ABSTRACT A STUDY OF PROJECTIONS OF 2-BOUQUET GRAPHS. A new field of mathematical research has emerged in knot theory,

Transcription

1 ABSTRACT A STUDY OF PROJECTIONS OF -BOUQUET GRAPHS A new field of mathematical research has emerged in knot theory, which considers knot diagrams with missing information at some of the crossings. That is, the observer does not know which strand lies over or under the other at a crossing, and this new type of crossing is known as a precrossing. Pseudodiagrams are knot-like diagrams that contain precrossings and crossings, while projections are knot-like diagrams that only contain precrossings. The trivializing number (or knotting number, respectively) of a pseudodiagram is the number of precrossings that need to be changed to a crossing to obtain a diagram that represents the unknot (or a nontrivial knot, respectively) regardless of how the remaining precrossings are resolved. Spatial graph theory is a subfield of knot theory that focuses on embeddings of graphs in three-dimensional space. In this thesis, we extend the concepts of trivializing and knotting numbers to spatial graph theory, where we focus on -bouquet graphs. Specifically, we calculate the trivializing and knotting number for projections and pseudodiagrams of -bouquet spatial graphs based on the number of precrossings and the placement of the precrossings in the pseudodiagram of a spatial graph. Elaina Aceves May 016

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3 A STUDY OF PROJECTIONS OF -BOUQUET GRAPHS by Elaina Aceves A thesis submitted in partial fulfillment of the requirements for the degree of Master of Arts in Mathematics in the College of Science and Mathematics California State University, Fresno May 016

4 APPROVED For the Department of Mathematics: We, the undersigned, certify that the thesis of the following student meets the required standards of scholarship, format, and style of the university and the student s graduate degree program for the awarding of the master s degree. Elaina Aceves Thesis Author Carmen Caprau (Chair) Mathematics Tamás Forgács Mathematics Oscar Vega Mathematics For the University Graduate Committee: Dean, Division of Graduate Studies

5 AUTHORIZATION FOR REPRODUCTION OF MASTER S THESIS I grant permission for the reproduction of this thesis in part or in its entirety without further authorization from me, on the condition that the person or agency requesting reproduction absorbs the cost and provides proper acknowledgment of authorship. X Permission to reproduce this thesis in part or in its entirety must be obtained from me. Signature of thesis author:

6 ACKNOWLEDGMENTS I would like to thank my advisor, Dr. Carmen Caprau, for introducing me to knot theory and for all of her time, support, and contagious enthusiasm for mathematics. I also want to thank Dr. Tamás Forgács for convincing me to change my major to mathematics from engineering (sorry Uncle Luke). Additionally, I want to thank Dr. Oscar Vega for introducing me to mathematical research and becoming my biggest fan at all of my presentations. Everyone in the math department has been so supportive and I will always be grateful for being a part of such a great group of people. Lastly, a big thanks to my family for always being there for me through all of the math you had to endure, even if you did just smile and nod.

7 TABLE OF CONTENTS Page LIST OF TABLES vi LIST OF FIGURES vii INTRODUCTION Mathematical Knots Pseudodiagrams of Knots Rigid Vertex Embeddings of Spatial -Bouquet Graphs Examples of Trivializing and Knotting Numbers INITIAL EXPLORATION SINGLE STACK OF PRECROSSINGS FINITE NUMBER OF STACKS OF PRECROSSINGS ADDITIONAL RESULTS INVOLVING PRETZEL PROJECTIONS.. 35 PRIME -BOUQUETS Trivializing and Knotting Numbers of Prime -Bouquets Weighted Resolutions Set CONCLUSION REFERENCES APPENDIX

8 LIST OF TABLES Page Table 1. Trivializing and Knotting Numbers of -Bouquets of Type K 43 Table. Trivializing and Knotting Numbers of -Bouquets of Type L 44

9 LIST OF FIGURES Page Figure 1. The Reidemeister Moves Figure. Pseudodiagrams Figure 3. Pseudodiagrams that are Trivial, Knotted, or Neither... 3 Figure 4. The Extended Reidemeister Moves Figure 5. Trivial -Bouquets of Type K and Type L Figure 6. Vertex with Labels Figure 7. Pretzel Projection (n 1, n,..., n k ) Figure 8. Pretzel Projection (1) Figure 9. Resolutions of (1) Figure 10. Pretzel Projection () Figure 11. Knotted Resolution of () Figure 1. Resolutions of () Figure 13. Alternate Version of the RV Move Figure 14. Proof of a Case of the Alternate Version of the RV Move. 14 Figure 15. Two Stacks Figure 16. Diagram T (p, q) Figure 17. Labeled Edges on a Vertex Figure 18. R + and R Figure 19. K-trivial Diagram Figure 0. L-trivial Diagram Figure 1. The Projection 5 k Figure. The Projection 5 l Figure 3. The Extended Pseudo-Reidemeister Moves

10 Page Figure 4. Proof of Invariance under PRII move Figure 5. Proof of Invariance under PRIII move Figure 6. The Projection 4 k Figure 7. Prime -Bouquets of Type K up to Six Crossings Figure 8. Prime -Bouquets of Type K up to Six Crossings (continued) 55 Figure 9. Prime -Bouquets of Type L up to Six Crossings viii

11 INTRODUCTION Mathematical Knots A knot K is an embedding of a circle into R 3. A diagram of a knot K is a projection of K onto a plane. Knot diagrams contain double points which illustrate which strand lies over or under the other when they interact. This type of double point is known as a crossing. Two knots are ambient isotopic if and only if their diagrams are related by a finite sequence of the Reidemeister moves given in Figure 1. RI: RII: RIII: Figure 1. The Reidemeister Moves A knot can be represented by more than one diagram. However, any two diagrams representing the same knot are related via a finite sequence of the Reidemeister moves. A knot is called trivial, or unkotted, if any diagram of it can be transformed via a finite sequence of the Reidemeister moves to a simple closed loop in a plane. Otherwise, a knot is called nontrivial, or knotted. We note that there is only one trivial knot up to ambient isotopy, which we call the unknot. A more detailed introduction to knots can be found in [1] and [5].

12 Pseudodiagrams of Knots For application purposes, one may want to allow the possibility that no information is known about which strand lies over the other at a double point in a diagram. We refer to this type of double point as a precrossing. A projection P is a knot diagram without over/under information at every double point. That is, all of the double points in a projection P are precrossings. A pseudodiagram Q of a knot is a projection P in which over/under information may be known at some of the precrossings of P. To be precise, a pseudodiagram can contain both crossings and precrossings. With these definitions in place, all projections are pseudodiagrams but not all pseudodiagrams are projections. Resolving a precrossing in a projection or a pseudodiagram is the action of replacing the precrossing with a crossing of either type. By resolving a precrossing of a pseudodiagram Q, we obtain two new pseudodiagrams Q 1 and Q, one for each way that we can obtain a crossing from a precrossing. A resolution of Q is a knot diagram that is obtained by resolving all of the precrossings of Q into crossings. In Figure, diagrams (a) and (b) are both pseudodiagrams but only diagram (a) is a projection. A pseudodiagram Q is trivial if any resolution of Q represents the trivial knot. Conversely, a pseudodiagram Q is knotted if any resolution of Q represents a nontrivial knot. Notice that a pseudodiagram can be trivial, knotted, or neither. In Figure 3, pseudodiagram (a) is trivial, (b) is knotted, and (c) is neither.

13 3 (a) (b) Figure. Pseudodiagrams (a) (b) (c) Figure 3. Pseudodiagrams that are Trivial, Knotted, or Neither We note that the above definitions apply to projections P, as well. The trivializing number of a knot projection P (if it exists), denoted tr(p ), is the minimum number of precrossings needed to be replaced by a crossing in order to obtain a trivial pseudodiagram. The knotting number of a knot projection P (if it exists), denoted kn(p ), is the minimum number of precrossings needed to be replaced by a crossing in order to obtain a knotted pseudodiagram. If the trivializing or knotting number does not exist,

14 4 we simply denote it as. For further information about trivializing and knotting numbers for knot projections, see Hanaki s work []. Remark 1. Given a pseudodiagram P with tr(p ) = n, if we resolve k precrossings into crossings, where k < n, there will always be a way to resolve the remaining n k precrossings to obtain a nontrivial pseudodiagram. Otherwise, the trivializing number would be k instead of n. Similarly, given a pseudodiagram P with kn(p ) = n, if we resolve k precrossings into crossings, where k < n, there will always be a way to resolve the remaining precrossings to obtain a trivial pseudodiagram. The majority of this work is based on ideas from Hanaki s paper [], which focuses on calculating the trivializing and knotting numbers for projections of knots. The main results from [] are listed below: Theorem ([], Theorem 1.6). For any knot projection P, the trivializing number of P is even. Theorem 3 ([], Proposition 1.9). For any non-negative even number n and any positive integer l 3, there exists a knot projection P with tr(p ) = n and kn(p ) = l. Hanaki s paper also includes results about the type of projections that are required to achieve equality between the number of precrossings and the trivializing or knotting numbers. We will adapt Hanaki s approach to projections of spatial -bouquet graphs instead of projections of knots.

15 5 Rigid Vertex Embeddings of Spatial -Bouquet Graphs A spatial graph is an embedding of a graph in R 3, while a diagram of a spatial graph G is a projection of G onto a plane. One of the simplest graphs to investigate are the bouquet graphs, which are graphs with one vertex and only loops as edges. The -bouquet graph is the graph with one vertex and two loops, as shown in Figure 5 on page 7. Throughout this thesis, we will refer to the loops of the -bouquet graphs as petals. We consider only rigid-vertex embeddings of graphs. Specifically, we regard a spatial graph as an embedding in R 3 of a 4-valent graph whose vertices have been replaced by rigid disks. Each disk has four strands attached to it, and the cyclic order of these strands is determined via the rigidity of the disk. Two spatial graphs, G 1 and G, with rigid vertices are called ambient isotopic if there exists an orientation-preserving homeomorphism of R 3 onto itself that maps G 1 to G. It is well-known that G 1 and G are ambient isotopic if and only if there is a finite sequence of extended Reidemeister moves transforming a diagram of G 1 into a diagram of G. The extended Reidemeister moves are depicted in Figure 4; the solid dot in moves RIV and RV represent the 4-valent vertex of the -bouquet graph. These moves introduce an equivalence relation on the diagrams, and as a consequence, we can view a spatial graph as the equivalence class of a spatial graph diagram. We refer the reader to Kauffman s work [4, 5] for more details on rigid-vertex embeddings of graphs.

16 6 RI: RII: RIII: RIV: RV: Figure 4. The Extended Reidemeister Moves We call two spatial graph diagrams equivalent if one can be transformed into the other via a finite sequence of the extended Reidemeister moves. Specifically, equivalent diagrams belong to the same equivalence class. In this paper, we restrict our attention to rigid-vertex embeddings of the -bouquet graph, and we call such an object a -bouquet for short. Moreover, we will consider pseudodiagrams and projections of -bouquets. There are two trivial rigid-vertex embeddings of the -bouquet in R 3, up to cyclic order of the edges meeting at the vertex, as shown in Figure 5 on page 7. We will refer to these as the unknotted -bouquet of type K and

17 the unknotted -bouquet of type L, respectively (see [6]). For simplicity, we will also refer to these as the trivial -bouquet of type K or of type L type K type L Figure 5. Trivial -Bouquets of Type K and Type L To identify a -bouquet diagram as type K or type L, we exit the vertex along a given edge. As we travel along the petal, we will return to the vertex at a different edge. If the two edges are adjacent, we have a -bouquet diagram of type K. Otherwise, we have a -bouquet diagram of type L. More precisely, we can label the edges around the vertex as in Figure Figure 6. Vertex with Labels If we exit the vertex via the edge labeled 1 and by traveling along the petal we return to the vertex via the edge labeled or 4 (the edge labeled 3, respectively), the diagram is of type K (type L, respectively).

18 8 We call a pseudodiagram of a -bouquet K-trivial (or L-trivial) if any diagram obtained by resolving all of its precrossings is equivalent to the standard diagram of the unknotted -bouquet of type K (or type L). Otherwise, a pseudodiagram of a -bouquet is called knotted. The trivializing number of a pseudodiagram Q of a -bouquet (if it exists), denoted tr(q), is the minimum number of precrossings one needs to replace by a crossing to obtain a K-trivial or L-trivial pseudodiagram. The knotting number of a pseudodiagram Q of a -bouquet graph (if it exists), denoted kn(q), is the minimum number of precrossings one needs to replace by a crossing to obtain a knotted pseudodiagram. If the trivializing or knotting number does not exist, we denote it as. It is clear from the previous definition that the trivializing and knotting numbers are non-negative integers or infinity. We will frequently consider the diagram shown in Figure 7, which we will refer to as a pretzel projection, where n i N for all 1 i k and where k is finite positive integer. The leftmost dashed circle contains only a 4-valent vertex, and the remaining k dashed circles from left to right contain n 1, n,..., n k precrossings stacked vertically. We will denote such a diagram as (n 1, n,..., n k ) n 1 n n k n i Figure 7. Pretzel Projection (n 1, n,..., n k )

19 9 Remark 4. We want to emphasize that only one of the dashed circles contains the 4-valent vertex. Also, the location of the 4-valent vertex among the k dashed circles does not matter, since we can always use planar isotopy to transform a diagram into another diagram where the 4-valent vertex is in the leftmost circle. Throughout the remainder of the thesis, we will refer to the n i precrossings that are stacked vertically in each dashed circle as a stack for each i. We will also label the first (top) precrossing in the stack of n i precrossings as p i,1. Similarly, we write p i,ni for the last (bottom) precrossing in the stack of n i precrossings. Remark 5. To simplify notation, we will use tr(n 1, n,, n k ) and kn(n 1, n,, n k ) to denote the trivializing and knotting numbers, respectively, of the pretzel projection (n 1, n,, n k ). We proceed to our first result. Lemma 6. Given a pretzel projection (n 1, n,..., n k ) of a -bouquet, at most one n i is even. Proof. We prove this result by contradiction. To this end, we suppose that n i and n j are both even for some i j. Without loss of generality, let i < j. Notice that we have another component in our diagram that joins stacks n i and n j by traveling through the top right hand corner of p i,1, the bottom right hand corner of p i,ni, through the stacks n i+1,, n j 1, through the bottom left hand side of p j,nj, and the top left hand side of p j,1. Since we have an additional component in union with our -bouquet, we do not have a

20 10 projection of a -bouquet. Thus, we can only allow one of our stacks of precrossings to contain an even number of precrossings. Examples of Trivializing and Knotting Numbers In this subsection, we provide two examples of projections of -bouquets and their trivializing and knotting numbers. More specifically, we consider pretzel projections of the form (n 1 ), for some n 1 N. Example 7. Figure 8 is the pretzel projection (n 1 ) with n 1 = 1: (1) = Figure 8. Pretzel Projection (1) After we resolve the precrossing in one of the two possible ways, we obtain the following diagrams in Figure 9. Figure 9. Resolutions of (1) Notice that both diagrams represent the unknotted -bouquet of type L. Thus, we have that tr(1) = 0 and kn(1) =. Example 8. The next pretzel projection that we consider is (n 1 ) with n 1 = which is shown in Figure 10 on the next page.

21 11 () = Figure 10. Pretzel Projection () First, we calculate the trivializing number of the diagram (). To begin, note that tr() 0 because a nontrivial diagram can be obtained by resolving the precrossings in the manner shown in Figure 11. Figure 11. Knotted Resolution of () If tr() = 1, we only need to resolve a single precrossing so that any resulting pseudodiagram obtained by resolving the remaining precrossing is trivial. Since diagram () is symmetrical, the precrossing we choose to resolve is arbitrary and the manner in which we resolve the precrossing is arbitrary as well. Thus, without loss of generality, we will choose the top precrossing to be resolved by the crossing where the overstrand will extend from the top left hand side to the bottom right hand side of the double point. Once we have done so, the following two pseudodiagrams in Figure 1 are obtained by resolving the remaining precrossing.

22 1 Figure 1. Resolutions of () Notice that the first diagram is nontrivial, since the two petals are linked together. On the other hand, the second diagram is K-trivial via a Reidemeister II move. Consequently, we must have that tr() =, because we need the precrossings to be resolved in such a way so as to allow one petal to lie above the other. This ensures that we can perform a Reidemeister II move to unlink the two petals, which results in a trivial -bouquet. Using similar reasoning, but instead attempting to force all the resulting diagrams to represent a knotted -bouquet, we obtain that kn() =.

23 INITIAL EXPLORATION In this section, we begin the exploration of calculating the trivializing and knotting numbers of our diagrams. Lemma 9. Let P be a projection of a -bouquet. Then the trivializing number of P is even. Proof. Recall that our goal with calculating the trivializing number is to find the minimum number of precrossings that we need to change to crossings, in order for our diagram to become a trivial pseudodiagram. Thus, we need to resolve our precrossings in such a way so that there exists a sequence of extended Reidemeister moves transforming our diagram into a trivial pseudodiagram. First, we consider the Reidemeister I move. If we want to use the Reidemeister I move, we need to have a precrossing in P that has two adjacent strands joined together. Regardless of how we resolve the corresponding precrossing (we only have two resolutions), we can always use a Reidemeister I move to simplify the diagram. Thus, we do not need to resolve the precrossing in a certain manner to make the diagram trivial. As a result, utilizing the Reidemeister I move does not affect our trivializing number. Next we consider the Reidemeister II move. To implement this move, we need to resolve two precrossings in P that are adjacent to each other (and involve two parallel strands of the projection) in such a way as to force one strand to be above the other. Therefore, the trivializing number will increase by two every time we apply a Reidemeister II move.

24 14 Consider the Reidemeister III move. To accomplish a Reidemeister III move, we need to resolve two precrossings that correspond to the strand that we are attempting to slide over or under the remaining crossing. As a result, the trivializing number will increase by two when we implement a Reidemeister III move. Using similar reasoning, the trivializing number will increase by two for an RIV move as well, to ensure that the strand will slide over or under the 4-valent vertex involved in the move. Lastly, we will consider the RV move. The RV move can be depicted as in Figure 4, as well as its equivalent representation given in Figure 13. RV: Figure 13. Alternate Version of the RV Move Indeed, Figure 14 shows that the first two diagrams depicted in Figure 13 are equivalent via an RIV move and a Reidemeister II move. The other cases of the alternate version of the RV move are verified similarly. RIV RII Figure 14. Proof of a Case of the Alternate Version of the RV Move

25 15 To employ the RV move, we need to resolve two precrossings in P so that two of the adjacent strands exiting the vertex involved in the move will become overstrands at the crossings while the remaining two adjacent strands exiting the vertex will become understrands at the crossings. Thus, to apply the RV move, the trivializing number will increase by two. Regardless of which one of the extended Reidemeister moves we employ, the trivializing number will either remain the same or increase by two. Consequently, the trivializing number is even. Remark 10. The trivializing number for projections of knots and -bouquets are both even by Theorem and Lemma 9. Before we can move on to the next result, we need to define what it means for a stack to be knotted, when it contains both precrossings and crossings. For our purposes, a stack must contain at least two double points, whether these are precrossings or crossings, and must be formed by linking exactly two parallel strands through the use of the double points. A stack of crossings and precrossings will be considered a knotted stack if, regardless of how the precrossings of the stack are resolved, the two strands are linked and cannot be separated with a series of Reidemeister II moves. Example 11. Of the two stacks in Figure 15, only the left stack is knotted, because regardless of how the precrossing is resolved, the two strands will always be linked together. On the other hand, we can resolve the two precrossings in the right stack with the resolution where the overstrand of the resulting crossings will have negative slope, so as to allow two consecutive

26 16 Reidemeister II moves (starting in the middle of the diagram) to separate the two strands. Thus, the right stack is not knotted, because there exists a resolution of the precrossings that results in the separation of the two strands. Figure 15. Two Stacks Now, we can proceed to our next result. Lemma 1. Given a pretzel pseudodiagram Q = (n 1, n,, n k ) where we allow crossings as well as precrossings in Q, if any stack of Q is knotted, then the pseudodiagram Q is knotted. Proof. Assume that the stack of Q that contains n i double points is knotted. In particular, n i. If n i is even, then the two petals only interact in that particular stack of n i double points. Since the stack containing n i double points is knotted, the two petals are linked together regardless of how the remaining precrossings are resolved, forcing Q to be knotted. Next, we consider the case when n i is odd. In an attempt to force our diagram to be trivial, we can resolve the remaining precrossings in the other stacks to create as many instances of the Reidemeister II move as we see possible. However, with the rigidity of the 4-valent vertex and the way in

27 17 which we construct our pretzel diagrams, a Reidemeister II move cannot be implemented by using one crossing in the stack of n i double points and another crossing in an adjacent stack, unless both stacks only contain a single crossing. Since the stack of n i double points contains at least two double points, we do not have a diagram where a Reidemeister II move can be applied between two adjacent stacks. As a result, the stack of n i double points will remain knotted after all of the remaining precrossings in the other stacks have been resolved. Since n i is odd and the stack of n i double points remains knotted regardless of the way the remaining precrossings are resolved, the two petals are linked together within the stack of n i double points. Therefore, we have that Q is knotted. Before we can progress to the next result, we will define the figure T (p,q) as the diagram in Figure 16 which contains p and q precrossings where p, q 0. T (p, q) = p q Figure 16. Diagram T (p, q) Lemma 13. Let P be a pretzel projection of a -bouquet and let Q be a pseudodiagram obtained from P by resolving some of the precrossings of P into crossings. Then tr(q) = 0 if and only if Q = (1) or Q = T (p, q) for some p, q 0.

28 18 Proof. ( ) We know that tr(1) = 0 because (1) is L-trivial by Example 7. Notice that any resolution of the diagram T (p, q) will result in multiple applications of the Reidemeister I move, so we have that T (p, q) is K-trivial. Hence, both (1) and T (p, q) have a trivializing number of 0 as diagrams of type L and type K, respectively. ( ) Let P = (n 1, n,, n k ) be a pretzel projection and Q be a pseudodiagram obtained from P by resolving some of the precrossings of P. Suppose that tr(q) = 0. By Lemma 1, we know that if one of the stacks is knotted, then the entire diagram is knotted. Since tr(q) = 0 and two precrossings will already provide a nontrivial diagram (see Example 8), we must have that each stack contains one precrossing, or every stack contains a single precrossing except for one stack which does not contain any precrossings to obtain diagrams that are trivial of type L and type K respectively. If n i = 1 for all i, then we have a single horizontal stack of precrossings. If i = 1, then we have the diagram (1) which is L-trivial, so Q = (1) is a possible diagram with tr(q) = 0. However, if i, we can resolve all of the precrossings in the same manner and have no instances where the Reidemeister II move can be applied. This resolution is a knotted diagram, contradicting the fact that tr(q) = 0. As a result, the diagram (1) is the only possible trivial diagram in this case. If n j = 0 for some 1 j k while n i = 1 for all i j, we have the diagram T (p, q) for some p, q 0 which we know to be K-trivial. Therefore, we must have that Q = (1) or Q = T (p, q) for some p, q 0.

29 19 For the next proof, we will require an additional definition. The mirror image diagram of a -bouquet diagram D is the diagram D obtained from D by changing overcrossings into undercrossings and vice-versa, for all crossings of D. The precrossings of D will remain unchanged. Lemma 14. Let P be a projection of a -bouquet. Then kn(p ). Proof. We prove the statement using contradiction. Suppose first that kn(p ) = 0. If P contains zero precrossings, then P = (0), which we know to be K-trivial. If P has a single precrossing, we can have either the diagram (1) or T (p, q) with p = 0 and q = 1 (or vice versa), which we know are trivial. Thus, P must contain at least two precrossings. However, we can resolve the precrossings to force one petal to lie above the other, forcing the diagram to be trivial. This is similar to the method of unknotting a knot via the crossings-change operation so as to obtain an ascending knot diagram. With the assumption that kn(p ) = 0, at least one resolution of P will always be trivial, a contradiction. Therefore, kn(p ) 0. Suppose now that kn(p ) = 1. If kn(p ) = 1, only one precrossing needs to be resolved in order to obtain a knotted pseudodiagram. Let Q be the knotted pseudodiagram obtained from P by resolving a precrossing, call it c, into a crossing. Now, let Q be the pseudodiagram obtained from Q by changing the type of the newly created crossing. Let D be a diagram obtained from Q by resolving all of the precrossings in Q. Notice that the mirror image diagram D can be obtained from Q, and thus D is knotted. Since the mirror image of a knotted spatial graph is also knotted, we have that D is a knotted -bouquet. Thus, Q is knotted. Since both Q and Q are knotted, we

30 0 do not need to resolve the precrossing c in any particular way. This implies that kn(p ) = 0, which we already know is a contradiction. Therefore, kn(p ).

31 SINGLE STACK OF PRECROSSINGS In this section, we begin calculating the trivializing and knotting number of our pretzel projections by considering a single stack of precrossings. Before we can perform the calculations, we need to determine which -bouquet diagrams are of type K and of type L. Lemma 15. Given a pretzel projection P = (n), P is of type K if and only if n is even and is of type L if and only if n is odd. Proof. To avoid repetition, we prove only one implication of each of the statements since the other implication follows similarly. We label the edges around the vertex as shown in Figure Figure 17. Labeled Edges on a Vertex Case 1: n is even. Without loss of generality, let the first petal begin by exiting the vertex along edge. We have that the edge labeled enters the precrossing p 1,1 through the top left hand strand. Since n is even, the edge labeled will pass through all of the precrossings and exit the precrossing p 1,n through the bottom left hand strand. Thus, edge returns to the vertex through edge 3. Since the edge labeled and 3 are adjacent to each other, we have that P is of type K. Case : n is odd. Using similar reasoning as in the previous case, the edge labeled will exit the vertex via p 1,1 through the top left hand strand,

32 pass through all of the precrossings, and exit precrossing p 1,n through the bottom right hand strand, since n is odd. Therefore, strand returns to the vertex via strand 4. Since strands and 4 are not adjacent to each other, we have that P is of type L. In an effort to simplify our explanations when calculating trivializing and knotting numbers, we will introduce the following notation. We assign R + to the crossing where the overstrand has a positive slope, and R to the crossing where the overstrand has a negative slope. R + = R = Figure 18. R + and R Proposition 16. Given a pretzel projection P = (n), where n N 0 = N {0}, we have that: n tr(p ) =. Proof. Suppose first that n is even. Then P is of type K and we need to reduce our pretzel projection into Figure 19. (0) = Figure 19. K-trivial Diagram Since n is even, we have that n = k for some k N 0 and k 0. If we resolve k precrossings with R and k precrossings with R +, we can implement

33 3 k Reidemeister II moves. Thus we have reduced our diagram to (0), which we know to be K-trivial. We conclude that tr(p ) n. Now, we need to show that tr(p ) = n. To do that, we must show tr(p ) n, which is the next possible value since the trivializing number is even, by Lemma 9. We can resolve k precrossings to perform k 1 Reidemeister II moves. We have two precrossings left, but we can resolve them using two R + crossings and the resulting diagram is nontrivial. As a result, tr(p ) n because we can resolve the precrossings in such a way as to obtain a nontrivial diagram. We conclude that tr(p ) = n, when n is even. Suppose next that n is odd. Then P is of type L, and we need to reduce our pretzel projection into Figure 0. Figure 0. L-trivial Diagram Since n is odd, we have that n = k + 1, for some k N 0 and k 0. Using similar reasoning as the previous case, we can reduce our diagram to (1) by implementing k Reidemeister II moves. Since (1) is L-trivial by Example 7, we have that tr(p ) n 1. To show that tr(p ) = n 1, suppose that we resolved k precrossings so that we can apply k 1 Reidemeister II moves. There are three precrossings left, but we can resolve each of them using R + and create a knotted diagram. Thus, we have that tr(p ) = n 1, when n is odd. The result follows by combining the two cases.

34 Now we calculate the knotting number of a single stack of precrossings. Proposition 17. Given a pretzel projection P = (n), where n N 0, we have 4 that: if n = 0, 1 kn(p ) = n + 1 if n > 1. Proof. If n = 0, we know that tr(0) = 0, since (0) is K-trivial. As a consequence, we must have that kn(0) =. If n = 1, then tr(1) = 0, since (1) is L-trivial, which forces kn(1) =. We now consider the case n > 1. Suppose first that n is even. Then n = k, for some k N 0, and P is of type K. Without loss of generality, if we resolved k + 1 precrossings with R, even if we maximized the number of Reidemeister II moves by resolving k 1 precrossings with R +, we can apply at most k 1 Reidemeister II moves. Then regardless of how we resolve the remaining precrossings, we have at least two R crossings which results in a nontrivial diagram, implying that kn(p ) n + 1. Now, we need to show that kn(p ) = n + 1 = k + 1. Suppose that we resolved k precrossings with R. To create the most Reidemeister II moves, we can resolve the remaining k precrossings with R +, to apply k Reidemeister II moves. By doing so, we reduce our diagram to (0), which we know to be K-trivial. As a consequence, kn(p ) = n + 1. Suppose next that n is odd. Then n = k + 1, for some k N 0, and P is of type L. If we resolved n + 1 precrossings with R, even if we maximized the number of Reidemeister II moves by resolving n 1 precrossings with R +, we would have at most n 1 Reidemeister II moves.

35 5 Therefore, we have at least three R crossings, which results in a nontrivial diagram, implying that kn(p ) n + 1. To show that kn(p ) = n + 1, suppose that we resolved n precrossings with R. To create the most Reidemeister II moves, we can resolve the remaining n precrossings with R + and apply n Reidemeister II moves. By doing so, we reduce our diagram to (1), which we know to be L-trivial. Thus, kn(p ) = n + 1. Therefore, regardless of the parity of n, where n > 1, we have that kn(p ) = n + 1.

36 FINITE NUMBER OF STACKS OF PRECROSSINGS Now that we have finished calculating the trivializing and knotting numbers for pretzel projections that contain a single stack of precrossings, we can proceed by considering pretzel projections that contain a finite number of stacks. In this section, we will make the additional assumption that given a diagram (n 1, n,, n k ), n i, for all 1 i k. Lemma 18. Given a pretzel projection P = (n 1, n,, n k ), the diagram is of type K if and only if one n j is even or all n i are odd and k is even. The diagram is of type L if and only if all n i are odd and k is odd. Proof. We prove only one implication of each of the statements to avoid repetition, as the other implication follows similarly. We prove the lemma by cases and use the same labeling of the edges at the vertex as in Lemma 15. Without loss of generality, let the first petal begin by exiting the vertex along edge. Case 1: The stack n j is even for some 1 j k. We have that the edge labeled enters the jth stack either via the top left hand corner of precrossing p j,1 or in the bottom left hand corner of precrossing p j,nj. Since n j is even, the strand through the top left hand corner of precrossing p j,1 is connected to the strand in the bottom left hand corner of precrossing p j,nj. When strand exits the jth stack, it will connect with edge 3 upon entering the vertex. Since edges and 3 are adjacent to each other, we have that P is of type K. Case : Suppose that all n i are odd and k is even.

37 7 Since n 1 is odd, strand will travel through the first stack of precrossings and exit the stack through the bottom right hand strand of precrossing p 1,n1 and enter the second stack through the bottom left hand strand of precrossing p,n. Since n is odd, edge will travel through the second stack of precrossings and exit the stack through the top right hand strand of precrossing p,1. This process will continue through the k stacks of our diagram. Since k is even, strand will exit the final stack through the top right hand strand of precrossing p k,1. Hence, the edge labeled will enter the vertex through the edge labeled 1. Since these edges are adjacent to each other, we have that P is of type K. Case 3: Suppose that all n i are odd and k is odd. Using the same reasoning as in the previous case, we know that strand will exit the last stack of precrossings through the bottom right hand strand of precrossing p k,nk. Thus, strand will enter the vertex through strand 4. Since strands and 4 are not adjacent to each other, we have that P is of type L. Theorem 19. Given a pretzel projection P = (n 1, n,, n k ) where n i for each i, the trivializing number of P is given by k i=1 n i ( k ) tr(p ) = n i 1 if k is odd and each n i is odd i=1 n j + if k is even and each n i is odd k (n i 1) if one n j is even for some 1 j k. i=1,i j

38 8 Proof. We prove this theorem by taking cases. Case 1: If k is even and each n i is odd, then P is of type K. By Lemma 1, we know that if one of the n k is knotted, the entire diagram is knotted. Recall that even a stack of two precrossings contains a nontrivial diagram, so we must perform as many Reidemeister II moves as possible in each stack of precrossings to result in a trivial pseudodiagram. Since each stack contains an odd number of precrossings, after we perform all of the Reidemeister II moves in each stack of precrossings, we have one precrossing remaining in each stack. We have reduced our diagram to the diagram (k). Since k is even, by Proposition 16, we have that tr(k) = k. Since we were forced to resolve every precrossing to obtain diagram (0) which is K-trivial, k tr(p ) = n i. i=1 Case : If k is odd and each n i is odd, then P is of type L. With the same reasoning as in the previous case, we perform all of the Reidemeister II moves in each vertical stack of precrossings, and obtain the diagram (k). Since k is odd, Proposition 16 states that tr(k) = k 1. Thus we have reduced our diagram to (1), which we ( know to be L-trivial. Since we did not k ) resolve the last precrossing, tr(p ) = n i 1. Case 3: If n j is even for one 1 j k, then P is of type K. Using the same reasoning as in the previous case, we will resolve pairs of precrossings in each stack to create as many instances of the Reidemeister II move as we see possible. Notice that this eliminates all of the precrossings in the n j stack to ensure that the two petals are not linked together. Then we have reduced our i=1

39 9 diagram to T (p, q) for some p, q 0. Since the diagram T (p, q) is K-trivial, k tr(p ) = n j + (n i 1). i=1,i j Theorem 0. Given a pretzel projection P = (n 1, n,, n k ) where n i for each i, the knotting number of P is given by kn(p ) = min { n1 + 1, n + 1,, nk } + 1. Proof. By Lemma 1, we know that if one of the stacks of precrossings is knotted, the entire diagram is knotted. Thus, we require the minimum number of precrossings to be resolved, so as to ensure that one of the stacks of precrossings is knotted. By Proposition 17, we know that the knotting number of a stack of n i precrossings is n i + 1. The result follows. Theorem 1. Given a pretzel projection P = (n 1, n,, n m, n m+1,, n k ) where m 1, n i = 1 for all 1 i m, and n j for all m + 1 j k, the trivializing number of P is given by m + tr(p ) = m 1 + n j + k i=m+1 n i k i=m+1 k i=m+1,i j n i if all n i are odd, m and k have the same parity if all n i are odd, m and k have different parity (n i 1) if one n j is even. Proof. We prove this theorem by cases. Case 1: Let k be even and all of the n i be odd. Since each n i is odd, by k Theorem 19, tr(n m+1, n m+,, n k ) = n i. In an attempt to make our i=m+1

40 30 k diagram trivial, we resolve the n i precrossings as described in i=m+1 Theorem 19 to eliminate the n m+1, n m+,, n k precrossings. As a result, we have reduced our diagram to (m). By Lemma 16, we know that diagram (m) has trivializing number m if m is even, and m 1 if m is odd. Case : Let k and all of the n i be odd. Using similar reasoning as in the previous case, Theorem 19 implies that ( k ) tr(n m+1, n m+,, n k ) = n i 1. i=m+1 ( k ) When we resolve the n i 1 precrossings as described in Theorem 19, i=m+1 the n m+1, n m+,, n k precrossings are reduced to a single precrossing. Overall, our diagram has been transformed into (m + 1). By Lemma 16, we know that diagram (m + 1) has trivializing number m + 1 if m is odd, and m if m is even. Case 3: Let one n j be even for some 1 j k. Since one n j is even, Theorem 19 states that k tr(n 1, n,, n k ) = n j + (n i 1). i=m+1,i j k After we resolve the n j + (n i 1) precrossings according to i=m+1,i j Theorem 19, we have reduced our diagram to T (p, q) for some p, q 0. By Lemma 13, we know that the diagram T (p, q) is K-trivial.

41 The result follows by combining all of the cases together, according to the parity of m and k. 31 Theorem. Given a pretzel projection P = (n 1, n,, n m, n m+1,, n k ) where m 1, n i = 1 for all 1 i m, and n j for all m + 1 j k, the knotting number of P is given by kn(n m+1,, n k ) if m k+1 or one n j is even kn(p ) = min { } kn(n m+1,, n k ), k+ if m k+ and P is of type K min { } kn(n m+1,, n k ), k+3 if m k+3 and P is of type L Proof. So far, when calculating the knotting number of a diagram, we obtained the results by only considering the stacks of the diagram. However, we also need to consider whether it is possible to reduce our knotting number by resolving some, or all, of the m initial precrossings in our diagram. Consider the case when one n j is even for some m + 1 j k. We want to prove that regardless of how we resolve the initial m precrossings in our diagram, a K-trivial diagram can be created by resolving the remaining precrossings. By contradiction, suppose that we resolve the initial m precrossings and the resulting diagram is knotted. We will resolve the remaining precrossings in the n m+1, n m+,, n k stacks to create the most possible Reidemeister II moves in each stack. This will ensure that the two strands that interact in the jth stack will not be linked and that each stack in our diagram is not knotted. Subsequently, we have transformed our diagram into the diagram T (p, q) for some p, q 0. Since T (p, q) is K-trivial by Lemma 13, we have a contradiction to our diagram being knotted. Therefore,

42 the m initial precrossings have no effect in obtaining a knotted diagram and only by knotting a stack of P will we obtain a knotted diagram. Thus, kn(p ) = kn(n m+1, n m+,, n k ), when one n j is even for some m + 1 j k. For the remainder of this proof, we assume that n l is odd for all m + 1 l k. Consider the case when m k+1. Similarly to the previous case, we will use a proof by contradiction. The most knotted diagram that we can obtain with m precrossings is created by replacing all of the precrossings with crossings of the same type. Without loss of generality, we resolve all of the m initial precrossings with R + and, towards a contradiction, suppose that the resulting diagram is knotted. We resolve the remaining precrossings in the stacks of P by creating as many Reidemeister II moves as possible in each stack and resolve the remaining precrossing in each stack with R. After applying all of the possible Reidemeister II moves, our diagram has been reduced to a diagram containing m (k m) = m k crossings of type R +. 3 Notice that ( ) k + 1 m k k = k + 1 k = 1. Therefore, we have at most a single instance of R + if P is of type L, and zero instances of R + if P is of type K. Regardless of the type, we have created a trivial diagram. This contradicts the fact that our diagram is knotted. As a result, the only way to create a knotted diagram is to create a knotted stack in P. Hence, kn(p ) = kn(n m+1, n m+,, n k ) when m k+1. Next consider the case when P is of type K and m k+. By Lemma 15, k is even, so k+ N. We claim that k+ is the knotting number of P.

43 33 Without loss of generality, resolve k+ many precrossings of the m initial precrossings with R +. In an attempt to make the diagram trivial, we can resolve the remaining precrossings of P by creating as many Reidemeister II moves in each stack as possible. Furthermore, all of the remaining single precrossings will be resolved with R. After completing all other instances of the Reidemeister II moves between the R + and R precrossings, we have exactly two R + crossings, as seen by the calculations below: k + ( k k + ) = k + k + k + = k + k = We know by Example 8 that the diagram containing two R + crossings is knotted. Thus, if we resolve k+ many precrossings, we can ensure that P will always be knotted, regardless of how we resolve the remaining precrossings. Therefore, we have that { kn(p ) = min kn(n m+1,, n k ), k + }. Lastly, consider the case when P is of type L and m k+3. By Lemma 15, k is odd, so k+3 N and we want to show that kn(p ) = k+3. Similar to the previous case, without loss of generality, resolve k+3 many precrossings of the m initial precrossings with R +. The remaining precrossings will be resolved to create as many Reidemeister II moves within each stack and all other precrossings will be replaced with R. After all of the Reidemeister II moves have been completed, we will have exactly three R + crossings remaining, as seen in the following calculations.

44 34 k + 3 ( k k + 3 ) = k + 3 k + k + 3 = k + 3 k = 3 Note that the diagram containing three R + is knotted, so if we resolve k+3 many precrossings, we can ensure that P is knotted. Therefore, we conclude that { kn(p ) = min kn(n m+1,, n k ), k + 3 }.

45 ADDITIONAL RESULTS INVOLVING PRETZEL PROJECTIONS In previous sections, we were given a pretzel projection and then we calculated the trivializing and knotting numbers. In this section, we will instead start with a number and attempt to create a pretzel projection with that number as its trivializing or knotting number. Proposition 3. For any non-negative even number n, there exists a pretzel projection P with tr(p ) = n. Proof. By Proposition 16, we know that the pretzel projection P = (n) has tr(p ) = n, since n is even. Proposition 4. For any n N, where n, there exists a pretzel projection P with kn(p ) = n. Proof. By Theorems 0 and, we have that the pretzel projection P = (1, (n 1)) has knotting number equal to n, since (n 1) kn(p ) = + 1 = n = n. The next result establishes the relationship between the trivializing and knotting numbers. Hanaki stated that the trivializing and knotting numbers for knot projections are independent of each other (see Theorem 3). However, in our case of projections of -bouquets, we find that the trivializing number depends on the knotting number of the projection.

46 36 Proposition 5. For any n N, where n, there exists a pretzel projection P with kn(p ) = n and tr(p ) = l, where l = (n 1) or l = 4(n 1) + (k 1), for some k N 0. Proof. Consider the pretzel projection P 1 = ((n 1)). Using Proposition 17, (n 1) kn(p 1 ) = + 1 = n = n. Moreover, Proposition 16 states that tr(p 1 ) = (n 1). For the pretzel projection P = ((n 1) 1, (n 1) + k 1), where k N 0, we use Theorem 0 to conclude that (n 1) 1 kn(p ) = + 1 = n = n = n. For all k N 0, by Theorem 19, we have that tr(p ) = (n 1) 1 + (n 1) + k 1 = 4(n 1) + (k 1). This completes the proof. For the remaining results in this section, we introduce an additional definition. Given a pretzel projection P, let p(p) be the number of precrossings in P. Proposition 6. Given a pretzel projection P, tr(p ) = p(p ) if and only if P = (n 1, n,, n k ), where k is even and each n i is odd, or

47 37 P = (n 1, n,, n m, n m+1,, n k ), where m 1, n i = 1 for all 1 i m, n j for all m + 1 j k, and m and k have the same parity. Proof. ( ) We have this implication from Theorems 19 and 1. ( ) Suppose tr(p ) = p(p ). Then P = (n 1, n,, n k ) or P = (n 1, n,, n m, n m+1,, n k ), where m 1, n i = 1 for all 1 i m, and n j for all m + 1 j k. First consider the case P = (n 1, n,, n k ). Since tr(p ) is even, p(p ) must be even. Thus, in the case when all n i are odd, we must have an even number of stacks for p(p ) to be even, which implies that k must be even. Then k tr(p ) = n i = p(p ), by Theorem 19. If one n j is even for some 1 j k, we know i=1 k tr(p ) = n j + (n j 1) < i=1,i j k n i = p(p ), i=1 by Theorem 19. Then in the case that one n j is even, we have that tr(p ) p(p ). Therefore, tr(p ) = p(p ) when P = (n 1, n,, n k ), where all n i are odd and k is even. Consider the case P = (n 1, n,, n m, n m+1,, n k ) where m 1, n i = 1 for all 1 i m, and n j for all m + 1 j k. First notice that, if exactly one n j is even for some m + 1 j k, then by Theorem 1, k tr(p ) = n j + (n j 1) < m + i=1,i j k n i = p(p ). i=1

48 38 Thus, tr(p ) p(p ) when one n j is even. Next, suppose that all n i are odd. If m and k have different parity, we have that p(p ) is odd. This particular case must be discarded, since the trivializing number is always even, by Lemma 9. Thus, we have that tr(p ) p(p ) when m and k have different parity. If m and k have the same parity, by Theorem 1, we have that tr(p ) = m + k n i = p(p ). i=1 Hence, tr(p ) = p(p ) when P = (n 1, n,, n m, n m+1,, n k ), where m 1, n i = 1 for all 1 i m, and n j for all m + 1 j k, and where each n i is odd, and m and k have the same parity. Proposition 7. kn(p ) = p(p ) if and only if P = () = (1, 1) or P = (3) = (1, 1, 1). Proof. ( ) This implication is a consequence of Proposition 17. ( ) Suppose kn(p ) = p(p ). Then P = (n 1, n,, n k ) or P = (n 1, n,, n m, n m+1,, n k ), where m 1, n i = 1 for all 1 i m, and n j for all m + 1 j k. First, we consider the case when P = (n 1, n,, n k ). Then kn(p ) = min { n1 + 1, n + 1,, nk } + 1, by Theorem 0. If k, we conclude that min { n1 + 1, n + 1,, nk } + 1 = ni + 1