Homomorphisms of Knot Groups on Finite Groups

Homomorphisms of Knot Groups on Finite GroupsAuthor(s): Robert RileySource: Mathematics of Computation, Vol. 25, No. 115 (Jul., 1971), pp. 603-619+s1-s22Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/2005224 .

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MATHEMATICS OF COMPUTATION, VOLUME 25, NUMBER 115, JULY, 1971

Homomorphisms of Knot Groups on Finite Groups

By Robert Riley

Abstract. We describe trial and error computer programs for finding certain homomorphisms of a knot group on a special projective group LF(2, p), p prime, and programs to evaluate H1(=; Z) where M is a finitely sheeted branched covering space of S3 associated with such a homomorphism. These programs have been applied to several collections of examples, in particular to the Kinoshita-Terasaka knots, and we state numerous conjectures based on these experiments.

About forty years ago a universal method for obtaining algebraic invariants of knot type was proposed and became standard. The method, as applied to a knot k of type K and having group 7rK = 7r1(S3 -k; *), begins with the determination of the homomorphisms of wrK on a given group G. These homomorphisms fall into equivalence classes under the action of the automorphisms of G, and a crude pre- liminary invariant of K is the number of homomorphism classes. In the next stage of the method, we fix a transitive permutation representation of G, perhaps of infinite degree. Each homomorphism class of 7rK on G is associated with a covering space c% of S3 - k such that the number of sheets in the covering is the degree of the permutation representation, and the group H&(Ut; Z) is an algebraic invariant of the knot type K. In this paper, we shall discuss the means and results of implementing the universal method on a computer when the group G is chosen to be one of the special projective groups L, = LF(2, p) = PSL(2, p), where p is a prime integer.

The universal method has been most thoroughly examined in the case where the group G is cyclic. The determination of the homomorphism classes becomes completely trivial, and all the homology invariants can be deduced from a single matrix, the Alexander matrix. These "cyclic invariants" have been applied with good effect to just about every problem in knot theory, but, alas, when the Alexander polynomial A(x) of the knot reduces to the constant 1 these invariants degenerate and are worthless. It is no good choosing a solvable group for G in such a case; to get useful results from the universal method, we must use nonsolvable groups. Of course, simple groups receive first consideration in this context, and of the families of classical finite simple groups, the family IL,4 is the most manageable. As further encouragement for this study, the group L4 is isomorphic to the alternating group AS, and R. H. Fox has shown in several papers ([5] is a good example) that the homomorphisms of a knot group on A5 have some interesting applications. There are, however, very few clues to suggest what the best method for finding homomorphisms on the special projective groups could be. In addition, almost nothing is known about the homology invariants of homomorphisms on noncyclic groups G. It therefore seems reasonable

Received July 17, 1970, revised December 2, 1970. AMS 1969 subject classifications. Primary 5520. Key words and phrases. Homomorphisms of knot groups on finite groups, homology of covering

spaces.

Copyright ? 1971, American Mathematical Society

603



604 ROBERT RILEY

to explore the subject initially by applying the most humble methods to a large number of examples. This leads directly to the development of computer programs to do the work and to large tables of output data. These are the subject of this paper.

Our method for finding the homomorphisms of a knot group irK on a finite group G is simple trial and error. Suppose that irK is given to us in the form

(0) 7rK = IxI, * n, x r1(x) = r2(x)= a = i.

Let a = (al, * * *, aJ) be a set of elements of G which together generate G. The assignment xi -* ai, for i = 1, * * *, n, can be extended to a homomorphism 0: 7rK - G such that (x,)@ = ai if and only if r,(a) = r2(a) = = 1. Hence the determination of the homomorphisms (and later their arrangement into classes) is a finite process. But before we actually carry out this process even on a computer, we should find ways of reducing the number of experiments to the irreducible minimum. When the presentation (0) is an over-presentation (see [5] or [7]), the generators xi are all conjugate in 7rK. This is such a great aid to the classification that we use only over- presentations for knot groups. In Section 4, we discuss a version of the over- presentation that is especially well adopted to our methods and programs.

In Section 2, we determine the irreducible set of experiments needed to determine the homomorphism classes of 7rK on A, when n in (0) is 2 or 3. It turns out that there are three possible kinds of such homomorphisms, and that the search for one kind is much easier to describe and implement in a computer program than the others. The characteristic property of these homomorphisms is that the image of an over- generator of (0) is an element of order 5 in A,. In addition, we found that such homomorphisms are more numerous than the others for our test cases. Hence, when we generalize the A, case to L, in Section 3, we restrict ourselves to the homomorphisms which carry an over-generator of irK to an element of order p in L,. We call such homomorphisms "reps" for brevity, and note that the property of being a rep is invariant, that is, it does not depend on the over-presentation (0) used in the definition. In Section 5, we discuss how to implement the second stage of the universal method on a computer when G is an arbitrary finite group. Our tools are now ready.

In Section 6, we discuss the outcome of applying our programs to various collections of test cases. We have concentrated most of our effort on the groups G = L, where p < 11, with the greatest emphasis on p = 5 because the results are all new and the computation time per knot is still reasonable. The first set of test knots that we discuss are the prime knots with at most 9 crossing points, conveniently drawn on pp. 70-72 of Reidemeister [13]. We call these the classical knots and we have two tables in the microfiche section for them. The first table is of over-presentations for their groups which may be used as a standard reference by anyone doing computation in knot theory. The second table is of the invariants of these knots associated with their homomorphisms on A, = L. We state a large number of conjectures about the homology invariants associated with reps of an arbitrary knot group on L, that are backed up by numerous tables of data which are not all discussed here. We hope that the reader shares the author's faith that only a small amount of wit and dis- crimination is needed to extract good conjectures from modest amounts of numerical data in this subject. The homology invariants for a certain special sort of reps are also discussed in Conjecture B, and we believe that this may be the best clue to the



HOMOMORPHISMS OF KNOT GROUPS ON FINITE GROUPS 605

whole subject. The section ends with a discussion of three particular knots having Alexander polynomial A(x) 1, namely Seifert's knot, the 11-crossing Kinoshita- Terasaka knot [10], and a third 11 -crossing knot discovered by J. Conway [2]. These three knots are shown to have different types.

In Section 7, we demonstrate the power and limitations of our programs by trying them on an infinite collection of knots that can be described by two integral parameters. These knots were discovered by Kinoshita and Terasaka [10] and shown to be nontrivial with A(x) _ 1. We shall show that a homomorphism of the group of one of these knots on A, determines a corresponding homomorphism for the groups of an infinity of other knots whose integral parameters vary in certain arith- rnetic progressions. If we vary one of these integral parameters in its progression, we get interesting formulas for the corresponding homology invariants. If these formulas were proven (but it is not worth the effort to do this), they would show that these knots determine infinitely many types. As it is, we have only proven that they determine at least 200 types.

The work described in this paper has led to a later paper, "Parabolic representations of knot groups" that we are submitting for publication elsewhere. In it, we prove that many of the knot groups discussed here have representations on subgroups of LF(2, C), from each of which one can derive an infinity of reps on the groups L,.

I am indebted to Dr. G. Edmunds for his useful comments and constructive criticism, and to Professor W. Magnus for his kind encouragement of this project at a time when its continuation was in doubt.

1. Throughout this paper we shall use a standard notation for groups and homomorphisms taken from Huppert [9]. In particular, we write operators on the right so that products of permutations are read from left to right, when x, y are elements of a group xv 3 y-xy, and G = (x1, * , x") means that G is the group generated by x1,.* * *, X".

All knots and knot types discussed in this paper are tame. We shall use the notation k for a knot in S3, K for its isotopy type, 1?' for the isotopy type of a mirror image of k and K for the type of k, i.e. two knots k1, k2 in S3 have the same type if there exists an autohomeomorphism of S3 such that a(kO) = k2. This notation is obviously consistent with subscripts and we use it with the understanding that mentioning any of ki, Ki, K' or Ki makes clear what the other symbols mean. The group of a knot, 7r1(S3 - k; *), depends only on K and we denote it irK. If T is the boundary of a nice tubular neighbourhood of k and *T is a point on T the group 7r1(T; *T) determines a conjugacy class of subgroups of rK, and 7rK with this conjugacy class specified will be called the marked group of K. The simplest way to mark 7rK is to name a longitude commuting with a chosen over-generator of rK, in practice always with the first named over generator.

Consider an over-presentation

(1) irK = lxl, * * , x. : r *

of the knot group rK. Each relation r, = 1 can be written as x, = W 1x, W where W E rK. We can use such a relation to eliminate the generator x8 whenever W can be written without using x.. By this means the original presentation (1) can be changed to a new presentation of the same form as (1), but with a smaller number of



606 ROBERT RILEY

generators n. Of course, the remaining generators are still over-generators of 7rK so they are still conjugate. We shall call such a presentation a normal presentation and we shall use only normal presentations in this paper. In Section 4 we will describe a modified version of a normal presentation which is more convenient in practice.

Among all the normal presentations of rK, there must be some written on the least number of generators. If this least number is v, then v depends only on K and we denote it by vK = v. It is not hard to see that vK is the minimal bridge number [14] of K, but we make no use of its known properties. We do not know how to calculate vK; however, it is usually possible to find its true value by inspection even when there is no proof available. vK must not be confused with the minimal number of generators for 7rK. For example, when K is a torus knot, this minimal number of generators is 2, but vK can be arbitrarily large [14, Satz 10].

The only knot K with vK = 1 is the trivial knot K = 0. The class of knots K with vK = 2 has been completely classified in [15] and differently in [2]. These knots are also known as rational knots (Viergeflechte) and they are all of alternating type. There is probably no reasonable scheme to classify the knots with vK = 3. A wild knot K would have vK = co.

Let rK be a knot group, G a group, and consider the various homomorphisms 01, 02, * * * of 7rK on G. We shall say that 01 is equivalent to 02, in symbols 0_= 02,

when there exists an automorphism X of G such that 02 = 01w. We shall also say that 0, is weakly equivalent to 02 if there is an automorphism a of rK such that 02 a01. Because the determination of the automorphism group of a knot group is an unsolved and presumably very knotty problem, we are forced to use equivalence much more than weak equivalence.

Let G be a group of permutations of the numbers 1, *., n. We will write the elements g of G as permutations in two different ways. In the first way, we write [al, .. , aJ for the element g E G such that i-g = a1, for i = 1, *., n. In the second way, we write g as a product of disjoint cycles.

(2) g = (b1l, b12, * Do, bi t1)(b2l, b22, . , b2 r.) (bai, b.2, *, bs ,,)

where r, + r2 + * * * + r, = n, b jig = bi j+1 (second index (mod r,)), and each of 1, * * *, n occurs in exactly one cycle. We shall omit cycles of length 1 from (2), and when s = 1 we shall also omit the parentheses. In fact, we shall omit the commas from both representations wherever possible, e.g. [23451] = 12345. We shall write E for the identity [123 * * n].

2. Let 7rK be a knot group presented by a normal presentation (1) on n = 2 or 3 generators. In this section, we shall reduce the problem of classifying the equivalence classes of homorphisms of 7rK on A, to a reasonable number of experiments, viz. 3 when n = 2, and 93 when n = 3. We begin with some assertions about A, which are best proven by direct calculation.

The group A, has 60 elements, viz. 1 of order 1, 15 of order 2, 20 of order 3 and 24 of order 5. All the elements of a given order except 5 are conjugate and the elements of order 5 lie in two conjugacy classes. If C E A, has order 5, then a is conjugate to C, C` but not to C2, C-2. The automorphism group of A, is the sym- metric group S, acting by conjugation, viz. let a E S5, g E A,, then ga = g'. All the elements of A, of a given order are equivalent under the action of S5. The maximal




subgroups of A, are the stabilizers of a symbol 1, *., 5 (these subgroups are isomorphic to A4), and dihedral groups of order 10. If two elements of A, commute, they either lie in a cyclic subgroup or they have order 2 and lie in the stabilizer of a symbol.

In this section, 0 always means a homomorphism of 7rK on A,. Since the over- generators of 7rK in any normal presentation (1) are all conjugate, the elements xi 0 C A, all have the same order, say r. If y, is a generator in a second normal presentation of rK, then y, is conjugate to xl or xl 1, therefore yO has order r. This means that the number r depends only on 0 and not on the presentation, and this allows us to define the order of 0 as r. Equivalent homomorphisms obviously have the same order, so that the problem of classifying the various homomorphisms is split into three cases. From now on, we assume that the number of generators n in the normal presentation (1) is 2 or 3.

If 0 has order 5, then x1 0 is a 5-cycle in A,. For some a E S5, a&`(x 0)a =

12345 = C so that 0 is equivalent to a homomorphism 01 such that xI 0 = C. We may assume 0 = 01. Next, x20 is one of the 12 elements of A, conjugate to C. These are C, C-', and 10 other elements which do not commute with C. These 10 elements lie in two sets, each of 5 elements, on which conjugation by C is a cyclic permutation. If one of these elements is a, then {CaCd I ]j = O .. , 4} is one set and IC'a1C' I I = 0, * * , 4} is the other. Thus, x:20 = C, C-1, or the automorphism of A,, defined by conjugation by a suitable power of C, carries x28 to a or a-. Hence, we may assume x20 = C, C-1, a or a-'. No automorphism of A, that leaves C fixed carries a to a1, so that two homomorphisms 01, 02, thus normalized with x201 F X20

are inequivalent. If n = 2, then (C, x20) = As so that x20 = a or a-1. We shall fix the element a to be (13254) = [35214].

Suppose n = 3 and x20 = C"l. We can apply the above argument to X30 and get X30 = a or a-1. However, if x20 = atl then (C, x20) = A. so that the only automorphism which leaves x1 and x20 fixed is the identity. This means that X30 can be any of the 12 elements conjugate to C and that two homomorphisms normalized so x1 01 = xI 02 = C, x201 = x202 = CZaI, but x301 5 x302, are inequivalent. Hence, when n = 3 the original homomorphism 0 is equivalent to a homomorphism defined by one of 2.2 + 2.12 = 28 choices for the images of xI, x2, X3.

Now let 0 be a homomorphism of order 3. We may assume x1 0 = 123. We have two cases, either 123 and x20 generate A5, or they do not. If they do, then x20 cannot leave 4 or 5 fixed and therefore x20 = a45 or a54 where a = 1, 2 or 3. When x20 = a54, the automorphism defined by (45) C S5 leaves 123 fixed and transforms a54 to a45, so that 0 may be replaced by an equivalent homomorphism (still called) 0 where x28 = a45. If a is a suitable power of 123, then a45' = 145 so that we may finally assume x20 = 145. Note that (123, 145) = As and that this argument shows that if ,3 is a 3-cycle moving both 4 and 5 then (123, A) = A,.

When n = 2 we must have (xI0, x20) = A5 so that the normalized homomorphism can only be x10 = 123, x20 = 145. However, when n = 3 and 01, 02 are two homomorphisms thus normalized, then 01-- 02 * X3 01 = X3 02. Therefore, our original 0 is equivalent to a homomorphism defined by one of 20 choices for the images of X1, X2, X3 when (xl0, x20) = A5.

Next, suppose that 0 is a homomorphism of order 3 such that xi 6 = 123 and (123) < (x1@, x20) < A,. Then x20 must move at least one of 4, 5, but not both.



608 ROBERT RILEY

If x20 moves 5, the automorphism defined by (45) E S5 transforms 0 to an equivalent homomorphism such that x20 moves 4, therefore, we can assume x20 moves 4. Now x?0 = ab4 where a and b are 1, 2, 3. We wish to normalize 0 so that a = 1, in which event x20 = 124 or 134. If a $ 1 already, the appropriate automorphism is defined by lac E S5 where c = 2 or 3 as a = 3 or 2. It is easy to check that if x1G1 = x282 =

123, x201 = 124, x202 = 134, then 01 $ 02. Furthermore, (123, 124) = (123, 134) = the subgroup of A, leaving 5 fixed. This subgroup is maximal so that if r is a 3-cycle moving 5 then (x1i, x20, 0) = A5. When n = 3, x30 must be one of these 12 3-cycles r = ab5. There is no freedom in the choice of 0 in its equivalence class left to move X3@, so that two homomorphisms 61, 02 normalized so that x101 = x202 = 123, x201 = x202 = 124 or 134, but x301 5 x302, are inequivalent. Therefore, our original 0 is equivalent to a homomorphism defined by one of 2 12 = 24 choices for the images of x1, x2, x3 when n = 3, 0 has order 3, and (x10) < (x10, x20) < A,.

Finally for order 3 assume that x20 = 123 or 132. Then (123, x38) = A, and the normalization argument above shows that we may replace 0 by a new 0 such that x30 = 145. Our original 0 is equivalent to a homomorphism defined by one of 2 choices for the images of X1, X2, X3.

Last and least we consider homomorphisms of order 2. It is well known (and easy to check for A,) that a nonabelian group generated by two elements of order 2 is dihedral, hence such homomorphisms do not exist for n = 2. We omit the case by case argument for n = 3 and simply tabulate the results in Table 1 at the end which summarizes the discussion of this section.

Now that we have a normal form for each equivalence class of homomorphisms we can test for the existence of each equivalence class of homomorphisms by testing the assignments x1 0, * *, x>0 of its normal form to see if 0 actually defines a homomorphism, as remarked in the introduction.

3. We now generalize in a naive manner some of the results of the last section to a classification of homomorphisms of knot groups on the linear projective groups Lp - LF(2, p) _ PSL(2, p) where p is a rational prime. For L4 = As, we found that the easiest case to handle was the homomorphisms of order 5. Accordingly, we shall fix our attention on the homomorphisms of order p on L,, and for brevity, we shall call such a homomorphism a "rep". We refer to Chapter XIV of Burnside [1] for the relevant background. The group L2 is somewhat exceptional so we shall assume p is an odd prime in Ehe classification arguments that follow.

The group L, has order p (p2 - 1)/2 and is simple except when p = 3 where L3 ~ A4. It contains p2 _ 1 elements of order p and these lie in two conjugacy classes. If a E L. has order p, then a is conjugate to a' iff n is a square (mod p). If f3 also has order p, either $ = a' for some n or (a, j3) = L,. All elements of order p are either powers of a or lie in one of p - 1 orbits of p elements under conjugation by a. The only automorphisms of L, that leave a fixed are conjugations by powers of a. For any pair fB,, 02 of elements of order p there is an automorphism r of L, such that /2 = 1.

Let (1) be a normal presentation of the knot group 7rK and assume the number of generators n is 2 or 3. To begin the classifications of the reps of 7rK on L,, we fix a EE L, of order p and alter each rep by an automorphism of L, so that the image of x1 is a. Next, consider the (p - 1)/2 orbits of elements of order p which are conjugate




to a. If we choose an element a1i in the ]th of these for j = 1, * * *, (p - 1)/2, then every element f3 of order p which is not a power of a has the form fi = a-'acaj for some j and some s. This means that every rep 0 can be altered so that x1 0 = a, x2 0 = a' for some s or x20 = ai for some j. When n = 2 the equivalence class of 0 is completely determined by j.

When n = 3, we have two cases. If x20 = a& for some j, then (x10, x20) = L, so that the equivalence class of 0 is completely determined by j and x3 0. However, if x20 = a' for some s, (s/p) = 1, then we can alter 0 by an automorphism of L, so that X3 0 = ai for some j. In this event, the equivalence class of 0 is determined by s and j.

This shows that when n = 2 the equivalence classes of reps can be found by (p - 1)/2 experiments and when n = 3 by

p -1p2 -I I I a-IMP -I +( 2) (P +2)(

experiments. The method generalizes immediately to n > 3 generators and the general formula for the number of experiments is

-p 1 (p + 1 - 1

2 p

We can determine the existence of reps on the solvable groups L4 and L3 from the Alexander polynomial AK(X) of K. Since L2 is dihedral of order 6, the argument of Section 10 of [6] shows that 7rK has a rep on L2 iff 3 divides AK(- 1). By a variation of this argument, we can show 7rK has a rep on L3 iff 2 divides RK(W) AK(U) where W is a nonreal cube root of 1. (This strange number is the product of the "Torsionszahlen drifter Stufe" in Reidemeister [13] which are tabulated there on p. 25 for the classical knots. We note that his table is wrong for the knots 91 and 96, the entries in both cases under h = 3 should read 2, 2. The rest of his table is correct.)

As a first hint that there is a better way to find reps, we will show that some knots have a rep on L, for every prime p. Let M be the group SL(2,Z), then every L, is a homomorphic image of M. Furthermore, the image of (1 1) in L, is an element of order p. This means that it is sufficient to find a homomorphism 0 of 7rK on M such that xi 0 = (1 1). The elements A = (' 1) and B = (_ ?) are conjugate and generate M, in fact BAB = C = (_? 1), AB = (_? 1) = D, and we know from Section 1.4, Exercise 19, of [11], that M = (C, D). Using the presentations in Table 2, we find the following homomorphisms 0 of 7rK on M such that x1 0 = (1 1).

K X20 X30

31, 91, 96, 923 (V 2) 21 1)

8,,( 818, 91 co nI inue

81sp 821# 924 0O 1/ V1 /

(continued)



610 ROBERT RILEY

K X20 x30

820 \-1 1J \- 01 0J 820(2 ) (2 )

818, 928 (2 I) (-I - )

81op 815, 818, 940 (-I I) (-1 I )

819 \- \1 2}

The above list is complete up to equivalence for the classical knots. Because a neces- sary condition for such a homomorphism 0 on M is the existence of a rep on L" for every p, the nonexistence of other 0 follows from the above mentioned table in Reidemeister or from Table 3 in the microfiche section in this issue.

4. If (1) is a normal presentation of 7rK on a small number n of generators, the relators are frequently very long words in xi, * * *, x". For example, for some rational knots with 9 crossings, the relator is a word in xl, x2 of length > 100. This difficulty can be avoided by introducing new generators x,+1, * , x", and new relations expressing x,,+i as a word in xl, * * *, x,+i-,, for i = n + 1, * , m. The resulting presentation of irK now has the form:

generators xi, - . -, xm,

(3) relations x"+, = word in xl, *, x+ li = 1, * *, m -n, relators rl(x1, I, Xm) , rn l(x1, , Xm).

This is still a normal presentation on n generators xl, * * *, xn and much more convenient in practice. Note that the subordinate generators Xn+19 ... , xm are arbitrary elements of irK and not necessarily over generators of 7rK.

If one is given a model knot k for K, a presentation (3) is defined implicitly by orienting k and selecting n arcs of the picture to correspond to x1, . .. , xn. Next, the other arcs are numbered n + 1, * * * , m in such a way that at one end arc(j) is crossed by an arc(i) with i < j and on the other side of arc(i) the continuation of arc(i) is arc(s) with s < j, j = n + 1, * , m. (This rule is stated for the Wirtinger presentation, but for an over-presentation for a nonalternating knot the rule is only slightly more complex. See the example below.)

Presentation 7rK = 1x1, X2, X3: r, r21

_-1 X4 X3 X2 X3,

-1 -1 X5 = X3 X2 XI X2 X3,

-1 -1 =i x5 xix2 X1

-1 -1 -1 r2= X4 X1 X2 X3 X2 X1

longitude X2 X3 X1 X3 X1 X2 X4 X5 X1




5

FIGURE 1

Table 2 in the microfiche section of this issue is a list of normal presentations of the marked groups of the classical knots. These presentations have been checked in numerous ways and are almost certainly correct. Table 3 of homomorphisms of these knot groups on A5 is based on Table 2.

5. Associated with a homomorphism 0 of a knot group 7rK on a finite group G are certain topological spaces whose homotopy types are invariants of the knot type K. In particular, suppose G is a transitive group of permutations of the symbols 1, - - *, s, and let H be the subgroup of rK whose image in G keeps the symbol 1 fixed. Let cu be the covering space of S3 - k corresponding to the subgroup H, then ca is a noncompact s-sheeted covering manifold of S3 - k. Furthermore, as explained in Section 8 of [6], ca has a certain natural closure MZ which is a compact manifold known as the s-sheeted cover of S3 branched over k (belonging to H). The spaces qt 9Mz are the topological spaces in question, they depend (up to homeomorphism) only on irK, the permutation representation of the group G, and the weak equivalence class of 0. In particular, their integral homology groups HlCu 8 H,(cl; Z), H1r - H1(M; Z) are invariants of K. A general algorithm to compute integral presentation matrices for these homology groups is informally derived in Section 8 of [6] as a consequence of a presentation for the fundamental groups 7rclU, 7rMZ.

We have translated this algorithm into a Fortran subroutine, Homology, which goes into action given a presentation (2) and the images of the generators {x, , i = 1, *. , n} expressed as permutations. It returns the groups H191, H19Z as Betti numbers and torsion coefficients. The subroutine has the same generality as the original algorithm (except for practical limitations) and so can be used to study more general groups than knot groups. A more complete description and a listing of Homology have been deposited in the UMT file; cf. the review in the review section of this issue.

To apply Homology to a homomorphism 0 of irK on L3, p prime, we need a permutation representation for L,. The different permutation representations of L3 will give rise to different spaces %a, MZ, but the smaller the number of sheets s, the smaller the presentation matrices for Hca, H1MIZ. It is known ([1] or [9]) that L3 always has a permutation representation of degree p + 1 but none of lower degree, except for p = 5, 7, 11, where there are representations of degree p. So for all p Z! 5, 7, 11, we restrict our attention to the representation of degree p + 1 but for p = 5, 7, 11 we shall use both representations. When irK, p and 0 are understood, we shall write 91,,+1, R'U, %l2Z3 for the corresponding covering spaces.



612 ROBERT RILEY

For the exceptional primes 5, 7, 11, we took particular permutation groups of degree p for the groups L4. For p = 5, we used A, as discussed in Section 2. For p = 7, we took the elements a, al, a2, a3 of Section 3 to be

2 4 a = (1234567), al = (1675243), a2 = 01, a3 = a1

This representation of L7 is taken from p. 208, line 5 of [1]. We took the elements a, {a,, j = 1, , 5} for LI, to be

a = (123456789ab), $ = (la)(37)(56)(89)(2)(b), a; =3a, = = 1, *- 5.

This representation came from 7.8 of [3]. In addition, for p = 5, 7, we wrote out specific isomorphisms between the above permutation groups and permutation groups of degrees 6, 8 resp. We may someday do the same for p = 11.

For the nonexceptional primes p, we find it convenient to deal with the groups L4 directly as matrix groups and derive our permutations from the matrices. Our work in this direction is not complete except for the primes p < 19.

Given the permutation group G of degree s and the homomorphism a of 7rK on G, suppose that x1 has c disjoint cycles. Then, as Professor Fox shows, the number of generators for 7r1Z (and consequently for HMlZ) is bounded above by ns - nc -

(s - 1); cf. [7]. When this bound is 0, MZ is a homotopy sphere. When PK = 2 this is true for Ml3 when 0 is a rep on L2 [7] and for ZS when 0 is a homomorphism of order 3 on A,. Consequently, we do not compute Hllt, H1MZ in these cases. When the upper bound is 1, ir15iz = H1r is cyclic and this happens for G = L2, vK = 3, and for G = Ls, vK = 2. (The number c is always 2 for the permutation representation of L, of degree p + 1.)

The space cu can be considered as the complementary space of the link I = -

in MZ. The link type of I in MZ is an invariant of the knot type K and the group Hcu. is only the crudest of a set of link invariants of I analogous to the Alexander polynomials and the Hosokawa polynomial [8]. We have not pursued this matter beyond the calculation (by hand) of one specific example. The rational knot 5, has a homomorphism of degree 3 on A, (Table 3), consequently ,5 is a homotopy sphere and I is a link in this sphere. We found that the Alexander matrix of I (in one variable x) reduces to

af3 0 a(X) = 0 ay

where a = 1- x, 3 = 1 + x5 and -y = 1 + x + X2 + X3 + x4. Consequently, the Hosokawa polynomial is V(x)

6. The algorithms of the preceding sections have been completely or partially coded as Fortran programs and put to work on up to 250 knots. In this and the next section we consider the results. They fill quite a few tables, some rather lengthy, so we will include just one in microfiche and merely discuss the conclusions to be drawn from the rest.

In this and the next section, we describe a homology group HAX H,(X; Z) by the diagonal entries of a square presentation matrix for HlX. If convenient, some of these entries may be 1. When the coefficient group is not Z, it will be Z, = Z/nZ for




some n. For example, H1X = 0, 1, 3 means H1X = Z 0) Z3, and if H1X = 3 then H,(X; Z2) = 1.

We begin with the reps on L. After this paper was submitted for publication, the paper "Metacyclic invariants of knots and links" by R. H. Fox appeared in Canad. J. Math., v. 22, 1970, pp. 193-201, and parts of that paper are very relevant to our work. In the first place, Fox points out that the longitudes of a knot group 7rK lie in the second commutator subgroup of rK, hence if 0: rK -+ G where G is a group whose second commutator subgroup is trivial, the longitudes are in the kernel of 0. This applies when G = L4 or L3 and supersedes a more cumbersome argument of mine. Secondly, Fox includes a table (which agrees with mine) of the groups H1=13 associated with the reps of the groups of the classical knots on L4. More interesting is the linking number v of the two components of z3- -cl3 for each rep which Fox includes in his table. Although I do not know how to calculate v, I found on comparison with one of my tables that in certain cases there seems to be a strict relation between v and a number I can calculate. Namely, if 0 is a rep of 7rK on M = SL(2, Z) where x10 = (G '), and By E 7K is a longitude commuting with xl, then zy0 necessarily has the form i:( 9) for some g E Z. In fact, g 0 (mod 6) because the image of So in L2 or L3 is E. Then, in the 15 cases where Fox's homomorphism is the mod 2 reduction of such a homomorphism on M, we have

A= 3v.

The sign depends on the normalizations and we cannot settle it here. This relation is probably at the heart of the reason why in all known cases v is a rational fraction with an even numerator.

For As = L,, we tabulate the detailed results for the classical knots in Table 3 in microfiche. All permutations in that table are in the bracket [ ] notation. There are altogether 51 reps, 26 homomorphisms of order 3, and 9 of order 2, for a total of 86 homomorphisms. There are 30 knots with no homomorphism of which only 932, 933, 934 are not rational. Five knots, viz. 91, 6 922, 925, 93,, 936, are proven nonrational by the number or types of their homomorphisms but not by any simpler method that I know. Most of the nonamphicheiral knots are proven to be such by Fox's argument [5].

The cases p = 7 and p = 11 have been run on the computer for the classical knots. We found 102 reps on L7 for these knots and that 27 knots have no rep, of which 6 are not rational. The maximal number of reps for any knot is 13 for 940 and these appear to lie in 6 weak equivalence classes. There are 100 reps on L11 and 29 knots with no reps, 8 of these are not rational. This time the maximal number of reps per knot is only 5, attained for K = 941, 947 and 948

This is as far as we have gone for the block of classical knots and as far as we have computed the homology. We have gone further to find at least one rep on some L4 for every classical knot. The largest prime needed for the first rep was 23 which was required for 75 and 812. All the nonrational knots, save 93, have a rep on L, for p < 11, but for 933 we needp = 17.

Out of the detailed results of the calculation of around 1000 homology groups Hlcuh, HlM,, we have found three general conjectures that appear certain, and two more which are very likely. The discussion splits naturally into the cases h = p + 1, h = p, and we exclude p = 2 because 2 is quite exceptional in this context. As



614 ROBERT RILEY

notation, write B('1t), B(YI) for the Betti numbers of Au,, (fml, respectively. We begin with h = p + 1.

Conjecture A. If 0 is a rep and p ? 3,

B(MW,+i) + 2 > B(cUl+1) _ B(Mz,+1) + 1,

and when B(Mz,+1) = 0, B('a,,+1) = 2. Furthermore H1MY+1 has a torsion generator of even order.

The usual value for B(Mlt,+1) is zero. We know that HMZ,+1 is a factor group of H1l,,+1, but apart from that and Conjecture A there seems to be no further universal assertion to be made about the comparison of Hlctt,+i with Hm%,Z+ 1. However, when o happens to be the reduction of a homomorphism on M the situation is much clearer. We have:

Conjecture B. For each prime p > 2, there are integers b., and d,, where d., is a divisor of p + 1, which have the following properties. Let 0: irK -> L, be the reduction modulo p of a homomorphism A5: 7rK -+ SL(2, Z) such that xlO = (1 1). Let the torsion subgroups of the corresponding homology groups HlqcU+1, HlM+1

have orders u, m, respectively. Then, B(n,+1) > b., and m/u is an integer which divides dX,. If B(Mz,+1) = b,, then mru = dz,.

We have verified this when p < 19 for the homomorphisms on M listed in Section 3. The numbers bX,, d., for p < 19 are

p 2 3 5 7 11 13 17 19

b, 0 0 0 0 2 0 2 2

d, 1 2 2 4 2 14 6 10

It should be emphasized that Conjecture B only concerns the orders u, m of the torsion subgroups, not the relations between the actual torsion coefficients. The tables show that the passage from cU to M will do the wildest things to these torsion numbers, subject to the known or conjectured restrictions. We have also considered the question of patterns in the groups HicU,+1, H1MTL+1 for a fixed K and varying p (corresponding to a homomorphism on M), but the results are too vague to be commented on here.

There is a conjecture similar to conjecture A for the groups HNc11, H1M, corresponding to a homomorphism of order 2 or 3 on L, = A,. For order 3, the conjecture is the same as Conjecture A, and for order 2 the only change is

B(M16) + 1 < B(c116) ? B(MZ6) + 4.

In all known cases, B(MnZ6) = 0, B(cu16) = 4, but this is not a conjecture. Now for the groups HAuD, HMNzE when p = 5, 7 or 1. We only discuss the cases

where 0 is a rep because the other cases are very different. Conjecture C. If 0: 7rK-> L. is a rep andp = 5,7 or 11 then B(C1L,) = B()zD) + 1.

If H1clt, = OS al, * * * an with l ? a, ? a2 < < an then for some l, ,

H1(iZ, = al, * ,paj, ... , an. In Conjecture C the torsion number a, may or may not be already divisible by p.

The hypothesis B(c9t) = 1 is essential. The two minor conjectures are that if H1ca. = 0, a,, * * * , an as in Conjecture C,

then 3 divides an when p = 5 and 2 divides a. when p = 7. The evidence for these is quite strong, but because there is no corresponding rule for p = 11, we hesitate to




be too dogmatic about them. As an example, the knot 74 has a unique rep on L11, and Hlqt, = 0, H1Nlr = 11.

To conclude this section, consider the three knots of Fig. 2, of interest because they have Alexander polynomial A(x) 1. The knot k, is

N14 ~ ~ ~ ~ ~ ~

ki k2 'ka

FIGuRE 2

Seifert's classical example of such a nontrivial knot. The knot k2 was constructed by Kinoshita and Terasaka [10] and further studied in [12]. The knot k, was discovered by J. Conway [2] in his classification of the nonalternating 11 crossing knots. He has shown that k, and k., are the only I 1-crossing knots with A(x) 1. (In Conway's notation, k2 = .-(2, 3). 2 and k = . -(3, 2). 2.) We first try the homomorphisms on A,.

K1 xl@ = (12345), x28 = (15432), x30 = (14523).

H1c1ls = O 2, 6. H1iZ5 = 2, 30. H1'1L, = 0, 0, 2, 4. H1W6 = 2, 2, 16.

K2 xl0 = (123), x20 = (145), x30 = (123), (longitude) 0 = E.

K3 x10 = (123), x20 = (145), x30 = (142), (longitude) 0 = E.

For both K2 and K3:

H1Il, = 0, 0, 0. Hi Mlt = 7. H1'L. = 0, 0, 12. H1lT6 = 180.

This shows that all three knots are nontrivial and that K1 is different from the other two. We have not yet shown K2 $ K3. It is possible that the deeper Alexander invariants of the link Z - it in ml can do that, but I bet not.

We now try reps on L4 for K2 and K3. For convenience, write C = (1234567). Each rep satisfies xi 0 = C and each knot has two reps.

X201 = (1675243), x301 = (1452736), (longitude) 0l = C.

Hl117 = 0, 4, 28. H1OT7= 28,. 28. Hl'8 = 0, 0, 3. H1lZ8 = 1920. K2IHC2

x202 = C, x382 = (1675243), (longitude) 02 = C2.

H 117 = 0, 2, 238. H1lZ7 = 14, 238. Hll8 = 0, 0, 6. H1rZ8 = 3, 228.

x201 = (1675243), x301 = (1723654), (longitude) 01 C.

H117 = 0, 2, 2, 4, 12. HjYZ7 2, 2, 4, 84. H1'1L8 = 0, 0, 4, 24.

HlZ8 = 4, 8, 24.

3 X202 = C, X302 = (1264735), (longitude) 01 = C2.

H1'117 = 0, 2, 2, 8, 40. H1X%7 = 2, 2, 8, 280. H1'1L8 = 0, 0, 2, 36.

HjMZ8 = 2, 6, 324.



616 ROBERT RILEY

This shows that K1 # K3 and that neither knot is amphicheiral. The results for L11 are very similar: two reps per knot but the groups Hictll1, H1W11 prove the knots are different.

We tried to prove that K2 $ K; by counting the reps on L, for p = 13, 17, 19, 23, 29 and 31 to see if there was ever a difference between the corresponding numbers for K2, K3. (This experiment took more than 3- hours on the Southampton computer.) The results suggest the rK2 and rK3 have the same number of reps on L, for every prime p and that these knots cannot be distinguished this way. In [12], Magnus and Peluso proved that irK2 has reps on L, for an infinity of p, but because they only netted a small proportion of the true set of reps their results cannot help us. We hope to return to this question in a later paper.

7. One of the best ways to understand the behavior of our knot invariants is to calculate them for a family of knots where the presentations of the knot groups depend on integral parameters. As a final collection-of examples, we consider the rather complicated family of Kinoshita-Terasaka (KT) knots. The KT knot K(p, n) is drawn in Fig. 3 in the case

2n

P <! P+ ~ ( p+I_/70 3\fWe

FIGURE 3. K(p, n)

where p and n are positive integers. When n < 0, the sense of rotation in the integral tangle (winding) Sy is reversed. In [10] Kinoshita and Terasaka proved that K@, n) is a nontrivial knot with Alexander polynomial A(x) =-1 if p > 2 and n # 0.* Obviously, K(p, n) = 0 if p = 0 or 1 or if n = 0. The knot K( of the last section is K(2, 1).

We may also define Kp, n) for p < 0 by reversing the sense of rotation in the tangles a, ,B, 3, e in Fig. 3. (Beware that a and e now have pI - 1 = I(p + 1)1 crossings.) If we turn the knot over and reflect in a mirror, we see K(p, n) = K(p, -n). Furthermore, we can turn the knot over and apply the Edmunds Flip to show K(J, n) = K(-p - 1, n) for all p, n. This allows us to assume p is even and n > 0 from now on.

Let ir(p, n) be the group of K(P, n). Since we have taken p even, we may write p = 2pog p =po + 1. Then 7r(p, n) has a normal presentation (4) of the form xi, x2, x3: r1, r2j where

* Our K(p, n) is written K{p, 2nz) in [10].




aX2 xi x4 a~'0 Xic~vo P~ao -Po~ a! = X2 X 1 X4 =1 0fXI Ci X.5 = Ci X2 C0

-1 x- X= o X4 1 X = Po X #-Po HX4 3 X6 X X86= o7IP1IV

-1 -n

(4) 'Y=X6 X x8=7 XI Y

6 = X2 X8 x= APO x2 5V-0 X10 = 8 x8 a

-1 -1 x o xi e-o e- Xg x3 Xii~ xC e

-1 - ri = X5 X10 r2 = X7 Xll

Suppose for some fixed (p, n) that K(p, n) has a homomorphism 0 of order s on A6. Say that the elements aQ, j80, y0, 60, eO of A, have orders a, b, g, d, e, respectively. Let q be the l.c.m. of a, b, d, e, and let x, 0 = ri, i = 1, 2, 3. From (4) it is easy to see that the assignment xi = ri, i = 1, 2, 3, defines a homomorphism 4. of K(p', n') of order s on A5 whenever p _ p' (mod 2q) and n n' (mod g). Because K(p, 0) =0 we must have g > 2, and because K(O, n) = K(- 2, n) = 0 we must have 2q _ 6 (in fact 2q is 30 or 60 in every case). We shall say that the homomorphisms 0 and 0 are in the same clan of homomorphisms 8.

Because the numbers g and q for a clan of homomorphisms must be divisors of 30, we know in advance that infinitely many KT knots have no. homomorphisms on A5. Clearly, the same difficulty will occur whatever finite (nonabelian) group G we may use in place of A6. Therefore, if we wish to resolve all the different types among the KT knots, we must either use an infinite group G or an infinite set of finite groups {G,,V= I,2,*..}.

We can show however, that infinitely many of the knots K(p, n) are different by considering the homology groups Hjlch, HlMZh, h = 5, 6, associated with a clan of homomorphisms 8. We examined a number of these clans and found by looking at 10 to 40 cases that these homology groups can usually be predicted by simple formulas. We must admit we have not tried to prove these formulas-the groups are complicated and the presentation matrices are rather large. However the method of proof (a giant calculation!) is clear and the results stated below should be con- vincing as they are. (Sceptics might try a simpler case of a clan 8E where the knots are alternating torus knots.) We present our formulas for five cases which illustrate most of what we should expect.

I. xi0 = (123), x28 = (123), x,0 = (145) when p = 2 (mod 30), n 1 (mod 3). Since K&, n) = K(p, -n), this clan also allows n _ 2 (mod 3), but using the standard normalization of 0 in its equivalence class we then write it xi 0 = (123), x20 = (145), X30 = (235). Whenp = 2 and n = 3s + 1:

H1'L, = 0, 0, 0. HPjD5 7 7 + 24s1. Hlts =0, 0, 12. HMlZs = 1180 + 648s[.

II. xI0 = (12345), x20 = (13254), X38 = (15324) when p 2 (mod 30), n _ 2 (mod 5). When n _ 3 (mod 5), we normalize 0 to x20 = (14523), X30 = (15324). (This clan is one of three clans of reps which exist for the same set of (p, n).) When p = 2 write n = 5s + 2, r = 2 if 11 4' s, = 22 if1 I s,

Hi% = 0, 2, 18. HjW5 = 2, 90. H1jcU. 0, 0, O. H1,L6 11100 + 2616s1.

III. x10 = (12)(34), X20 = (12)(35), X30 = (13)(25) when p -4 (mod 30), n I (mod 3). For n 2 (mod 3) we write x28 = (13)(24), x38 = (12)(35). When



618 ROBERT RILEY

p = 4, n = 3s + 1 and T = 178 + 240sl we get:

HlL5= 0, 0, 0,T. HI1Z5=T. H1cBL = O, OO ,O ,2. HIW6 = 4.

IV. x10 = (12)(34), x20 = (13)(24), X30 = (14)(25) when p 2 (mod 30), n 2 (mod 5). For n _ 3 (mod 5) we write X30 a (13)(25), X30 = (15)(34). When p = 2 the groups are:

H1U5c= 0, 0, 0, 3. H15 = 3. H1 = OO, OSO. HI1 6 = 2.

V. x10 = (123), x20 = (134), X30 = (135) when p -4 (mod 60), n -1 (mod 2). (This clan is one of four such which lie in pairs, the corresponding homology groups for the homomorphisms in a pair being the same.) When p = 4 write n = 2s + 1, p = 18 1s + I1, andletr = 9 if 31s, T = 3 if 3 4' s. Then:

HiUt5 = O.,O ,O ,r. HI L5 = 2, p.

Hial = 0, 0, a(s), b(s) where a(s) = 4r(s), b(s) = 1 when s is even, b(s) = 4 if s 1 (mod 4), and b(s) = 2 if s 3 3 (mod 4). HiS, = c(s), d(s), t(s) where c(s) = 3 if s is even, c(s) = 2, otherwise, and d(s) = 1 if s is even, d(s) = 3 * b(s) otherwise. The final summand t(s) varies so wildly that I cannot guess a formula for it.

Table 1. The images x1 0, X20, X30 of the generators of a normal presentation on three generators for the equivalence classes of homomorphisms on A,. See Section 1 for the [ ] notation.

I. Order 5. x10 = [23451]. 28 combinations

x20 = [35214] or [43152]. X30 = [23451], [51234], or one of [35214], [41532], [54213], [31524], [54132],

[43152], [25413], [43521], [24153], [35421].

x20 = [23451] or [51234]. X30 = [35214] or [43152].

II. Order 3. xl0 = [23145]. 46 combinations

x20 = [42351]. X30 = any 3-cycle.

x20 = [24315] or [32415]. X30 = any 3-cycle which moves 5.

X30 = [23145] or [31245]. X30 = [42351].

III. Order 2. xI0 = [21435]. 19 combinations

x20 = [21543]. X30 = one of [35142], [32154], [34125], [42513],

[45312], [52431].

x20 = [35142]. X30 = one of [21543], [21354], [34125], [32154],

[43215], [45312], [53241], [52431],

[14523], [15432].

x20 = [34125]. X30 = one of [21543], [35142], [45312].

Department of Mathematics The University of Southampton Southampton, England




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Table Thiw.

. r h ism on of the kot urpw in table tvo,

topU nr itii ab corr"pcoding hooqo groups

ati bomrpblem is d*fiaed by the permutation. wbih are\

th. itige. of, the tuAdpendet" over geeratorw. The lare

f a io irnfwred Tr th orter or the hoastisa.



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5 S"rKS b91111 1 lt4S tN VA I u

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5 SkIlTS 19Slt!I Too1ls3I S P,

DRACHED BITT! S t431(SI P 30 A SO VIlS BITT I a TiI"SW I l" iRAWCS1) flTT 0 TORS I N P

KNOT 6.1@ I * OR)UtS 5 M 3%$4)A X3 3%S14 Liu STlt Till

% SHwft1 ['11l 1 I Ulfl34U 45 RAnCHWID PTlS 0 TO ISION

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Stan, Ryra, ewd Gelberd recently $abliahed table. ot weight.

m *ecieee for the eveluatiem by (.eusien qutdreture ot Integrals

as tile f or:

(1) fft(aeapt1s )da

vhich onre In ducleer isactor deign celculet lna. Slelar integrals,

except with e *fljjf vetghting fuictios w(x)f2s ep(-s ) rather

them their Rolcm wett. oceut frequently in atomic *ed nloculer

co11it4u pro*li Althouh thee integrale could be regtrded aa

spectel efled of (1),0 better ceerge?ace L he eb ted trft 'uadratur. ~1

fetwle" explicitly tektag into eccomat the additional factor 23 as

part ef the veight faction. TI61 I of this note givae absciosse end

veihte for 1 polit C lasom qadrature, of the fre:

(2) fawh ePl-2 )f(s~da 2 2~~~

Theewere obte in the usual W tfr ;h (nth nleaic)

polyoslsla 13(x) orthogonal on (,O,') 0ith reapoct to w(z),

f-weI? daf4t by tek I ?0(a)fl fud te f*olmtq aes"ttion

for recursion

(} *~~x (rn.X)P3(a) * R/| l(a)

Of eawerol nericel tochaiquo in letigatod for determining

thsee Polytel the N*o ufferit let troa lo*e of marical



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Homomorphisms of Knot Groups on Finite Groups