1 Introduction
The HararyHill Conjecture states that the crossing number of the complete graph is given by the formula
Some of the known families of drawings of achieving crossings have the geometric character of being spherical: a spherical drawing of a graph is a drawing in the unit sphere in which the vertices of are represented as points—no three on a great circle—and the edges of are shortestarcs in connecting pairs of vertices.
Examples of such families are: Hill’s Tin Can Drawings [15]; Kynčl’s general spherical drawing in his posting [17]; and, most recently, the family of Ábrego et al. [3] in which every edge is crossed at least once. Even though there are spherical drawings of with crossings, the HararyHill Conjecture remains unknown for this special class of drawings. It is not clear to us whether a 2pagedrawing of having crossings is necessarily spherical. See [4].
A surprising result by Moon [20] states that the number of crossings in a random spherical drawing of has, as goes to infinity, crossings. Thus, spherical drawings are linked with the asymptotic version of the conjecture:
A rectilinear drawing of a graph is a drawing of in the plane so that its edges are straightline segments. One of the most important recent accomplishments in the study of crossing numbers is a result of Lovász et al. [18] and, simultaneously and independently, Ábrego and FernándezMerchant [1], showing that rectilinear drawings have at least crossings. (It is now known that, for , rectilinear drawings have strictly more than crossings.)
The proofs that rectilinear drawings of have at least of crossings use machinery for studying arrangements of pseudolines, based on the property that the edges in a rectilinear drawing can be extended to such an arrangement. An analogous property is satisfied by spherical drawings: each edge can be extended to a great circle (a circle in of maximum diameter, and the edge is the shorter side joining the two vertices). This means that the edges can be extended into an arrangement of pseudocircles, defined as a set of simple closed curves in such that every two intersect at most twice, and every intersection is a crossing between two curves.
The following questions are principal motivations for this work.

[label=Question 0., ref=,leftmargin=75pt,topsep=0pt,itemsep=0pt]

Is every optimal drawing of spherical?

Is the spherical crossing number of equal to ?
The success of the geometric approach for the rectilinear crossing number of suggests trying an analogous approach for spherical drawings, replacing arrangements of pseudolines with arrangements of pseudocircles. A principal goal of this work is to introduce pseudospherical drawings as a generalization of spherical drawings.
There are nice characterizations of pseudolinear drawings of . Aichholzer et al. [5] prove that a drawing of in the plane is pseudolinear if and only if every crossing has the facial 4cycle bounding the infinite face of the . Arroyo et al. [7] have an equivalent characterization based in their notion of a convex drawing of . Arroyo et al. [6] characterize when a drawing of a general graph in the plane is pseudolinear.
Therefore, the other principal goal of this work is to characterize pseudospherical drawings of ; this is also based in the notion of convex drawings mentioned in the preceding paragraph.
Motivated by the remarks above, a drawing of a graph in the sphere is pseudospherical if, for any distinct edges and :

[leftmargin=60pt,label=(PS0), ref=(PS0),itemsep=0pt]

is contained in a simple closed curve such that no vertex other than an end of is contained in ;

and all intersections are crossings; and

has at most one point.
A drawing is weakly pseudospherical if it satisfies 1, 3, while 2 is relaxed to

[leftmargin=60pt,label=(PS0w), ref=(PS0w),itemsep=0pt,start=2]

and all intersections are crossings.
Our main result is the following characterizations of pseudospherical drawings of . (The definition of hconvex will be given in Section 4.)
Theorem 1.1.
It is not difficult to prove that the Blazěk and Koman 2page drawing of having crossings is weakly pseudospherical [4, 11]. Theorem 1.1 shows that it is, therefore, pseudospherical and hconvex.
The implication 12 is trivial. Although the reader cannot see it now, the implication 23 is quite easy. The hard part is 31. We do not see how to prove 21 directly.
The proof of 31 proceeds by iteratively finding a pseudocircle for the next edge to extend by one the set of pseudocircles satisfying the conditions 1–3. There are two principal steps involved. The first step is to find two initial approximations for the next pseudocircle (when added to , either of these will satisfy 1), while the second is to repeatedly shift one of the initial approximations, gradually increasing the number of pseudocircles in that it intersects until it intersects them all, at which point it is a transversal.
Each of these steps has its challenges. To find the initial approximations, we require an extensive study of hconvex drawings; this is done in Section 4. Crucially, each edge of an hconvex drawing partitions the vertex set of into two pseudolinear drawings of (typically smaller) complete subgraphs; the initial approximations are near the outer boundary of each of these pseudolinear subdrawings. (Motivated by this study, we tried to show this partitioning holds for pseudospherical drawings of general graphs. In a personal communication, Xinyu Lily Wang has shown that it is false in the more general context.)
Producing the pseudocircle transversal requires shifting one of a pair of initial approximations from the preceding paragraph towards the other using analogues of Reidemeister II and III moves. The core of the shifting turns out to require a characterization of arrangements of pseudocircles.
Let be an arrangement of simple closed curves in the sphere; that is, a set of simple closed curves, any two of which have finitely many intersections, all of which are crossings. A spiral of is an arc in the union of the curves in that always makes the same—either all left or all right—turn in changing from one curve to another (see Figure 1). In Section 2, we give a more precise definition of spiral and define coherent spirals. The auxiliary result that we need is the following.
Theorem 1.2.
Let be an arrangement of simple closed curves. Then is an arrangement of pseudocircles if and only if has no coherent spirals.
Our study of spirals led us to two drawings, one for each of and . The former has no extension of its edges to an arrangement of pseudocircles. The latter has such an extension, but no extension has all pseudocircles crossing exactly twice. These examples are exhibited in Section 7.
Independently, Aichholzer at the 2015 Crossing Number Workshop in Rio de Janeiro and Pilz at the Geometric Graph Week in Berlin (2015) asked if every drawing of has an extension to an arrangement of pseudocircles. The drawing of answers this question in the negative. The drawing of answers negatively the related question of improving a pseudocircular extension to a pseudospherical extension.
In the next section, we introduce spirals and prove Theorem 1.2. Section 3 contains the proof that an initial pair of approximations for implies the existence of a transversal. Section 4 introduces hconvex drawings and proves the easy implication 23. Section 5 contains the necessary discussion of hconvex drawings for the initial approximations. The proof of Theorem 1.1 is completed in Section 6 by showing how to inductively obtain in an hconvex drawing the initial curves for the sweeping. The interesting drawings of and are in Section 7, while Section 8 has concluding remarks.
2 Spirals and Coherence
A spiral is a special arc, illustrated in Figure 1 and defined precisely below, in the union of an arrangement of simple closed curves that has all its crossings facing the same side of the arc. (Alternatively, a spiral always makes the same—either all left or all right—turn in changing from one curve to another.) The main result in this section is a characterization of arrangements of pseudocircles as either (a) not having any “coherent” spirals or, equivalently, (b) every spiral has an “external segment”. That an arrangement of pseudocircles has no coherent spiral is the point required for the proof of Theorem 1.1.
In an arrangement of simple closed curves (defined just before Theorem 1.2), three or more of the curves in may cross at the same point. This is necessary for extensions of drawings of , where curves pairwise cross at each vertex.
A nontrivial arc in the union of the curves in with ends and has a unique decomposition sequence of subarcs of , as depicted in Figure 2, such that:

[label=(),leftmargin = .50 truein,ref=()]

is an end of , is an end of ;

for each , there is a such that ; and

for , the curves and in are distinct and is a crossing of and .
The number is the weight of . Figure 2 shows an arrangement of simple closed curves with an arc of weight 3. For , the crossing of with is denoted . For convenience, we set and .
For , and are the closures of the components of incident with and , respectively. Four such are illustrated in Figure 3. Evidently, consists of the continuation of through either () or () up to the next meeting with .
The continuations and both leave on the same side of . This is the side of that faces. In Figure 2, and face different sides of the dotted arc. We are now prepared for the definitions of spiral, external segment, and coherence.
Definition 2.1 (Spiral, External Segment, Coherence).
Let be an arrangement of simple closed curves and let be an arc in with decomposition sequence . For each , let be such that . (Only consecutive are required to be distinct; if , then could be the same as .)

The arc is a spiral if all of face the same side of .

For , the segment is external for if ; otherwise, is internal for .

For every and , the arc is a coherent extension if: (a) is internal for and (b) has both ends on the same side (see discussion of “side” below) of the interior of . (In Figure 3, is not a coherent extension; the other three are.)

is coherent if at least one of and is a coherent extension. In Figure 3, both and are coherent.

is coherent if, for each , is coherent.
As we traverse from one end to the other, there are naturally left and right sides. The two ends of the arc are in . The issue in the definition of coherence is: are the points near each end of on the same side of or not. “Left” and “right” depend on an orientation of and are irrelevant to us.
To be on the same side, the ends of must be in the interior of . As the points and are not in the interior of , we have the following.
Remark 2.2.
Neither nor is coherent.
The following characterization of an arrangement of pseudocircles in terms of its spirals seems to be quite interesting in its own right. We only need 13 for the proof of Theorem 1.1.
Theorem 2.3.
This characterization fits in well with much recent work on arrangements of pseudocircles. The paper of Felsner and Scheucher [12] is one example; their references [2], [7], [10], [15], [19], and [26] are others. Felsner and Scheucher have a web page devoted to pseudocircles [13]. Another recent work about unavoidable configurations in the sense of Ramsey’s Theorem is Medina et al. [19].
The example to the left in Figure 4 has a spiral in an arrangement of pseudocircles. This spiral is not coherent because is external for .
In addition to its use in the proof of Theorem 1.1, Theorem 2.3 is useful for constructing drawings of graphs that cannot be extended to arrangements of pseudocircles. For example, the righthand diagram in Figure 5 cannot be extended to an arrangement of pseudocircles because the arc indicated by the dashed curve induces a coherent spiral in any such extension. A similar idea is used in Section 7 to construct a nonextendible drawing of .
Proof of Theorem 2.3. The implications 43 and 32 are trivial. We prove 21 and 13 to complete the proof that 1, 2, and 3 are equivalent and finish the proof with 14.
21. Suppose by way of contradiction that are distinct curves in such that . Then . Let and, traversing in one direction starting at , let be the first three points of encountered.
Let be the arc obtained by starting at , continuing along through to and then following through to a point just beyond . The decomposition of is , with, for , . Since every arc in with weight at most 1 is a spiral, is a spiral. The fact that the points and of are consecutive in imply that is coherent.
On the other hand, the subarc of intersects the subarc of just in . This shows that is coherent, completing the proof that is coherent, the required contradiction.
13. To obtain a contradiction, suppose is a coherent spiral with least weight. An arc with weight 0 is not coherent, so the decomposition of has .
The first claim imposes constraints on what happens at points “under” a forward jump such as, in Figure 3, under . For an extension of , denotes the other end of .
Claim 1.
Suppose that is a coherent extension of . If there is an such that is in , then:

[ref=(0)]

for each , is disjoint from ; and

either: for some , ; or intersects .
Likewise, suppose is a coherent extension of . If there is an such that is in , then:

[start=3,ref=(0)]

for each , is disjoint from ; and

either: for some , ; or intersects .
Proof. We prove the first statement; the “likewise” is the same, but for the traversal of in the reverse direction.
By way of contradiction, suppose first that, for some , intersects ; let be the first intersection with as we traverse from . The interior of the subarc of from to is on the side of the unique simple closed curve contained in that is opposite to the side that contains .
See Figure 5 for an illustration of this proof.
Let be the subarc of consisting of and the portion of from to . Likewise, let be the subarc of consisting of and . Then the arc consisting of segments has smaller weight than . Also, even if , is a spiral. (In case , then , , and all cross at . This ensures that the cyclic rotation of these three curves at is , where is the continuation of from .)
To see that is coherent, first let be the simple closed curve . For each of the segments of , contains some segment of . Let be a coherent extension of for . Follow from its end in (or, if , from ). If we never encounter , then we arrive at on the same side. On the other hand, if we encounter , it is not at a point in and so it is in .
Label as the outside of the side of containing . Since starts on the outside of , its first intersection with is from that side. Thus, the portion of up to that first intersection with is a coherent extension of , as required.
To complete the proof that is coherent, we note that, if, for the segment of , both and are coherent, then they are contained in coherent extensions of the corresponding segment of . Since these extensions for are distinct, as extensions for they are also distinct. Thus, there is a coherent spiral with weight less than , a contradiction.
For 2, if, for each , and is disjoint from , then and (using 1) each is a coherent extension in . The same argument as in the preceding paragraph shows that is a coherent spiral with smaller weight than , a contradiction.
The next claim considers “reverse” coherent extensions. This claim rules out the possibility of an extension such as in Figure 3.
Claim 2.
There do not exist such that and is a coherent extension with an end in .
Likewise, there do not exist such that and is a coherent extension with an end in .
Proof. We only prove the first statement. Choose the least for which such an exists. Suppose first that . Thus, . Because , . Thus, Hypothesis 1 implies . Let be the simple closed curve contained in . Except for , is disjoint from .
Note that is not in . Therefore, and the start of from are on different sides of , which is impossible. Thus, .
The choice of implies that either is not coherent or it does not intersect . Therefore, must intersect at a point other than . This gives the two crossings of with .
An intersection of with yields a third intersection of with , and the theorem is proved. Therefore, we assume is disjoint from . On the other hand, the choice of implies that, for each with in the interior of , . Therefore 2 of the preceding claim shows is not disjoint from , the final contradiction.
We may now suppose that there does not exist an that is a coherent extension with an end in any such that . Similarly, we may assume that there does not exist an that is a coherent extension with any end in any such that .
The final claim combines the first two to completely determine the nature of a coherent extension. Before we get to it, we require one more detail.
Claim 3.
Let . Suppose is not a coherent extension and that has an intersection with . Then is not coherent.
Likewise, if is not coherent and has an intersection with , then is not coherent.
Proof. We only prove the first statement. Because is coherent and is not a coherent extension, is a coherent extension of . Let be the intersection of with ; it follows that . Since , is neither nor . Hypothesis 1 implies .
Since is disjoint from ; in particular, it is disjoint from . For , the preceding claim implies that is not coherent. Thus, we suppose .
The union of and is a simple closed curve . Let be a point of near . From trace an arc alongside , across and, continuing beside , on to a point near the end of . Thus, is along the side of faced by all the . Because is not coherent, it does not return to on this side and, therefore, crosses only once. Consequently, crosses only once.
Suppose by way of contradiction that is a coherent extension of . Because does not intersect , it cannot cross . Therefore, it does not intersect the portion of from its crossing with to its end near . In particular, has no end in . The first paragraph shows is also disjoint from . Thus, is in , contradicting the preceding claim.
We are now ready to get the fine detail of the coherent extensions of .
Claim 4.
For , if is a coherent extension of , then .
Likewise, for , if is a coherent extension of , then .
Proof. We only prove the first statement. Suppose that is least such that is a coherent extension of and . We show by induction that, for each , is not a coherent extension of . For , Remark 2.2 shows that is not a coherent extension of . Now let and suppose that is not a coherent extension of .
Since is coherent and is not a coherent extension of , we have that is a coherent extension of . By the choice of , . The preceding claim implies that is not a coherent extension of , completing the proof that, for each , is not a coherent extension of .
Let be such that . The second claim and the choice of show that . Part 1 of the first claim implies that, for each , is disjoint from , while the second claim asserts that . Part 2 of the first claim now implies that intersects at a point .
We showed that is not a coherent extension of and that intersects at and . Consequently, the coherence of and the preceding claim show that is a coherent extension. The second claim shows that, for some , . The first two claims show that, for each , is disjoint from .
Let be the subarc and let be the arc consisting of , . Just above, we showed that is disjoint from . Therefore, is a coherent extension of with respect to . The second claim shows that has both ends in .
Since intersects in and , we see that ; therefore . Moreover, it follows that is disjoint from . This, together with the fact that, for each , is disjoint from , contradicts the first claim.
Because is coherent and Remark 2.2 shows is not a coherent extension of , is a coherent extension of . Likewise, is a coherent extension of . It follows that there is a such that is a coherent extension of and is a coherent extension of . The fourth claim implies that both ends of are in and both ends of are in . This implies that , completing the proof that 13.
Lemma 2.4.
Let be arcs in the sphere such that and both have their ends in , but otherwise are disjoint from . We assume and have finitely many intersections and these are all crossings.
Assume that short subarcs of starting at each end of are on different sides of the unique simple closed curve in . Then has at least one point.
By way of contradiction, let be a leastweight spiral in having no segment external for . Since the only segment in an arc of weight 0 is external for that arc, has positive weight . Let be the decomposition of .
If any spiral in is coherent, then 13 implies that some two curves in have at least four crossings, a contradiction. In particular, we may assume is incoherent; let be an incoherent segment of . By definition, both and are both incoherent extensions. Thus, short subarcs near their ends and are on the side of that is opposite the side faced by all the crossings . We remark that one or both of and might be in . If , then only the subarc near is forced by incoherence to be on the side of not faced by the crossings; the fact that is a simple closed curve implies is as well. An analogous statement applies if .
As we traverse from to , we first encounter and then (these could be equal). Therefore, either or . As these are symmetric up to reversal of , we assume the latter.
Let be the spiral ; evidently, its weight is less than that of . Therefore, has a segment that is external for . Notice that the side of faced by all its crossings is separated from by the simple closed curve .
Since no segment of is external for , intersects at a point outside . This implies that and intersect at points and , respectively, that are not in . Since is external for , and are in .
Claim 5.
Traversing and from and , respectively, they each have an intersection with and these intersections are distinct points of .
Proof. In an extreme case, ; here is the required common point between and . In the other extreme case, and ; here is the common point between and . (These could happen simultaneously.) Otherwise, and are interior points of .
If we take , and to be either or , then our trivial lemma implies there are distinct points, one in each of and .
Since , the only points in are the points in each of and . In particular, and are both disjoint from .
Because and is disjoint from , . On the other hand, the disjointness of with implies that is not further from in than is. In turn, this implies that is in . Therefore, reversing the direction of traversal of , may play the role of in the preceding argument.
Thus, there is a such that is a segment of , external for , but not external for . The argument above for and applies to and . Thus, and have their endpoints and , respectively, in . The trivial lemma implies that each of and has an intersection with each of and . In particular, has at least four points, as required.
3 A pseudocircle transversal
We recall that the proof of Theorem 1.1 has two parts. For our current set of pairwise intersecting pseudocircles, we must (i) find a pair of initial approximations to the pseudocircle for the next edge and (ii) show that the pair of approximations imply the existence of the desired curve crossing all the curves in . In this section, our focus is on the second of these.
In particular, in Section 6, we will show how to find two initial approximations that together intersect all the curves in our current . The main result of this section is to use these two approximations to find the single curve that intersects all the curves in .
We are reminded of the theorems that are:

[topsep=0pt, itemsep=4pt,leftmargin=11pt]

Hellytype: if a collection of sets is such that every of the sets admits a transversal, then the whole collection admits a transversal; and

Gallaitype: if a collection of sets is such that every of the sets admits a transversal, then the whole admits a small set of partial transversals whose union is a transversal.
Our theorem has the following different character: if a collection of sets admits a small set of partial transversals whose union is a transversal, then it admits a transversal. We do not know of an another example of this type of theorem.
Definition 3.1.
Let be an arrangement of pseudocircles.

A set of simple closed curves is a transversal if every curve in intersects at least one of the curves in .

A simple closed curve is a pseudocircle if is an arrangement of pseudocircles. [Text removed.]
Theorem 3.2.
Let be an arrangement of pseudocircles. Let and pseudocircles such that is a transversal. Suppose

is a nontrivial arc and

if are such that and , then .
Then there exists a pseudocircle containing and is contained in the closure of the face of not incident with .
Our proof shows that one can sweep either of or to the required . In their classic paper [22], Snoeyink and Hershberger show how to sweep one curve through the others in an arrangement of pseudoarcs and pseudocircles. In particular, up to sweeping, is unique.
Proof. Let and, for , let . Because is a transversal, . Let . Define as the number of crossings in included in the face of not incident with . We proceed by induction on .
We can assume that neither nor is empty, else we pick to be either equal to or .
If there is an arc of some incident with a face of that is included in and incident with , then by shifting some part of to cross via a Reidemeister Type II move, we obtain a curve such that the pair satisfies the same hypothesis as . Since , the result follows by induction.
In the alternative, there exists an arc with ends in , but otherwise contained in , such that separates from . Let be the closure of the component of that is incident with both and . Among the finitely many choices for , we choose so that is minimal under inclusion. See Figure 6.
We apply Theorem 2.3, specifically 13, to see that there is a crossing in facing . The proof is by contradiction, assuming that every crossing faces the other side of . In particular, is a spiral; the contradiction follows from the following.
Claim 1.
There is a crossing in facing .
Proof. By way of contradiction, suppose every crossing faces the other side of ; in particular, is a spiral. We show that is coherent, contradicting Theorem 2.3 (specifically, 13). It suffices to show that if is an arc in the decomposition of , then is coherent. Let be the curve in containing .
Let be the closed arc in such that is a simple closed curve. Consider the continuation of from one end of . Since crosses twice, the continuation must eventually reach ; in particular, it must have a first intersection with .
We show below that it is impossible for both continuations to have these first intersections in . Therefore, for one of them, the first intersection is in the interior of , showing that is coherent.
So suppose both continuations intersect for the first time at and . Since , . Therefore, is contained in the region bounded by and intersects the boundary of this region only at . This shows that , the required contradiction.
Claim 2.
Let . Then every arc in has one end in the interior of and one end not in the interior of .
Proof. Let be an arc . If has no end in , then contains an arc that separates from such that is properly contained in , contradicting the choice of .
If has no end in the complement of the interior of in the boundary of , then is contained in and has both ends in the interior of . Since has exactly two points, these are the two ends of . The preceding paragraph shows that is the only arc in , and hence is included in the side of disjoint from . Thus, for any
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