CCrossing Number Bounds in Knot Mosaics
Hugh Howards, Andrew KobinNovember 9, 2018
Abstract
Knot mosaics are used to model physical quantum states. The mosaic number of a knot is thesmallest integer m such that the knot can be represented as a knot m -mosaic. In this paperwe establish an upper bound for the crossing number of a knot in terms of the mosaic number.Given an m -mosaic and any knot K that is represented on the mosaic, its crossing number c isbounded above by ( m − − m is odd, and by ( m − − ( m −
3) if m is even. In theprocess we develop a useful new tool called the mosaic complement. In [7], Lomonaco and Kauffman introduce a standard system of knot mosaics as a model ofphysical quantum states. Since then mosaics have been used in this original manner, but alsoas a tool by which to better understand knots themselves.Much like crossing number and other easy to define invariants of knots it is not hard to getvery loose bounds for mosaic number, but strong bounds can be difficult to compute. Many ofthe recent known results can be found at the Online Encyclopedia of Integer Sequences webpagedevoted to the number of n × n knot mosaics [9].Mosaics contain 5 distinct tiles, up to rotation, and all 11 orientations are shown below. Welabel the tiles with roman numerals for the 5 types and when applicable the letters a though d for the distinct rotations of those types. We also introduce a type 0 tile which consists of asquare with a dot in the center. Type 0 tiles are not a part of a mosaic, and have not previouslybeen used in the literature but will be used when we introduce a new tool called the mosaiccomplement in Section 2.For a positive integer n , define an n -mosaic M n as an n × n matrix composed of mosaic tiles.As is typical when studying mosaics, we are only interested in mosaics which represent knotsand links so we restrict to those cases in the standard way as follows. A connection point is themidpoint of a tile edge which is also the endpoint of an arc drawn on that tile. Thus type I Figure 1.1:
The tiles of a mosaic and its mosaic complement. a r X i v : . [ m a t h . G T ] M a y igure 1.2: Pictured above are two 4 × tiles have no connection points, type II and III tiles have 2 connection points and type IV andV tiles have 4 connection points as seen in Figure 1.1. Contiguous tiles are any tiles which falldirectly next to each other in the same column or row. We can then say that a tile within amosaic is suitably connected if each of its connection points is identified with a connection pointof a contiguous tile.A link n -mosaic is an n -mosaic with all of its tiles suitably connected so that after all thetiles are placed on the mosaic, the result is a projection of a link. In such a mosaic, tiles of typesII, III, IV, and V must have adjacent tiles that are also of one of these four types to extend thearcs started on those tiles.A knot n -mosaic is a link n -mosaic that corresponds to a projection of a one component link(a knot). Thus every knot mosaic is a link mosaic, but not every link mosaic is a knot mosaic.Define the mosaic number of a knot (or link) to be the smallest natural number m such that theknot (link) is able to be represented as a knot (link) m -mosaic. See Figure 1.2 for an example ofa 4-mosaic that is not suitably connected as well as a knot 4-mosaic (that is suitably connected).An important motivation for studying knot mosaics is the Lomonaco-Kauffman Conjecture,which states that knot mosaic theory is equivalent to classic (tame) knot theory. This wasproven by Kuriya and Shehab in [4], so as a result, we can treat knot equivalence and crossingnumber of knot mosaics in the usual way.At times we will want to examine specific tiles within a mosaic. When specifying a giventile, we model subscripts after the entries in a matrix so the tile R i,j will refer to the tile in the i th row and j th column, where columns are counted from the left and rows are counted fromthe bottom (we diverge from matrix notation slightly here to allow row numbers to reflect aheight function). If we think of M n as an n × n square disk, the 4 n − ∂M n will be called boundary tiles and the other ( n − tiles will be referred to as the interior of themosaic and denoted S . See Figure 1.3 for a depiction of S and the boundary tiles in a mosaic.We will often be focused on S . Figure 1.3:
In a 4 × S is the shaded 2 × In this paper we are explore the relationship between crossing number and mosaic number.We begin by listing some of the known results on crossing number and mosaics. We start with heorem 3.1 in [3]. Theorem 1.1 (Upper Bound on Crossing Number) . [3] Given an m -mosaic, if a knot is rep-resentable on the mosaic then its crossing number is bounded above by: c ≤ ( m − This upper bound follows immediately from the well known observation that all crossing tileson an n -mosaic occur on the interior S of a mosaic (see Lemma 3.1) and S has exactly ( n − tiles. Using this bound, one can quickly find a lower bound for the mosaic numbers for manyknots. For a few simple knots such as the trefoil it yields the exact mosaic number. However,it is clear that this bound is of limited use when it comes to determining the mosaic number ofmore complex knots. Ludwig, Evans, and Paat, for example, show that the relationship betweencrossing number and mosaic number can be subtle in [8] by building an infinite family of knotswhere each knot can only achieve its mosaic number in a projection on a mosaic that does notrealize its crossing number.While we will focus on an upper bound, Lee, Hong, Lee, and Oh give a lower bound oncrossing number in [5] showing that the mosaic number of a non-trivial knot is always less thanor equal to the crossing number of a knot plus 1. In this paper, we sharpen the upper boundon crossing number by proving Theorem 8.2 . Given an m -mosaic and any knot K that is projected onto the mosaic, thecrossing number c of K is bounded above by the following: c ≤ (cid:40) ( m − − m = 2 k + 1( m − − ( m −
3) if m = 2 k .In the next section we introduce a new tool, the mosaic complement , denoted C , togetherwith T , an ordered triple, associated to C . C and T anordered triple associated to C Given a mosaic M n for link L (or knot K ), we define the mosaic complement on S , the interiortiles of M n . The mosaic complement C is obtained by replacing the tiles on S in the followingmanner: every type V is replaced with a type I, type IV with type 0, type IIIa with type IIIb,type IIIb with type IIIa, type IIa with type IIc, type IIc with type IIa, type IIb with type IId,type IId with type IIb, and finally type I with either a type IVa or IVb (although this meansthe mosaic complement is not uniquely defined, either type IVa or IVb is fine).Intuitively, the mosaic complement is complementary to the link in the mosaic in the sensethat the union of the mosaic complement and the mosaic will intersect the boundary of each tilein S exactly four times - once on each of the tile’s edges. The name follows from the fact thatif we put a link on a mosaic together with its mosaic complement, the union of the two types ofarcs on the interior of the mosaic S will consist of entirely type V tiles, type IV tiles, and typeIV tiles together with a dot (the last set comes from when the link has a type IV tile and themosaic complement contributes a type 0 tile). One set of arcs on these tiles belongs exclusivelyto the link and the complementary set of arcs belongs only to the mosaic complement.Although the definition of the mosaic complement is not quite unique since we could chooseeither type IV tile to replace each type I tile, there is an inverse function associated to thedefinition that is unique (reverse the orders in the definition above), so the portion of the linkin the mosaic contained in S may be deduced from C . See Figure 2.1 for a picture of a knotmosaic together with its mosaic complement (the outline of S is also pictured).In general, the mosaic complement consists of loops, type 0 tiles, and arcs with both end-points on ∂S which we call edges. The term arc will be used to refer to a subset of a loop oran edge. The constant | C | is defined to be the total number of tiles in C (excluding the blanktype I tiles, of course). Throughout the paper R i,j refers to the tile representing the not and T i,j refers to the corresponding tile for the mosaic complement. Although T i,j is defined by looking at R i,j , we will be focused on the mosaic complement in most of ourarguments so we will almost always refer directly to T i,j . Figure 2.1:
An example of a knot mosaic and its mosaic complement.
We next define an ordered triple T , which is computed from the mosaic complement. Recallthat C is the set of all tiles (that are not blank) in the mosaic complement. Let C (cid:48) be the setof all tiles that form the complement in C of all of the type 0 tiles in C . Note that the tiles of C (cid:48) together form a set of loops and properly embedded edges on the square disk S . Let C (cid:48)(cid:48) bethe set of all tiles T i,j in C that are of type IV. Let l = | C | , l (cid:48) = | C (cid:48) | and l (cid:48)(cid:48) = | C (cid:48)(cid:48) | . We definethe ordered triple T = ( l, l (cid:48) , l (cid:48)(cid:48) ). Notice that although the mosaic complement is not unique fora given mosaic, the only choices were which of the two type IV tiles to pick and so any choiceof mosaic complement for a given mosaic will give the same ordered triple T . In Figure 2.1,for example, T = (5 , ,
1) because the mosaic complement has 5 nontrivial tiles, 3 of those tilesare not type 0, and 1 of those tiles is type IV. In Figure 3.2 , T = (3 , ,
0) because there are 3nontrivial tiles in the mosaic complement, but 0 of those tiles are not type 0 and, of course, 0of those tiles are type IV.Of all possible ways to embed a specific knot K on an n × n mosaic, let M n be a mosaicthat minimizes the ordered triple T lexicographically, and say that in such a case that T is minimal . For example, if K can be built with with a mosaic complement yielding T = (7 , , T = (8 , ,
0) we pick the first embedding since T < T lexicographically. A mosaic is said to be saturated if every tile of S , the interior of the mosaic, is a crossing tile.In this case C = ∅ . Conversely, if a link mosaic has a nonempty mosaic complement, it is notsaturated. Theorem 8.2 stated above shows that the even and odd knot mosaic boards haveradically different properties regarding how “nearly saturated” they can be. Lemma 3.1.
In a link mosaic, boundary tiles cannot be crossing tiles. Therefore all crossingsof a link must fit on S , the interior of the mosaic.Proof. This Lemma is easy to prove and well known. By definition for a mosaic to be suitablyconnected, connection points cannot occur on the boundary of the mosaic. Every edge of acrossing tile contains a connection point, so therefore boundary tiles cannot be crossing tiles(Tile R , in the mosaic on the left in Fig. 1.2. Theorem 3.2.
A saturated mosaic cannot contain the projection of a knot that achieves itscrossing number. igure 3.1: An odd mosaic with | C | = 2. Figure 3.2:
There are three type 0 tiles in this mosaic complement on M . One could think of the mosaic as asaturated mosaic with three crossings smoothed reducing the number of components in the link from 4 to 1. Figure 3.3:
A projection of Solomon’s link on a saturated mosaic ( M ). roof. Start by filling S with type V tiles. The proof will not depend on if we choose typeVa or Vb so at this stage we have not lost generality of the argument no matter which type Vtiles we choose. Now we notice that since edges intersecting ∂S can only connect to one of itstwo adjacent edges in ∂S there are only two choices of how to connect up the strands throughthe boundary tiles to get a knot or link. The vertical strand in R , , for example, must eitherconnect to the vertical strand in R , as it does in Figure 3.5 or R , as in Figure 3.4. In thefirst case this means tile R , is a type IId and R , is type IIc and in the second case tile R , is a type IIc and R , is a type IId, again both matching the examples in Figures 3.5 and 3.4respectively. On a saturated board once this single choice has been made the rest of the choiceson the outside are uniquely determined.Suppose K is a mosaic representation on M n which is saturated. If n is odd, both choicesof how to connect up along the boundary tiles leads to a pair of nugatory crossings in oppositecorners of the board like the crossing shown in Fig. 3.4. If n is even, one choice will result in alink (not a knot) and the other will result in a knot with a nugatory crossing in all four corners.In each of the cases where we have a knot instead of a link type I Reidemeister moves will reducethe number of crossings. Therefore a saturated mosaic cannot contain the projection of a knotthat achieves its crossing number. Figure 3.4:
A nugatory crossing in the corner.
Figure 3.5:
A link of n − We next consider mosaics with mosaic complements consisting of a single tile, that is | C | = 1.Almost every knot will fail to achieve its crossing number on such mosaics, and the followinglemma precedes a general theorem for mosaics with | C | = 1. igure 3.6: Odd mosaic with corners easily removed by type I Reidemeister moves..
Lemma 3.3.
The trefoil knot has mosaic number 4.Proof.
The trefoil knot has crossing number 3. Since a 3 × × M and achieve its crossing numberof 3. Therefore the trefoil knot has mosaic number 4. Theorem 3.4.
Given a knot K with crossing number c , suppose its mosaic number m is odd.Then c ≤ ( m − − .Proof. We showed above that an odd mosaic represents a knot with crossing number at most( n − −
1. A mosaic with two interior tiles that are not crossing tiles (type V) will have c ≤ ( m − − K is achieved with only one interior tile that is not a crossing tile, then thismeans | C | = 1. This can only happen if C is either a single type 0 tile or if C is a single type IItile in one of the corners of S . If C is not in one of the corners of S , then M n has two oppositecorners with crossings that can be reduced by a type I Reidemeister move just as in a saturatedodd board. Placing the mosaic complement in one of the corners can at most remove one ofthese nugatory crossings. Thus even though this embedding of K has ( n − − K does have an embedding with ( n − − M n where we assume n is even and T is minimal withrespect to all n × n knot mosaics giving knot K . We now construct an argument showing that we can assume C contains no loops while keeping T minimal. Lemma 4.1.
Let M n be a mosaic for a knot K for which T is minimal and the number of loopsin mosaic complement C is minimized over all such possible mosaics and for which | C | ≤ n − . et { c , c , . . . , c k } be the set of loops in the mosaic complement. Then if the set of loops is notempty, at least one of the c (cid:48) i s contains a type II tile.Proof. Since each loop has at least 4 corners, the only way for a loop to avoid a type II tileis if each corner is part of a type IV tile. If none of the corners are type II then the loop hasat least 4 type IV corners that meet other components of the mosaic complement. If one ofthese corners is type IVa replace it with one that is type IVb; if not do the opposite. This swapyields the connect sum of the loop in the mosaic complement with another component of themosaic complement, decreasing the number of loops in the mosaic complement by one. Since itdoes not change T and since it still yields a mosaic complement for K (the type IV corners arechosen arbitrarily) we see that the original mosaic complement did not meet the hypothesis ofthe lemma, a contradiction. Thus the corners of each loop may be assumed to be type II, aneven stronger conclusion than the one type II tile in the lemma. Lemma 4.2.
If some loop in the mosaic complement contains a type II tile then T = ( l, l (cid:48) , l (cid:48)(cid:48) ) is not minimal.Proof. If there is such a loop, call it c . Any time c crosses the knot, K , we may dictate thatit goes under K . By virtue of the definition of the mosaic complement, c never crosses itself.Now remove c from the mosaic complement and add it instead to the knot, replacing the knotmosaic with a link mosaic containing K and an unknot. Next place a type V tile into the mosaic(type V in the link, not the mosaic complement) where the type II tile of c had been. Because c was entirely below K , and c had no crossings with itself, we have just taken the connectsum of K with an unknot. Thus we have a new mosaic representing K , but | C | has droppedcontradicting the minimality of the ordered triple T in the original mosaic.In the proof above we found an unknot in the mosaic complement that contained a type IItile, swapped it out of the mosaic complement and into the mosaic and changed the type II tileto a type V, yielding a connect sum of K with an unknot resulting in a new embedding of K and lowering T . We will repeat this process multiple times in different contexts throughout thepaper and we call the process corner conversion .Lemmas 4.1 and 4.2 imply Corollary 4.3.
Let M n be a mosaic for knot K with | C | ≤ n − and for which T achieves itsminimum over all such mosaics. We may then choose M n so that simultaneously C containsno loops and T is minimal. We now want to look at a particular class of loops. These are the shortest possible loops:ones of length 4 coming from a combination of 4 tiles of types II and IV. We call these shortloops bubbles . We see a bubble in each of the pictures in Figure 4.1.The proof above showed that the mosaic complement may be chosen to contain no loops if T is minimal, but it did not show that a mosaic complement containing loops could not also beminimal. Later we may start with a mosaic complement that contains no loops and use movesthat create bubbles, which we then want to argue is impossible, so we need a stronger resultsaying that if T is minimal the mosaic complement cannot contain bubbles. In the argumentit will be useful to have the following lemma that gives us some flexibility in where a bubblemight be positioned. Lemma 4.4.
Given a knot mosaic M n for knot K with mosaic complement C and orderedtriple T , if M n contains any × n or n × subset of tiles, n ≥ , that consists of exclusivelytype IV tiles in the complement, then we may pick any × subset of these tiles and form a(possibly) new mosaic complement for M n and K in which there is a bubble in the × subsetso that T is unchanged for the new mosaic complement.Proof. The proof is easy. Pick each of the four tiles to be IVa or IVb so that they have a bubblewithin them. Leave the other type IV tiles unchanged. Since swapping type IV tiles in the osaic complement doesn’t affect K or T , the new mosaic complement has shifted the bubbleto the desired location and satisfies our requirements on K and T . Figure 4.1:
A bubble percolates up through a 3 × This process allows us to shift the location of an existing bubble through nearby type IVtiles. We call this process of shifting a bubble to a new location percolation . A 3 × Theorem 4.5.
Let M n be a mosaic for knot K for which T is minimal. C cannot contain abubble.Proof. If we ever have a bubble which contains a type II tile, then we can do a corner conversionas we did in Lemma 4.2, yielding a new embedding of K but lowering T . This contradicts thefact that T was minimal. Thus we proceed with the argument under the assumption that thebubble is entirely contained in type IV tiles.Next we show the mosaic complement is not minimal if the knot intersects either a row ora column of S containing the bubble. In this case, the ability to rotate the mosaic assures usthat we may assume that the intersection is in a column above the bubble. If there are any typeIV tiles below the knot in those columns, but above the bubble we use Lemma 4.4 to shift thebubble up to the four tiles directly below the knot.Explicitly, if K intersects T i +1 ,s ∪ T i +1 ,s +1 we let the bubble be contained in tiles T i − ,s , T i,s , T i − ,s +1 , and T i,s +1 . Without loss of generality let K intersect T i +1 ,s (and possibly T i +1 ,s +1 ,too). Because it is directly above a type IV tile, but it contains part of the knot, T i +1 ,s is eithertype IIb, IIa, or IIIb (the knot cannot intersect the bottom edge of the tile). T i +1 ,s +1 is alsoabove a type IV tile and must pair with T i +1 ,s . This means T i +1 ,s +1 must be a type IIa tile ora type IV tile if T i +1 ,s is type IIb, and T i +1 ,s +1 must be IIb tile or a type IIIb if T i +1 ,s is IIa orIIIb.We show moves in Figure 4.2 for six possible combinations that allow us to connect sumthe bubble with the knot and reduce T contradicting minimality – in the case of a type IV in T i +1 ,s +1 we show only type IVb case since the IVa case is nearly identical.Finally we are left with the case where neither the columns nor the rows of S containing thebubble intersect the knot. Thus they are exclusively full of type IV tiles in the complement.Since K exists and in any interesting case has at least one crossing, we know that some row of S intersects K . (Of course the unknot fits on a 2 × S = ∅ so we are only interestedin knots with positive crossing number.)By Lemma 4.4 we can percolate the bubble within the columns containing it to make itintersect the row that already contains part of K . Now as before we have not changed K or T . We have, however, reduced to the previous case, which shows T can be reduced withoutchanging K , contradicting minimality. Since we now know that we can get rid of loops in a mosaic complement without increasing T we turn our attention to edges. Lemma 5.1. If | C | ≤ n − then no edge in C runs from one side S to the opposite side. igure 4.2: A bubble can never appear in the mosaic complement when T is minimal. Here we see that if a tileabove the bubble contains an arc of K , there is always an embedding of K that decreases T and ‘bursts’ the bubble. Proof.
The proof is trivial since such an edge must be of length at least n − e is an edge of the mosaic complement, then the endpoints will have toeither be on adjacent sides, as in edges f , f , f and f in Figure 7.3, or on the same side of S ,i.e. starting and ending in the exact same row or column as in edges e , e and e in the samefigure. If the endpoints are on the same side as each other we call it an XX-edge and if adjacentsides we call it an
XY-edge .Both XX and XY-edges cut S into two disks. The smaller side is considered the outside ofthe edge. In topology we may not have a metric so we often avoid talking about the smallerpart of a disk, but a mosaic as an n × n subset of the plane has a natural metric on it so we arefree to use the term.Because the edges in the mosaic complement are relatively short, if e is an XX-edge then theboundary of the outside (smaller) disk consists of e together with part of one side of S . Likewiseif e is an XY-edge the outside consists of e together with parts of two sides of S . Thus for anyarc e we have a notion of outside . An edge e ⊂ C (cid:48) is called outermost in C (cid:48) if there are no edgesoutside of e on S in C (cid:48) .Note that if C contains no loops – which Corollary 4.3 allows us to assume – and C (cid:48) (cid:54) = ∅ , sothere is at least one edge, then there must be an outermost edge e . Also note that our definitionis not quite the same as the traditional definition of outermost arcs on disks in topology. If e isoutermost in our context it is outermost in the traditional definition, but not every traditionallyoutermost arc is outermost in our definition because it might be outermost, but on the wrongside (the side of its larger disk). An edge can still be outermost even if there is a type 0 tile of C outside of it. We establish a set of moves that when applied to the arcs of the mosaic complement will reducethe ordered triple T without changing the isotopy class of the knot. One primary use of themoves is to lower an arc a of the mosaic complement that represents a local maximum (possiblyafter rotating the mosaic). This will eventually lead to the conclusion first that no such movescan be made to the edges of C , and then after a further argument that C contains no edges atall if T is minimal. We define the moves starting with the more elementary moves. .1 The type IV moves: bubble release and XX-through-XY moves In the proof of Lemma 4.1 we swapped type IV tiles to reduce the number of loops in a mosaiccomplement by taking the connect sum of a loop with another part of the mosaic complement.We now consider the inverse operation on the mosaic complement when it would create a bubble.We define a bubble release move when we swap a type IVa tile for IVb or vice versa to yield abubble without altering K or T . Such a move is pictured in Figure 6.1. Figure 6.1:
Neither the embedding of the knot nor T are altered when we break a bubble off of an arc of the mosaiccomplement using the bubble release move. Note that the move is identical if any of the type II mosaic complementtiles are swapped for type IV tiles. We know, however, by Theorem 4.5 that a minimal mosaic complement can never containa bubble and T is unchanged by a bubble release move so we see immediately the followinglemma. Lemma 6.1. If T is minimal then C cannot contain tiles on which we can perform a bubblerelease move. The other type IV move is the XX-through-XY move. Let C contain e an XY-edge thathas e an XX-edge outside of it such that the two edges share a type IV tile. Switching the tilefrom IVa to IVb or vice versa will, of course, have no effect on T or K , but will replace e and e with a new XY-edge and a new XX-edge. Call the XY-edge e (cid:48) and the XX-edge e (cid:48) . Themove reduces the overall number of XX-edges outside of XY-edges. In particular at the veryleast e (cid:48) is not outside of e (cid:48) and e (cid:48) has fewer XX-edges outside of it than e did.Iterating this process will eventually terminate since the number of type IV tiles is boundedby | C | .We define a knot mosaic together with a mosaic complement C and associated ordered triple T to be a minimal embedding for a knot K if T is minimal, C contains no loops and in C no XX-through-XY moves are possible. Corollary 4.3 together with the process we have just describedassures that every knot K that has a knot mosaic with | C | ≤ n − minimal mosaic complement . We describe this move in terms of an arc that acts as a local maximum for an edge and is moveddownward. By symmetry, we can rotate the mosaic any multiple of 90 degrees or reflect alonga horizontal or vertical line so the move is equally valid if the arc is a minimum and is movedupward, or one that is concave right and is moved to the right or concave left and is moved left.
Corner-corner move:
Let e be an edge in C that intersects row i in an arc a that representsa local maximum for e . A local maximum must run directly across row i, i >
2, from a typeIIb (or IVb) tile in column s to a type IIa (or IVa) tile in column t with s < t as in Figure 6.2.We want to reduce the ordered triple T = ( l, l (cid:48) , l (cid:48)(cid:48) ) by moving part of the knot up across a andshorten a by moving it down. Figure 6.3 shows the basic move.We pay close attention to any portion of the mosaic complement in tiles T i − ,w with s ≤ w ≤ t (row i − a ). Since each of the tiles of a of the form T i,w with s < w < t consistsexclusively of type IIIa tiles, clearly those tiles of the form T i − ,w cannot ever be type IIc, IIdIIIb, IV or V.Certainly T i − ,w can be a Type 0 tile as shown, together with the corresponding corner-corner move in, Figure 6.3. On the other hand, if there is a tile in the mosaic complement
11 * * * * *
Figure 6.2:
Each * denotes one of four types of corners possible in a corner-corner arc (two up to reflectivesymmetry).
Figure 6.3:
We see a basic corner-corner move. T i − ,w with s < w < t that is type IIa, IIb, or IIIa then the corner-corner move is undefinedon a . Such examples are seen in the nested arcs in Figure 7.3. Another obstruction to thedefinition we can encounter is that if T i − ,t is type IId, IVa, or IVb. Symmetrically it is alsoundefined if the mosaic complement tile T i − ,s is type IIc, IVa, or IVb. We see arcs of thisform in Figure 6.4. We will never need to use the corner-corner move in any of the undefinedcontexts, so the lack of definition here will not be a problem.We now focus on the definition of the move in the situations where it can be applied. Themove at its core just takes an arc a that is a local maximum for the mosaic complement andpushes it down one row when there is nothing from the mosaic complement already below it toblock it. The exact prescription is given in two parts. For each w , s < w < t we switch T i − ,w with T i,w . This tells us how we apply the move to tiles that are between the corners of the arc,but not in the corners themselves. We now specify the move on the two corner tiles and the twotiles directly below them ( T i,s , T i − ,s , T i,t and T i − ,t ). The corner tiles T i,s and T i,t are eithertype II tiles or type IV. When the move is defined tile T i − ,s must be type IIIb tile or type IIc.On the other corner, T i − ,t is either type IIIb or IId. As mentioned earlier, the move is notdefined if either of the tiles below the corners are type IV; we address this situation later.The swap for a typical situation is pictured in Figure 6.3. If the tile T i,s or T i,t is type IIwe replace it with a type 0 tile. If it is type IVa it is replaced with a type IIc tile. A type IVbis replaced with type IId. If the tile T i − ,s is type IIIb it is replaced by a type IIb tile. If it istype IId, it is replaced by IIIa. If T i − ,t is type IIIb then it is replaced by type IIa. If it is IIc,it is replaced by type IIIa. Lemma 6.2.
A corner-corner move causes a planar isotopy of K and reduces T . Hence therecannot be an arc on which a corner-corner move can be applied in a mosaic that minimizes T .Proof. The lemma follows directly from the definition of the move. The tiles between columns a (cid:48) a a (cid:48) Figure 6.4:
The corner-corner move is not defined on the arcs a and a at the top of the edges because of thebottom tile in a (cid:48) and a (cid:48) blocking the move, but this is not a problem because it is defined on the two-tile arcs a (cid:48) and a (cid:48) . s and t swap places, but pairwise remain identical and thus cannot change T . As seen in thefigures no matter which configuration appears in column s and t , the ordered triple T decreasesin these columns. Specifically, the contributions to | C | remains the same, but the contributionin column s to either | C (cid:48) | or | C (cid:48)(cid:48) | is reduced by one and the same is true in column t .We now turn our attention to the two cases in which the corner-corner move was not definedto see that neither of these is a problem. The following lemma states that the first one can neveroccur in a reduced mosaic complement. Lemma 6.3.
A corner-corner arc a with T i − ,t either type IId or type IVb or with T i − ,s typeIIc or IVa cannot occur if T is minimal.Proof. Given a in the mosaic complement running from T i,s to T i,t if T i − ,t in the mosaiccomplement is either type IVb (meaning R i − ,t is type I) or type IId then the corner-cornermove is not defined on a . Let the portion of e in tiles T i,t and T i − ,t be called a (cid:48) . Arcs a (cid:48) and a (cid:48) in Figure 6.4 are examples of such arcs. If t = s + 1 and T i,s is type IV then we are in a situationsuch as Figure 6.1, but this is impossible since T is minimal and the existence of a bubble releasemove would contradict minimality. Given the structure of a corner-corner arc a together withadjacent two-tile corner-corner arc a (cid:48) , this is the only case in which the corner-corner move isnot defined on a (cid:48) Therefore we push it to the left so in all other cases a corner-corner move canbe applied to a (cid:48) reducing T just as it can be to a (cid:48) and a (cid:48) in Figure 6.4. By Corollary 6.2 weknow that the move cannot happen if T is minimal, so T i − ,t cannot be type IId or IVb. Theanalogous argument holds by reflective symmetry if T i − ,s is type IIc or IVa.Thus it is not a problem that the corner-corner move was not defined in this context. Weare left only with the following situations in which the corner-corner move was not defined. Wecould have a corner-corner arc a in the mosaic complement running across row i from T i,s to T i,t and if t > s + 1 we have a mosaic complement tile T i − ,w with s < w < t that is type IIa,IIb, or IIIa. If t = s + 1 then T i,s is type IVb and T i − ,t is type IVa; T i,s may also be type IVband T i − ,t may be type IVa if t > s + 1, too, of course. Lemma 6.4. If T is minimal, then the only way we can have a corner-corner arc a in row i is if a is part of a nested series of corner-corner arcs { a , a . . . a i − } with each a j contained inrow j for ≤ j ≤ i − .Proof. We have seen already that the definition of the mosaic complement dramatically limitsthe choices for tiles beneath a in row i − T i − ,w must be a corner-corner arc a i − from T i − ,s (cid:48) to T i − ,t (cid:48) for some s (cid:48) and t (cid:48) with s ≤ s (cid:48) < t (cid:48) ≤ t .Iterating the process we either find a corner-corner arc that does not have a corner-corner arcbelow it in some row j with 2 < j ≤ i contradicting minimality or there are nested corner-cornerarcs extending in every row from i down to 2 as in the edges e , e and e in Figure 7.3. .3 Corner-edge moves We again for simplicity choose to describe this move as it moves an arc a of the mosaic comple-ment down, but as before, symmetrical moves to the right, left, or up are all valid by rotationsor reflections of the mosaic. The move is very similar to the corner-corner move as are thearguments about it.Our goal in applying the corner-edge move is to reduce T , and we will always do any availablecorner-corner moves before doing any corner-edge moves, so we need not worry about definingthe corner-edge move on an edge for which a corner-corner move is possible. Corner-edge move:
Let e be an edge of the mosaic complement that intersects row i > a running directly across i in columns 2 through t and turning down in column t, t ≥ e intersects row i in an arc a such that each tile in the mosaic complement T i,w , w < t is a type IIIa tile and tile T i,t is a type IIa or IVa tile as in Figures 6.5 and 6.6, respectively.As in the corner-corner move, we pay close attention to any portion of the mosaic complementin tiles T i − ,w with 2 ≤ w < t (row i − a ). Again T i − ,w cannot ever be typeIIc, IId IIIb, IV or V, but can be a type 0 tile without causing any problems.As before, if there is a tile in the mosaic complement T i − ,w with s < w < t that is type IIa,IIb, or IIIa then the corner-edge move is undefined on a . Such examples are seen in the nestededges f , f , f and f in Figure 7.3.We may have an obstruction where t = 2, T i − ,t is type IId or type IVb and T i,t ∪ T i − ,t forms a two tile outermost XX-edge, but we never apply a corner-edge move in this context sowe do not mind this obstruction. With this exception we do not encounter an obstruction to thedefinition where T i − ,t is either type IId or type IVa because it would lead to a reduction via acorner-corner move of tiles T i,t ∪ T i − ,t to the left which already contradicts the minimality of T for the mosaic complement. Figure 6.5:
The corner-edge move may have a type II tile in its corner.
We again reduce T by moving part of the knot across the mosaic complement. Figures 6.5,6.6, and 6.7 show the basic move.Because there can be no corner-corner moves in a minimal mosaic complement and there arenever type V tiles in a mosaic complement, the tile T i − ,t must be either a type IIIb, type IIcor type IVa. If t = 2 and T i − ,t is type IVa then the corner-edge move is not defined, but againthis is an obstruction that we do not mind as we will never need to apply it in this context.Instead examine the case where t >
2. If T i − ,t is type IVa we could swap the type IVa mosaiccomplement tile with a type IVb tile, replacing mosaic complement C by mosaic complement B without affecting the knot. Since the only thing we have changed to go from C to B is onetype IV tile for another, the ordered triple T C for C is clearly identical to the ordered triple T B for B . T B is reduced by a corner-corner move, showing it was not minimal for the knot K whose mosaic complement is B (and C ) and therefore T C also was not minimal for K . Since wealways choose our embedding of K so that T is minimal we may assume that tile T i − ,t is nottype IVa when t > igure 6.6: Alternatively, the corner-edge move may have a type IV tile in its corner. The resulting move is onlyslightly different.
We are now left with the possibilities that T i − ,t must be type IIIb (Figure 6.5) or IIc asdepicted in Figure 6.7 and we define the corner-edge move accordingly. The exact prescriptionfor the move is that for each w < s we switch tile T i − ,w with tile T i,w . T i − ,t and tile T i,t are treated exactly as they were in the corner-corner move: if T i,t is type II we replace it witha type 0 tile. If it is type IVa it is replaced with a type IIc tile. If T i − ,t is type IIIb then itis replaced by type IIa. If it is IIc, it is replaced by type IIIa. Typical corner-edge moves aredepicted in Figures 6.5 through 6.7. Figure 6.7: T i − ,t may be a type II tile oriented as pictured instead of a type III tile. The resulting move stillreduces T . Again this move was described in terms of a row, but it can be rotated or reflected to movecorner to edge row arcs up and down and corner to edge column arcs right and left.
Lemma 6.5.
A corner-edge move causes only a planar isotopy of K and reduces T . Thereforein a minimal mosaic, there cannot be an arc on which a corner-edge move may be applied.Proof. The lemma follows directly from the definition of the move. The argument is analogousto Lemma 6.2.We now have the moves defined and in the next section will turn our focus to the XX-edges (edges with both endpoints on the same edge), showing that they cannot exist withoutcontradicting minimality. Then once we know there are no edges of this type we will eliminateXY-edges, too. Reduction steps towards the main theorem
Lemma 7.1. If E = { e , e , . . . , e n } is the set of all edges in C and | C | ≤ n − then there issome e i containing a type II tile in the mosaic complement. If the XX-edges do not share a typeIV tile with the XY-edges then at least one of the XX-edges contains a type II tile or the set isempty. The same is true for the XY-edges.Proof. Each XX-edge has at least two corners which are either a type II tiles or else Type IVtiles where they meet another edge of the mosaic complement. Similarly each XY-edge musthave at least one such corner. To avoid any type II tiles, the mosaic complement would have tostretch from one side of S to the opposite side, but this would mean | C | ≥ n − e is an XX-edge with both endpointson the bottom in a minimal mosaic complement then e contains exactly one corner-corner arc,and that arc can only be concave down. We note that as always, symmetrical arguments canbe made by rotation and reflection for edges with endpoints on the other sides of S . Lemma 7.2. If e is an XX-edge in C with both endpoints on the bottom side of S or an XY-edgewith one endpoint on the bottom of S and the other on the left side and | C | ≤ n − and T isminimal then e cannot contain a corner-corner arc a that is concave up. By symmetry this alsomeans the XY-edge cannot have a corner-corner arc that is concave right.Proof. By Lemma 6.4, T can be reduced via a corner-corner move applied to a unless there is anested set of corner-corner arcs inside of a including one in each of the rows of S above a . This,however, cannot happen since it would imply that there are mosaic complement tiles in everyrow of S , contradicting | C | ≤ n − Lemma 7.3.
Suppose | C | ≤ n − and T minimal. Let e be an XX-edge in C with both endpointson the bottom edge of S or an XY-edge with one end point on the bottom of S . If a is a corner-corner arc of e in row i that is concave down, then a is the only corner-corner arc on e that isconcave down.Proof. A second concave down corner-corner arc would require a concave up corner-corner arcbetween the two: an edge in the plane with endpoints at the same height may not have twolocal maxima without a local minimum. We know we cannot have a concave up corner-cornerarc on e by Lemma 7.2. Lemma 7.4.
Let e be an XX-edge or XY-edge in C where | C | ≤ n − and T is minimal.Then e cannot contain a corner-corner arc concave to the left (representing a maximum in thedirection right) and also a corner-corner arc concave right.Proof. Each corner-corner arc would need nested corner-corner arcs going all the way to theedge of S . This would, of course, require at least one tile in each column of S , contradicting thefact that | C | ≤ n − Lemma 7.5.
Let e be an XX-edge in C with both endpoints on the bottom edge of S with | C | ≤ n − and T minimal. Let the left endpoint of e be in column s and the right endpointin column t , s < t . We cannot encounter a corner-corner arc concave to the left intersectingcolumn w for w < t . The same is true for corner-corner arcs concave right in columns w with s < w roof. If a is such a corner-corner arc in e , to connect with T ,t in the first case and T ,s in thesecond, e would need to turn around via a corner-corner arc concave in the opposite directioncontradicting Lemma 7.4. Lemma 7.6.
Let e be an XX-edge in a minimal mosaic complement C with both endpoints onthe bottom edge of S with | C | ≤ n − and T minimal. Then e cannot have any corner-cornerarcs concave to the left or right.Proof. We show the proof for corner-corner arcs concave to the right since the proof to the leftis identical up to symmetry. By Lemma 7.5 the concave right corner-corner arc a must startand end in column w , w ≤ s . However by Lemma 6.4, a must have nested corner-corner arcsinside of it extending all the way to the right side of S . Since a is on the left side of e , thisimplies that at least one of the nested corner-corner arcs for a is also in e . For e to contain twocorner-corner arcs that are concave to the right it must also contain a corner-corner arc concaveto the left between them. This contradicts Lemma 7.4 . Thus there were no corner-corner arcsconcave to the right.These lemmas imply Corollary 7.7.
Let e be an XX-edge in minimal mosaic complement C with both endpoints onthe bottom edge of S with | C | ≤ n − and T minimal. Let the left endpoint of e be in column s and the right endpoint in column t , s < t . Let row i contain the maximum of e . Then e isstrictly contained between columns s and t (inclusive) and below row i (inclusive). We now apply this result to outermost XX-edges to build an argument that they must consistof only two tiles exemplified by edge e in Figure 7.3. Lemma 7.8. If e is an outermost XX-edge in minimal mosaic complement C with | C | ≤ n − then e consists of two adjacent tiles in the second layer. Each of the tiles is either type II ortype IV.Proof. Without loss of generality let both endpoints of e be on the bottom edge of S . First weargue that if e is an outermost edge then e is totally contained in row 2 and consists of T ,s atype IIb or IVb tile, T ,t a type IIa or IVa tile and type IIIa tiles T ,w for s < w < t (if t = s + 1then this last set of tiles is not used). Up to rotation, h and e are examples of such arcs inFigure 7.3. We then strengthen the result to show that e not only is in the second row, but itcontains only two tiles.Let a be the corner-corner arc of e in row i , the highest row that contains a tile of e . If i = 2 we are done with the first step of the proof. If i > i + 1 just above a cannot contain any portion of e since it is above the global maximum of e and this dictates thatthe tiles in row i + 1 are inside (not outside) of e . In turn this means that the row i − a must either contain part of e or be outside of e . Since T is minimal, there must be acorner-corner arc a (cid:48) ⊂ C below a or we could do a corner-corner move on a pushing it downand reducing T . Then a (cid:48) cannot be part of an edge other than e since that would imply it wasoutside of e and e is outermost. Thus a (cid:48) ⊂ e , but this contradicts Lemma 7.3. This implies that a is in row 2 and this can only happen if e = a and consists of T ,s a type IIb tile, T ,t a typeIIa tile and T ,w type IIIa tiles for s < w < t .Now we know that e is entirely contained in row 2. If e contains more than 2 tiles, then theleftmost tile of e can be moved to the right using a corner-edge move to reduce T contradictingminimality. Thus e must be just 2 tiles long and we are only left with the desired type ofoutermost XX-edges (see arcs e , c , c , g and g in Figure 7.3). Lemma 7.9. If e is an XX-edge in a minimal mosaic complement C with both endpoints on thebottom edge of S and | C | ≤ n − and the maximum of e occurs in column i , i > then thereis a set of nested edges { e , e . . . e i − } outside of e with the maximum of each e j in row j . roof. We must have nested corner-corner arcs outside of e in each row and each edge has onlyone corner-corner arc. Lemma 7.10. If e is an XX-edge in minimal mosaic complement C with both endpoints on thebottom edge of S and | C | ≤ n − and the left endpoint of e occurs in column s at T ,s , and e is not outermost, then there is a nested set of edges { e s +1 , e s +2 . . . e k } outside of e with the leftendpoint of e j in column j for each j , s + 1 ≤ j ≤ k and e k an outermost edge in C .Proof. Because the XX-edges contain no corner-corner arcs that are concave left or right wecan always use a corner-edge move to reduce T unless there is an edge with its endpoint in theadjacent column blocking the move. Lemma 7.11. If e is an XX-edge in C and | C | ≤ n − and all the edges outside of e are nestedwith each other, and e contains a type II tile, then C is not minimal.Proof. Examine the subset of M n representing K . If the arcs of K contained in S are connectedusing the tiles under e in row 1, as it does in the top mosaic in Figure 7.1, then the edge in row1 also runs under an outermost edge e (cid:48) outside of e ( e = e (cid:48) if e is outermost). We can add e (cid:48) to K , connecting it up to the original knot as in the bottom picture in Figure 7.1, and obtaina knot isotopic to K , but we have reduced the ordered triple contradicting minimality.If K does not use the tiles in row 1 under e , then K is disjoint from all of the tiles inrow 1 between the endpoints of e . If necessary change crossings between the knot and mosaiccomplement – but not the knot with itself, thus not changing the knot at all – to make sure e always goes under K , and connect e to itself through row 2 giving a loop. Remove the loop fromthe mosaic complement and add it to the mosaic creating a link mosaic with two components, K and an unknot. This takes us to the middle picture in Figure 7.2. Then use a corner conversionby placing a type V crossing tile where the type II tile of e had been going from the middle tothe bottom picture in Figure 7.2. As in the proof of Lemma 4.2, a corner conversion takes theconnect sum of K with an unknot giving another version of K on an n × n mosaic, but with areduced ordered triple contradicting the minimality of T . Figure 7.1: If K passes outside of an outermost arc of the mosaic complement, the move pictured shows the mosaiccomplement is not reduced. Lemma 7.12. If e is an XX-edge in a minimal mosaic complement C and | C | ≤ n − then allthe edges outside of e are nested with each other.Proof. If not, then examine the outermost edge e which has edges outside of it which are notnested with each other. Let e and e be the innermost non-nested edges outside of e (so e is not outside of e and vice versa and e and e are just outside of e ). Up to rotation, thissituation is depicted by arcs g , g , g and by c , c , c in Figure 7.3. None of the corners at thetop of e ∪ e can be type II or T is not minimal by the previous lemma, but if they are all typeIV tiles then e has a corner-corner arc that is concave up contradicting Lemma 7.2. igure 7.2: If a nested XX-edge in the mosaic complement has a type II tile in the mosaic complement and K doesnot pass outside the edge, we can alter the mosaic complement reducing T . f f f f e e e h c c c g g g Figure 7.3:
An example of a knot mosaic and its mosaic complement. hese lemmas imply Theorem 7.13. C contains no XX-edges containing a type II tile if C is chosen minimally. Now that we know that all the corners in an XX-edge are type IV we exploit this fact to getrid of all XX-edges.
Theorem 7.14. If | C | ≤ n − and C is chosen minimally, then C contains no XX-edges.Proof. If C does contain an XX-edge then there is at least one that is not outside of any of theother XX-edges – the innermost edge from any of the nested sequences would suffice. Call thatedge e . Without loss of generality let e have both end points on the bottom of S , specificallyin tiles T ,s and T ,t with s < t . We know e has no type II tiles by Theorem 7.13, but it mustcontain two type IV tiles at its maximum. Let f be an edge that meets e in one of these type IVtiles. Note that the tile coming from a maximum for e implies that f is not outside of e . Since e has endpoints on the bottom of S and | C | ≤ n − f cannot have an end point on thetop edge of S . If there is a second edge g that shares the other type IV tile from the maximumof e , it cannot be the case that one of these edges had an end point on the right side of S andthe other on the left side, since such an edge would stretch across S , forcing | C | ≥ n −
2. Sowithout loss of generality we may assume that the end points of f and g are contained in atmost the left and bottom sides of S .We next argue that both end points of f (or g ) cannot just be on the bottom of S . If f hasboth end points on the bottom it is by definition an XX-edge, but recall that we chose e so itwas not outside of any XX-edges so we know that e is not outside of f . As a result both endpoints of f have to be contained in rows that are on the same side of rows containing the endpoints of e . In particular f must have end points in tiles T ,u and T ,v with u < v since they areon the bottom of S , but then since the two edges are not nested and do not intersect we musthave u < v < s < t or s < t < u < v . Either way by Corollary 7.7 f never intersects any of thecolumns between s and t and e never leaves these columns so they cannot contain a commontype IV tile. This implies that f does not have both end points on the bottom of S .Now without loss of generality f (and any edge sharing a type IV tile with e ) either has bothend points on the left side of S if it is an XX-edge or one on the left side of S and the other onthe bottom side if it is an XY-edge. In the latter case, since e is not outside of f we also knowthat the end point on the bottom of S is in some tile T ,u with u < s < t .Examine again the top right type IV tile from the maximum of e and the edge f that sharesthis tile. The portion of f within this tile looks like a IIc tile. Because both of its end points areto the left of e and it does not intersect e we know f must contain a concave up corner-cornerarc to contain this tile. This, however, contradicts Lemma 7.2.Thus e cannot exist so there are no XX-edges in the mosaic complement.From here we proceed by eliminating XY-edges. Theorem 7.15. If | C | ≤ n − and C is chosen minimally, then C contains no XY-edges.Proof. We know all edges must be XY-edges, and that if the set of edges is nonempty then atleast one of the XY-edges contains a type II tile by Lemma 7.1.Let e be such an XY-edge. We can rotate the entire mosaic if necessary until it encloses thebottom left corner, so let e have one endpoint in tile T s, on the left edge of S and the otherin T ,t on the bottom edge of S enclosing the bottom left corner on its outside. First observethat e cannot contain a corner-corner arc. By Lemma 7.9, any corner-corner arc requires a setof nested corner-corner arcs terminating in an XX-edge on the boundary of S . However byTheorem 7.14, C contains no XX-edges so this is impossible.The lack of any corner-corner arcs implies that e is completely contained in the rows belowrow s (inclusive) and to the left of column t (inclusive). In turn this implies that if s > { e , e . . . e w . . . e s − } for each w , 2 ≤ w < s . If not wecould apply a corner-edge move to reduce T . The analogous result holds for the endpoints onthe bottom of S with respect to the columns. ow we close the argument in the same manner as before. The outermost XY-edge mustbe a type II tile or one of the arcs from a type IV tile in T , . The next most outermost hasone endpoint in tile T , and the other in T , etc. Examine the tiles representing K . If K runsthrough row 1 and column 1 past the ends of these arcs, then we add the outermost XY-edgeto K and connect it up to give a planar isotopy of K and reducing T , contradicting minimality.If not, recall that e contains a type II tile. Since K does not go through the tiles in row 1or column 1 outside of e we can turn e into an unknot by hooking the endpoints of e to itselfthrough these tiles. There is no effect on K if we assume that e always passes under K . Asin previous theorems we use a corner conversion to replace the hypothesized type II tile from e with a type V tile to connect sum the new unknot with K , yielding another embedding of aknot isotopic to K for which T has decreased contradicting minimality. Thus there can be noXY-edges.Since the mosaic complement contains no XX-edges, no XY-edges and no loops, we may nowconclude the following Corollary. Corollary 7.16. If M n = M k is an even knot mosaic yielding knot K , with minimal mosaiccomplement C and | C | ≤ n − , then l (cid:48) = | C (cid:48) | = 0 . Therefore we may assume C consistsexclusively of type 0 tiles. We use Corollary 7.16 to show our main theorem below.
Theorem 8.1. If M n = M k is an even knot mosaic yielding knot K , then the crossing numberof K is less than ( n − − ( n − .Proof. A mosaic with | C | = l and C (cid:48) = ∅ has l type 0 tiles in the mosaic complement andnothing else. It therefore is obtained from a saturated mosaic by smoothing l crossings. In thelanguage of tiles, we are replacing l type V tiles in the link with l type IV tiles. Each time thisis done the number of components in the mosaic changes by at most one.If l > n − K has at most ( n − − ( n −
5) crossings failing to exceed our bound.As we saw in the section on saturated mosaics, a saturated even mosaic has n − n − S are connected in the boundary tiles ofthe mosaic. Since we are only smoothing l crossings, we see that if l < n − l ≤ n − l ≥ n − l = n − n − n − n − n − S as smoothing one of these crossings fails to lower thenumber of components in the link. This means that the knot that results from smoothing n − n − − ( n −
4) crossings, but it also still has the 4 trivial loops in the cornersthat can be removed with type I Reidemeister moves. Thus K could be embedded with 4 fewercrossings, showing its crossing number is at most ( n − − ( n − − < ( n − − ( n −
4) so thisknot does not exceed our bound on crossing number on even mosaics. Therefore a knot mosaicon an even board cannot have crossing number greater than or equal to ( n − − ( n − M showinga knot of crossing number 3 can be built on a 4 × heorem 8.2 (New Upper Bound for Crossing Number) . Given an m -mosaic and any knot K that is projected onto the mosaic, the crossing number c of K is bounded above by the following: c ≤ (cid:40) ( m − − if m = 2 k + 1( m − − ( m − if m = 2 k . At the beginning of this paper we used Theorem 1.1 to relate crossing number and mosaicnumber. In a similar fashion, Theorem 8.2 may be used to bound a knot’s mosaic number frombelow. First we define B = √ c + 2 B = 5 + √ c − Corollary 9.1 (New Lower Bound for Mosaic Number) . Let K be a knot with crossing number c and mosaic number m . Then m ≥ min { B , B } . This will prove useful in future computations of mosaic number. For now, we will brieflyexplore the behavior of B and B . It is easy to see that B and B are asymptotic, aslim c →∞ B B = 1 . Informally, this means that B and B , as functions of c , grow at relatively the same rate. Astronger result is that the difference | B − B | is bounded by , and although this difference isalways increasing, it turns out that lim c →∞ | B − B | = . This result may be somewhat surprising: our work has shown that the even and odd casesrequire different approaches, but in reality the estimates for each are actually quite similar andthe predictive power of one never strays too far from the other.
10 Future research directions
Our research has provoked several questions about knot mosaics which are left open to furtherinvestigation.
Question.
How does mosaic number behave for the connect sum of knots?
Question.
Are there any knots whose mosaic number is 2 greater than the number predicted byTheorem 8.2?
Question.
What are the mosaic numbers for all knots of 10 crossings or fewer?
We note that Lee, Ludwig, Paat, and Peiffer compute the mosaic number of all prime knotsof 8 crossings or fewer [6] so progress is currently being made in this direction.It is also natural to look at a more general class of mosaics where instead of insisting theboard be n × n we allow it to be n × m . One might then define the rectangular mosaic numberin terms of the number of tiles in the mosaic or perhaps even better the number of tiles on itsinterior. Mosaics that need not be square should allow for more efficient embeddings especiallyin the case of knots that are not prime. Question.
How does crossing number relate to rectangular mosaic number? he authors in [5] establish an upper bound on the mosaic number of a knot using arc index,and use this to prove stronger bounds for several classes of knots. Their results, together withTheorem 8.2 and Corollary 9.1 of this paper, provide a clearer picture of the relation betweenthe mosaic number and crossing number of a knot. However, in general a more precise formulathat relates the mosaic number and crossing number of a knot would be desirable. The mosaicnumber of some relatively simple knots remains unknown including many prime knots with 9crossings and some composite knots with fewer than 9 crossings.Note that a 5 × × × standard cube as an analog of a mosaic tile: a cube contains 0, 1, 2 or 3 strands, andeach face of the cube intersects at most 1 strand in the center of the face. In addition, foreach strand within a cube there is at most one critical point in any direction on the interiorarc of the strand. Define an n-cubic knot as an n × n × n array of suitably connected standardcubes. Furthermore, define the grid number of a knot (analog of mosaic number) to be thesmallest natural number g such that the knot is representable as a g -cubic knot. Note thatmosaic number is a (bad) upper bound for grid number. Further research questions on the topicof cubic knots can be asked such as the one proposed below. Question.
For a knot K with mosaic number m and grid number g , g ≤ m . Find a sharperupper bound for g . The authors would like to thank Sam Lomonaco and Lou Kauffman for use of many of thefigures in this paper, as well as Joe Paat and Lew Ludwig for inspirational discussions resultingfrom their paper with Erica Evans [8].
References [1] Adams, Colin C.
The Knot Book: An Elementary Introduction to the MathematicalTheory of Knots . Henry Holt and Company, New York (2002).[2] Burde, G., and Zieschang, H.
Knots, nd rev. ed. Berlin: de Gruyter (2002).[3] Gardu˜no, Irina T. Virtual Mosaic Knots.
Rose Hulman Undergraduate MathematicsJournal . Journal ofKnot Theory and Its Ramifications vol. 23, Issue 1 (2014): 1450003.[5] Lee, H.J., Hong K., Lee H., Oh S., Mosaic number of knots
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Vol. 23 (2014) No. 13[6] Lee, HJ, Ludwig, LD, Paat JS, and Peiffer, Knot Mosaic Tabulation.
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Vol. 11(2018), No. 1, 1326[7] Lomonaco, Samuel J., and Kauffman, Louis H. Quantum Knots and Mosaics.
Journalof Quantum Information Processing . vol. 7, nos. 2-3 (2008). pp. 85-115.[8] Ludwig, Lewis, Evans, Erica, and Paat, Joseph. An infinite family of knots whose mosaicnumber is realized in non-reduced projections.
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The Online Encyclopedia of Integer Sequences webpage devoted to the numberof n × n knot mosaics : https://oeis.org/A261400 (updated Aug 19, 2015).: https://oeis.org/A261400 (updated Aug 19, 2015).