KKNOT COLORINGS: COLORING AND GOERITZ MATRICES
SUDIPTA KOLAYA
BSTRACT . Knot colorings are one of the simplest ways to distinguish knots, dating back to Reide-meister, and popularized by Fox. In this mostly expository article, we discuss knot invariants likecolorability, knot determinant and number of colorings, and how these can be computed from ei-ther the coloring matrix or the Goeritz matrix. We give an elementary approach to this equivalence,without using any algebraic topology. We also compute knot determinant, nullity of pretzel knotswith arbitrarily many twist regions.
1. I
NTRODUCTION
The purpose of this note is to give an exposition of knot colorings through coloring andGoeritz matrices, and to discuss of n -colorings of pretzel knots. Although the coloring andGoeritz matrices can be combinatorially defined given a diagram of the knot, the classical proofsof this equality of their determinants, and related results in the literature [4, Chapter 9] requiresome background on the part of reader. We give an elementary proof of these results withoutusing any algebraic topology. Our approach here is somewhat similar to that of [7] (buildingon [3]), although more elementary. While most of this article is expository, the results (and ourmethods of computation) in the last section for the determinant and number of colorings ofpretzel knots with arbitrarily many twist regions are new. It is hoped that this article would behelpful to people at all levels interested in learning about knots and knot colorings.A knot is a smooth embedding of a circle S in three-space R . Here are two examples:F IGURE
1. The trefoil and the figure eight knotsThe fundamental question in knot theory is given two knots, can we tell them apart? One ofthe simplest ways of telling knots apart is tricolorability , whether we can non-trivially (i.e. usingat least two colors) color the knot with three colors so that at any crossing the three strandscoming together has either all the same colors, or three different colors. This is the simplestinvariant which tells the above two knots, the trefoil and the figure eight knot, are different (notisotopic, meaning one cannot be deformed into another).The notion of tricolorability goes back at least to Reidemeister, at the beginning of the nine-teenth century. Ralph Fox generalized this notion to n -colorability, and popularized this way ofstudying knots. The property of being n -colorable (and related invariants, like total number of n -colorings) is a fairly powerful knot invariant, which can be developed with minimal amountof mathematical machinery, namely linear algebra.There is a related integer valued knot invariant called the determinant of a knot whose divi-sors n are exactly those for which the given knot is n -colorable. It is well known that this invari-ant determinant is the absolute value of the determinant of any coloring matrix of the knot andit is also the determinant of any Goeritz matrix for the knot. We will give an elementary proofof this result in Sections 6 and 7, after reviewing necessary background materials up to Section5. We will use ideas from Section 6 to set up an equivalent linear system for pretzel knots in thefinal section. a r X i v : . [ m a t h . G T ] O c t SUDIPTA KOLAY
Acknowledgements . The author would thanks John Etnyre for making helpful comments onearlier drafts of this paper. This work is partially supported by NSF grants DMS-1608684 andDMS-1906414. 2. K
NOTS AND LINKS
A link is a smooth embedding (or in other words an injective map which is differentiable ateach point) of disjoint union of circles in R (so a knot is just a link with one component). Givenany such embedding, we can orthogonally project everything with respect to the z -axis, and weget a projection of the knot in the xy -plane, which we will call the link diagram. Generically , wemay assume that the only non-embedded points in the diagram are isolated double points. Theadvantage of knot diagrams is that it is much easier to draw on a piece of paper and manipulateit. In fact, all the examples of knots and links given here (or any other paper based format) is adiagram. While it is clear that given any link in R we have a diagram for it, one may wonderif we can say when two diagrams represent the same knot. The answer is yes, as proven byAlexander-Briggs [1] and independently Reidemeister [6].Any two knot diagrams of the same link, up to planar isotopy, can be related by a sequenceof the following three moves (called the Reidemeister moves). Each move operates on a smallregion of the diagram and is one of three types (see Figure 2):(1) Twist (or untiwst) a strand.(2) Isotope a strand over (or under) another strand.(3) Isotope a strand completely over (or under) a crossing.F IGURE
2. Reidemeister movesThe above result shows it is enough to understand links diagrams to understand links. Moreover,if we can define an object from a diagram, and show it remains unchanged under any of theReidemeister moves, it follows that it is in fact an invariant of the link (and not the diagram).3. C
OLORING OF KNOTS AND LINKS
Now we are ready to introduce n -colorings of links. Given any knot diagram D , an n - coloring of a link is is an assignment to each strand an element of Z /n Z , such that whenever we havea crossing with the overstrand associated to y , and the understrands associated to x or z asillustrated by Figure 3, we have x + z = 2 y in Z /n Z .F IGURE
3. Coloring equation: y = x + z . Exercise 1.
Show that given any knot diagram with such a coloring, if a Reidemeister move isapplied the new diagram has a unique coloring which is the same outside the small region wherethe move was applied. Conclude that an n -coloring is in fact an invariant of the link (and notjust the diagram). It may happen that some knot diagram has triple (or higher order) crossings, however we can isotope some of thestrands so that the only crossings that remains are double points.
NOT COLORINGS: COLORING AND GOERITZ MATRICES 3 If n is small it is customary to color the strands with n distinct colors, and this is where thename colorability comes from. For example when n = 3 , we use three colors customarily red,blue and green; and -colorability is also known as tricolorability. When n is large, it is not F IGURE
4. Examples of a 3-coloring and a 5-coloringpractical to use different colors, and we simply use the labelling (perhaps this invariant couldalso have been called n -labelling) on each of the strands. Note that if we color (or label) eachstrand by constant element s ∈ Z /n Z , then all the above constraints are trivially satisfied. Such acoloring is called a trivial coloring, and any link has n such trivial n -colorings. Hence we will saya link is n -colorable if it has a non-trivial n -coloring (because otherwise every link has a coloringand this notion would not be able to distinguish between links). We will see in the next sectionthat the set of all n -colorings (including the trivial ones) has additional structure, and this isalso an invariant (see Exercise 5). Had we removed the trivial colorings from this collection, theremaining set does not canonically get such a structure.For the rest of this article, we will restrict to the case of knots, because some of the statementsthat follow would get very technical otherwise. The technical issues come from the fact that onehas more trivial colorings for split links (i.e. links which can be split apart to lie inside disjointthree balls in R ), and to get appropriate count of number of different colorings we need todivide out by the number of trivial colorings. All of the statements hold for non-split links (i.e.links which not split links). If we had a split link, we could separate them into different regionsand work out coloring invariants for each of the separate non-split links. This would suggest thatthere is no loss of generality to restricting to the case of non-split links, except for the caveat thatgiven a link diagram, it is not easy to determine if it is a diagram of a split link or not; or if it issplit, how many components it has. 4. C OLORING MATRIX
Notice that the constraints we had at each crossing for n -colorings is a linear equation, whichsuggests that we can consider all such equations together and use tools from linear algebra togain a better understanding of what is going on.Given a knot diagram D with c crossings, we have exactly c strands (with the only exceptionbeing the standard diagram of the unknot, which has one strand and no crossings). For eachstrand let us assign a variable x i , and for each crossing we can write down an equation x i + x k =2 x j , which has to be satisfied if the x i ’s gave rise to a valid coloring. If we consider the set oflinear equations, and write out the matrix form, we get ˜ C(cid:126)x = (cid:126) , where ˜ C is a c × c matrix, and (cid:126)x is a column vector with i -th entry x i . We define ˜ C to be the pre-coloring matrix for the diagram D . Example . In this example we work out the pre-coloring matrix of the figure eight knot, seeFigure 5. The system of equations we obtain are: x − x + x = 0 − x + x + x = 0 x − x + x = 0 x + x − x = 0 SUDIPTA KOLAY F IGURE
5. Coloring equations for the figure eight knot.Hence the pre-coloring matrix ˜ C equals − − − − . Notice that we made some choices, we could have labelled the strands in a different order,or permuted the equations in the linear system, and we would have obtained a different pre-coloring matrix.We observe that the set of all possible n -colorings is nothing but the solution space of theabove equation reduced modulo n , i.e. the Null Space of the pre-coloring matrix, and as suchgets the structure of a vector subspace. Exercise 3.
Go back to your solution of Exercise 3 and check that the linear structure of the setof all n -colorings is preserved under each of the Reidemeister moves. Conclude that the solutionspace is an invariant of the knot. We note that the sum of the columns of these matrices are (cid:126) , and this is in fact true more gen-erally (possibly with a signed sum, depending on if we write the equation as x i + x k − x j = 0 or − x i − x k + 2 x j = 0 ). In other words the constant vector (cid:126)s (with all entries s ) is a solution tothe system ˜ C(cid:126)x = (cid:126) , and we will call such solutions semitrivial . Hence if (cid:126)x is any solution, (cid:126)x + (cid:126)s isagain a solution for any scalar s . We can fix this redundancy, by requiring one of the x i ’s is zero.Since the pre-coloring matrix always has a non-trivial solution, it is singular, we know that thereis some linear dependence among the rows as well. So let us delete some row and some columnfrom the pre-coloring matrix, and we will call this resulting square matrix C a coloring matrix for the diagram D . While the matrix C is not an invariant (even for the knot diagram D ), it turnsout the absolute value of the determinant of C is an invariant of the knot (not just a diagram!),which we will call the determinant of the knot.We digress to some linear algebra to see why various cofactors of pre-coloring matrices arethe same. Recall, for any square matrix A , the ( i, j ) cofactor C i,j is ( − ij times the determinantof the matrix A i,j obtained by removing the i -th row and j -th column from A. Proposition 4.
Suppose A is an n × n matrix with real entries (or more generally with entries insome commutative ring) so that sum of all the columns is the column vector of all zeros, and sumof all the rows is the row vector of all zeros (equivalently, the sum of entries of each row and eachcolumn add up to zero). Then all the ( i, j ) cofactors of A are the same. Exercise 5.
Work out the following outline to prove the above proposition:
NOT COLORINGS: COLORING AND GOERITZ MATRICES 5 (1) For any such matrix A , show that you can reconstruct A if you know the submatrix A i,j ,and the indices i, j .(2) Using (cid:126)R + ... + (cid:126)R n = (cid:126) , and properties of the determinant, relate C , and C , .(3) Relate C , with C ,j .(4) Relate C , with C i,j . In order to use this result from linear algebra, we need to know that for an arbitrary knot dia-gram we can choose a pre-coloring matrix in the specified form, just as we saw in the example.
Exercise 6.
Given any knot diagram D , show that one can number the crossings and strands insuch a way so that hypothesis of Proposition 4 holds. Another approach is to find a modified version of Proposition 4 which can be applied to anypre-coloring matrix.
Exercise 7.
Suppose A is an n × n matrix with real entries so that a linear combination (witheach weight being ± ) of every column is the column vector of all zeros, a linear combination(with each weight being ± ) of every row is the row vector of all zeros. Then all the absolute valuesof cofactors (or equivalently, minors) of A are the same. The aforementioned results from linear algebra tell us that the determinant is indeed an in-variant of the diagram, but we still need to justify that it does not depend on the diagram. Oneway (which is commonly used in the literature) to go about this is to show that the knot determi-nant is invariant under the Reidemeister moves. We will take a slightly different approach here,by invoking the fact (as seen in the last section) that n -colorability is a knot invariant. Proposition 8.
The following are equivalent:(i) A pre-coloring matrix ˜ C has a non-semitrivial solution in Z /n Z .(ii) A coloring matrix C has a non-trivial solution in Z /n Z .(iii) det C = 0 in Z /n Z .(iv) n divides det C .Proof. Suppose C is obtained from ˜ C by deleting the i -th row and j -th column. ( i ) ⇐⇒ ( ii ) : Given a solution (cid:126)x of the pre-coloring matrix ˜ C , we can add a constant vector (cid:126)s to get another solution (cid:126)y of ˜ C with y j = 0 (note that this operation preserves if the solutionvector was semitrivial or not). The column vector obtained by deleting y j from (cid:126)y is a solution tothe coloring matrix C , and moreover one can go backwards by adding to the j -th spot. Thisprocess describes a bijection between all solutions of ˜ C with j -th entry zero, and all solutions of C , and further the non-semitrivial solutions of ˜ C correspond exactly to the non-trivial solutionsof C . ( ii ) ⇐⇒ ( iii ) is a standard fact in linear algebra; and ( iii ) ⇐⇒ ( iv ) follows from the definition ofthe quotient set Z /n Z . (cid:3) Suppose D and D are two diagrams for a knot K , with coloring matrices C and C . Thenfor every n dividing det( D ) , we see by Proposition 6 that the diagram D is n -colorable, andhence (since n -colorability is a knot invariant) the diagram D is n -colorable as well, and againby Proposition 6 we have n divides det( D ) . So we see that det( D ) divides det( D ) , and bysymmetry det( D ) divides det( D ) . Thus the absolute values of det( D ) and det( D ) are thesame, so knot determinant is invariant of knot.The determinant of knot K determines for which n , the knot K is n -colorable. However thereader may have noticed that in order to find the knot determinant directly from definition, weneed to compute the determinant of a matrix of size ( c − × ( c − , which in some cases maynot be computationally efficient. In the next sections, we will talk about an alternate, easierapproach to compute the knot determinant and related invariants.5. G OERITZ MATRIX
Given a knot diagram we can checkerboard color the regions, for example see Figure 7. Sup-pose we assign to each crossing c the sign η ( c ) ∈ {± } according to Figure 6:Suppose the shaded regions are enumerated R , ..., R k . Let us define the k × k pre-Goeritzmatrix ˜ G for the diagram D by the following: For the off diagonal entries ( i (cid:54) = j ), we set ˜ G i,j := (cid:88) η ( c ) , SUDIPTA KOLAY -1+1 F IGURE
6. Signs at crossings. Note that these signs are independent of an ori-entation on the knot, but will flip if we switch the shaded and unshaded regionsin the checkerboard coloring.where the sum ranges over all crossings c where regions R i and R j come together. Let us nowdefine the diagonal entries by ˜ G i,i := − (cid:88) j (cid:54) = i ˜ G i,j . We now note that ˜ G is a square symmetric matrix whose columns (respectively rows) sum tozero. R R R -1 -1-1 -1-1 -1 F IGURE
7. A knot diagram with a checkerboard coloring.
Example . The pre-Goeritz matrix for the knot diagram in Figure 7 with the given checkerboardcoloring is: − − − − − − Just as we saw earlier for the pre-coloring matrices, the determinant of ˜ G is zero, however ifwe form a Goeritz matrix by deleting any row and any column, the absolute value of determinantthe resulting matrix is well defined for the diagram, which we will call the Goeritz determinant of the diagram. We get a similar result regarding the Goeritz matrices as we had for the coloringmatrices, and the same proof works.
NOT COLORINGS: COLORING AND GOERITZ MATRICES 7
Proposition 10.
Let us pick any pre-Goeritz matrix ˜ G for a knot diagram D , and obtain a Goeritzmatrix G by deleting some row and some column. The following are equivalent:(i) ˜ G has a non-semitrivial solution in Z /n Z .(ii) G has a non-trivial solution in Z /n Z .(iii) det G = 0 in Z /n Z .(iv) n divides det G . As the reader can probably expect, the Goeritz determinant does not depend on the knot di-agram. In fact, we will show it is exactly the same as the determinant of the knot (coming fromcoloring matrix)! This result also means that determinant of a Goeritz matrix obtained from theshaded regions is same in absolute value to the one coming from the unshaded regions. Whileit is not too difficult to see that the Goeritz determinant is invariant under the Reidemeistermoves, it is not quite clear from this perspective if the determinants corresponding to the un-shaded and shaded regions are related, or if they are related to the determinant of a coloringmatrix. In the next two sections, we will describe another approach where we will create a bi-jection between the solution spaces of the pre-coloring and pre-Goeritz matrices, and this willgive us the above result about determinants, and a bit more.6. D
IFFERENCE
Let us note that if we have a colored knot diagram with a collection of half twists as illus-trated in Figure 8, the difference between the colors on top and the bottom is the same for every x y yz z x F IGURE
8. Differencegeneric vertical slice (i.e. does not pass through a crossing or touch the knot diagram), sincethe coloring equation is equivalent to x i − x j = x j − x k . In other words, if we drew any genericarc between the regions R and R in the part of the diagram illustrated above, and took the dif-ference of the colors, then we would get the same value. One may wonder if something similarworks for any two crossing-adjacent shaded (or unshaded) regions. Indeed, this will be the casemore generally (that is for non adjacent regions, even with different checkerboard colors), oncewe formulate the right generalization for this difference.Given a knot diagram D with regions R , ..., R r (ignoring checkerboard coloring for now), ofa knot K with an n -coloring, let us define the difference d ( R i , R j ) ∈ Z /n Z between any tworegions by taking any generic (i.e. not passing through a crossing, and intersecting D trans-versely) oriented simple arc between points in the interior of regions R i and R j , and computingthe alternate (beginning with positive) sum of the colors on the strand the arc crosses. Claim 11.
The difference d ( R i , R j ) is well defined, that is it does not depend on the arc we choseto compute the alternating sum. We sketch two proofs of this well definedness below, after we discuss an example.
Example . Let us work out the differences in 5-coloring of the figure eight knot we saw earlier,see Figure 9. Notice that for the arc γ between regions R and R , there are three intersectionpoints, and the alternating sum is − − in Z / Z , where as there is only one By two curves touching at a point we will mean there is a point of intersection where the tangent lines agree.
SUDIPTA KOLAY intersection for the arc γ and here the alternating sum in 4. Let us assume Claim 11 for nowand find the various values of the differences d ( R , R j ) for different j . d ( R , R ) = 2 , d ( R , R ) = 3 , d ( R , R ) = 0 , d ( R , R ) = 4 , d ( R , R ) = 4 . Exercise 13.
Compute the other differences d ( R i , R j ) for the knot coloring in Figure 9. F IGURE
9. Differences in a coloring of the figure eight knot.
First proof of Claim 11.
Note that in order to show this alternating sum is same for any two arcswith same endpoints, it is equivalent to show the alternating sum along any generic closed loopin the plane is zero (concatenate one arc and the other’s reverse). Let us begin by consideringa generic simple closed loop γ (i.e. no self intersections, it can intersect D ). We can think of D together with the loop γ as being a diagram of a link of two components, one of them beingthe original knot K , and the other knot lying entirely below K . The latter is easily seen to bethe unknot (as the diagram has no crossings), unlinked from K , by the way we chose crossings.Hence we can isotope the unlinked unknot away from the original knot diagram, we can colorit arbitrarily by some color. Since n -colorability is an invariant of the link, we know that the linkdiagram we started off with also has an n -coloring. Suppose the color of the strand of this newunknot in some region R k is c , if we use the coloring equations check that the color we end upwith once we traverse along γ is the sum of c and the alternating sum of the colors the γ crosses,and so this alternating sum has to be zero. Now if γ was not simple, if γ is a subloop of γ whichis simple then by our above discussion we see that the alternating sum along γ is zero. Hence,we can we can start deleting innermost loops from γ without changing the alternating sum, andhence we get the alternating sum along any generic closed loop is zero. Note that we can varythe endpoints of the arcs in the same region by concatenating with an arc that does not cross D , and hence the alternating sum remains the same. Hence, it follows that d ( R i , R j ) is welldefined. (cid:3) Second proof of Claim 11.
An alternate approach to check well definedness would be to use thefact that any two such arcs are related by the following two moves: going over a crossing, andencountering a birth/death (think about the non-generic instances and then the various wayswe can perturb them to get to a generic arc), as illustrated in Figure 10. We leave it to the readerto verify that the alternating sum remains the same under these moves. (cid:3)
Let us now consider some checkerboard coloring of the knot diagram. For any triple i, j, k ;we can concatenate an arc from R i to R j ; and an arc from R j to R k to obtain an arc from R i to R j . It follows that for any triple i, j, k , the differences satisfy d ( R i , R j ) + d ( R j , R k ) = d ( R i , R k ) if R i and R j have same checkerboard coloring, and d ( R i , R j ) − d ( R j , R k ) = d ( R i , R k ) if R i and R j have different checkerboard coloring.Conversely, if we are given such a collection { d ( R i , R j ) } for all pairs of regions satisfying theabove relations, then we obtain an n -coloring on the knot K : pick any two regions adjacentalong a strand of the knot, and the difference between the two regions tells us the coloring onthe strand. In other words, the data of an n -coloring is equivalent to the data of the differences. NOT COLORINGS: COLORING AND GOERITZ MATRICES 9 x y yz xx F IGURE
10. Moves relating a pair of isotopic arcs, when the knot diagram re-mains fixed.Since there are relations among the various d ( R i .R j ) (cid:48) s , there is some redundancy; we notethat having d ( R i .R j ) (cid:48) s for adjacent regions is enough to recover the coloring. In fact, as we willsee in the next section, to recover the coloring (up to translation by constant colorings), it isenough to only know the differences d ( R i .R j ) (cid:48) s among the crossing-adjacent shaded regions;and the nullspace of the (pre-)Goeritz matrix stores the information of the differences in a com-pact way. 7. B IJECTIONS BETWEEN SOLUTIONS OF COLORING AND G OERITZ MATRICES
In this section we demonstrate a bijection between solutions of the coloring matrix and Goeritzmatrix of a diagram D . Suppose the shaded regions of D are enumerated R , ..., R s . Suppose wehave an element (cid:126)v in the nullspace of the pre-Goertiz matrix ˜ G , i.e. ˜ G(cid:126)v = (cid:126) . Let us define d ( R i , R j ) := v j − v i , and we see that d ( R i , R j ) + d ( R j , R k ) = v j − v i + v k − v j = v k − v i = d ( R i , R k ) . If these numbers d ( R i , R j ) are differences coming from an actual coloring (this is true aswe shall see) of the knot diagram, if we know a color on one of the strands, we can use thesedifferences to figure out the colors on all the other strands of the knot diagram. We will use thisobservation and assign a color to one strand, and given the numbers d ( R i , R j ) , we will assigncolors on all the other strands, and verify that this actually gives rise to a valid coloring. Just likein our discussion of the difference, we will need to make some choices and we need to checkthat they are all give consistent coloring on a strand.To this end, let us define an auxiliary knot diagram to be a modification of a knot diagramwhere we break up the overstrands at each crossing, and we will draw a small rectangle at eachcrossing to keep track of which strand was the overcrossing, see Figure 11.We will define an auxiliary n -coloring on a knot diagram to be an assignment of colors toeach of strands of the auxiliary knot diagram, where at each crossing if the overstrands are la-beled w and y ; and the understrands are labelled x and z , then w + y = x + z , i.e. the overstrandsneed not have the same labelling, see Figure 12. If it turns out that at each crossing both the seg-ments of the overstrands have the same labelling, then this auxiliary coloring gives us an actualcoloring of the knot diagram. Note that we can define the difference for an auxiliary knot col-oring exactly the same way we did for a knot coloring, and the second proof of Claim 11 carriesover to show that this difference is well defined.Let us pick a strand α in the auxiliary knot diagram and color it x . It will be part of theboundary of exactly one shaded region, let us call it R . As we go around the boundary of R in the counterclockwise direction, if we come across a crossing c with the other shaded regionbeing R j ; we will add η ( c ) d ( R , R j ) to the coloring of the subsequent strand, see Figure 13. We F IGURE
11. Auxiliary knot diagram for the knot diagram in Figure 7.F
IGURE
12. Auxiliary coloring equation: w + y = x + z . -1+1 R R R R F IGURE
13. Assigning colors to each strand bounding a region in the auxiliaryknot diagram.need to make sure that when we go all the way across and come back to the strand α we stillget the color x . Note that the color we get by going all the way across is x + (cid:88) c is a crossing of R η ( c ) d ( R , R j ( c ) ) , where R j ( c ) denotes the other shaded region which has crossing c . We observe that (cid:88) c crossing of R η ( c ) d ( R , R j ( c ) ) = (cid:88) c crossing of R η ( c )( v − v j ( c ) ) = k (cid:88) j =2 ˜ G ,j ( v j − v ) NOT COLORINGS: COLORING AND GOERITZ MATRICES 11 = ( k (cid:88) j =2 − ˜ G ,j ) v + k (cid:88) j =2 ˜ G ,j v j = ˜ G , v + k (cid:88) j =2 ˜ G ,j v j = k (cid:88) j =1 ˜ G ,j v j = ˜ G(cid:126)v = (cid:126) since (cid:126)v is in the null space of the pre-Goeritz matrix ˜ G .We can do the exact same thing for each of the regions by picking one strand and arbitrarilyassigning a color, and coloring all the other strands by adding η ( c ) d ( R i , R j ) to the coloring as wepass across a crossing c with shaded regions R i and R j , and we obtain an auxiliary coloring ofthe knot diagram.Let us now choose a collection of s − crossings joining the s shaded regions R , ..., R s , i.e.given any i and j , we can draw an arc from a point in the interior of region R i to the a point inthe interior of region R j so that the arc passes between regions only in arbitrarily small neigh-borhoods of the chosen crossings. Starting with a color x on a strand α in the region R , weobtain an auxiliary n -coloring on the entire knot diagram by choosing to color the overstrandsat the chosen crossings by the same element of Z /n Z , and extending this to each of the strandspartially bounding a region by the procedure explained in the preceding two paragraphs. Byconstruction, the colorings on each of the overstrands agree on the chosen s − crossings, andwe show below they agree on every other crossing as well, thereby giving us an actual knot col-oring.Let us denote by δ ( R p , R q ) the difference between the regions R p and R q of this auxiliary n -coloring. Suppose c is any crossing in the knot diagram, with the shaded regions R i and R j incident on it. We have an arc γ from a point in the interior of region R i to the a point in theinterior of region R j , traversing the regions R l = R i , R l , ..., R l r = R j , so that it goes betweenregions R l k and R l k +1 in an arbitrarily small neighborhoods of the chosen crossings. Then wehave δ ( R i , R j ) = r − (cid:88) k =1 δ ( R l k , R l k +1 ) (by choosing the arc γ to compute the difference) = r − (cid:88) k =1 d ( R l k , R l k +1 ) (by construction of the auxiliary coloring) = d ( R i , R j ) (telescoping sum since d ( R p , R q ) = v q − v p ).It follows that at the crossing c the colorings on the overstrands agree, and since c is arbitrary,this is true at any crossing. Thus, we have obtained a valid knot coloring with its difference forshaded regions δ ( R i , R j ) agreeing with d ( R i , R j ) we defined earlier. We should note that thisimplies we would obtain the same knot coloring independent of the choice of which of the s − crossings we chose earlier. Also, if instead of coloring the strand α with x , we colored α withthe color x + a , then the above procedure would give a new coloring where every strand wouldget the color a plus the original coloring.Conversely, we note that if we start with a valid coloring, we can look at the differences d ( R i , R j ) among the shaded regions, and they would satisfy the Goeritz relations (cid:88) c crossing of R i η ( c ) d ( R i , R j ( c ) ) = 0 for each region R i , which in turn would correspond to a solution of the pre-Goeritz matrix, oncewe choose a value for a region, say R .Hence, under this procedure semitrivial (respectively non-semitrivial) solutions of the pre-Goeritz matrix correspond to the semitrivial (respectively non-semitrivial) solutions of the pre-coloring matrix. More concretely, we have: Theorem 14.
Suppose we have a knot diagram D for a knot K with c crossings and s shadedregions, let us choose a strand α of the knot diagram partially bounding a shaded region R .Let’s suppose we delete the first column and first row of the pre-coloring matrix ˜ C (respectivelypre-Goeritz matrix ˜ G ) to obtain coloring matrix C (respectively Goeritz matrix G ). Then for any n ∈ N there is a bijection among the collection of:(i) Non-trivial solutions ˜ v ∈ ( Z /n Z ) c of ˜ C with first entry 0.(ii) Non-trivial solutions v ∈ ( Z /n Z ) c − of C .(iii) Non-trivial solutions w ∈ ( Z /n Z ) s − of G .(iv) Non-trivial solutions ˜ w ∈ ( Z /n Z ) s of ˜ G with first entry 0. Moreover the above bijection between Nul( C ) and Nul( G ) is linear, and so we get an isomorphismof these null spaces of Z /n Z -modules (for n prime, this means a vector space isomorphism). Inparticular, for prime n , the mod n nullity (i.e. dimension of the null space) of both these matricesare equal. Exercise 15.
Complete the proof of the above theorem by checking linearity of the bijection be-tween the solution spaces.
For a prime n , the mod n nullity of either the coloring or Goeritz matrix completely deter-mines how many n -colorings there are of a knot K , and is an invariant of the knot, which wewill refer to as the n - nullity of K . We remark that if n is composite, then knowing the cardinalityof Nul( C ) is not enough to understand the structure of Nul( C ) as a Z /n Z -module. But when n isprime, Nul( C ) is a vector space, and we completely understand it once we know the dimension.We note a few consequences of Theorem 14 (combined with earlier propositions):(1) n divides det C ⇐⇒ C has a non-trivial solution in Z /n Z ⇐⇒ K has a non-trivial n -coloring G ⇐⇒ G has a non-trivial solution in Z /n Z ⇐⇒ n divides det G .(2) The Goeritz determinant is an invariant of the knot, and moreover equals the coloringdeterminant.(3) For any n , the null-space of a Goeritz matrix as a Z /n Z -module is a knot invariant, whichis isomorphic to the null space of any coloring matrix of K .Thus, we can find the determinant of knot; the space of n -colorings determinant from a Goeritzmatrix, which is frequently smaller than a coloring matrix.8. C OLORINGS OF P RETZEL KNOTS
In this final section, we will use ideas from the last few sections and find determinant and n -nullity of pretzel knots. They have been previously computed pretzel knots with up to fourtwist regions [2, 5] using coloring matrices.Recall that a pretzel knot P ( q , q , ..., q m ) has m twist regions joined up as illustrated in Figure14, where there are q i (which can be both positive and negative) half-twists in the i -th region.See Figure 15 for the diagram of the pretzel knot P (3 , , − .F IGURE
14. Pretzel Knot P ( q , ..., q m ) .Note that for the pretzel knot P ( q , q , ..., q m ) the coloring matrix from the above diagramwill be of the order Q × Q , where Q = | q | + ... + | q m | , where as a Goeritz matrix will be a ( m − × ( m − matrix. It turns out that it seems it is easier to compute the determinantand nullity of pretzel knots from another linear system, constructed with the differences, asindicated below. An interested reader may work out determinant and n -nullity from a Goeritzmatrix (and even from a coloring matrix, if feeling particularly adventurous).Let d i denote the difference d ( R i , R i +1 ) between shaded regions R i and R i +1 . Note that foreach twist region if there are q half-twists, and the difference is d , then we can figure out thecolors on all the strands if we know a color on the leftmost strand. We observe that the differencebetween the colors of the top left (respectively right) and the bottom left (respectively right) NOT COLORINGS: COLORING AND GOERITZ MATRICES 13 strand in the i -th twist region is q i d i . Also, note that the top (respectively bottom) right strandof the i -th twist region is exactly the same as the top (respectively bottom) leftt strand of the ( i + 1) -th twist region. Thus, for adjacent twist regions we must have the increase in the verticaldirection must be the same, so we have q i d i = q i +1 d i +1 for all i . Moreover if we drew an archorizontally traversing just below (see figure) each of the twist region regions we see that − ( d + d + ... + d m ) = 0 , since the leftmost and rightmost strands are the same and must have thesame color. We illustrate the above discussion with the explicit example of of the pretzel knot P (3 , , − in Figure 15. F IGURE
15. Pretzel Knot P (3 , , − .Thus we obtain the following collection of linear equations in Z /n Z : d + d + ... + d m = 0 − q d + q d = 0 ... − q d + q m d m = 0 Let us write out the above system in matrix form (with d i ’s being the variables) A(cid:126)d = (cid:126) , wherethe matrix A is: . . . − q q . . . − q q . . . . . . . . .. . . . . .. . . . . . − q . . . q m Exercise 16.
For a pretzel knot diagram of P ( q , q , ..., q m ) , pick a strand and name it α . Givena solution (cid:126)d to the linear system described above, and any labelling of the strand α by some x ∈ ( Z /n Z ) , show that there is a unique extension to coloring on the entire knot diagram byusing the solution (cid:126)d . Moreover show that this defines a linear bijective correspondence betweenthe nullspace of the matrix A reduced modulo n , and the space of n -colorings of the knot diagram(i.e. nullspace of any coloring matrix). It follows that we can use the matrix A to compute the determinant and n -nullity of P ( q , q , ..., q m ) ,and this method seems to be the easiest way of computing them. Claim 17.
The determinant of the m × m matrix A is given by q q ...q m ( 1 q + 1 q + ... + 1 q m ) Proof.
We will prove the claim by induction on m . Exercise 18.
Check the base cases for m = 2 , . In order to compute the determinant of A , we use Laplace expansion along the last row. det A = ( − m +1 q det . . . q . . . q . . . . . . . .. . . . .. . . . . . . q m − + q m det . . . − q q . . . − q q . . . . . . . . .. . . . . .. . . . . . − q . . . q m − We observe that by expanding along the last column, we can compute the determinant of thefirst matrix easily since we get a diagonal matrix: det . . . q . . . q . . . . . . . .. . . . .. . . . . . . q m − = ( − m − det q . . . q . . . . . . .. . . .. . . . . . . q m − = ( − m − q ...q m − Inductively, we know what the second determinant in the above expansion is. det . . . − q q . . . − q q . . . . . . . . .. . . . . .. . . . . − q . . . q m − = q q ..q m − ( 1 q + 1 q + ... + 1 q m − ) Combining them we obtain det A = ( − m +1 q ( − m − q ...q m − + q m q q ...q m − ( 1 q + 1 q + ... + 1 q m − )= q q ...q m − + q q ...q m ( 1 q + 1 q + ... + 1 q m − ) = q q ...q m ( 1 q + 1 q + ... + 1 q m ) (cid:3) Let us now assume n is a prime and compute mod n nullity of A . We make a few observations: Claim 19.
If all the q i ’s are coprime to n then the mod n nullity of A is either 1 or 0, depending onwhether n divides det A = q q ...q m ( q + q + ... + q m ) or not.Proof. Note that the submatrix A , obtained by deleting the first row and first column is a fullrank diagonal matrix and so has mod n rank m − , since we are assuming n does not divide anyof q , ..., q n . It follows that the mod n rank of A is either m − or m (which by the rank nullitytheorem is equivalent to saying the mod n nullity is 1 or 0). The proof is completed by using theInvertible Matrix Theorem, that mod n nullity of A is 0 iff det A = 0 modulo n . (cid:3) Claim 20.
If some of the q i ’s are divisible by n , then the mod n nullity of A is the total number of q i ’s divisible by n , minus . If some of the q i ’s are divisible by n , we may assume n divides q (note that there is cyclicsymmetry for pretzel knots, P ( q , q , ..., q m − , q m ) is isotopic to P ( q , q , ..., q m , q ) ). In this case,the claim is equivalent to:If n divides q ,then the mod n nullity of A is the number of q i ’s (apart from q ) divisible by n . NOT COLORINGS: COLORING AND GOERITZ MATRICES 15
Proof.
When A is reduced modulo n , it becomes a upper triangular matrix, and the rank is thenumber of pivots (i.e. the number of q i ’s not divisible by n , plus 1 ), and the nullity is the numberof 0’s in the main diagonal (the number of ”other” q i ’s divisible by n ). (cid:3) Exercise 21.
Give a direct proof Claim 20 (without using that n -nullity is invariant of the knot),i.e. show the result is true when q is coprime to n . (Hint: Use row operations). To summarize, if none of the q i ’s are divisible by n , the pretzel knot can have at most onecoloring and the determinant determines whether there is one. In cases that some of the q i ’s aredivisible by n , for any n -coloring(1) the colors of all the strands in the i -th twist region will be the same when the corre-sponding q i is coprime to n ;(2) for those q i which are divisible by n , the colors of the strands in the i -th twist regioncan be different, and these sort of correspond to the free variables, except one has toremember that the colors of the very last such twist region is determined (by the coloringon the first twist region), and this constraint is where we get the ”minus 1” in the formulafor n -nullity. R EFERENCES[1] J.W. Alexander and G.B. Briggs. On types of knotted curves.
Ann. of Math. , 28 : 2 (1927/28) pp. 563–586.[2] K. Brownell, K. O’Neil and L. Taalman. Counting m-coloring classes of knots and links.
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The American MathematicalMonthly , Vol. 121, No. 6 (June–July 2014), pp. 506-514.[4] W. B. R. Lickorish. An Introduction to Knot Theory.
Springer GTM , 1997.[5] Robert Ostrander. P-Coloring of pretzel knots. Masters Thesis, 2014.[6] K. Reidemeister. Elementare Begrundung der Knotentheorie.
Abh. Math. Sem. Univ. Hamburg , 5 (1927) pp. 24–32[7] Lorenzo Traldi. Link colorings and the Goeritz matrix.
Journal of Knot Theory and Its Ramifications , Vol. 26, No. 08,2017.S
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