Strong Hanani-Tutte for the Torus
SStrong Hanani-Tutte for the Torus
Radoslav Fulek
Department of MathematicsUC San DiegoLa Jolla, California 92093, USA [email protected]
Michael J. Pelsmajer
Department of Applied MathematicsIllinois Institute of TechnologyChicago, Illinois 60616, USA [email protected]
Marcus Schaefer
Department of Computer ScienceDePaul UniversityChicago, Illinois 60604, USA [email protected]
December 2, 2020
Abstract
If a graph can be drawn on the torus so that every two independent edges cross an even number oftimes, then the graph can be embedded on the torus.
A pair of edges in a graph is independent if the edges in the pair do not share a vertex. A drawing of a graphon a surface is independently even if every pair of independent edges cross an even number of times in thedrawing. Independently even drawings of graphs on surfaces are important relaxations of graph embeddingswith a wide array of applications, which we will discuss in detail in Section 7.In the plane, there is a beautiful characterization of planar graphs known as the Hanani-Tutte theoremwhich says that a graph is planar, that is, it has an embedding in the plane, if and only if it has anindependently even drawing in the plane. Equivalently, any drawing of a non-planar graph in the plane mustcontain two independent edges that cross oddly.There are several proofs of the Hanani-Tutte theorem, including the original 1934 proof by Hanani andthe 1970 proof by Tutte, see [27] for more references. We also know that the result remains true for theprojective plane [26, 4]. On the other hand, counterexamples were found recently which show that theHanani-Tutte theorem does not extend to orientable surfaces of genus more than 3 [8]. This opened up apath to the study of approximate versions of the Hanani-Tutte theorem [9, 11].We complement these results by proving that the Hanani-Tutte theorem does extend to the torus.
Theorem 1.1.
Let G be a graph. Suppose that G can be drawn on the torus so that every two independentedges cross evenly. Then G can be embedded on the torus. Among orientable surfaces, this leaves only the double and triple torus, for which we do not know whetherthe Hanani-Tutte theorem holds. For non-orientable surfaces, all cases starting with the Klein bottle areopen.Our approach extends and refines techniques developed in [4, 26] and other papers, in particular inSections 3 and 4. After that, in Sections 5 and 6 our proof requires new tools. A sphere with a crosscap. We assume that the reader is familiar with the basic terminology of drawings and embeddingsin surfaces. For background see [23, 6]. a r X i v : . [ c s . D M ] N ov e carry out the proof of Theorem 1.1, by first, carefully defining a partial order on drawings of graphs onthe torus, and then showing that no minimal counterexample to an appropriate strengthening to Theorem 1.1(cf. Theorem 3.2 in Section 3) with respect to our partial order exists. Our reductions are mostly, but notalways, minor preserving. In the base case of our proof, we work with drawings of subdivisions of K ,t , t ≤ K , possibly with some attached subvidided edges in the case of K ,t . The implementation ofthe reduction steps goes through relatively smoothly by using fairly natural redrawing tools up to the pointwhen the graph is formed by a subdivision of K ,t or K with additional simple bridges, in which case anextensive case analysis seems to us the only way to proceed. The bulk of the paper is dealing with how tohandle such graphs. The main difficulty in this part of the proof lies in performing the reduction steps sothat the case analysis becomes manageable. The Hanani-Tutte theorem [2, 36] has been known with many proofs for the plane for a while, for a surveyof known results see [33]. It was shown to be true for the projective plane by Pelsmajer et al. [26] using theexcluded minors for the projective plane, and later directly, without recourse to excluded minors by ´E. Colinde Verdi`ere et al. [4]. Working with excluded minors to establish Hanani-Tutte style results for surfacesstops being useful at that exact point, since we do not know the complete list of excluded minors for thetorus or any higher-order surfaces.It is known that a weak version of the Hanani-Tutte theorem is true for all surfaces.
Theorem 1.2 (Weak Hanani-Tutte for Surfaces [1, 29]) . If a graph can be drawn in a surface S (orientableor not) so that every pair of edges crosses an even number of times, then the graph can be embedded in S ,without changing the rotation system. Unfortunately, the available proofs of the weak version do not give any insight on how to establish thestrong version on a surface whenever this is possible. It is too weak, for example, to settle the deceptivelynon-trivial problem of showing that a graph which can be drawn in a surface so that the only crossings arebetween dependent edges, can actually be embedded in that surface.There is also work on applying Hanani-Tutte to different planarity variants, see, for example, [13, 14, 15,17, 32]. A variant of the (strong) Hanani-Tutte theorem in the context of approximating maps of graphs,which works on any surface, was recently announced in a paper co-authored by the first author [7].
For a particular drawing D of a graph G on a closed surface S , we make the following definitions: Two edgesform an odd pair if they cross each other an odd number of times. A subgraph (or single edge) is even ifnone of its edges are in any odd pairs. A subgraph (or single edge) is iocr- C on a closed surface S is non-essential if it forms the boundary of a (not necessarilyconnected) closed sub-manifold of S , or equivalently, if its complement in S can be two-colored so that pathconnected components sharing a non-trivial part of C receive opposite colors. In terms of the homology,a closed curve C on S is essential if its homology class does not vanish over Z . A closed curve separatesthe surface if removing the curve from the surface disconnects the surface. A curve is simple if it is free ofself-intersections. A closed simple curve C separates the surface if and only if C is non-essential. An essentialsimple closed curve is therefore non-separating .We apply the same terminology to cycles in graph drawings, since a cycle determines a closed curve. Wesay a subgraph in a drawing is essential if it contains an essential cycle.Every closed curve in the plane is non-essential. The existence of essential curves on every other closedsurfaces poses the main challenge in proving variants of the Hanani-Tutte theorem on higher genus surfaces.The reason is that a pair of essential curves can cross an odd number of times, for example an equator and ameridian on the torus, see, for example, [11, Section 5.1]. A non-essential curve, on the other hand, crosses2very other curve an even number of times, which is easy to see using the two-coloring of the complement ofthe curve from the above definition. We will use these two facts extensively.To make this more precise, suppose we are given a drawing D of a graph G . We associate with everyclosed walk W of G a 2-dimensional vector H ( W ) over Z representing the 1-dimensional homology class of W in D , intuitively, we count how often W crosses an equator and meridian of the torus, modulo 2. Thevectors H ( W )’s satisfy the following properties, where ⊕ denote the symmetric difference applied to edges.(i) H ( W ) = (0 ,
0) if and only if W is non-essential;(ii) H ( W ⊕ W ) = H ( W ) + H ( W ), for every pair of closed walks W and W ; and(iii) the number of crossings over Z between closed curves that are drawings of closed walks W and W in D (possibly after a small perturbation to achieve a generic position) is H ( W ) T (cid:18) (cid:19) H ( W ),that is, W and W cross an odd number of times if and only if they belong to different non-vanishinghomology classes.The following lemma rephrases a basic property of essential curves in a surface, see, for example, [11,Section 5.1] or [22, Proposition 4.3.1]. Lemma 2.1.
The family of essential cycles in a graph drawn in a (compact) surface satisfies the -pathcondition: given three internally disjoint paths with the same endpoints, if two of the cycles formed by thepaths are non-essential then so is the third.Proof. Let γ , γ , γ be three curves from point x to point y in the surface, and for each i let γ − i be thethe reversal of γ i (i.e., γ i taken from y to x ). If γ γ − (the concatenation of γ and γ − ) and γ γ − arenon-essential then γ γ − γ γ − is also non-essential. The claim follows since γ γ − γ γ − is in the same1-dimensional homology class over Z as γ γ − . Edge-vertex move. An edge-vertex ( e, v )- move (also known as van Kampen’s finger-move , or edge-vertexswitch ) is a generic deformation of the edge e in a drawing of G changing the parity of the number ofcrossings between e and all the edges incident to v , without changing any other crossing parities. An edge-vertex switch is performed as follows. We cut e by removing a small portion that is not crossed by any edgefrom its relative interior. Let a and b denote the severed ends of e . We draw a small circle C around v andwe cut C in the same manner as e . Let a (cid:48) and b (cid:48) denote the severed ends of C . We recover the edge e byconnecting a and b with a Jordan arc obtained by concatenating C and a pair of non-crossing arcs joining a with a (cid:48) , and b with b (cid:48) drawn close to each other. Edge flip. An edge flip , or flip , in a drawing is a redrawing operation that happens near a vertex, whichtakes two consecutive edges in the rotation at that vertex and exchanges their position in the rotation atthat vertex. As a result, the parity of crossing between the two flipped edges changes, without affecting anyother crossing parities. Edge contraction and vertex split. A contraction of an edge e = uv in a topological graph is anoperation that turns e into a vertex by moving v along e towards u while dragging all the other edgesincident to v along e . If e is even, then contracting e in this fashion does not change the parity of crossingbetween any pair of edges. Contraction may introduce multi-edges or loops at the vertices.The three moves we have just described may introduce self-crossings of edges. Such self-crossings areeasily resolved [16, Section 3.1]: remove the crossing, and reconnect the four ends so that the edge consistsof a single curve. This redrawing does not affect the topology of the cycles in the drawing, that is, essentialcycles remain essential and non-essential cycles remain non-essential.We will also often use the following operation which can be thought of as the inverse operation of theedge contraction in a topological graph. A split of a vertex v in a drawing of a graph G is the operation3hat replaces the vertex v by two vertices v (cid:48) and v (cid:48)(cid:48) drawn in a small neighborhood of v joined by a shortcrossing free edge so that the neighbors of v are partitioned into two parts according to whether they arejoined with v (cid:48) or v (cid:48)(cid:48) in the resulting drawing, the rotations at v (cid:48) and v (cid:48)(cid:48) are inherited from the rotation at v ,and the new edges are drawn in the small neighborhood of the edges they correspond to in G . -Drawings on the Torus In this section we establish some of the basic redrawing tools for iocr-0-drawings in Section 3.1, some fromearlier papers, show how to work with disjoint essential cycles, and give a first application of these tools tothe 1-spindle.
In this section, we collect some redrawing tools which (mostly) work on all surfaces, and may be useful forestablishing Hanani-Tutte type results for surfaces other than the torus. Our terminology is particularlyfocused on the orientable surfaces.We say that a vertex v in a drawing D of a graph on a surface is even if every pair of edges incidentto v cross each other an even number of times; otherwise, it is odd . A drawing D of a graph G in anorientable surface S is compatible with a drawing D of G in S if every even vertex in D is even in D ,and the rotation at even vertices in D is preserved in D . Note that the compatibility relation is transitive,but not necessarily symmetric. We can define a notion of connectivity on even vertices: two even vertices u and v in a drawing of a graph are evenly connected if there exists a path connecting u and v consistingonly of even vertices. An evenly connected component in a drawing is a maximal connected subgraph in theunderlying abstract graph induced by a set of even vertices.A drawing D of a graph G in an orientable surface S is weakly compatible with a drawing D of G in S if every even vertex in D is even in D , and for every evenly connected component C in D either therotation in D at every vertex v ∈ V ( C ) is the same as the rotation of v in D , or the rotation in D atevery vertex v ∈ V ( C ) is the reversal of the rotation of v in D . Note that the weak compatibility relationis transitive, but not necessarily symmetric, similarly as compatibility.With these notions we can state a version of the Hanani-Tutte theorem on the plane that implies boththe weak and the strong version. While the result itself follows directly from the proof of the Hanani-Tuttetheorem in [27], it was first explicitly stated, and given a new proof, in [10]. Theorem 3.1 (The Unified Hanani-Tutte theorem [10]) . If G has an iocr- -drawing in the plane, then G has an embedding in the plane compatible with the iocr- -drawing. Theorem 3.1 does not generalize to arbitrary surfaces [8, Theorem 7]. However, this no longer the case ifwe replace in the statement of the theorem the notion of compatibility with the notion of weak compatibility.Indeed, we prove the following strengthening of our main result Theorem 1.1.
Theorem 3.2. If G has an iocr- -drawing in the torus, then G has an embedding in the torus that is weaklycompatible with the iocr- -drawing. Theorem 3.2 follows by Lemma 5.5 and the strengthened version of Theorem 1.1 stated at the beginningof Section 6.3, where we prove our main result.The following lemma is one of our main redrawing tools.
Lemma 3.3.
Let D be an iocr- -drawing of a graph G on a surface S , and let C be a finite collection ofdisjoint, closed simple curves so that S − (cid:83) γ ∈C γ is connected. If every edge of D crosses every curve in C an even number of times and every curve in C contains at most one vertex of G , which we do not count This will not conflict with the usual degree-based definition of even and odd for vertices, since we will not be using thatterminology. s a crossing, then G has a compatible iocr- -drawing D (cid:48) on S in which D (cid:48) is disjoint from any curve in C except for at the vertices of G which still line on the same γ ∈ C , they belonged to in D . Every cycle of G is essential in D if and only if it is essential in D (cid:48) . We apply a 1-dimensional variant of the Whitney trick from [16, Lemma 2].
Proof.
We process simple closed curves in C one at a time. Let γ ∈ C be such a curve. By assumptionevery edge e crosses γ evenly. We cut every such e at every crossing with γ and reconnect every pair ofconsecutive severed ends of e along γ on both sides of γ by arcs closely following γ and avoiding, if it exists,the vertex of G on γ (here we need that each curve contains at most one vertex). This operation splits e into a set of arcs consisting of finitely many closed arcs and an arc connecting its former end vertices. Since C does not separate the surface, we can reconnect the components of e by pairs of parallel arcs which do notintersect any curve in C , in particular γ itself, thereby turning e into an arc joining its end vertices, makingthe drawing proper again. Repeating this for every e crossing γ and every γ ∈ C , we obtain a drawing D (cid:48) of G which is compatible with the original drawing and in which all curves in C are free of crossings with edgesof G ; if γ ∈ C contains a vertex of G in D (cid:48) , it contained the same vertex in D .The next lemma, Lemma 3.4, originally stated for the projective plane, already occurs in [26]; essentiallythe same proof (based on ideas from [27, Theorem 3.1]) allows us, for any compact surface w/o boundary,to “clear” an essential cycle—or even a set of disjoint cycles that forms a vanishing homology class over Z . Lemma 3.4.
Let D be an iocr- -drawing of a graph G on a surface S . Suppose that C is an essential cyclein D . Then there exists an iocr- -drawing D (cid:48) of G compatible with D on S in which C is crossing free, andevery cycle is essential in D (cid:48) if and only if it is essential in D .Proof. We start by modifying the drawing so that every edge of C is even. This can be achieved by flippingedges incident to odd vertices of C : For any odd vertex on C first ensure that the two C -edges incident tothe vertex cross evenly, and then move the remaining edges at the vertex so they cross both C -edges evenly;this can be done using edge-flips; we do not need to change the rotation at even vertices.Let e ∗ be an edge of E ( C ). We then contract all edges in E ( C ) − e ∗ one by one as follows: each edge uv ∈ E ( C ) − e ∗ is contracted by moving one of its end vertices, say u , along uv to v , extending other edgesincident to u (as in [27]). Since all edges cross uv evenly, the resulting drawing is still iocr-0. At the end ofthis process, C has turned into a loop e ∗ . We eliminate any self-crossings of the loop if necessary. Clearly,this does not change the property of C being essential.We now apply Lemma 3.3 with C = { C } to clear C of all crossings ( C contains exactly one vertex).Finally, we recover edges of C by successively performing splits that are inverse to the contractions weperformed on E ( C ) − e ∗ . Clearly, the homology class of every cycle over Z remains unchanged during thewhole redrawing. Corollary 3.5. If G has an iocr- -drawing on the torus containing an essential cycle C consisting of evenvertices only, then G has an embedding on the torus that is compatible with the iocr- -drawing. In Lemma 3.11 we will show that the result remains true even if at most one vertex of C is odd. Proof.
Use Lemma 3.4 to free C of all crossings; the embedding is compatible, and all vertices of C remaineven. We can then cut the torus along C , creating two duplicate copies of C in a sphere with two holes;let the new graph be G ∗ . By filling in the holes, we obtain an iocr-0-drawing of G ∗ on the sphere, in whichboth copies of C are free of crossings (and each bounds an empty face). Apply the Unified Hanani-TutteTheorem, to obtain a compatible embedding of G ∗ ; we can then reidentify the two copies of C by adding ahandle to the sphere, obtaining a compatible embedding of G on the torus.The following lemma shows, roughly speaking, that for an iocr-0-drawing it is not a single vertex thatmakes the difference between planarity and non-planarity. (It appeared, with a slightly different proof, forthe projective plane in [26].) 5 v u vC ∈ C C ∈ C Figure 1: Contracting the edge uv by pulling u along uv to v . Lemma 3.6.
Let x ∈ V ( G ) and H = G − x . Suppose there is an iocr- -drawing of G in a surface S suchthat H is non-essential. Then G is planar. The planar embedding of H induced by G is compatible with theoriginal iocr- -drawing of H inherited from the iocr- -drawing of G . In other words, under the assumptions of the lemma, G has a planar embedding in which only the rotationat x may change. Lemma 3.6 also implies that if a graph H has a non-essential embedding on a surface,then H has a compatible plane embedding. Proof.
We consider the surface S as a sphere with handles and crosscaps. For each crosscap we can choosea 1-sided closed curve cutting through the crosscap; for each handle, we can pick two 2-sided closed curves(sharing a single point) so that cutting the surface along the two curves results in a single (square) boundaryhole. For the given surface S , we can choose a set C of disjoint 1-sided and 2-sided essential curves so thatcutting the surface along these curves in C results in a planar surface (with a single hole for each crosscapand handle). Given an edge uv , we may contract the edge by pulling u toward v until uv no longer crosses any curveof C . For any edge e that crosses uv , when u reaches e then e will be deformed so that instead of u crossing e , e will get pulled along to just stay in front of u , see Figure 1. This may cause e to add new crossings withcurves of C , two crossings at a time, whenever u crosses a curve of C . Thus the crossing parity between eachedge of G not incident to u and each curve of C will be unchanged by this operation.Let F be a rooted maximum spanning forest in H := G − x . For each component of F , perform a breadth-first search transversal, contracting each edge as described towards the root of the component. After an edge e of F is contracted it has zero crossings with every curve of C ; later contractions may add crossings between e and curves of C but without affecting the parity of crossing. Thus, at the end of this process, every edgeof F crosses every curve of C evenly.Consider any edge e ∈ E ( H ) − E ( F ). Since we assumed that H is non-essential (and that did not changeduring the redrawing we did), e must cross each C ∈ C evenly (if it did not, the (unique) cycle in F ∪ { e } isessential, a contradiction). In summary, every edge of H crosses each C ∈ C evenly.Now cut the surface along the curves in C , and fill in each boundary hole with a disk, resulting in asphere. For each edge-crosscap intersection, reconnect the ends with a curve passing straight through thedisk; for each edge broken at a handle, reconnect it straight across the square-shaped disk. Since each edgeof H intersects each curve of C evenly, its parity of crossing with other edges does not change. In particular:if a pair of edges did not intersect oddly before the redrawing, but they do now, then both edges are incidentto x , and even vertices, except possibly x , remain even.We can then apply Theorem 3.1 to obtain a plane drawing of G in which the subdrawing of H is compatibleto the original drawing.The following result shows that an even tree (all its edges are even) can always be cleaned of crossings.The result remains true for forests, but we will not need that stronger version. Mohar [21] calls these “planarizing system of disjoint curves”. If we deformed the curves of C so that they all shared asingle point, we’d get a set of generators of the fundamental group of the surface. emma 3.7. If G has an iocr- -drawing D containing an even tree T , then there exists an iocr- -drawingdrawing of G that is compatible with D in which T is free of crossings. Only edges incident to T are redrawn.Proof. Fix a root of the tree and orient all edges towards the root. Process the edges in a depth-first traversal(any order in which ancestor edges are processed first is fine) as follows: partially contract each edge uv by moving the child v along uv towards u until uv is free of crossings. This does not change the parity ofcrossings between any pair of edges, and an edge, once contracted, remains crossing-free, since it cannot beincident to an edge contracted later on. This clears the tree of crossings. A pair of edge-disjoint cycles C and C touch at a vertex v ∈ V ( C ) , V ( C ) if in the rotation at v the edgesof C and C do not interleave. Otherwise, we say the cycles cross at v .A pair of weakly disjoint cycles is a pair of edge-disjoint cycles for which any shared vertices are evenand touching. Lemma 3.8. If G has an iocr- -drawing on the torus containing two weakly disjoint essential cycles, then G has an embedding on the torus that is compatible with the iocr- -drawing. This lemma implies that we can assume that a counterexample to the Hanani-Tutte theorem on the torusdoes not contain two vertex-disjoint essential cycles; we extend this result in Section 5.
Proof.
Let C and C be two weakly disjoint essential cycles in an iocr-0-drawing of G .It is sufficient to prove the result for two vertex-disjoint essential cycles: If C and C touch at an evenvertex v , we can split v so that C and C are vertex disjoint. We can then apply the result for vertex-disjointessential cycles, and then merge the split vertices by contracting edges added in the splitting. Since eacheven vertex v splits into two even vertices, whose rotations will be preserved, the final contraction recoversthe original rotation of v , and we have compatibility for even vertices.We can therefore assume that C and C are vertex-disjoint. Apply Lemma 3.4 to make C crossing-free.Next, we redraw C proceeding more or less as in the proof of Lemma 3.4 until the last step, which isdifferent: As before, we redraw edges of C near V ( C ) so that dependent pairs of edges cross evenly, andthen redraw the other edges incident to V ( C ) near the endpoints in V ( C ) so that edges of C are even.Let e ∗ be an edge of C . Contract the edges of C − e ∗ , so that e ∗ becomes a loop. Remove self-intersectionsfrom e ∗ . Now we can uncontract and recover C , giving us a drawing in which C and C are disjoint simple closed curves. Every edge of C ∪ C is crossing-free except possibly e ∗ , which is even. Here is where theproof differs from the proof of Lemma 3.4: C ∪ C separates the surface into two faces , each homeomorphic to a cylinder. For each edge f thatcrosses C (at e ∗ ), cut f at each crossing. Since f crosses e ∗ an even number of times, we can reconnectthe broken ends of f in consecutive pairs by drawing curves alongside e ∗ . This does not change the parityof any edge crossings, and f no longer crosses C . However, f may have a number of curve components,exactly one of which is a simple curve between the endpoints of f , while the others are closed curves, each ofwhich lies in one of the two cylinder-faces. The closed components in the same face as the simple curve canall be joined into a single curve, using pairs of “parallel curves”, while the closed components in the otherface will simply be removed. We do this for all edges f that cross C . Suppose this construction yieldedtwo edges f and f (cid:48) that cross oddly; then both f and f (cid:48) would have to lie in the same cylinder-face. Thenall their discarded components were closed curves in the other cylinder-face; since any two closed curves inthe plane cross evenly, removing those discards did not change the crossing parity, so f and f (cid:48) must havecrossed oddly to start with. We have thus obtained a compatible iocr-0-drawing of G on the torus such that C and C are both crossing-free.Let G ∗ be a plane graph obtained by removing one of the faces and replacing it with two disks D i eachcontaining a vertex v i and crossing-free edges to each vertex of V ( C i ), i ∈ { , } . By Theorem 3.1, G ∗ hasa planar embedding in which the orientation of C and C does not change. We can remove v , v from thedrawing and transfer it back to its face. Doing this with the other face as well completes the proof.7f the second cycle C (cid:48) is not essential, we cannot directly prove that G has a (compatible) embedding,but the following lemma allows us to clear both cycles of crossings. Lemma 3.9. If G has an iocr- -drawing on the torus containing an essential cycle C and a cycle C (cid:48) thatis weakly disjoint from C , then G has an compatible iocr- -drawing in which C and C (cid:48) are crossing-free.Proof. If C and C (cid:48) are both essential cycles, then the result follows from Lemma 3.8. So we can assume thatevery cycle that is weakly disjoint from C is non-essential, including C (cid:48) .As in the proof of Lemma 3.8, we can first split vertices where C and C (cid:48) touch to make them vertex-disjoint, and after proving it for that case we can recontract and recover the original graph, with rotationsof even vertices preserved. That is to say, we can assume that C and C (cid:48) are vertex-disjoint.Lemma 3.4 allows us to redraw so that C is free of crossings (without changing the rotation of any evenvertex). Next, make all edges of C (cid:48) even by flips at its vertices, then contract all but one edge of C (cid:48) so that itbecomes a loop. Remove self-intersections of the loop, and uncontract. We now have an iocr-0-drawing of G in which all crossings with C (cid:48) occur along a single edge (and that edge is even). Cut edges crossing that edgeof C (cid:48) and reconnect their ends pairwise (as usual). Edges which crossed C (cid:48) consists of multiple componentsnow, one of them connecting the endpoints of the edge, and the other components closed curves. Reconnectthe closed curves to the edge components if they are on the same side of C (cid:48) ; otherwise drop them. We claimthat the parity of crossing between pairs of independent edges remains zero: if there were two edges thatnow cross oddly, they must be on the same side of C (cid:48) to do so. But two closed components belonging tothese edges can only cross evenly: if they lie inside C (cid:48) , because C (cid:48) is contractible, if they lie outside C (cid:48) ,because they lie in a cylinder (as they cannot cross C ). -Spindle Before we move on to the torus, we show how to apply our tools so far in a much simpler context than thetorus (or the projective plane).We show that there is a strong Hanani-Tutte theorem for the 1-spindle. The 1 -spindle is a pseudosurfaceobtained from the sphere by identifying two distinct points (the pinchpoint ). Embeddings are defined asusual, for drawings we allow at most one edge to pass through the pinchpoint. In an embedding, we canalways assume that there is a vertex at the pinchpoint (if not, and there is an edge passing through thepinchpoint, pull the edge until an end-vertex lies in the pinchpoint, otherwise, the pinchpoint isn’t necessary,and the graph planar, in which case we can make an arbitrary vertex a pinchpoint).
Theorem 3.10 (Hanani-Tutte for 1-Spindle) . If a graph has an iocr- -drawing on the -spindle, it can beembedded on the -spindle. The embedding is compatible with the iocr- -drawing, except for (possibly) thevertex at the pinchpoint. It would seem that this result should follow from a Hanani-Tutte theorem for the projective plane or thetorus, but that does not seem immediately obvious.
Proof.
Fix an iocr-0-drawing of G on the 1-spindle. We can assume that there is a vertex v at the pinchpoint.If not, there must be an edge e passing through the pinchpoint (otherwise, we have an iocr-0-drawing onthe sphere, and we are done the Hanani–Tutte theorem in the plane). Consider the subarc γ of e betweenthe pinchpoint and a vertex v . For any edge f which crosses γ oddly, we perform an ( f, u )-move for everyvertex u . This does not change the parity of edge crossings, but it does ensure that γ is an even curve. Wecan then partially contract e by pulling v along γ onto the pinchpoint. Since γ is even, the drawing remainsiocr-0.Refer to Figure 2. Remove the pinchpoint (splitting v into two copies), giving us a drawing on a sphere,which is iocr-0, except for the two copies of v . Add a handle close to the two copies, and move them backtogether, merging them into a single vertex. Add an essential cycle, say a C , through v using the handle(and free of crossings), let G ∗ denote the new graph. If the drawing of G − v (as part of G ∗ ) does not containan essential cycle, then G is planar, by Lemma 3.6, and we are done. So the drawing of G − v contains an8 v v v Figure 2: Illustration of the surgery in the proof of Theorem 3.10. The elimination of the pinched pointsplits the vertex v into two vertices v and v that are subsequently merged using the handle. Afterwards C is introduced as a meridian of the handle passing through v .essential cycle, implying that the drawing of G ∗ contains two disjoint essential cycles. By Lemma 3.8 we canfind a compatible embedding of G ∗ in the torus. Note that in that embedding the essential cycle C is stillessential (and free of crossings), so we can contract it until it becomes a point and we have an embedding of G on the 1-spindle.The proof of Theorem 3.10 suggests the following result, which is useful by itself. Lemma 3.11.
If a graph has an iocr- -drawing on the torus containing an essential cycle C such that exactlyone vertex of C is odd, then the graph has a compatible embedding on the torus. Recall that Corollary 3.5 deals with the case that all vertices on C are even. Proof.
Fix an iocr-0-drawing of G on the torus. Use Lemma 3.4 to clear C of crossings. By assumption thereis a v ∈ V ( C ) so that all vertices in V ( C ) − v are even. Let γ be the open, simple curve from v to v along C (excluding v ). Cut the torus along γ , creating two copies of each edge in E ( C ) and all vertices in V ( C ) − v .Let G ∗ be the resulting graph. After filling in the hole, we have an iocr-0-drawing of G ∗ in the torus. Ifall essential cycles in the drawing of G ∗ pass through v , then G ∗ has a compatible planar embedding, byLemma 3.6. From that, we can reconstruct a compatible embedding of G in the torus as follows: both copiesof the original E ( C ) lie on the boundary of faces in the plane embedding of G ∗ (only the rotation at v canhave changed, since it is odd), so by adding a single handle, we can merge the two copies of V ( C ) − v , anddrop the edges E ( C ) of one of those two copies. This gives us a compatible embedding of G in the torus.Otherwise, there is an essential cycle in the iocr-0-drawing of G ∗ on the torus which avoids v . In thiscase, we can add a C through v which lies between the two copies of C , and thus us essential. This C and the essential cycle avoiding v are vertex-disjoint, and we can apply Lemma 3.8 to find a compatibleembedding of G ∗ ∪ C in the torus. By removing the C (except for v ) from this embedding, and mergingthe two copies of C , we obtain a compatible embedding of G in the torus, which is what we had to show. Fulek and Kynˇcl [11] showed that the Hanani-Tutte theorem is true for any surface if we restrict ourselvesto the graphs known as Kuratowski minors which include K ,t for any t ≥ Lemma 4.1 ([11]) . For t ≥ , K ,t does not admit an iocr- -drawing on the torus. Figure 3 shows a toroidal embedding of K , , so Lemma 4.1 is sharp, even, as we can see in the figure, ifwe add a 3-cycle on the vertices of degree 6. Therefore, Lemma 4.1 implies that the Hanani-Tutte theoremon the torus is true for all K ,t , t ≥ K , : there is aniocr-0-drawing of K , on the torus which does not have a compatible embedding. We can show, however,9 b ca b c Figure 3: A torus embedding of K , ∪ C , with the union taken over the three vertices of degree 6. The six K , -claws incident to c alternate connecting to a and b between ab and ba .that there is a weakly compatible embedding of K ,n up to n = 6 which is sharp, since K , has no embeddingon the torus.We establish a slightly stronger version. Let a , b , and c be the three vertices of degree n in K ,n (for n = 3 pick the vertices of one of the two sides). We call a graph a K ,n with spokes (at a, b, c ) if it is theunion of the K ,n with (any number of) interior-disjoint paths of length at most two between any two of { a, b, c } . Lemma 4.2.
Let G be a subgraph of a subdivision of K ,n with spokes at a, b, c . For every iocr- -drawing ofa subgraph of G on the torus there exists an embedding of the subgraph on the torus that is compatible withthe iocr- -drawing.Proof. We can assume that G contains no vertices of degree 1, since removing such vertices does not affectembeddability. If u is a vertex of degree 2 in G , and u is not part of a spoke s, u, t with s, t ∈ (cid:0) { a,b,c } (cid:1) , then u can be suppressed. This does not affect iocr-0-ness of the drawing, or embeddability of G , since a suppressedvertex can always be reintroduced in an embedding; finally, since u is not part of a spoke, suppressing itcannot lead to multiple edges.This allows us to assume that G is a subgraph of a K ,n with spokes at a, b, c . A proper subgraph of a K ,n with spokes at a, b, c is just another K ,n (cid:48) , n (cid:48) ≤ n , with spokes at a, b, c up to degree-1 vertices whichcan always be removed without affecting embeddability. We can therefore assume that G is a K ,n withspokes at a, b, c . If any of the edges ab , ac , or bc occur, we subdivide them so that all spokes have length 2.Let S = { a, b, c } . We can assume that all the vertices of degree at most 3 are even, by performingedge-flips if necessary. So any odd vertices must belong to S .If S contains no even vertices we are done, since K , ∪ S has an embedding on the torus, as shown inFigure 3, and we can embed G as part of it. (Any two rotations of a degree-3 vertex are weakly compatible,since each of these vertices is its own evenly connected component.) If S contains only even vertices, we aredone by Theorem 1.2, the weak Hanani-Tutte theorem for surfaces. So S contains either one or two evenvertices.Consider the case that S contains two even vertices. If there is an essential cycle that avoids the oddvertex in S , we are done by Corollary 3.5. Otherwise, all essential cycles pass through the odd vertex in S ,in which case we are done by Lemma 3.6.This leaves us with the case that S contains one even vertex. We will establish this case separately inLemma 4.3.The following lemma covers the missing case in the proof of Lemma 4.2. Lemma 4.3.
Let G be a K ,n with spokes at a, b, c , where n ≤ . If G has an iocr- -drawing in which c iseven, and a and b are odd, then G has an embedding on the torus that is compatible with the iocr- -drawing. roof. Using edge-flips, if necessary, we can ensure that all vertices of degree 3 are even. Since c is an evenvertex, we can use Lemma 3.7 to clear all its incident edges of crossings. Every edge incident to c belongs toeither a path of length 2 to a , a path of length 2 to b , or a K , with legs connecting to a and b (an ab -claw or claw ). In the last case we can distinguish between the legs occurring in order ab or ba , depending on therotation at the center of the K , (which is an even vertex). This allows us to summarize the rotation at c asa cyclic list, for example ab, a, ba, b, a, b describes the rotation at c for a K , with (four) spokes, see Figure 4(left) for an illustration. We assume for the moment that there are no spokes between a and b , so the cycliclist completely determines the rotation at even vertices of G . cb aba b a a b ca b a ca b aa Figure 4: The rotation at c is summarized by the cyclic list ab, a, ba, b, a, b (left). Introducing a claw if ab, a appears in the summary of the rotation at c (right).Suppose G is a counterexample to the statement of the lemma. So G has an iocr-0-embedding, but itdoes not have a weakly compatible embedding. Under this condition, let n be maximal, and the number ofspokes minimal. Then n ≤
6, since K , does not have an iocr-0-embedding by Lemma 4.1.Consider the cyclic list we constructed for c . That list cannot contain the patterns a, a or b, b : if it did,we could simplify to just a or b , removing one of the spokes. The resulting graph has a weakly compatibleembedding if and only if the original one does, since we can duplicate a spoke/delete a spoke. Hence, thiswould contradict the minimality of the number of spokes.The cyclic list also cannot contain the patterns a, b or b, a . Suppose it contained a, b (the other case issimilar). Let G (cid:48) be the graph in which a, b is replaced by ab, b . If G has a weakly compatible embedding,then so does G (cid:48) (duplicate the b -spoke and merge it with the a -spoke to get an ab -claw), and vice versa (inthe claw delete the connection to b to get an a -spoke). If G (cid:48) has more than six claws, it cannot be embeddedin the torus, so neither can G . If G (cid:48) has fewer than six claws, G did not maximize n .Refer to Figure 4 (right). A similar argument shows that the cyclic list cannot contain any of the foursub-patterns ab, a ; a, ba ; b, ab ; or ba, b . Suppose it did, without loss of generality, suppose there is an ab, a in the cyclic list. Recall that the center vertex of the ab -claw is even, so we can route a path of length2 from c to b between (and along) the ab -claw and a . We can then duplicate the a -spoke, and merge thenew b - and a -spokes into an ab -claw. That is, we go from ab, a to ab, b, a to ab, b, a, a to ab, ba, a . We stillhave an iocr-0-drawing, but there is a new claw, corresponding to the pattern. Since K , does not have aniocr-0-drawing, it must be the case that n + 1 ≤
6, so n ≤
5, which contradicts the choice of n being maximal(we saw how to add an n + 1-st claw into the iocr-0-drawing).Since the cyclic list does not contain any of the sub-patterns we identified, any a -spoke must be precededby ba and succeeded by ab , and b -spokes must be preceded by ab and succeeded by ba . In other words, clawsalternate between ab - and ba -claws. We can then take the standard embedding of K , shown in Figure 3,for which the claws alternate as well, and add the a - and b -spokes into the embedding as required by thecyclic list; note that this correctly reproduces the rotations at all even vertices. Finally, note that we canadd any number of spokes between a and b into the embedding, since a and b lie on the same face (and bothare odd, so their rotation does not matter). Remark 4.4.
We will apply Lemma 4.3 in a situation where a and b are part of an essential cycle. Theconstruction in Lemma 4.3 makes this cycle plane. It is possible to show, however, that the cycle can bekept essential. This requires a slightly sharper analysis, using that n can be at most 4, and not 6 in thiscase. 11e also need a Unified Hanani-Tutte theorem for K . Lemma 4.5.
For every iocr- -drawing of a subgraph G of a subdivision of K on the torus, there exists anembedding on the torus that is compatible with the iocr- -drawing.Proof. Suppressing a subdivision vertex does not change the iocr-0-ness of a drawing, and a suppressedvertex can always be reintroduced in an embedding, so we can assume that there are no subdivision vertices.We can also remove any vertices of degree 1, since they do not affect embeddability. It follows that G is asubgraph of K n , for some n ≤
5. For n ≤
4, the result is easy, so we assume that G is a subgraph of K .Let S be the set of even vertices in the iocr-0-drawing of G . We distinguish cases by the cardinality of S . Case | S | ∈ { , , } . If S is empty we can embed G as part of a toroidal embedding of K ; if S contains allvertices we are done using Theorem 1.2; if S contains a single even vertex we take a toroidal embedding of G and map the vertex and its neighbors to the embedding to get a compatible embedding. Case | S | = 4 . We distinguish two subcases: if there is an essential cycle consisting of vertices in S , weare done by Corollary 3.5. Otherwise, all essential cycles in the drawing must contain the fifth vertex notbelonging to S . In that case, we are done by Lemma 3.6. Case | S | = 2 . In this case we assume without loss of generality that G = K , since we can delete the edgesnot present in G in the end.Let the two even vertices be u , v and label the remaining vertices a , b , and c , so that the rotation at u is vabc . Draw u and v in the plane with half-edges to a , b , and c attached (this will look like a (cid:54) = sign). Unlessthe rotation at v is uabc , there are at least two half-edges at v which can be connected to the half-edgesemanating from u without crossings. The (at most one) remaining pair of half-edges can be connected usinga handle. This leaves us with the three edges between a , b , and c . Edges ab and bc can be drawn close to u ,since half-edges for ua and ub as well as ub and uc are consecutive at u . Edge ac can be added at v , if va and vc are consecutive in the rotation at v . If not, then the rotation at v must be ucba , since we excludedthe rotation uabc in this case. Then half-edges could be connected in the plane, and we use the handle forthe edge ac . See Figure 5 (left) for an illustration of this case. (The square with the cross inside representsthe handle in the figures.) cb a u v cb a u v Figure 5: (Two even vertices)
Compatible embedding of K in the case that v has rotation ucba (left). (Twoeven vertices) Compatible embedding of K in the case that v has rotation uabc (right). We use (cid:1) for the(orientable) handle.This leaves us with the case that the rotation at v is uabc . Figure 5 (right) shows a compatible embeddingof the K for this case. Case | S | = 3 . Let u , v , w be the even vertices, and label the two remaining vertices a and b . Let us firstassume that G contains the cycle uvw . If the cycle on uvw is an essential cycle, we are done by Corollary 3.5.We can therefore assume that the cycle is non-essential or one of its edges is missing in G . If the cycle uvw exists in G we use Lemma 3.7 to clear uv and vw of crossings. It follows that uw does not cross either uv and vw , so the cycle on uvw is a simple closed curve which, since it is not essential, bounds a plane region.If the edge ab exists it crosses the cycle an even number of times, so both a and b must lie on the sameside of the cycle. Since each edge incident to a and b also crosses the cycle an even number of times, alledges attaching to the cycle attach on the same side of the cycle (all inside the plane region, or all outside).12 au vw b au vwb a baab b aa b Figure 6: (Three even vertices)
Compatible embeddings of K in the case that half-edges leaving the triangleare abbaab and ababab .This allows us to build a compatible embedding of the G as follows. In the plane draw a crossing-free cycle uvw , and add half-edges, as determined by the rotation, starting at the cycle towards a and b . As we arguedabove, we can assume that a and b , and the half-edges lie on the same side, without loss of generality, insidethe cycle. If we write down the half-edges in the order in which they occur along the cycle uvw , there areonly two possible results (up to renaming of a and b and shifting the list cyclically), namely ababab , and abbaab . Figure 6 shows a compatible embedding for each case, completing the proof.If the edge ab does not exist and a and b are not on the same side of the cycle, the previous argumentdoes not apply only when the two edges attaching at a vertex of the cycle are on its opposite sides, in whichcase we can easily construct a compatible embedding of G . u v w u v w u v wab ab ab Figure 7: (Three even vertices)
Compatible embeddings of G in the case uw (cid:54)∈ E ( G ).Finally, if uvw is not a cycle the above argument applies (by an edge deletion in the end) unless uvw isa path in G , uw (cid:54)∈ E ( G ), v is a degree-4 vertex, and its two remaining edges attach on the opposite sides ofthe path uvw (otherwise, we can add an even edge uv into the drawing). In this case, up to symmetry thereare three cases to consider depending on the rotations at u and w , each of which can be easily completedinto a compatible embedding of G , see Figure 7. In this section we collect properties of a minimal counterexample to the Hanani-Tutte theorem on the torus.We order graphs by their number of isolated vertices ( in creasing), number of edges ( in creasing), and then bynumber of vertices ( de creasing). We denote by ≺ the strict partial ordering based on these three numbers,and refer to a ≺ -minimal graph (with certain properties). In a later section, we will refine ≺ by an ordering ≺ (cid:48) ; that is, whenever G ≺ H , then G ≺ (cid:48) H . In particular, any ≺ (cid:48) -minimal graph with a certain property isalso ≺ -minimal with that property. So the properties proved for ≺ -minimal graphs in this section are alsotrue for ≺ (cid:48) -minimal graphs, which is why we write minimal, rather than ≺ - or ≺ (cid:48) -minimal in this section.13ince isolated vertices do not affect embeddability on a surface, or the Hanani-Tutte criterion, a minimalcounterexample contains no isolated vertices. Graphs without isolated vertices are then ordered by numberof edges (increasing), and number of vertices (decreasing); in other words, we use the (strict, partial) lexico-graphic order on ( | E ( G ) | , | E ( G ) | − | V ( G ) | ); since graphs without isolated vertices satisfy | V ( G ) | ≤ | E ( G ) | ,this order is well-founded. Note that if H is a proper subgraph of G , then H ≺ G , simply because it hasfewer edges (since there are no isolated vertices).Section 5.1 presents some basic properties of minimal counterexamples with respects to cycles and cuts.In Section 5.2 we identify what we call an X -configuration, which must occur in a minimal counterexampleto the Hanani-Tutte theorem. The main proof requires a strengthened assumption on (weakly) compatibleembeddings, for which we cannot immediately show that the X -configuration occurs. Hence, Section 5.3extends the proofs of Section 5.1 to show that they (mostly) still hold if the presence of an X -configurationis required. The following lemma is true when read with the bracketed weakly, or without it.
Lemma 5.1. If G is a minimal graph that has an iocr- -drawing D on the torus, but does not have a[weakly] compatible embedding on the torus, then D does not contain a pair of weakly-disjoint cycles at leastone of which is essential. For the weakly compatible version, every bracketed weakly in the proof needs to be read. For thecompatible version, all bracketed occurrences of weakly have to be dropped.
Proof.
Let C and C (cid:48) be weakly disjoint cycles in D . By Lemma 3.8 not both of them can be essential, solet us assume that C is essential, and C (cid:48) is non-essential.Using Lemma 3.9, we can clear both C and C (cid:48) of crossings. Let H be the part of G drawn within thedisk bounded by C (cid:48) , and including C (cid:48) . Then H has a compatible planar embedding, by Theorem 3.1, inwhich C (cid:48) bounds the outer face (to see this, add a new vertex outside C (cid:48) and connect it to every vertex on C (cid:48) before applying Theorem 3.1, the resulting drawing is still iocr-0 in the plane). C CC (cid:48)
C CH G G ∗ v Figure 8: Construction of the drawing of the graph G ∗ in the proof of Lemma 5.1.Refer to Figure 8. Construct G ∗ from G as follows: remove the inner vertices of H , that is V ( H ) − V ( C (cid:48) ),as well as E ( C (cid:48) ), the edges of C (cid:48) ; in other words, we clear the disk bounded by C (cid:48) , including the edges of C (cid:48) , but leaving its vertices. Add a new vertex v in the disk, and connect it to all vertices of C (cid:48) . Then v iseven, since all its incident edges are free of crossings.The drawing of G ∗ is iocr-0, and G ∗ ≺ G in our graph ordering: if H contains a vertex apart from V ( C (cid:48) ),because the number of edges decreased; otherwise, H = C (cid:48) , and the number of edges remained the same,but the number of vertices increased.By minimality, G ∗ has a [weakly] compatible embedding on the torus.All our redrawing operations and reductions Lemma 3.4, Lemma 3.8, and Lemma 3.9 yield compatibledrawings. We can therefore remove v and edges incident to it from the drawing, and reinsert H (in the14eakly compatible case we may have to flip all of H ) to obtain a [weakly] compatible embedding of theoriginal graph G on the torus, completing the proof.We use the previous lemma repeatedly to thin out G in order to show that no minimal counter-exampleto the Hanani-Tutte theorem on the torus exists.The following lemma allows us to focus on 2-connected (for the compatible and weakly compatible case)or 3-connected counterexamples (in general). Lemma 5.2. If G is a minimal graph that has an iocr- -drawing on the torus, but does not have a [weakly]compatible embedding on the torus, then G is -connected and no 2-cut consists of two odd vertices in thedrawing.If we do not require the embedding to be compatible, then a minimal counterexample has to be -connected.Proof. If G is disconnected, and some component K of G does not contain an essential cycle, then K isplanar and has a compatible embedding (apply Lemma 3.6 with a new x (cid:54)∈ V ( K )). By minimality of G ,there also is a [weakly] compatible embedding of G − V ( K ) on the torus, and the two embeddings can becombined into a single [weakly] compatible embedding of G . If every component of G contains an essentialcycle, we can apply Lemma 3.8 to obtain a compatible embedding. Therefore, G is connected.Suppose G has a cut-vertex x . Then G − x consists of at least two components; if each component of G − x contains an essential cycle, then Lemma 3.8 implies that G has a compatible embedding, which is acontradiction. Hence, some component K of G − x does not contain an essential cycle. If x is odd, we useLemma 3.6 to argue that L = G [ V ( K ) ∪ x ] is planar. The graph L can be embedded in a disk with x on itsboundary, and G − V ( K ) has, by minimality of G , a [weakly] compatible embedding on the torus. Since x is odd, we can change the rotation at x , and combine the two embeddings into a single [weakly] compatibleembedding of G on the torus.If x is even, we need to argue more carefully, since we cannot change the rotation at x . This means wecannot make use of Lemma 3.6. By Lemma 5.1 we know that D does not contain two weakly disjoint cycles atleast one of which is essential. There must be at least one component K of G − x for which L = G [ V ( K ) ∪ x ]contains an essential cycle C (otherwise G contains no essential cycle, and therefore has a compatible planarembedding, as implied by Lemma 3.6, for example). If C is contained in K , then T = G [ V ( G ) − V ( K )] has tobe a tree (by Lemma 5.1) with x as a leaf of the tree. Clearly, the tree T has a compatible plane embedding.Since G was minimal, L has a [weakly] compatible embedding on the torus; to that we can add the planeembedding of T to obtain a compatible embedding of G (reversing all rotations of T if the rotation of x in L is reversed). Therefore, the essential cycle C must pass through x in L . By a similar argument, there canbe no components K (cid:48) of G − x for which L (cid:48) = G [ V ( K (cid:48) ) ∪ x ] does not contain a cycle. (Such an L (cid:48) is a tree,which has a compatible plane embedding, which can be added at x , so such an L (cid:48) does not exist in a minimalcounterexample.) Therefore, if K (cid:48) is a component of G − x other than K , then L (cid:48) = G [ V ( K (cid:48) ) ∪ x ] mustcontain a cycle C (cid:48) . Because of Lemma 5.1 again, this cycle cannot be weakly disjoint from C , so it must passthrough x and cross C (cid:48) in x . This forces C (cid:48) to be essential (the underlying curves of C and C (cid:48) cross oddly,since x is even, so both curves must be essential). We clear C of crossings using Lemma 3.4. We then usethe same procedure described in Lemma 3.4 to clear C (cid:48) of crossings. This is possible in this particular case,since C and C (cid:48) intersect in an even vertex x (so the rotation of x does not need to be modified for applyingLemma 3.4), and C does not separate the torus, and neither does C ∪ C (cid:48) once the crossings along C (cid:48) havebeen removed. We thus obtain a compatible drawing in which both C and C (cid:48) are free of crossings. We knowthat K (cid:48) is a tree (if K (cid:48) contained a cycle, it would be vertex disjoint from C , contradicting Lemma 5.1.The leaves of K (cid:48) must be adjacent to x , otherwise they are leaves in G and would not occur in a minimalcounterexample. Hence, there must be some path P between a vertex y on C (cid:48) − x and x . Then P togetherwith the path from y to x on C (cid:48) that ends on the same side of C as P yields a cycle weakly disjoint from C and touching it (at x ) which is a contradiction. Therefore K (cid:48) is just C (cid:48) − x , so L (cid:48) = C (cid:48) . Since x is even,and we can make all other vertices of C (cid:48) even (they have degree 2), we can apply Corollary 3.5 to obtain acompatible embedding of G in the torus.Hence, we may assume that G is two-connected. Let { x, y } be a cut-set with both x and y being oddvertices. If K and K (cid:48) are components of G − { x, y } , by Lemma 3.8 we may assume that either G [ V ( K ) ∪ { x } ]15r G [ V ( K (cid:48) ) ∪ { y } ] contains no essential cycle: say, the former. Let L ( K ) = G [ V ( K ) ∪ { x, y } ] + xy where xy is an edge from x to y , drawn along an x, y -path in G − V ( K ), added only if G [ V ( K ) ∪ { x, y } ] doesn’talready contain such an edge. This gives us an iocr-0-drawing of L ( K ) in which every essential cycle in ithas to pass through xy . By Lemma 3.6, L ( K ) is planar, and has a compatible planar embedding. We candraw L ( K ) in a small disk, with xy along the boundary of the disk. Let G ∗ = G − V ( K ) + xy , with the edge xy drawn along an x, y -path in G [ V ( K ) ∪ { x, y } ]. G ∗ has an iocr-0-drawing, so by minimality of G , we havea [weakly] compatible embedding of G ∗ in the torus. We remove a small disk from the torus minus G ∗ , suchthat the boundary of disk intersects G ∗ at xy , and replace it with the disk containing L ( K ) so that the two xy ’s match up. Since x and y are odd, rotations at these vertices do not matter. We then delete xy from thedrawing if it does not belong to G and we have obtained an embedding of G on the torus. This contradictsthe assumption that a minimal counterexample contains a 2-cut consisting of two odd vertices.If we do not require the minimal counterexample to be compatible, or weakly compatible, the sameargument shows that there is no 2-cut in a minimal counterexample, so it is 3-connected. X -Configuration The Hanani-Tutte theorem holds for cubic graphs on arbitrary surfaces; this is because vertices of degree 3can be made even by edge-flips, at which point Theorem 1.2, the weak Hanani-Tutte theorem for surfacescan be applied. Edge-flips are no longer sufficient to deal with vertices of degree 4, and this case is the mainobstruction as we will see presently.Consider four edges e , e , e , e incident on the same vertex v with ends in order e , e , e , e . Supposethat e and e cross oddly, and every other pair of edges crosses evenly. No matter what edge-flips areperformed at v , there will always remain at least one pair of edges crossing oddly. The configurationdescribed above is the unique obstacle to a vertex being even, up to edge-flips: suppose v is incident to fouredges. Using edge-flips we can make three consecutive edges at v cross evenly with each other. Say the edgesare e , e , e , e , in this order, and every two of e , e and e cross evenly. If e crosses exactly e oddly, weare done. If it crosses e and e oddly, we move the end of e once around v , so that e crosses exactly e oddly. In all other cases, e can be made to cross all of e , e and e evenly using edge-flips. Lemma 5.3.
If a vertex in a drawing cannot be made even by flips, then it is incident to four edges whichcannot be made even by flips.Proof.
Let v be a vertex in a drawing so that v cannot be made even by edge-flips. Then v has degree atleast 4. If it has degree exactly four, we are done by the argument preceding the lemma. So we can assumethat the degree of v is larger than 4. Looking at all possible sequences of edge-flips, find a drawing whichcontains the longest block B of consecutive ends so that all edges in the block cross each other evenly. Fixthe rotation of that drawing. We can assume that B contains at least four ends (otherwise, we can pickany four edges at v , since no four edges can be made to cross pairwise evenly). Let the block B start withedge e and end with edge f , and let the next edge after f be g . Then g (cid:54) = e , since otherwise v is even. If g crosses both e and f oddly, we rotate g once around v flipping the parity of g with all other edges incidentto v . In particular, we can assume that g does not cross both e and f oddly. If g crosses both e and f evenly, it has to cross some edge h in B − { e, f } oddly, since otherwise we could have added g to the block B of edges crossing pairwise evenly. In this case, { e, f, g, h } are the four edges we were looking for. We cantherefore assume that g crosses exactly one of e and f evenly, without loss of generality let us say g crosses e evenly, and f oddly (in the other case, we move g in the rotation next to e ). Then B can be written as eB g (cid:48) B f , where g crosses all edges in B f oddly, and g (cid:48) evenly. Move g past f and B so the new rotationis eB g (cid:48) gB f . It is not possible that g (cid:48) = e or that g crosses every edge in B evenly, since in those cases eB g (cid:48) gB f would have been a longer block than B . Hence, there must be an edge h ∈ B which crosses g oddly. Then e, h, g (cid:48) , g are the four edges we are looking for.The core of the inductive proof will be working with a specific configuration in iocr-0-drawings: Twoessential cycles C and C which intersect in a single vertex so that the ends of C and C at v cannot bemade even with respect to each other by flips. We call such a pair ( C , C ) an X -configuration , see Figure 9.16 C Figure 9: A pair of essential cycles C and C meeting in a single vertex v so that the rotation at v cannotbe “corrected” to make the edges of C and C cross each other evenly. Claim 5.4.
If ( C , C ) is an X -configuration, and C and C are two cycles that intersect in the same vertex u as C and C , are otherwise disjoint, and u is incident to the same four edges in C ∪ C as in C ∪ C ,then ( C , C ) is an X -configuration. Proof.
We only have to show that C and C are essential. Using edge-flips, make all edges of C even.Make both of the edges of C incident to u even with respect to C . If these two edges were on the same sideof C , they could be made even with respect to each other, contradicting the assumption that they cannot(since ( C , C ) is an X -configuration). So they are on opposite sides, implying that C and C cross eachother oddly as curves, implying they are both essential. Lemma 5.5. If G is a minimal graph that has an iocr- -drawing D on the torus, but does not have anembedding on the torus (not necessarily compatible), then D contains an X -configuration ( C , C ) . Figure 9 serves as an illustration.
Proof.
By Lemma 5.2 we can assume that G is 3-connected.If every vertex of G in an iocr-0-drawing of G on the torus can be made even by flips, we obtain anembedding of G . This follows from Theorem 1.2, the weak version of the Hanani-Tutte theorem. Therefore,there exists a vertex v for which this is not the case. By Lemma 5.3 there are four edges e i = vu i , 1 ≤ i ≤ v which cannot all be made even with respect to each other by edge-flips.Since G is 3-connected, we can find edge-disjoint cycles C , C intersecting in v as follows: Consider aminimal path in G connecting two u i vertices that avoid v ; without loss of generality this path is P between u and u in G − { v, u , u } . Let C denote the cycle obtained as the edge union of E ( P ) , { e } and { e } .Let P denote a path in G − { v, u } between u and V ( C ) − { v } . Let w denote the end vertex of P on C − { v } . Finally, let P denote a path in G − { v, w } between v and ( V ( P ) ∪ V ( C )) − { v, w } . Clearly,the edge union of E ( C ) , E ( P ) , E ( P ) , { e } and { e } contains a pair of cycles C and C meeting exactlyin v .As explained before Lemma 5.3, we can assume that the e i cross each other evenly, with the exception ofone pair of ends which cross oddly, and are not consecutive. If the ends of C and C do not alternate at v ,then the two edges crossing oddly cannot both belong to the same C i , since they would then be consecutive;therefore one belongs to C and the other to C , which implies that C and C cross each other oddly (alledges of the cycles not incident to v cross evenly, and the two cycles do not cross at v ), hence both cyclesare essential. If the ends of C and C do alternate at v , then the two edge crossing oddly must belong tothe same cycle (since otherwise they would be consecutive), so again C and C cross each other oddly (at v , no other odd crossings), and both are essential. X -Configuration In this section we show that Lemma 5.1 still works when restricted to graphs with iocr-0-drawings containingan X -configuration. With some restrictions, the same is true for Lemma 5.2. We need these variants for ourmain induction. 17 emma 5.6. If G is a minimal graph that has an iocr- -drawing D on the torus containing an X -configuration ( C, C ) , but does not have a [weakly] compatible embedding on the torus, then D does not contain a cycleweakly-disjoint from C (or C ).Proof. It is sufficient to verify that the iocr-0-drawing of the graph G ∗ constructed in the proof of Lemma 5.1contains an X -configuration. Let ( C, C ) be the X -configuration in G . Since C and C (cid:48) are weakly-disjoint,they can only touch in even vertices. It follows that C cannot intersect the interior of H and is thereforepresent, unchanged, in G ∗ . The vertex v ∈ V ( C ) ∩ V ( C ) is odd, by the definition of X -configuration, soit cannot lie on C (cid:48) . If C contains no edges of H ∪ C (cid:48) , then C exists unchanged in G ∗ , and we are done.So C (cid:48) must contain a first and a last vertex of C as we traverse it from v to v . We modify C to directlyconnect these two vertices via the newly added vertex in G ∗ . This gives us a new cycle C (cid:48) which intersects C exactly in v . Therefore, C (cid:48) must still be essential, and we have found the X -configuration ( C, C (cid:48) ) in G ∗ we needed.Lemma 5.2 showed that a minimal counterexample (with compatible embedding) has to be 2-connected,and placed restrictions on a 2-cut. The 2-cut part needs further restrictions for compatible and weaklycompatible embeddings, and we state that result separately. Lemma 5.7. If G is a minimal graph that has an iocr- -drawing on the torus containing an X -configuration,but does not have a [weakly] compatible embedding on the torus, then G is -connected.Proof. Let G contain an X -configuration ( C , C ) with C and C intersecting in v . The proof of Lemma 5.2still shows that G is connected.Suppose that G has a cut-vertex x . If x (cid:54) = v , then C ∪ C is contained in G − K for some component K of G − x . If L = G [ V ( K ) ∪ x ] contains a cycle, we are done by Lemma 5.6, since either C or C isvertex-disjoint from that cycle (not both of them can pass through x (cid:54) = v ).We conclude that x = v , so x is odd. In that case, the original proof in Lemma 5.2 works, as long as C ∪ C lies in G − K for some component K of G − v . If that is not the case, then C − v and C − v lie indifferent components of G − v . Let K be the component of G − v containing C − v and let L = G [ V ( K ) ∪ v ].By Lemma 3.4, we can assume that C is free of crossings. By Lemma 5.6, K cannot contain a cycle (itwould be vertex-disjoint from C ), all essential cycles in L pass through v , and so, by Lemma 3.6, L has aplane embedding, which is compatible, except for, possibly, at v (in L , v may be even; we can ignore this,however, since in the original drawing it is odd). Let G ∗ be the subgraph of G obtained by removing alledges of E ( K ) − E ( C ), and all (resulting) isolated vertices. Lemma 3.11 then applies to G ∗ and C : Odddegree-2 vertices of C can be easily made even, and hence, v is the only odd vertex of C in G ∗ . Therefore,we know that G ∗ has a [weakly] compatible embedding (except for at v ) on the torus. We can then reinsertthe compatible embedding of L along C to obtain a [weakly] compatible embedding of G : Every connectedcomponent of K − C is connected by an edge with at most one other vertex of C besides v . This contradicts G being a counterexample. (Note that we did not apply Lemma 5.7 inductively in this case; we could nothave, because removing v destroyed the only X -configuration we knew about.)In particular, G is 2-connected. Lemma 5.8. If G is a minimal graph that has an iocr- -drawing on the torus containing an X -configuration,but does not have a [weakly] compatible embedding on the torus, then G has no 2-cut consisting of two oddvertices, unless both vertices belong to the X -configuration, with one of them belonging to both cycles. The lemma does not cover the case in which the 2-cut consists of the vertex in which the two cycles ofthe X -configuration intersect together with another odd vertex belonging to one of the cycles. We will dealwith this situation later when investigating bridges. Proof.
By Lemma 5.7, we know that G is two-connected. Let { x, y } be a cut-set for which both x and y areodd. Let ( C, C ) be the X -configuration, with v being the vertex in which C and C intersect. If C ∪ C contains at most one of x and y , say y , and that vertex is not v , then the proof in Lemma 5.2 can be used:As K (cid:48) we choose the connected component of G − { x, y } containing C ∪ C − { y } , so that G [ V ( K (cid:48) ) ∪ { y } ]18ontains C ∪ C which forces G [ V ( K ) ∪ { x } ] to not have an essential cycle. The proof can then be completedas described, since we induct on a graph G ∗ = G − V ( K ) + xy which contains the X -configuration ( C, C ).Next, let us consider the case that both x and y belong to C ∪ C , both of them different from v . If x and y lie on the same cycle, C or C , then we can choose K and K (cid:48) so that the other cycle belongs to G [ V ( K (cid:48) ) ∪ { y } ]. The rest of the proof is as in Lemma 5.2, but we observe that G ∗ = G − V ( K ) + xy containsan X -configuration based on ( C, C ) in which one of the cycles may have been shortened by replacing thepath between x and y on that cycle by xy . If x and y lie on different cycles, C ∪ C − { x, y } is connected, sobelongs to the same component, and we let K (cid:48) be that component. The rest of the proof is then as before.We are left with the case that the 2-cut has the form { v, y } for some vertex y (cid:54)∈ V ( C ∪ C ). If C − { v } and C − { v } both belong to G [ V ( K )] for the same component K of G − { v, y } , we can argue as in Lemma 5.2,since the G ∗ constructed in that proof contains the X -configuration ( C, C ). Otherwise, C −{ v } and C −{ v } belong to different components of G − { v, y } . In that case, we can argue that if G [ V ( K ) ∪ { y } ] contains anessential cycle for any component K of G − { v, y } , we are done by Lemma 3.8, since that cycle is vertexdisjoint from either C or C . Therefore, any essential cycle in a G [ V ( K ) ∪ { v, y } ] must pass through v , andthis remains true if we add an edge vy drawn along a path, call it vy , connecting v and y in G − K . Hence, G [ V ( K ) ∪ { v, y } ] + vy has a compatible embedding in the plane,by Lemma 3.6, for any component K of G − { v, y } . We can then combine all of these plane embeddings to obtain a [weakly] compatible embeddingof G in the plane. In this section we prove Theorem 1.1. We explain how to view iocr-0-drawings of graphs on a torus ascylinder-drawings of an essential cycle with bridges in Section 6.1, and then introduce some tools to simplifybridges in Section 6.2. The main proof is in Section 6.3, with the exception of a lemma which we proveseparately in Section 6.5, after presenting a small set of customized tools in Section 6.4.
In this section we introduce a different way of looking at an iocr-0-drawing on the torus. We assume thatthe drawing contains an essential cycle C . Cylinder.
We can assume that C is crossing-free (by Lemma 3.4). If we cut the torus along C we obtain:Drawing D (cid:48) of the graph G (cid:48) on a cylinder C , in which C is replaced by two cycles of the same length, C L and C R . Indeed, ( S × S ) − C ∼ = I × S , where I is an open interval and S is 1-sphere. In the current section G (cid:48) always denotes such a cylindrically drawn graph. The boundary of C consists of the disjoint union of C L and C R . D (cid:48) can have independent odd pairs of edges touching the boundary. We fix a drawing D (cid:48) of G (cid:48) onthe cylinder as described above. Bridge.
Given a subgraph F of G (possibly just a subset of V ( G )), an F -bridge is a subgraph of G thatconsists of either ( i ) an edge of E ( G ) − E ( F ) with both endpoints in V ( F ), or ( ii ) a component of G − V ( F )together with all edges between the component and V ( F ) and their endpoints in V ( F ). A subgraph F and the F -bridges together partition E ( G ). The vertices of F in an F -bridge are its feet , the edges of thecomponent incident to its feet are its legs , and, for type ( ii ), its component of G − V ( F ) is its core . In G (cid:48) , C -bridges correspond to ( C L ∪ C R )-bridges, identical except for their feet. Upper Indices L and R . Let v ∈ V ( G ) be a vertex on C . We denote by v L and v R its copy on C L and C R , respectively, in G (cid:48) . Let e ∈ E ( G ) be an edge intersecting C (graph-theoretically, C is free of crossings).We denote by e L , e R and e LR , respectively, its corresponding edge in G (cid:48) intersecting only C L , C R and both C L and C R , respectively, in G (cid:48) . Only edges of C can have more than one corresponding edge in G (cid:48) . Then e LR must be an edge joining C L with C R , and e L an edge with at least one end vertex on C L and none on19 R . Similarly, for e R . (Note that e L , e R , and e LR may be undefined.) If P is a path in C we denote by P L and P R its copy on C L and C R , respectively, in G (cid:48) .Let H ⊆ G be a subgraph of a C -bridge. We denote by either H L , H R or H LR the subgraph of G (cid:48) corresponding to H , depending on whether the subgraph corresponding to H in G (cid:48) intersects (in its feet)either only C L , only C R , or both C L and C R , respectively.For other subgraphs of G (cid:48) the upper indices cannot always be used, for example, a subgraph containingparts of C L and C R . In these remaining cases, we write the subgraph of G (cid:48) without using the upper index. Supporting Independent Odd Pairs
If the two edges of an odd pair in G attach to the same vertex u ∈ C on opposite sides of (the crossing-free) C , then the two edges turn into an independent odd pair ofedges in G (cid:48) , attaching to u L and u R . We say that u (or u L and u R ) support(s) an independent odd pair(relative to C ) . If both edges are legs of the same C -bridge H , we say that H supports an independent oddpair at u (or u L and u R ). Cycles and Paths on a Cylinder.
The complement of a closed curve drawn on a cylinder is partitionedinto interior and exterior according to the two-coloring of the connected components. Whenever we talkabout an interior or exterior of a curve (or a cycle) in the drawing of G (cid:48) we specify which components areunderstood to be the interior and exterior.An L -diagonal is a path P L connecting a pair of distinct vertices on C L internally disjoint from both C L and C R . Similarly we define an R -diagonal. Let P L be an L -diagonal. The L -foundation of P L (similarlywe define a R -foundation) is a path P (cid:48) L contained in C L connecting u L with v L such that the counterpartin G of the cycle C (cid:48) = P L ∪ P (cid:48) L is non-essential. By viewing C R from the perspective of C L as being “atinfinity”, for non-essential cycles C (cid:48) obtained in this way the interior is defined as the union of regions in thetwo-coloring of the complement of C (cid:48) having the same color as the component bounded by P (cid:48) L , and similarlyif the roles of C L and C R are exchanged. If we concatenate P L with the path that is complementary to P (cid:48) L on C L , we get an essential cycle by Lemma 2.1.An LR -diagonal is a path P LR in G (cid:48) joining a vertex on C R with a vertex on C L internally disjoint from C R ∪ C L . Three-Stars. A three-star is a K , which occurs as a bridge of three vertices { u, v, z } . Our goal will beto show that a counterexample to Hanani-Tutte can be reduced to a set of pairwise edge-disjoint three-starsthat are { u, v, z } -bridges, where none of u, v, z are even. (For this we need some additional tools developedin the following sections.) The underlying graph then must be a K ,t , which is not toroidal for t ≥
7. Wecan then complete the proof by using the results on Kuratowski minors from Section 4.
The easiest case occurs when all independent odd pairs in a cylindrical drawing are incident on the samevertex of the X -configuration. Lemma 3.11 deals with this case, but we restate it here from the cylindricalpoint of view. This is one of the few results which does not require minimality. Lemma 6.1.
Suppose G contains an X -configuration ( C, C ) . If there is a a vertex on C which supportsall independent odd pairs of edges in G (cid:48) , then G has a compatible embedding in the torus.Proof. Immediate consequence of Lemma 3.11.In the following lemma, a non-trivial C -bridge is a bridge that is not an edge, so it must contain a vertexin its core. Lemma 6.2. If G is a minimal graph that has an iocr- -drawing on the torus containing an X -configuration ( C, C ) , but does not have a weakly compatible embedding on the torus, then there is at most one non-trivial C -bridge that has exactly two feet on C both of which are odd, and that C -bridge contains C . If the C -bridgecontaining C has two feet on C , it is not possible that it supports independent odd crossings at both feet. C and C . Proof.
Let u be the intersection of C and C . Let H be a non-trivial bridge with (exactly) two feet on C . If u is not one of the feet, we can apply Lemma 5.6, since in this case H ∪ C − { u } contains a cyclevertex-disjoint from C . So u must be one of the feet of H , let the other foot be v . Suppose H does notcontain C . Then H − u cannot contain an essential cycle (since it would be disjoint from C , therebycontradicting Lemma 3.8). Therefore, all essential cycles (if any) in H pass through u . The same is true for H + uv , where we draw the edge uv along C . By Lemma 3.6, H + uv has a compatible plane embedding(the rotation at u may change). Let G ∗ = G − ( V ( H ) − { u, v } ) + uv , where uv is drawn along a path in H . We have that G ∗ ≺ G , since H is non-trivial, and G ∗ contains ( C, C ). By induction, G ∗ has a weaklycompatible embedding in the torus in which the plane embedding of H + uv can be embedded alongside uv (both are odd, so it is ok if their rotation changes). We conclude that H must contain C , in other words itis the unique non-trivial C -bridge with two odd feet.To see the second claim, suppose, for the sake of contraction, that H supports independent odd crossingsat both its feet u and v . Let P LRv be the path in H from v L to v R which starts and ends with the pair ofedges involved in the independent odd crossing. Then P LRv corresponds to an essential cycle C in G . Wehave two X -configurations then: ( C, C ) and ( C, C ). By Lemma 5.6,(*) there exists no cycle in G disjoint from C .We claim that there is no non-trivial C -bridge besides H . Consider a C -bridge H (cid:48) other than H , trivialor not. Then H (cid:48) must have at least two feet on C if H (cid:48) is trivial, and the same is true if H (cid:48) is non-trivial, byLemma 5.7. The bridge H (cid:48) cannot have a foot different from both u and v : if it did, there would be a cyclein H (cid:48) ∪ C − { u } vertex-disjoint from C , or in H (cid:48) ∪ C − { v } vertex-disjoint from C , contradicting Lemma 5.6.Hence, H (cid:48) must have exactly two feet, which are u and v , on C . By the first part of the lemma, which wealready proved, H (cid:48) is trivial and consists of the single edge uv . It follows that H is the only non-trivial C -bridge.Since u and v are the only feet on C , minimality implies that C is the cycle uvw , with a degree-2 vertex w . Let G = G − { w } . Then G ≺ G since G has fewer edges than G , and G has an iocr-0-drawing on thetorus, inherited from G . Since u and v are odd, a weakly compatible embedding of G could be extendedto a weakly compatible embedding of G (which does not exist), by embedding the path uwv along uv . Byminimality of G , we have(**) G does not admit a weakly compatible embedding on the torus, and hence, does not contain an X -configuration by the minimality of G .We claim that every C -bridge in G except the one containing v , which we denote by H , is a path oflength at most 2 ending in u . To see this, let H ∗ be a C -bridge other than H . Since H ∗ cannot contain acycle weakly disjoint from C , it has at most two feet on C , one of which is u , since C ∩ C = { u } . By thesame token, every cycle in H ∗ must pass through u . Therefore, all cycles in H ∗ are disjoint from its secondfoot on C . Hence, all cycles in H ∗ are disjoint from C . By (*), there is no cycle in H (cid:48) , and since there isalso no cut-vertex in G , by Lemma 5.7, H ∗ is a (subdivided) edge. By the minimality of G , H ∗ is a path oflength at most 2.To summarize, C is a cycle in G which has one non-trivial C bridge containing v , and all other C bridges are paths of length at most 2 ending in the odd vertex u .We now choose C as the shortest cycle in G which contains a non-trivial C -bridge H with an odd foot u so that all other C -bridges are induced paths of length at most 2 ending in u , and so that the number ofedges in H is maximized.We show by induction of the number of C -bridges other than H that G admits a weakly compatibleembedding on the torus, which contradicts (**), and therefore concludes the proof.21n the base case, we are done by the second part of (**): In this case there is only one C -bridge, H .By (**) there is no X -configuration, so all of the vertices on C can be made even. Then the cylindricaldrawing of G (cid:48) is in fact an iocr-0-drawing in the plane (an annulus) and we are done by an application ofTheorem 3.1.In the inductive case, G contains a C -bridge P different from H that is a (subdivided) edge incidentto u , which is odd, and a second vertex u (cid:48) on C . Suppose there is such a C -bridge P for which u (cid:48) is alsoan odd vertex. If P joins two consecutive vertices along C we can remove P , apply induction and insert P back into the embedding along the edge of C . If P does not join two consecutive vertices of C , by thechoice of C and H , a part of C between the end-vertices of P is a subdivided edge of the same lengthas P , which must be 2. Therefore we can again remove P , apply induction, and insert P back in theembedding along the edges of C .It remains to deal with the case that all the C -bridges different from H are paths whose second foot,the foot different from u , is an even vertex. We claim that u is the only vertex on C that supports anindependent odd pair in G (cid:48) , the cylindrical drawing of G . Suppose there were a foot v (cid:54) = u on C whichsupports an independent odd pair. Then v can only be a foot of H , since all other bridges have u or aneven vertex as a foot. It follows that the independent odd pair at v must belong to H , which forces an X -configuration in C ∪ H , contradicting (**). We can therefore assume that any two edges attaching at v ∈ C − { v } on opposite sides of C cross evenly. This allows us to split off two vertices from v , one on eachside with edges incident to v on that side attached, and attached to v , turning v into an even degree-4 vertex;the drawing remains iocr-0. At this point, u is the only odd vertex on C , and we can apply Lemma 3.11 (notinductively) to obtain a compatible embedding of the modified G . Contracting the edges which resultedfrom the splits, we obtain a compatible embedding of G .The following lemma is the heart of the reduction; we will state it and prove it in Section 6.5. Beforethat, we will see how to use it to complete the proof of the main result. Lemma 6.3. If G is a minimal graph that has an iocr- -drawing on the torus containing an X -configuration ( C, C ) , but does not have a weakly compatible embedding on the torus, and there is a C -bridge H whichattaches to C in at least two vertices u and v that support independent odd pairs of legs in H , then ( i ) if there is more than one C -bridge, then there is an X -configuration ( C, C ) (not the same C and C as the original) for which there is only one C -bridge, and ( ii ) if there is only one C -bridge, then G is a subgraph of a subdivision of a K or a K ,t with spokes. Assume, for a contradiction, that Theorem 1.1 fails. Then there is a graph G with an iocr-0-drawing D on the torus, so that G cannot be embedded on the torus. Let G be a minimal such counterexample, withiocr-0-drawing D . By Lemma 5.2 we know that G is 3-connected, and by Lemma 5.5 that D contains an X -configuration ( C, C ). So there is a counterexample to the following strengthened version of Theorem 1.1:Let D be an iocr-0-drawing of a graph G containing an X -configuration ( C, C ), then G has aweakly compatible embedding on the torus.If the condition is met, let us say that G —or more precisely, ( G, D, C, C ), satisfies HT-XWC , acronymfor Hanani-Tutte–X-Weakly-Compatible. So we know that there is a (
G, D, C, C ) which does not satisfyHT-XWC.Fix a ( G, D, C, C ) violating HT-XWC for which G is minimal with respect to first, ≺ , second, thenumber of C -bridges, and third, | E ( C ) | + | E ( C ) | . We write ≺ (cid:48) for the strict partial ordering defined in thisway; by definition ≺ (cid:48) refines ≺ , so a ≺ (cid:48) -minimal counterexample is also a ≺ -minimal counterexample.By Lemma 6.3( i ) if there is a C -bridge which supports independent odd pairs at two (or more) verticesof C , then we can choose C so that there is only one C -bridge. By Lemma 6.1 there cannot be a single22ertex on C supporting all independent odd pairs in G (cid:48) , so there must be at least two such vertices, and,since there is only a single C -bridge, the independent odd pairs are supported by that single C -bridge. Wecan then apply Lemma 6.3( ii ) to conclude that G is a subgraph of a subdivision of K or a K ,t with spokes.The first case is not a counterexample, by Lemma 4.5. So G is a subgraph of a subdivision of a K ,t withspokes; let t be minimal. Then G contains a subdivision of a K ,t , so G is not a counterexample for t ≥ t ≤ G is not a counterexample, because it has aweakly compatible embedding on the torus by Lemma 4.2.We conclude that for every C -bridge H there is at most one vertex on C so that H supports an independentodd pair at that vertex. By definition, the two C -edges incident to u form an independent odd pair at u ,so there is a path P LR from u L to u R in G (cid:48) , following C , which contains an independent odd pair at u . Let H LR be the C -bridge containing P LR . We already argued that H LR does not support an independent oddpair at any vertex v ∈ V ( C ) − { u } . Also, recall that u is not even.There must be a v ∈ V ( C ) − { u } so that v L and v R are incident to edges forming an independent oddpair l and l (in G (cid:48) ), since otherwise we are done by Lemma 6.1. Since H only supports independent oddpairs at u , it is not possible that both l and l belong to H , so at least one of them must belong to adifferent C -bridge H (cid:48) .First, suppose that H (cid:48) is not just a single edge. Since v and u are odd, by Lemma 6.2 H (cid:48) must have afoot w (cid:48) on C L or C R different from u L , u R , v L and v R . Then H (cid:48) contains a path from w (cid:48) to v (on C L or C R )disjoint from P LR . Adding to that a subpath of C connecting v to w (cid:48) while avoiding u gives us a cycle C (cid:48) in G which is vertex-disjoint from P LR , and, therefore, C . This contradicts the choice of G by Lemma 5.6(note that C (cid:48) may be essential or not).Second, H (cid:48) consists of a single edge e . We distinguish two cases: e is incident to one side of C only, or e is an LR -bridge.Suppose first that e is incident to C L only. This contradicts the choice of C : Let C (cid:48) be the result ofshortening C by using e to replace the L -foundation of e on C L . Then C (cid:48) is essential, and C (cid:48) cannot havemore bridges than C , since V ( C (cid:48) ) ⊆ V ( C ). The path P LR remains an LR -diagonal for C (cid:48) , since its endsremain on the opposite sides of C (cid:48) . (This is not necessarily true if e is an LR -diagonal.) The two new cyclesstill intersect in u which can still not be made even by flips. Since C (cid:48) is shorter than C , and C did notchange, this contradicts the minimality of the X -configuration ( C, C ). u L u R v R C L C R v L e u L u R v R C L C R v L eH LR H LR f fw L P (cid:48) R P (cid:48) R w R Figure 10: A C -bridge H (cid:48) is a single edge e (left) that can be embedded along the path P (cid:48) R after all theother C -bridges H different from H are embedded along C R (right).In the remaining case, e is an LR -diagonal. We want to redraw C -bridges other than H LR so that e canbe drawn along C R or C L , see Figure 10.By symmetry, we can assume that e is incident to v L . The other endpoint of e must be u R : otherwise wehave an essential cycle disjoint from C , which passes through u , consisting of e followed by the path fromthe endpoint of e on C R to v R avoiding u . This contradicts Lemma 3.8.Let P (cid:48) be a path between u and v on C . This path corresponds to two paths P (cid:48) L and P (cid:48) R on C R and C L ,respectively, in G (cid:48) . We claim that it is not possible that H LR has feet in the interior of both both of thesepaths. If it did, we could use a path in H LR connecting the two feet (and otherwise avoiding C ) combinedwith a subpath of P (cid:48) L (or P (cid:48) R ) to obtain an essential cycle which is vertex-disjoint from the essential cycle23ormed by e and the path connecting u and v on C which is not P (cid:48) . This contradicts Lemma 3.8.Without loss of generality then, we can assume that P (cid:48) R contains no feet of H LR (the case for P (cid:48) L canbe dealt with symmetrically). Consider a C -bridge H other than H and H (cid:48) = e . Then H must have atleast two feet in C by Lemma 5.7. On the other hand, H cannot have two feet on C L (or C R ) which aredifferent from u , since this would contradict Lemma 5.6 by giving us two vertex-disjoint cycles, the essentialcycle C and a cycle in H ∪ C avoiding u . Similarly, H cannot have a foot on both C L and C R bothdifferent from u L and u R , since such a H would lead to an essential cycle in G vertex-disjoint from C ,contradicting Lemma 3.8.Therefore, each such bridge H has exactly one foot on C L or C R other than u L or u R (and at leastone of these). We only consider bridges with a foot in P (cid:48) R . If there are no such bridges, we can redraw H (cid:48) = e close to P (cid:48) R , crossing-free. The rotation at u and v changes, but both are odd vertices, so that isfine. Moreover, ( C, C ) is still an X -configuration (since the rotation of the edges in H LR did not changewith respect to C , and those edges could not be made even by flips). We are now back in the previous case,where e can be used to shortcut C .Otherwise, there is a bridge with a foot w R on P (cid:48) R . We pick w R as close to u R (on P (cid:48) R ) as possible. If w is odd, we pick any bridge H incident to w R . If w is even, we pick the bridge H first in the rotation at w R after w R u R (anti-clockwise); note that by Lemma 5.6, there is a single edge between w R and the core of H in this case. Similarly, a cycle in H avoiding u would contradict Lemma 5.6 due to the X -configuration( C, C ). Therefore all cycles in H pass through u , and H ∪ uw has a compatible embedding (except for,possibly, at u ), by Lemma 3.6.We can then insert the plane embedding of H close to w R u R ; in the case that w is even, this reestablishesthe rotation at w , since we picked the bridge with the closest leg to w R u R in the rotation at w , and thebridge connects to w R via a single edge.Note that this last argument is a local induction on the number of bridges (other than H and H (cid:48) ) witha foot in P (cid:48) R ; within this case, we are not using minimality of G (cid:48) , we only redraw the bridges so that e canbe redrawn without crossings.We have shown that there is no minimal counterexample to HT-XWC, and, therefore, no counterexampleto Theorem 1.1, completing the proof. In preparation for the proof of Lemma 6.3 in the next section, we present three reduction lemmas. Foreach lemma, we assume that G is a minimal graph that has an iocr-0-drawing on the torus containing an X -configuration ( C, C ), but does not have a weakly compatible embedding on the torus, Lemma 6.4.
Let v ∈ V ( G ) be a vertex of degree , and u and w two of its neighbors. If both u and w areodd, then uw (cid:54)∈ E ( G ) unless G − uw does not contain an X -configuration.Proof. Otherwise we would violate the minimality of G . Indeed, removing uw results in a graph that is nota counterexample, and uw can be inserted into a toroidal embedding of G − uw without introducing an edgecrossings and while maintaining the rotations at even vertices in the given independently even drawing. Lemma 6.5.
Suppose that G contains a 2-cut { w (cid:48) , x } such that w (cid:48) belongs to C , but not to C , and x doesnot belong to C . If the edges incident to w (cid:48) L (or w (cid:48) R ) in G (cid:48) − E ( C ) connect w (cid:48) with at least two connectedcomponents of G [ V − { w (cid:48) , x } ] or at least one such component if w (cid:48) L x (or w (cid:48) R x ) is an edge of G (cid:48) , then thevertex x is even.Proof. If w (cid:48) is even, then the union of two connected components of G [ V − { w (cid:48) , x } ] or the union of one suchcomponents together with the edge w (cid:48) L x or w (cid:48) R x contains a cycle C (cid:48) weakly disjoint from C , contradictingLemma 5.6. We conclude that w (cid:48) must be odd. Since w (cid:48) does not belong to C and x does not belong to C ,Lemma 5.8 applies, and x must be even. 24 emma 6.6. Suppose that G has a 2-cut { u, v } such that v (cid:54)∈ V ( C ) is even and u ∈ V ( C ) . Let H = G [ V ( G ∗ ) ∪ { u, v } ] , where G ∗ is the union of all components of G − { u, v } excluding the component containing C − { u } . If H does not contain an essential cycle, then H is a path. In particular, H has only a single edgeadjacent to v .Proof. If H is an edge or path, we are done. Hence H must contain a cycle C (cid:48) : all internal vertices of H , thatis, vertices other than { u, v } , have degree at least two, otherwise there would be a cut-vertex. In particular, H contains at least three edges. We claim that u is odd. Suppose u were even. The cycle C (cid:48) in H cannot beessential, by assumption, so C and C (cid:48) cross an even number of times. Since u is even this means that C and C (cid:48) are either vertex-disjoint, or touch in u , contradicting Lemma 5.6. This shows that u is an odd vertex.The edges of H at v have to be consecutive in the rotation at v : If not, there has to be a pair of paths P and P between v and C − u internally disjoint from H , whose ends at v alternate with edges e , f of H .Since H and C only share u , the cycle in P ∪ P ∪ ( C − u ) and the cycle in H containing e and f only havethe even vertex v in common, and, since their ends alternate at v and the drawing is iocr-0, the two cyclescross an odd number of times, which implies both of them are essential, contradicting the assumption that H does not contain an essential cycle.In G , we now replace H with a path of length two between u and v (following a uv -path in H andsuppressing all but one interior vertex). Then the resulting graph G ∗ satisfies G ∗ ≺ G (since H containsat least three edges). Moreover, G ∗ still contains an X -configuration, with C unchanged, and C possibleshortened: if C ∩ H is non-empty, then that piece of C is replaced by the new uv -path of length two. Byminimality of G , there is a weakly compatible embedding of G ∗ in the torus.Let H ∗ be H , together with a path P between v and u internally disjoint from H (such a P exists, sinceotherwise u is a cut-vertex, contradicting Lemma 5.2. Then any essential cycle in H ∗ must use P (since H does not contain any essential cycles by assumption), and, therefore, some interior vertex x of P . We canthen apply Lemma 3.6 to show that H ∗ has a plane embedding, in which the embedding of H is compatible.Removing the internal vertices of P from the drawing of H ∗ , yields a compatible plane embedding of H inwhich u and v lie in the same (outer) face. We can the insert this embedding in place of the uv -path in G ∗ which replaced H , to obtain a weakly compatible embedding of G . Since u is odd, we do not have to recoverthe rotation at u .Again, we have reached a contradiction, so we can conclude that H is a path (possibly a single edge). We restate the lemma for reference:If there is a C -bridge H which attaches to C in at least two vertices u and v that supportindependent odd pairs in H , then( i ) if there is more than one C -bridge, then C and C can be chosen so that there is only one C -bridge, and( ii ) if there is only one C -bridge, then G is a subgraph of a subdivision of a K or a K ,t withspokes.Let the two independent odd pairs of H at u and v be ( l L , l R ) and ( l L , l R ) with l L , l R incident to u L , u R , and l L , l R incident to v L , v R . See Figure 11. ( l L , l R ), ( l L , l R )We denote by P L a path in H LR from u to v starting with l L and ending with l L . Similarly, P R is apath in H LR from u to v starting with l R and ending with l R . Both paths P L and P R can be assumed to P L , P R avoid vertices of C in their interior. We choose P and P so that the sum of the lengths of P L and P R isminimal; we will use this assumption in Case (B2) below at the very end of the proof. Then, P L and P R ,are L and R -diagonals. Let C and C , denote the cycles obtained by concatenating P L and P R with their L and R -foundations on C L and C R . C , C A brief outline of the remainder of the proof: We distinguish two cases depending on whether P L and P R intersect. In both cases we find, by the choice of G (cid:48) , a vertex z that together with u and v forms a set25f vertices that due to the minimality of G yields in G a collection of { z, u, v } -bridges that must be “almostlike” three-stars. The heart of the matter then is to reduce the instance so that Lemma 4.2, or Lemma 4.5apply.For goal ( i ), replacing C with an essential cycle having only a single bridge, we will use a specific way toshort-cut a path. Suppose P is a path in G (cid:48) which corresponds to a cycle in G . A P -shortcut is a path that P -shortcut,sc( P )is obtained from P by the following inductive procedure. Start with P = P . If there exists a P i -bridge B with feet u and v , that is a (subdivided) edge, and the subpath of P i between u and v is not a (subdivided)edge (in G ) then we construct P i +1 from P i by replacing the subpath of P i between u and v by B . If no such B exists then P k is a shortcut of P . We write sc( P ) for a P -shortcut; it may not be unique, but it alwaysexists. Moreover, sc( P ) is a path in G (cid:48) connecting the same endpoints as P , and corresponds to a cycle in G . When P or sc( P ) correspond to a cycle in G , we will write P in G and sc( P ) in G to refer to those cycles. Claim 6.7.
Suppose P is a path in G (cid:48) which corresponds to a cycle in G and both P ∩ C L and P ∩ C R arepaths, with at least one of them being a single vertex.( i ) P and sc( P ) are essential cycles in G .( ii ) If ( C, P ) is an X -configuration, and sc( P ) has a single bridge in G , then sc( P ) is part of an X -configuration.( iii ) Any bridge of P in G is part of a bridge of sc( P ) in G , or has become part of sc( P ). Proof. ( i ). Since P is a path in G (cid:48) and a cycle in G , it must connect x L on C L to x R on C R for some x ∈ V ( G ). By symmetry we can assume that x L is the only vertex of P on C L . Since P is a path in G (cid:48) , anda cycle in G that cycle must be essential in G . The shortcut sc( P ) also is a path between x L and x R in G (cid:48) .By the construction of sc( P ), x is also the only vertex such that x L is on sc( P ). Indeed, only vertices in V can be on sc( P ) and not on P . However, vertices in V ∩ C are not feet of any C -bridge, and therefore arenot adjacent to any vertex of P besides possibly x . Therefore sc( P ) is an essential cycle.( ii ). We can then make the edges of sc( P ) even (in G ), using Lemma 3.4. Let P ∩ C = { u } . We knowthat u cannot be made even by edge-flips; since it still lies on sc( P ), that means there are two edges incidentto u which cross oddly. Since sc( P ) only has a single bridge, there is a cycle containing both edges, and noother vertices of sc( P ). Then that cycle, together with sc( P ), forms an X -configuration.( iii ). The shortcut sc( P ) consists of (a subset of the) vertices of P and degree-2 vertices. Therefore, anybridge of sc( P ) must attach to sc( P ) in vertices of P , so it is, or contains, a bridge of P .We begin with the case that P L and P R do not intersect, illustrated in Figure 11. CASE (A): P L and P R do not intersect. There exists a path P in H that joins a vertex of P L with a vertex z of P R . Since all the vertices of P L P (Case A)are contained in the interior of C and all the vertices of P R are contained in the interior of C all the feetof H LR are on the cycles C and C . By Lemma 6.2, H LR has another foot on C L or C R . Let us say thatfoot is w L , (cid:54) = u L , v L on C L . Let P (cid:48) L be a path between w L and P R in H LR . w L (Case A) P (cid:48) L (Case A)For future reference, we give names to some of the LR -diagonals. Let Q u denote the path in G (cid:48) consisting Q u (Case A)of the subpath of P R between u R and z ; P (cid:48) L ; and the subpath of C L between w L and u L not containing v L .Symmetrically, let Q v denote the path in G (cid:48) consisting of the subpath of P R between v R and z ; P (cid:48) L ; and Q v (Case A)the subpath of C L between w L and v L not containing u L . Let R LRu denote the LR -diagonal in G (cid:48) obtained R LRu (CaseA)by concatenating the part of P R between u R and z ; P ; and the part of P L between the end vertex of P and u L . Symmetrically, we define R LRv . Refer to Figure 11. By Claim 6.7 all of Q u , Q v , R LRu and R LRv are R LRv (CaseA)essential cycles in G . Then each of the pairs ( C ⊕ C , Q u ), ( C ⊕ C , Q v ), ( C, R
LRu ), and (
C, R
LRv ) containsthe four edges at u (or v ) which cannot be made even, so each pair is an X -configuration, implying that allof Q u , Q v , R LRu , and R LRv are part of X -configurations.26 L C R H LR u L v L v R u R l L l L l R l R PP (cid:48) L R LRu C L C R H LR u L v L v R u R l L l L l R l R Q u w R w L w R z w L sc ( Q u ) z Figure 11: The case when P L and P R connecting u L , v L and u R , v R , respectively, in H LR do not intersectin G (cid:48) . The paths P, P (cid:48) L , R LRu highlighted on the left, Q u and sc ( Q u ) highlighted on the right.We collect some crucial properties of z, w L and P (cid:48) L which are needed in the remainder of the argument. Claim 6.8.
We have the following.(a) Vertex z is the end vertex of P (cid:48) L .(b) Vertices { w L , z } disconnects any interior vertex of P (cid:48) L from the rest of G (cid:48) .(c) Vertex z is odd.(d) Vertex w L has degree 1 in H LR .A statement analogous to the claim with the third foot being w R on C R holds by symmetry (we will usethat variant of the claim below). Proof.
We start with the proof of (a) Note that P (cid:48) L is disjoint from P L (since its vertices lie outside C which contains all vertices of P L in its interior). The end vertex of P (cid:48) L on P R must be z , the end vertexof P on P R : If P (cid:48) L had an end vertex z (cid:48) (cid:54) = z on P R , there would be two vertex-disjoint cycles: considerthe cycle starting with w L , following P (cid:48) L up to z (cid:48) , then the part of P R either to u R or v R whichever avoids z , and finally, the part of C R back to w R avoiding either u R or v R ; this cycle is vertex disjoint from either R u or R v , depending on which of these was avoided by the first cycle. Hence, P (cid:48) L ends in z on P R , whichconcludes the proof of (a).We continue with (b). Claim (b) is true for the vertices in the core of H , since the core of H is a tree,leaving only vertices of C L , u R and v R due to the fact that the interior vertices of P (cid:48) L are in the interior of C . If there were a path from an interior vertex of P (cid:48) L to a vertex of C L , or u R or v R , avoiding both w and z , then such a path, together with a subpath of C L , forms a path giving rise to a cycle in G vertex-disjointfrom an essential cycle corresponding to R LRu or R LRv , since the subpath of C L can be chosen to avoid at leastone of u and v . This would contradict Lemma 5.6. By Lemma 5.6, it also follows that the { w L , z } -bridge B containing P (cid:48) L is just a (subdivided) edge, i.e., all of its internal vertices are of degree two. Indeed, if thebridge B is not just a path it must contain a cycle disjoint from z or w L . The cycle is not contained in thecore of H , which is a tree. The cycle must then pass through w L and not through z . Hence, Lemma 5.6applies if we use R LRu or R LRv .Next, we prove (c). If z is even, or can be made even by flips, then Q u and R LRv would correspond toweakly disjoint essential cycles in G , contradicting Lemma 3.8. Hence, z is odd and cannot be made evenby flips.Finally, we prove (d). The parts (a) and (b), imply that any edge of H LR incident to w L must becontained in a { w L , z } -bridge that is a path (possibly just an edge). Hence, if there is another such edgeincident to P (cid:48) L besides the one in P (cid:48) L then { w, z } forms a 2-cut in G . Combining this with the oddness of27 , proved in part (c), Lemma 6.5 then implies that w L has degree 1 in H , that is, the core of H has onlyone leg ending in w L , which concludes the proof.We are now in a position to prove ( i ) for Case ( A ). CASE (A) - Establishing Property ( i ) . We would like to argue that Q u has only a single bridge; this need not be true, however. For example, ifwe subdivide the edge u L z in G (cid:48) in the left picture of Figure 11, the resulting path is a bridge of Q u . Wetherefore work with a Q u -shortcut, sc( Q u ). Claim 6.7 tells us that sc( Q u ) is an essential cycle in G and thatany bridge of Q u (in G ) is contained in some bridge of sc( Q u ), or has become part of sc( Q u ).Suppose then that sc( Q u ) contains a bridge H other than the bridge containing R LRv . If H has twofeet on sc( Q u ) different from z , then sc( Q u ) ∪ H contains an essential cycle avoiding z (passing through thetwo feet), which contradicts Lemma 3.8, since R LRv corresponds to a disjoint essential cycle, so this does nothappen. Since G is 2-connected, by Lemma 5.7, H must have exactly two feet on sc( Q u ), one of which is z .The other foot of H cannot be one of the degree-two vertices short-cutting Q u (being a foot), and it cannotbelong to the core of H , since that is a tree. Therefore, the other foot must lie on C . Suppose the foot lieson C L , and call it w (cid:48) L . We allow the case w (cid:48) L = u L , but note that w (cid:48) L (cid:54) = w L , since w L has degree 1 in H as we argued earlier.By Claim 6.8 (b) and (d), H is a subdivided edge (in case w (cid:48) L = u L it is an edge, by Lemma 5.8, sinceboth z and u are odd; we work with X -configuration ( C, Q v ) to apply the lemma). Now sc( Q u ) contains apath from z to w (cid:48) L ; if this path contained any vertex of Q u , it would have been replaced by H in sc( P );since, by assumption, that did not happen, the path from z to w (cid:48) L in sc( Q u ) contains no vertices of Q u ,which implies that it a (subdivided) edge. This contradicts Lemma 6.5, because z is not even; we work with X -configuration ( C, Q v ). The lemma applies, since w (cid:48) belong to C and not to Q v , and z does not belong to C . Hence sc( Q u ) has only one bridge, the bridge containing R LRv . Moreover, the cycles corresponding tosc( Q u ) and R LRv are essential and intersect in a single vertex z , which, as we saw, cannot be made even byflips. This concludes the argument of part ( i ) of Case ( A ). CASE (A)- Establishing Property ( ii ) . For part ( ii ) we can assume that C has only a single bridge.Suppose C has length at least 4. Then there must be a fourth vertex w (cid:48) on C . If w (cid:48) is not incident to a w (cid:48) (Case A)leg of H LR we can suppress w (cid:48) thereby violating the minimality of the counterexample. So w (cid:48) is incident toa leg of H LR , and this leg must be contained in a cycle vertex-disjoint from an essential one, contradictingLemma 3.8: If w (cid:48) L is incident to such a leg l , a cycle passing through l , and containing z and w L is vertex-disjoint from an essential cycle whose edge set is the symmetric difference of E ( C ) and E ( C L ). Otherwise,if l is incident to w (cid:48) R ; in this case, a cycle passing through l and a subpath of C , and containing v R , say, isvertex disjoint from Q u . In both cases, we contradict Lemma 3.8.Therefore, C has length 3. Consider the case that w has degree 3 in G . In that case we are done,by Lemma 6.4, since the edge uv violates the lemma: u and v are odd, and the essential cycle sc( Q u ) (orsc( Q v )), as we argued in part ( i ), has a single sc( Q u )-bridge, and removing the edge uv does not change that.If all vertices of sc( Q u ) can be made even by flips, then, by Corollary 3.5, G − uv has a weakly compatibleembedding in the torus, and to that embedding we can add back uv to get a weakly compatible embeddingof G in the torus. Hence sc( Q u ) contains an odd vertex which cannot be made even by edge-flips, and, sincesc( Q u ) has a single bridge, there must be a cycle C (cid:48)(cid:48) in G − uv so that (sc( Q u ) , C (cid:48)(cid:48) ) is an X -configuration,contradicting Lemma 6.4.We conclude that w has degree at least 4 and C consists of three vertices only, u, v and w . We arguedearlier that w L has degree at most one in a C -bridge. Claim 6.8 (b) and (d) applies also to w R . Hence, w has degree exactly 4, with both w L and w R being of degree 3 in G (cid:48) . We can then define Q u and Q v with w R playing the role of w L . It follows (by symmetry) that in G there exists an essential cycle avoiding z and28 , and an essential cycle avoiding z and u that can play the role of C similarly as the cycles correspondingto sc( Q u ) or sc( Q v ). Moreover, there is a path P (cid:48) R between w R and z (cid:48) , the endpoint of P on P L ; we have z (cid:54) = z (cid:48) since P L and P R do not intersect by assumption. As we did with P (cid:48) L , we can show that P (cid:48) R is asubdivided edge, and z (cid:48) is odd.Let us summarize what we know at this point: C consists of vertices u , v , and w ; w has degree 4 in G and is connected to u and v by an edge, and to z and z (cid:48) by subdivided edges. We claim that G has no other { u, v, z } -bridge (attached at all three vertices) besides the one containing w . If it did, we could replace zu L in Q u (for w L ) by a path from z to u through that bridge (avoiding v ), creating a vertex-disjoint cycle tothe Q v we constructed for w R (which passes through w R so belongs to the bridge containing w ). This wouldcontradict Lemma 3.8. Since z and z (cid:48) are odd and do not lie on C , Lemma 5.8 implies that { z, z (cid:48) } is not acut, and similarly { u, z } and { v, z } . Since there are no cycles disjoint from C , and we have accounted for alledges attaching to C , this means that z and z (cid:48) if they are connected, are connected by an edge. We concludethat G is a (subgraph of a) subdivision of K with all vertices, u , v , w , z , and z (cid:48) odd. By Lemma 4.5 sucha G is not a counterexample. CASE (B) P L and P R intersect. The intersection of P L and P R must be a path, possibly consisting of a single vertex: If x and y are interiorvertices of P L , then the unique path between x and y in the core of H LR (which is a tree) must belong to P L . The same is true for P R , showing that P L and P R intersect in a path.First, we consider the case that C and C are (edge)-complements of each other on C . CASE (B1) C and C are complementary on C . See Figure 13 (left) for an illustration; in the figure the red subpath of C L is the complement of the greysubpath of C R .We correct the rotation at the vertices of P L so that the edges of P L cross every other dependent edgein G (cid:48) an even number of times. Let H (cid:48) LR denote the union of P L and P R . H (cid:48) LR (CaseB1)Let R LRu and R LRv denote the paths joining u L and u R , and v L and v R , respectively, in H (cid:48) LR . Since C R
LRu , R LRv (Case B)is essential, and C and C are non-essential, C ⊕ ( C ⊕ C ) = R u ⊕ R v has non vanishing homology over Z . Hence, R u and R v as curves cross an odd number of times. Indeed, R u and R v are essential cycles in G , and therefore belong to different non-vanishing homology class over Z . It follows that the order of endvertices of P L ∩ P R along P L and P R when traversing P L and P R , respectively, from u L to v L , and from u R to v R , is reversed.Let w R denote a third foot of H LR (a third foot must exist by Lemma 6.2, we arbitrarily assume it lies w R (CaseB1)on C R , the case C L is symmetric).Let z denote the vertex in H (cid:48) , but not on C , joined by a path P R connecting H (cid:48) LR with w R (internally z , P R (CaseB1)disjoint from C R , C L and H (cid:48) LR ). Claim 6.9.
The vertex z lies in the intersection of P L and P R , and there exists no pair of vertex disjointpaths in G (cid:48) internally disjoint from H (cid:48) LR joining the intersection of P L and P R with C R , and the sameapplies to C L by symmetry. Proof.
Refer to Figure 12. If the first part of the claim does not hold we contradict Lemma 3.8, see thetop-left part of the figure. If the second part of the claim does not hold we are done by the same token, seethe top-right and bottom part of the figure.Recall that R LRu is the path in H (cid:48) LR joining u L with u R , and similarly for R LRv . By choice of u and v both ( C, R
LRu ) and (
C, R
LRv ) are X -configurations, but R LRu , and R LRv , may have more than one bridge.That is, why we work with sc( R LRu ), and sc( R LRv ), the R LRu - and R LRv -shortcuts.29 L C R u L v L v R l L l L l R l R w R w L C L C R u L v L u R l L l L l R l R w R w L y R P L P R zP R z (cid:48) P L P R y L C L C R u L v L v R u R l L l L l R l R w R w L P L P R zu R v R z (cid:48) z Figure 12: P L and P R intersect. A pair of gray essential cycles violating Lemma 3.8 if z is not in theintersection of P L and P R (left). A pair of disjoint paths joining the intersection of P L and P R with C R yielding a pair of gray essential cycles violating Lemma 3.8 (right,bottom). C L C R w L w R u R P R P L P R v R zu L v L C C C L C R w L w R u R P R P L P R v R u L v L zT v C L C R u L v L v R u R l L l L l R l R w R w L P L P R zQ v Figure 13: The interiors of C (red), and C (grey) in the regions incident to their vertices (left). The dashededges (paths) are not possible (right). The one between z and w L due to Lemma 3.8.30 ASE (B1)- Establishing Property ( i ) . We prove property ( i ) by showing that sc( R LRu ) is an essential cycle with a single bridge (and similarly forsc( R LRv )). By Claim 6.7, sc( R LRu ) is an essential cycle, so we have to show that sc( R LRu ) only has a singlebridge, which then implies that sc( R LRu ) is part of an X -configuration by Claim 6.7.We will work with an essential cycle C in G defined as follows: The cycle C combines P R , the subpath C (Case B1)of C R connecting w R with v R avoiding u R , and the part of P R between v R and z . See Figure 14. The cycle C is essential due to the fact that its small generic perturbation crosses R LRu an odd number of times. C L C R u L v L l L l L l R l R w R w L P L P R sc( R LRu ) C v R u R z C L C R u L v L l L l L l R l R w R w L P L P R sc( R LRu ) C v R u R zC (cid:48)(cid:48) Figure 14: Path sc( R LRu ) and the cycle C . The dashed path cannot exist if z is even since it yields a pairof weakly disjoint essential cycles, which in G (cid:48) form a pair of paths between u L and u R , and v L and v R ,respectively. We depict the two cases depending on the position of w R on C R with respect to C .Suppose there were a sc( R LRu )-bridge H different from the bridge containing C − u . Then H cannothave two feet in the interior of sc( R LRu ), since this would imply that H ∪ sc( R LRu ) − u contains a cyclevertex-disjoint from C , which contradicts Lemma 5.6. Since H must have at least two feet (by Lemma 5.7),those feet must be { u, u (cid:48) } , where u (cid:48) is an interior vertex of sc( R LRu ). Since vertices of sc( R LRu ) have eitherdegree 2 or belong to R LRu , we know that u (cid:48) , as a foot, must belong to R LRu .Since there are two sc( R LRu )-bridges, { u, u (cid:48) } must be a cut-set. We can then apply Lemma 5.8, with X -configuration ( C, R
LRv ), to conclude that u (cid:48) is even.Let F denote the union of connected components of G − { u (cid:48) , u } not containing v (this also includesan edge u (cid:48) u if it exists). Let H (cid:48)(cid:48) = G [ V ( F ) ∪ { u (cid:48) , u } ]. By Lemma 6.6, either H (cid:48)(cid:48) has only a single edgeadjacent to u (cid:48) or there is an essential cycle C (cid:48)(cid:48) in H (cid:48)(cid:48) . In the first case, H (cid:48)(cid:48) is a (subdivided) edge—byLemma 5.6—between u and u (cid:48) , which cannot be the case due to the existence of H .Hence, H (cid:48)(cid:48) contains an essential cycle C (cid:48)(cid:48) . C (cid:48)(cid:48) must pass through u (otherwise C (cid:48)(cid:48) and C are vertex-disjoint, contradicting Lemma 3.8). If u (cid:48) (cid:54) = z , then u (cid:48) lies either before or after z on the path R LRu from u L to u R . If u (cid:48) lies on the u L -to- z part of R LRu , then C (cid:48)(cid:48) is vertex-disjoint from the essential cycle Q v through v Q v (CaseB1)and z consisting of the part of P L from v L to z followed by P R and the part of C R from w R to v R avoiding u R ; if u (cid:48) lies on the z -to- u R part of R LRu , then C (cid:48)(cid:48) is vertex-disjoint from the essential cycle C . In bothcases, Lemma 3.8 applies, establishing a contradiction.We conclude that u (cid:48) = z . Recall that u (cid:48) is even. Therefore, if C (cid:48)(cid:48) touches either Q v or C in u (cid:48) = z , weare done by Lemma 3.8. Hence C (cid:48)(cid:48) crosses both of those cycles in z . Then C (cid:48)(cid:48) must be weakly disjoint from R LRv , see Figure 14. This contradicts Lemma 3.8.This completes the proof of ( i ) in case (B1).Before turning to ( ii ), we show that there are two more paths in G (cid:48) that can (potentially) play the roleof C , in the sense of being essential with a single bridge, and therefore part of an X -configuration.31efer to Figure 13 (middle-right). Let Q u be the path in G (cid:48) obtained by following P L from u L to z , then Q u (CaseB1)following P R from z to w R , and C R from w R to u R avoiding v R . We already defined Q v earlier ( P L from v L to z , then P R from z to w R and C R from w R to v R avoiding u R ). By Claim 6.7, both Q u and Q v areessential cycles in G . Let T v denote the cycle starting at v R following P R until it intersects P L ; from that T v (Case B1)intersection continue along P L to u L and, along C L from u L to v L , avoiding w L . Similarly, T u starts at v R T u (Case B1)and follows P R until it intersects P L , which it follows to u L ; then from u L to v L along C L avoiding w L .If z does not lie on T u , then ( Q u , T u ) forms an X -configuration; if z does not lie on T v , then ( Q v , T v ) isan X -configuration. If P L and P R intersect in at least one edge, then z cannot belong to both T u and T v ,so one of the two cases holds. CASE (B1)- Establishing Property ( ii ) . P L and P R contains at least one edge. We can now complete the proof of ( ii ) in case that the intersection of P L and P R contains at least one edge.We can use the fact that either ( Q u , T u ) or ( Q v , T v ) is an X -configuration (though we may not know whichof the two it is). First, we show that( ◦ ) w L cannot be a foot of H LR .Suppose for the sake of contradiction that w L is a foot of H LR . Let P Lw denote the shortest path between w L and H (cid:48) LR , that is completely contained in H LR . We distinguish two cases depending on whether P Lw ends in z . If P Lw ends in z , then the essential cycle consisting of P Lw , P R would be vertex-disjoint fromeither T u or T v (here we use that P L and P R have at least one edge in common), which is a contradiction.Therefore, P Lw ends in a vertex y (cid:54) = z, w L on H (cid:48) LR . Then y is in the interior of the cycle C Ru formed by P Ru (or P Rv in which case a symmetric argument applies) and its R -foundation. Indeed, the part of P L between u L and y crosses P R , and hence, P Ru , an odd number of times. If a pair of legs incident to w L and w R forman independent odd pair, w can play the role of v and we end up in CASE (A) . Otherwise, since y is in theinterior of C Ru , P Ru must intersect P Lw an odd number of times, which is not possible (contradiction). Thiscompletes the proof of ( ◦ ).If C consists of three vertices only, u, v and w then similarly as in CASE (A) we are done by Lemma 6.4:By ( ◦ ), w L is only incident to the two C -edges (since H is the only C -bridge now). In G , we can then split w to make it of degree three and remove the edge uv (this contradicts the minimality of G , the number ofedges remains the same, but the number of vertices increased).Hence, there must be a fourth foot of H LR . We want to show that this implies that z is odd. Suppose thefourth foot lies on C R , call it w (cid:48) R . By Claim 6.9, the shortest path S R in H LR between w (cid:48) R and P L ∪ P R w (cid:48) R (CaseB1)must attach in z . If z is even, then S attaches outside of C , and inside C , which means that w (cid:48) R lies on C , like w R . Then the cycle formed by P R , S R , and the path from w R to w (cid:48) R along C R avoiding u R isweakly disjoint from from sc( R LRu ), contradicting Lemma 5.6. We conclude that z is odd in this case. Weproved the following.(*) If w (cid:48) R is a foot of H LR different from u R , v R and w R then the shortest path S R (in H LR ) between w (cid:48) R and H (cid:48) LR = P L ∪ P R ends in z , which must be odd in this case.Suppose the fourth foot of H LR lies on C L , call it w (cid:48) L . Let S L be a shortest path between w (cid:48) L and w (cid:48) L (CaseB1) H (cid:48) LR = P L ∪ P R . Similarly as S R from the above, path S L must end in the intersection P L ∩ P R . Indeed,otherwise its concatenation with a part of P L or P R , and a part C L or C R is a cycle in G that is vertexdisjoint from R LRu or R LRv (contradiction by Lemma 5.6), in particular, it attaches to P L .We need the following analog of (*).(**) If w (cid:48) L is a foot of H LR different from u L , v L and w L then the shortest path S L (in H LR ) between w (cid:48) L and H (cid:48) LR = P L ∪ P R ends in z , which must be odd in this case.32uppose S L instead attaches at a vertex z (cid:48) (cid:54) = z . We distinguish four cases, based on whether z (cid:48) occursbefore or after z on P L on the way from u L , and whether w and w (cid:48) belong to the same, or different partsof C − { u, v } . Let us first assume that w and w (cid:48) belong to different parts of C − { u, v } . Let us also assumethat z (cid:48) occurs before z on P L . As we argued in part ( i ), in this case z does not lie on T v , and so Q v is partof a X -configuration ( Q v , T v ). But Q v is vertex-disjoint from W (cid:48) u starting at w (cid:48) L , following S L to P L , thenfollowing P L to u L , and C L from u L to w (cid:48) L avoiding v L . This contradicts Lemma 5.6. If, instead, z (cid:48) occursafter z on P L , we know that z does not lie on T u , so ( Q u , T u ) is an X -configuration. In this case, considerthe cycle W (cid:48) v starting at w (cid:48) L , following S L to P L , then following P L to v L , and C L from v L to w (cid:48) L avoiding u L . W (cid:48) v is disjoint from Q v , so we have a contradiction by Lemma 5.6.We can therefore assume that w and w (cid:48) belong to the same part of C − { u, v } . Suppose the orderingalong C is u, w, w (cid:48) , v . If z (cid:48) occurs before z on P L , consider W v starting at v R , following P R until it intersects P L , then following P L to z (cid:48) , then S to w (cid:48) L and C L to v L avoiding w L and u L . The second cycle is W u starting at u L following P L to z , then P to w R and C R to u R avoiding w (cid:48) R and v R . Cycles W v and W u arevertex-disjoint, and, both are essential. Therefore Lemma 3.8 applies and we are done in this case. If theordering along C is u, w (cid:48) , w, v , we can swap sides and apply the same arguments to obtain a contradiction.We conclude that S L attaches in z . Suppose z were even. If the foot of S L on C L belongs to C L − C ,then S L attaches to z from outside C (like P R ), and the cycle consisting of P R , S L , and the path from w R to w (cid:48) R along C R avoiding u R is weakly disjoint from from sc( R LRu ), contradicting Lemma 5.6. Hence,the foot of S L must belong to C ∩ C L . Then the cycle consisting of S (which must attach to z from inside C ) together with the subpaths of C connecting it to u L touches Q v and the cycle consisting of S togetherwith the subpaths of C to v L touches Q u . Since at least one of Q u or Q v belongs to an X -configuration,we can invoke Lemma 5.6 to conclude that this is a contradiction. Again, we conclude that z is odd therebycompleting the proof of (**).Together, (*) and (**) tell us that in the presence of a fourth foot, z must be an odd vertex. ByLemma 5.8, neither { z, v } , nor { z, u } are 2-cuts (work with ( C, R u ) and ( C, R v )); and neither is { u, v } (since there is only one C -bridge, and an additional foot). It follows that there are no { z, v } -, { z, u } - or { u, v } -bridges, except, possibly, the edges zu , zv , and (certainly) uv . Since there are no cut-vertices, thismeans that except for these three edges, G is the union of { u, v, z } -bridges, with each bridge attaching to allthree vertices. We would like to show that every { u, v, z } -bridge is (a subdivision of) K , , since this proves( ii ). We distinguish two cases: Subcase 1. The vertex w R is incident to exactly one leg of H LR . Without loss of generality, we assume that z does not lie on T v , so ( Q v , T v ) forms an X -configuration. Let H w denote the { u, v, z } -bridge containing w R , and thus P R . H w (CaseB1)We will show that that H w is a (subdivision) of K , , but first, using that w R is not incident to morethan one leg of H LR , we prove that(***) the two edges of P L incident to z belong to different { u, v, z } -bridges (or { u, z } , { v, z } -bridges) andboth bridges are different from H w .For a contradiction, assume two of those bridges are the same; then there is a cycle C z passing throughthe two P L -edges incident to z , or through one of those edges, and an edge belonging to H w . Since C z belongs to a { u, v, z } -bridge (or { u, z } , { v, z } -bridge), it avoids both u and v . Then the cycle C z must stillintersect C due to Lemma 5.6, and therefore we either violate (*) or (**).Next, we show that H w can have only one leg at z . If there were two legs, then there would be a cycle C z through z in H w . By (***), H w does not contain any of the two edges of P L incident to z , neither does C z ,so C z is disjoint from T v , contradicting Lemma 5.6; here we use that z does not belong to T v . Furthermore,a cycle avoiding z and u (respectively, v ) in H w is disjoint from R LRu (respectively, R LRv ), which is again acontradiction with Lemma 5.6. Hence, H w cannot contain any cycle. Finally, there are no cut-vertices byLemma 5.7 in G , and therefore H w is a (subdivision of) K , .33y (***), Q v passes through a trivial { v, z } -bridge or { u, v, z } -bridge H Q v and H w ; and T v is containedin a single { u, v, z } -bridge H T v which Q v meets only in v . Both H T v and H Q v have a very simple structurewhich we delve further into next. C L C R u L v L v R u R w R w L P L P R P R z C L C R u L v L v R u R w R w L P L P R P R zH LRQ v H LRT v Figure 15: The bridges H LRT v and H LRQ v (left). The doubling of the path in H LRQ v between z and v L (right).Refer to Figure 15. We show that(****) The vertex v is incident to a single edge in H Q v , and that u is incident to a single edge in H T v .If H T v contains a cycle passing through u and disjoint from v and z then this cycle is disjoint from Q v (contradiction with Lemma 5.6). If H Q v contains a cycle passing through v and disjoint from u and z then this cycle is disjoint either from R LRu or Q u depending on whether z is the end point of P L ∩ P R . Inparticular, if z is the end point of P L ∩ P R then it is disjoint from R LRu , and otherwise it is disjoint from Q u and also ( Q u , T u ) forms an X -configuration in this case, and hence, we can apply Lemma 5.6 (to derivea contradiction).By (***) and (****), we see that both v and z are incident to a single edge in H Q v . Therefore if H Q v contains a cycle passing through u and disjoint from v and z then this cycle must intersect P L . Since nocycle is disjoint from C , H Q v is a path with a single bridge B u which is a (subdivided) star with the center u .Unless H Q v contains two edges attached to the opposite sides of P L at an even vertex or each at a differenteven vertex we can remove edges from B u in order to convert H Q v into a (subdivision of) K , (apply theminimality of the counterexample and reinsert the deleted edges into the embedding). Otherwise, we removeall the edges from B u , except for the u -to- P L paths ending in the two special edges attached to the oppositesides of P L . Again, this is possible (apply the minimality of the counterexample and reinsert the deletededges into the embedding). Now, we split H Q v into two (subdivisions of) K , by doubling P L except atits end-vertices, which can be easily achieved while constructing an iocr-0-drawing of the resulting graph.Indeed, every interior vertex of P L is even at this point.Similarly, if H T v contains a cycle passing through v and disjoint from u and z then this cycle mustintersect P L and analogously as in the previous paragraph we reduce H T v to at most 2 subdivisions of K , .Now, any other { u, v, z } -bridge is easily seen to be (almost) a subdivided K , : a cycle through u in suchbridge that avoids z and v is disjoint from Q v . Now, a cycle through v in any other { u, v, z } -bridge thatavoids u and z is either disjoint from R LRu (contradiction), or we can again split the bridge into at most two K , , and we are done. 34 ubcase 2. The vertex w R is incident to more than one leg of H LR . If there exists at least three legs of H LR at w R they can be extended into pairwise interior disjoint pathsending on P L . By Lemma 6.5, they cannot all end in z , since z is odd. Hence, one of them ends in z (cid:48) (cid:54) = z .It follows that both ( Q v , T v ) and ( Q u , T u ) form an X -configuration if we appropriately redirect Q v or Q u through z (cid:48) . Now, we obtain a cycle through two of the legs disjoint from T v or T u (contradiction withLemma 5.6). Hence, using ( ◦ ) there exists at most two legs and therefore w R can be made even, but in thiscase we have a cycle weakly disjoint from C (contradiction with Lemma 5.6).We conclude that all { u, v, z } -bridges (except for the edges { uz, vz, uv } ) are subdivisions of K , attachedto { u, v, z } .This completes the proof of ( ii ) in the case that P L ∩ P R contains at least one edge. CASE (B1)- Establishing Property ( ii ) . P L and P R intersect in a single vertex. We are left with showing ( ii ) in the case that P L and P R intersect in a single vertex z . If z can be madeeven by flips, we are done by Lemma 3.8, since R LRu and Q v touch in the even vertex z . We conclude that z is odd, and cannot be made even by edge-flips.At this point we can complete the argument as in the case that P L and P R have an edge in common. Inthis case, the previous argument applies unless there exists a path S in H LR connecting z and w L ; or w (cid:48) R and w (cid:48) R such that w (cid:48) R is in C if and only if w (cid:48)(cid:48) R is in C .If there exits a pair of feet w (cid:48) L and w (cid:48)(cid:48) R of H LR such that w (cid:48) and w (cid:48)(cid:48) are both in C − C or both in C , and both w (cid:48) and w (cid:48)(cid:48) are even, possibly w (cid:48) = w (cid:48)(cid:48) , we remove all the legs of H LR incident to C R − C except for the pair attached at w L and w (cid:48) R . Then we double the part of C passing through C − C or C .Here, we again use the minimality of the counterexample when removing the edges. Hence, we obtain apair of { u, v, z } -bridges in G that are (subdivisions of) K , . Otherwise, we turn { u, v, z } -bridges containingedges of C (there can be at most two of them) into (subdivisions of) K , by edge removals. The rest of theargument goes through as in the previous case since we don’t need to use the X -configuration ( T u , Q u ) or( T v , Q v ) at this point. CASE (B2) C and C agree on C . C L C R u L v L v R u R zP R P L H z Figure 16: A { u, v, z } -bridge with both u and v as feet.We consider the case that the subpaths of C on C L and C on C R are the same as subgraphs of G .If the intersection of P L and P R is not a single vertex we will find a pair of vertex-disjoint paths in C ∪ C , contradicting Lemma 3.8. Let R LRu and R LRv denote the paths joining u L and u R , and v L and v R , R LRu , R LRv (Case B2)respectively, in P L ∪ P R . Now, C ⊕ C = R u ⊕ R v has vanishing homology over Z . Hence, R u and R v Z . It follows that the order of end vertices of P L ∩ P R along P L and P R when traversing P L and P R , respectively, from u L to v L , and from u R to v R , is the same. Hence, R LRu and R LRv are disjoint. Since the cycles in G corresponding to R LRu and R LRv are essential, this contradictsLemma 3.8.Therefore, we can assume that the intersection of P L and P R consists of a single vertex z . If z can be z (Case B2)made even by flips, we do so. In this case, R LRu and R LRv touch at z : The parity of crossing between R LRu and R LRv is the same as the parity of crossing between P L and P R . Now P L and P R (as part of two closedcurves C and C ) cross evenly, so R LRu and R LRv do too. The only edges of R u and R v which can crossoddly, are the edges incident to u and v , and the edges incident to z . The edges incident to u and v in R LRu and R LRv form two independent odd pairs, so together they do not affect the parity of crossing between R LRu and R LRv . Thus, R u and R v cross evenly overall. Now, if R LRu and R LRv “cross” in z , it would follow that thecycles corresponding to R u and R v are not in the same homology class over Z . However, we already showedthe opposite in the previous paragraph. Hence, R LRu and R LRv touch at z , which contradicts Lemma 3.8.We conclude that z is odd and cannot be made even by flips. Let the edges of R LRu incident to z be e and f . Correct the rotation at z so that e and f cross each other and every other edge at z evenly. If e and f are consecutive in the rotation at z (not separated by any edge in G ), then R LRv and R LRu touch at z , and we are done, as before. Therefore, there must be some edge g separating e and f . Since the graph is2-connected (by Lemma 5.7), there must be a path P (cid:48) starting with g at z , and connecting z to C ( P (cid:48) avoidsthe interior vertices of H LR , since otherwise, we would contradict Lemma 5.6). By inspecting the rotationat z (note that we made edges of P L and P R cross each other evenly at z , see Figure 16), the vertices ofthis path lie both in the interior of C and C , so P (cid:48) can only attach to C by crossing C or C oddly, whichit can only do if it connects to u L , u R , v L , or v R . So g is part of a { u, v, z } -bridge H z . If both u and v arefeet of H z (which can happen, see Figure 16), then H z contains a path from u to v avoiding z and disjointfrom P R . We can therefore treat this as CASE (A) or CASE (B1) instead depending on whether H LRz attaches at the both C L and C R . Hence, any such { u, v, z } -bridge is either a { u, z } -bridge or a { v, z } -bridge.By the choice of P L and P R minimizing the sum of their lengths, which we assumed at the beginning of thethe proof, the { u, z } -bridge or { v, z } -bridge H z is not an edge and therefore feet of H z form a 2-cut. Since z is odd, Lemma 5.8 implies that there is no such H z (since P L ∪ P R already contains paths from z to eachof u L , u R , v L , and v R ; we work with X -configurations ( C, R u ) and ( C, R v ) to make sure the lemma can beapplied). Hence e and f are consecutive after all, and that case we already dealt with. We survey results which rely on the Hanani-Tutte theorem, and discuss how Theorem 1.1 can lead to toroidalversions of some of these results; along the way, we also encounter some open questions, and make someconjectures. This extends the discussion in [33]).
Dependent Crossings
Theorem 1.1 implies that a graph which can be drawn on the torus so that the only crossings are betweendependent edges is toroidal. This would appear to be intuitively clear, but we only know it for the plane, theprojective plane, and now for the torus, by virtue of their respective Hanani-Tutte theorems; even for theplane no simpler proof is known. Since we know that the Hanani-Tutte theorem does not hold for orientablesurfaces of genus at least 4 [8], this begs the question whether we can always remove dependent crossings.
Conjecture 7.1.
If a graph can be drawn in a surface without any independent crossings, then the graph isembeddable in that surface.
Crossing Number Variants
The Hanani-Tutte theorem is related to the independent odd crossing number , iocr( G ) of a graph G , thefewest number of pairs of independent edges that have to cross oddly in a drawing of G . The Hanani-Tutte36heorem then states that iocr( G ) = 0 implies that cr( G ) = 0, where cr( G ) is the traditional crossing numberof G . It is even true that iocr( G ) = cr( G ) for iocr( G ) ≤ G forwhich iocr( G ) < cr( G ) [28]. The two crossing numbers cannot be arbitrarily far apart though, one can showthat cr( G ) ≤ (cid:0) G )2 (cid:1) [30]. It is not known whether similar bounds, or any bounds, for that matter, holdfor other surfaces, not even the projective plane. (The subscript in crossing number variants indicates thesurface we work on.) Conjecture 7.2. cr Σ ( G ) ≤ (cid:0) Σ ( G )2 (cid:1) , where Σ is the projective plane, or the torus. We also now have a crossing lemma for iocr on the torus. Since iocr lower-bounds the algebraic and paircrossing numbers, iacr, acr, pcr, and pcr − , see the survey [31] for the definition of these crossing numbervariants, we also obtain crossing lemmas for these crossing numbers on the torus. Corollary 7.3 (Crossing Lemma) . iocr T ( G ) ≥ cm /n , where T is the torus, c = 1 / , m = | E ( G ) | , and n = | V ( G ) | . The proof follows from the standard crossing lemma argument done carefully combined with Theorem 1.1(see the Section on crossing lemma variants in [31]). Since iocr lower bounds several other crossing numbervariants, such as iacr, pcr, and pcr − this also implies a crossing lemma for these variants. Also see [19] fora sketch of an argument that shows a crossing lemma for iocr for arbitrary surfaces, but without explicitconstant c . Pseudodisks and Admissible Regions
Smorodinsky and Sharir [34] showed that if P is a collection of n points, and C a collection of m pseudo-disks in the plane such that every pseudo-disk in C passes through a distinct pair of points in P , and nopseudo-disk contains a point of P in its interior, then m ≤ n −
6, where 3 n − m ≤ n − m ≤ n for pseudodisks in the torus.A family of simply connected regions is k -admissible if each pair of region boundaries intersect in aneven number of points, not exceeding k . Whitesides and Zhao [38] showed that the union of n ≥ k -admissible sets is bounded by at most k (3 n −
6) arcs, where, again, 3 n − k (3 n −
3) for regions in the projective plane [33], and 3 kn for regions inthe torus.Keszegh [18] gave a more uniform treatment of these types of results based on hypergraphs, which atthe core again uses the Hanani-Tutte theorem. While he only states results for the plane, all of his materialshould lift to the projective plane and the torus based on the availability of the Hanani-Tutte theorem onthose surfaces. Surfaces and Pseudosurfaces
We know that, at least for orientable surfaces, Hanani-Tutte characterizations will stop at genus 4. Thisstill leaves a fair number of open cases, and our general set-up may be useful in exploring those cases. Forexample, we may ask whether our proof can be adjusted to cover the projective plane (this would give athird proof, after [26] and [5]), or extend it to the Klein bottle (the first case where there are non-trivialsurface separating cycles), or the spindle pseudosurface with n pinchpoints.The graph minor theorem for surfaces implies that embeddability in a surface can be characterized by afinite set of forbidden minors. For the plane and the projective plane, proofs of the Hanani-Tutte theoremwere possible, because the list of forbidden minors is known explicitly [26]. For the torus, this is no longerthe case (though progress is being made on all cubic obstructions). One may ask, whether it is possible togo backwards: from the Hanani-Tutte theorem to the set of forbidden minors? For the plane this was doneby van der Holst [37]. Can this work be extended to the 1-spindle, the projective plane (where [5] gave aproof of Hanani-Tutte not using forbidden minors), or even the torus?37he proof of Theorem 1.1 is algorithmic in the sense that given an iocr-0-drawing of G we can find anembedding of G in the torus. This does not imply that testing whether a graph can be embedded in the toruslies in polynomial time. The planar Hanani-Tutte theorem can be turned into a linear system of equationsover GF
2, which is solvable in polynomial time. Unfortunately, modeling the handle of the torus requiresquadratic equations, which loses us the polynomial-time algorithm. This is similar to the situation for theprojective plane, where the projective handle also leads to quadratic equations.
Question 7.4.
Can the Hanani-Tutte criteria for the projective plane and the torus be turned into apolynomial-time test?For the 1-spindle this is easily possible, but not quite surprising.We have focussed on the Hanani-Tutte theorem over Z , that is, we count crossings by parity. We canalso count crossings algebraically, over Z , by assigning a crossing the value +1 or − Z -version [8]; does the Hanani-Tutte theoremhold algebraically, over Z ? Similarly, we saw that the Unified Hanani-Tutte theorem, Theorem 3.1, due to[10] fails for the torus. Does the algebraic variant, over Z , hold on the torus?Finally, the Hanani-Tutte theorem has some variants in the plane which have not yet been generalized.For example, if a graph can be drawn in the plane so that every cycle in the graph has an even numberof independent self-crossings, then the graph is planar [33, Theorem 1.16]. Is this result still true for theprojective plane, or the torus? The proof in the plane uses the Kuratowski minors. Arf and Approximating the Hanani-Tutte Theorem
The fact that the Hanani–Tutte theorem cannot be extended to all orientable surfaces has some positivepractical consequences. Some results about graphs embedded on a closed surface remain true if we onlyrequire that the graph has an independently even drawing on the surface, and this is a strictly larger classof graphs in general (starting at genus 4).One notable example is the Arf invariant formula for the number of perfect matchings in a graph embeddedon an orientable surface [20, Remark 1.4, Theorem 1] , see also [3, 24]. The proof of the similar resultby Tesler [35], which applies to all closed surfaces, still works for independently even drawings insteadof embeddings. In both cases, the complexity of the formula depends exponentially on the genus of theunderlying surface, so for graphs which have an independently even drawing in a surface in which theycannot be embedded, the complexity of the formula is improved; its length does not depend on the genus ofthe graph, but rather its Z -genus , the smallest genus of a surface on which the graph has an independentlyeven drawing.It is therefore interesting to study how large the gap between the genus and the Z -genus of a graph maybe. It is known that the gap can be at least linear [8, Corollary 4.1]. There also is an upper bound on thisgap, but it is not effective [12]. References [1] Grant Cairns and Yury Nikolayevsky. Bounds for generalized thrackles.
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