside the cell of R \ f i that has [v l j , v r j ] as part of the boundary. It follows from the connectivity of fragments that f i and f j have no crossings in a reduced diagram D R .
(c) If i j and f i and f j are conflicting, only crossings of one type can occur. Since the entire spatial graph G is an embedding in Z (as well as in T ), it is not possible that both crossing types between f i and f j occur (Figure 7). We remark that if i j and f i and f j are conflicting, they have at least one crossing in D R : Without affecting D R , a fragmentf can be added in (0, 1) × [−1, 0] ∪ {(0, 0)} ∪ {(1, 0)} so thatf conflicts with both f i and f j . Then, the subgraph consisting of l, f i , f j andf is abstractly nonplanar by Tutte's Theorem 3. The fragmentf does not lie in D R by construction. Therefore, there exist neither crossings betweenf and f i nor betweenf and f j . It follows from the abstract planarity that a crossing between f i and f j must exists.
. Assume that f i over cross f j1 in a non-empty set of points but under crosses f j2 in a non-empty set in D R . By the connectivity of fragments, there exists a path p : [0, 1] → f i ∪ v ia ∪ v ib , with v ib ∈ v( f i ), a b with endpoints p(0) = v ia and p(1) = v ib , that over cross f j1 in p(t 1 ) and under crosses f j2 in p(t 2 ). To change from an over to an under crossing, a path in D R has to intersect either the bottom or the top line of R . As p does only intersect the bottom line in v ai and v ib , p must have an intersection point p(t) with the top line of R , so that t 1 < t < t 2 . But as f j1 is nested in f j2 , it follows that v l j2 < v ia < v r j2 . Therefore, p starting from v ia overcrosses f j1 as well as undercrosses f j2 before it can intersect the top line of R . This is a contradiction.  The conclusions made in (3) and (4) allow rotations of the spatial fragments of G around the longitude l in R 3 which gives an ambient isotopy from G to a realisation G in which all fragments that are elements of S 1 lie on the torus T and all elements of S 2 lie on a second torusT . The torusT is glued to T in l along a longitude.
The rotations can be chosen as follows (compare Figure 9): Let { f 1 } be the set of all fragments of G with respect to l (Figure 9a) and order some variables as 1 ≤ k 11 ≤ · · · ≤ k t1 ≤ n. If a step during the procedure which is decribed below cannot be performed, continue with the next step.
(a) Fix f 1 , . . . , f k 11 −1 where f k 11 is the first fragment that conflicts with a fragment f i , 1 ≤ i ≤ k 11 − 1 (Figure 9a). Then rotate all spatial fragments that are elements of the set { f k 11 ,m 1 } rigidly in R 3 about π around the longitude l. They are now embedded onT . It is possible to choose the direction of the rotation so that no spatial fragments pass through each other since the spatial fragments in { f k 11 ,m 1 } have only over crossings (respectively only under crossings) with f 1 . . . f k 11 −1 by construction and by steps (3) and (4) (Figure 9f). This gives a realisation G of G which is ambient isotopic to G. By (3), a pair of spatial fragments of G has no crossings in a reduced diagram The diagram D R is the composition of two diagrams defined as in (2) (Figure 9.g). (7) Showing that a single spatial fragment f i has no crossings in a reduced diagram D R of G ( Figure 14): Each spatial subgraph f i ∪ [v li , v ri ] is embedded on a sphere S 2 i . To see this, let wlog f i be embedded on T and take two meridians of T intersecting l in v li and v ri so that the meridians do not intersect G except in v li and v ri . Then glue two meridional discs in, one in each meridian. S 2 i consists of the two meridional discs and the part of T where f i is embedded in that lies between the meridians. We now want to ambient isotope f i inside the ball bounded by S 2 i where we take the inside to be the component of R 3 \ S 2 i that does not intersect G . This isotopy will transform the diagram D R to a diagram in which the subdiagram corresponding to f i that is crossing free. Take the subdiagram D i R of the diagram D R that corresponds to f i ∪ [v li , v ri ]. Peform all reducing Reidemeister I & II moves on it (Figure 14, a-b). Simplify D i R by isotopy whenever possible. We can assume that the diagram D i R has edges crossing the top line of R ; otherwise it is crossing free and we are done. If one of those edges intersects the top line of R more then once or runs through a crossing in D i R , we subdivide the edge by adding vertices (Figure 14, a-b). Therefore, we can assume that an edge of D i R that intersects the top line of R intersects it only once and is crossing free in the diagram. (This does not affect our argument since if a subdivision of a spatial graph is planar embedded, the spatial graph itself is.) Denote the edges crossing the top line of R by {e 1 , . . . , e˜k}. Each edge e j has two endpoints, e ju and e jo (fat in Figure 14b). By the connectivity of G , there is at least one edge e j in {e 1 , . . . , e˜k} for which there exists a path in f i from an endpoint of the edge wlog e ju to an element v ∈ v( f i ) that does not intersect the top line of R . Denote such a path with endpoints e ju and v by p(e ju , v) (fat in Figure 14b). The set of all such paths is called {p( ju)}. The set {p( jo)} is analogously defined for the endpoint e jo of e j . If an edge e j ∈ {e 1 , . . . , e˜k} has an endpoint e ju or e jo which is not the endpoint of any path in {p( ju)} or {p( jo)} (e 3u in Figure 14b), e j can be deformed not to intersect the top line of R by moving e j away from the top-line while keeping D R − e j fixed. After this procedure, a subset of edges {e 1 , . . . , e k } ⊆ {e 1 , . . . , e˜k} remains in which every edge e j has endpoints e ju and e jo so that there exists at least two (possibly constant) paths p(e ju , v a ) and p(e jo , v b ) with v a , v b ∈ v( f i ). Such a path p(e ju , v a ) or p(e jo , v b ) cannot have both over Figure 9. The rotation of fragments described in (6). For clarity, each fragments is chosen not to have crossings with itself.
crossings and under crossings since in this case the path would intersect ∂R which it does not by construction. Also by construction, if a path in {p( jx)}, x = u, o has an overcrossing (or respectively undercrossing), no path in {p( jx)} has an under (respectively over) crossing. In addition, we can assume wlog that there is no edge e j ∈ {e 1 , . . . , e k } that has an endpoint e ju or e jo so that all paths {p( ju)} or {p( jo)} are crossing free (e˜k u in Figure 14b) as in this case we can deform e j away from the top line while keeping D R − e j fixed. Also, if {p( ju)} and {p( jo)} have only one type of crossings (e˜k −2 in Figure 14b), we can again deform e j away from the top line while keeping D R − e j fixed. Therefore, every edge e j has one endpoint e ju so that at least one path in {p( ju)} has crossings which all are undercrossings and one endpoint e jo so that at least one path in {p( jo)} has crossings which all are overcrossings. Furthermore, in {p( ju)} (respectively {p( jo)}) are no paths that have over crossings (respectively undercrossings) by definition of the paths. Assign to each edge e j ∈ {e 1 , . . . , e k } the set w(e ju ) ⊆ {w j1 , . . . , w jl } ( analogously w(e jo ) ⊆ {w j1 , . . . , w jl }) which is the set of points in v( f i ) that are endpoints of at least one element in {p( ju)} (respectively {p( jo)}). The union of the two sets is the endpointset of e j denoted by w(e j ) = {w j1 , . . . , w jl } := w(e ju ) ∪ w(e jo ). The sets in the example in Figure 14c are w(e ku ) = {w 7 , w 8 , w 9 } and w(e ko ) = {w 1 , w 2 , w 7 , w 10 , w 11 , w 12 } . Denote the union of w(e 1u ) ∪ · · · ∪ w(e ku ) by w(u) (and the union w(e 1o ) ∪ · · · ∪ w(e ko ) by w(o)). (In Figure 14c), w(u) = {w 3 , w 4 , w 5 , w 6 , w 7 , w 8 , w 9 }, w(o) = {w 1 , w 2 , w 7 , w 10 , w 11 , w 12 }) See that there exist no four points w u1 , w u2 ∈ w(e ju ) and w o1 , w o2 ∈ w(e jo ) that are interlaced as wlog w u1 < w o1 < w u2 < w o2 : the cycle (l − (w u1 , w o2 ), p(e ju , w u1 ), e j , p(e jo , w o1 ), [w o1 , w u2 ], [w u2 , w o2 ]) would have three pairwise conflicting fragments [w u1 , w o1 ], p(e ju , w u2 ) and p(e jo , w o2 ) ( Figure 10). This contradicts the bipartiteness of G which by Theorem 3 contradicts its abstract planarity. (Also, this graph forms K 3,3 where the points w u1 , e jo , w u2 , w o1 , e ju , w o2 are the vertices.) Therefore, w(e ju ) and w(e jo ) can only be arranged as the endpointsets in Remark 1 and it is allowed to restrict to those cases as done below.  (Figure 14e). There exists a path p j : [0, 1] → D R − {e j } from p j (0) ∈ w(e jo ) to p(1) ∈ w(e jo ) such that p j (0) ≤ w(e ju ) ≤ p j (1) and so that there exist two distinct points p j (t 1 ), p j (t 2 ) with t 1 , t 2 ∈ [0, 1] that have the following property: The diagram D R − {e j } − p j (t 1 ) − p j (t 2 ) splits such that the component C jo containing e jo does not contain any points of [v li , v ri ] (Figure 14e). Furthermore, after performing a Whitney 2-flip on C jo (Figure 14, e-f), the edges of {e j } can be reintroduced to the diagram D R − {e j } without introducing crossings (Figure 14g). A Whitney 2-flip replaces a component by its mirror image as shown in Figure 11, left. This corresponds to a rotation in R 3 about π that would not pass the spatial graph G through itself -even if all edges {e j } are left attached (Figure 11, right and Figure 14, d-g). Therefore, we have an ambient isotopy, that eliminates the crossings of {e j }. After continuing this procedure, all remaining edges of {e 1 , . . . , e k } have endpointsets so that for each edge all elements of w(e ju ) are smaller or equal than all elements of w(e jo ) (or all elements of w(e ju ) are greater or equal than all elements of w(e jo )). See with a contradiciton that it is always possible to find two points p(t 1 ) and p(t 2 ) with the required property as follows. If no two points p(t 1 ) and p(t 2 ) with the required property would exist, there existed a subgraph of G that is abstractly nonplanar (Figure 12): Wlog there would exist two points w(e ju ) w 1 < w 2 ∈ w(e jo ) such that there would exist an edge e w2 between w 2 and an inner point p w2 of a pathp(e jo , w) with w < w 1 . Let 0 < t s < 1 such that p(e jo , w)(t s ) = p w2 . This gives a cycle with non-bipartite conflict graph (alternatively, see K 3,3 ): The cycle consists of the following segments: l − (w, w r j ), p| [0,s] , e w2 , [w 1 , w 2 ], p(w 1 , e ju ), e j , p(e jo , w r j ) and the fragments are (w, w 1 ), p| (s,1) and (w 2 , w r j ). This contradicts abstract planarity by Theorem 3.
Case 2: All endpoints in w(e ju ) are smaller or equal to all endpoints in w(e jo ) (or respectively w(e ju ) ≥ w(e jo )). If {e 1 , . . . , e k } is empty or has one element only, it follows that the diagram of f i has no crossings. So consider the case that k ≥ 2. Wlog, assume that all elements of w(e ju ) are smaller or equal to all elements of w(e jo ). If all elements of w(e ( j+1)u ) are greater or equal to all elements of w(e ( j+1)o ), it follows that w(e jo ) = w(e ( j+1)o ) by construction and the connectivity of fragments (Figure 13, left). If all elements of w(e ( j+1)u ) are smaller Figure 11. Whitney 2-flip and corresponding rotation about π or equal to all elements of w(e ( j+1)o ) but there exists an element w II ∈ w(e ( j+1)u ) that is smaller than an element w III ∈ w(e jo ), there are four elements w I < w II < w III < w IV , w I ∈ w(e ju ), w IV ∈ w(e ( j+1)o ). The paths p(w I , e ju ), e j , p(w jo , e III ) and p(w II , e ( j+1)u ), e j+1 , p(w jo , e IV ) are connected via a path inR by the connectivity of fragments. The subgraph of G (fat in Figure 13, right) consisting of those three paths and l is abstractly nonplanar (it is K 3,3 ) which can again be shown with a similar argument as above by choosing any hamilton cycle of the subgraph (i.e. a cycle that runs through every vertex of the subgraph once) and seeing that its conflict graph is not bipartite.
Therefore, it is shown in step (7) that a reduced form of the diagram D R has no crossings. It follows together with step (6) that G and therefore G is planar embedded. The argument of step (6) relies on step (5) and step (4). By step (3), which can be performed because of step (2), G is also planar embedded. This proves the theorem by step (1). Corollary (Ravels do not embed on the torus). Every nontrivial embedding of θ n -graphs on the torus contains a nontrivial knot.
Proof. As there exist no pair of disjoint cycles in a θ n -graph, such a graph does not contain a nonsplit link. Since θ n -graphs are planar, the statement of the corollary follows directly from Theorem 1.

2.3.
Alternative proof of Theorem 1: The proof of Theorem 1 given above can be differently finished using Theorem 4 by Wu [31]. This is a shortcut in the argument but does not give an explicit deformation.
Theorem 4 (Criterion for planar embeddedness of an anbstractly planar graph [31]). An abstractly planar graph is planar embedded if and only if every cycle in the spatial graph bounds an embedded disc with interior disjoint from the spatial graph.
Alternative proof of Theorem 1.
Proof. Start with step (1) and step (2) of the proof that is given in the section above. Recall that the point P is defined as follows: Beside elementary cases, there exists a meridian of the torus on which the spatial graph G is embedded on so that the meridian intersects the spatial graph in only one point. Take this point to be P. To apply Theorem 4, observe that every cycle in G bounds a disc embedded in R 3 with interior disjoint from G. This is clearly true for any meridian and for any cycle that does not intersect P as G \ P is embedded on a sphere already. We are left to consider cycles C that run through P for which there exists a natural number n so that the cycle has knot type T (n, 1). Let C n be one of those cycles. We can find an ambient isotopy i n of G that transforms C n to a longitude l = i n (C n ) of a new torus i n (T ) (not isotopic to T ), by possibly performing another twist as described in step (3) in the proof above. Call the spatial graph that results from this twist by i n (G). As a longitude bounds a disc internally disjoint from the torus, it follows that the cycle i n (C n ) bounds a disc internally disjoint from i n (G). Since ambient isotopies preserve embedded discs and do not pass them through the graph, it follows that the cycle C n in G bounds a disc internally disjoint from G. For every n ∈ N, we can perform such an ambient isotopy of G. This shows that every cycle in G bounds a disc which is internally disjoint from the spatial graph. As G is abstractly planar by assumption, it follows from Theorem 4 that G is planar embedded.
Remark 2. It is not possible to weaken the assumptions of Theorem 1 as shown by giving counter examples in [25].