BRANCHED COVER HOMOLOGY AND Q EVALUATIONS

This is a draft. I would be grateful for any comments and corrections!

A. Stoimenow^{∗}

Max-Planck-Institut für Mathematik, Vivatsgasse 7, D-53111 Bonn, Germany, Postal Address: P. O. Box: 7280, D-53072 Bonn,

                                  e-mail: alex@mpim-bonn.mpg.de,
                     WWW: http://www.informatik.hu-berlin.de/~stoimeno

Current version: September 8, 1999 First version: May 18, 1999

Abstract. We show that the known evaluations of the Q polynomial are the only ones determined by the homology of the double branched cover of S^{3} over a link. Keywords: double branched cover, Q polynomial. AMS subject classification: 57M25 (primary), 57M12, 42A28, 05A15 (secondary).

1. Introduction

The Alexander polynomial [Al] is a classical invariant in the theory of knots and links in S^{3}, which is closely related to the topology of the knot (or link) complement [Ro].

After the appearance of the relatives of the Alexander polynomial [J, H, BLM, Ka], it has been tried to find a topological understanding of these invariants, too. This succeeded only for special evaluations of the above polynomials [LM, J2]. A summary on this can be found in [L]. In all cases these evaluations have been related to the homology modules of branched coverings of S^{3} over the link with values in some finite field. Of particular interest is the Q polynomial of Brandt-Lickorish-Millett-Ho [BLM, Ho], a polynomial invariant with values in ℤ[z,z^{−1} ], from which the rank of the homology of the double branched cover with values in ℤ_{3} can be obtained from the evaluation at z = −1 (see Theorem 8.4.8 (2) of [Kw]) or the one with values in ℤ_{5}, recoverable from the evaluation at z = (\sqrt{5}− 1)/2 [J2]. Thus, by [We, theorem p. 690] or by direct skein/Kauffman relation arguments as in [Tr], these evaluations give lower bounds for the unknotting number.

In this note we attempt to show that these are the only evaluations of Q, which are, in some reasonable sense, determined by the homology of the double branched cover with values in some finite field. A similar reasoning for the Jones polynomial we leave to an interested reader.

2. Special Q evaluations determined completely by dimH_{1}(D_{L},ℤ_{n})

We start with the following

Theorem 2.1 Let z ∈ ℂ \ {0} be so that Q_{L}(z) is determined by the rank of the homology H_{1}(D_{L},ℤ_{n}) of the double branched cover D_{L} of L with values in some finite field ℤ_{n}, n prime. Then z = −2,1, or z = −1 and n = 3.

∗Supported by a DFG postdoc grant.

2 2 Special Q evaluations determined completely by dimH_{1}(D_{L},ℤ_{n})

∞  0  −1  1  4

Figure 1: The Conway tangles.

Remark 2.1 The case z = (\sqrt{5}− 1)/2 does not occur here, because the evaluation is determined just up to a sign by dimH_{1}(D_{L},ℤ_{5}). Thus we will attempt to prove a stronger version of theorem 2.1, where Q_{L}(z) is replaced by |Q_{L}(z)|, but, unfortunately, our method will work with much more effort.

Proof. We use the result of Przytycki [Pr], that H_{1}(D_{L},ℤ_{n}) is unchanged by a n-move. A n-move is a move on a knot or link diagram, replacing a 0 tangle (in the Conway [Co] sense) by an n or −n tangle (where a −n tangle is the obverse of an n tangle):

−→ _{n half-twists}

Two links are called n-equivalent if there is a sequence of Reidemeister and n-moves transforming a diagram of the one link into one of the other link.

Recall, that the Q polynomial is a Laurent polynomial in one variable z for links without orientation, defined by being 1 on the unknot and the relation

A_{1}+ A_{−1} = z(A_{0}+ A_{∞}),  (1)

where A_{i} are the Q polynomials of links K_{i} and K_{i} (i ∈ ℤ ∪ {∞}) possess diagrams equal except in one room, where an i-tangle (in the Conway sense) is inserted, see figure 1.

Now take all K_{i}, except K_{∞}, to be knots, and consider the series

f(x,z) := ∑_{i=0}^{∞} A_{i}(z)x^{i}.

As the coefficients of Q are exponentially bounded in the crossing number, the series has a positive convergence radius around (0,0) and converges absolutely within this radius and defines an analytic function.

More precisely, the series converges absolutely for |x| < 1 and |z| < 1, and for |z| > 1 and |xz| < 1. Therefore, for any given z we can choose x small enough so as to perform the following calculations.

From (1) we obtain

f(x,z) = \frac{−f(x,z) + A_{0}(z) + A_{1}(z)x}{x^{2}} + z \frac{f(x,z) − A_{0}(z)}{x} + \frac{z}{1 − x} A_{∞} (z),

whence

f = \frac{(1 − zx)(1 − x)A_{0}+ x(1 − x)A_{1}+ zx^{2}A_{∞}}{(1 − zx + x^{2})(1 − x)} .  (2)

The dependency of f on A_{0,1,∞} we will not mark explicitly, but should implicitly keep in mind.

Assume now that for some concrete value of z, Q(z) depends just on H_{1}(D_{L},ℤ_{n}). Then f(x,z) would have a n-periodic Taylor expansion in x around 0 (for any z_{0}, f(x,z_{0}) converges absolutely for |x| < ϵ_{z_{0}}). So

((1 − zx)(1 − x)A_{0}+ x(1 − x)A_{1}+ zx^{2}A_{∞} )(1 − x^{n}) = (1 − x)(x^{2}− zx + 1)P(x,z),

for some polynomial P ∈ ℤ[z,x] of degree at most n − 1 in x.

Now, first we show that z = 2cos2πk/n for some natural number 0 ≤ k ≤ n − 1. Assume, that it is not the case. Then x^{2} − zx + 1 has zeros, which are not zeros of 1 − x^{n}. Then x^{2} − zx + 1 must divide the first factor on the left. But, already making the simplest non-trivial choice A_{∞} = A_{1} = 1, A_{0} = 2/z−1, the rest is x(2z^{2}−2)+(2−2z) ≠ 0, unless z = −1, which is uninteresting.

Therefore, z = 2cos2πk/n. Now, connected sum shows that the map rankH_{1} ↦ Q(z), should be multiplicative under addition, and hence must be an exponential. The only candidate for a base is found by examining the 2 component unlink. So we must have

Q(2cos2πk/n) = ( \frac{1}{cos2πk/n}− 1) ^{dimH_{1}(D_{L},ℤ_{n})}.

Consider the trefoil with polynomial Q = −3 + 2z + 2z^{2} and n > 3. The above equality specializes to

−3 + 4c + 8c^{2} = 1,

with c:= cos2πk/n, whence c = 1/2 or c = −1, which implies z = 1 or z = −2. So we are done checking the cases n ≤ 3 directly. □

3. The evaluations determined by norm

Now replace Q(z) by |Q(z)|. This makes life somewhat more complicated. The generating series is now

f̃(x,z) := ∑_{i=0}^{∞} |A_{i}(z)|^{2}x^{i}.

This series converges absolutely for |x| < 1 and |z| < 1, and for |z| > 1 and |xz| < 1. Therefore, for any given z we can choose x small enough so as to perform the following calculations.

f̃still can be expressed in terms of f by Fourier calculus, but rather complicatedly. One way is

f̃(x,z) = ∫_{0}^{1} f(\sqrt{x}e^{2πik},z)\overline{f(\sqrt{\overline{x}}e^{2πik},z)} dk = ∫_{0}^{1} f(\sqrt{x}e^{2πik},z)f(\sqrt{x}e^{−2πik} ,\overline{z})dk.  (3)

Now we need to examine for which z there is an n such that

\frac{∂^{n}}{∂x^{n}}|_{|x|<ϵ} ∫_{0}^{1} f(\sqrt{x}e^{2πik},z)\overline{f(\sqrt{\overline{x}}e^{2πik},z)} (1 − x^{n})dk ≡ 0

for any choice of A_{0,1,∞}.

The result is the one we like.

Theorem 3.1 Let z ∈ ℂ\{0} be so that |Q_{L}(z)| is determined by dimH_{1}(D_{L},ℤ_{n}), n prime. Then z = −2,1, or z = −1 and n = 3, or z = ^{\frac{−1±\sqrt{5}}{2}} and n = 5.

Proof. Again we are interested in a periodic Taylor expansion with regard to x around x = 0, this time of the integral in (3). This integral can be expressed as a curve integral

\frac{1}{2πi} ∮_{|k|=\sqrt{x}} f(k,z)f( \frac{x}{k},\overline{z}) \frac{1}{k}dk

= \frac{1}{2πi} ∮_{|k|=\sqrt{x}} \frac{[(1 − zk)(1 − k)A_{0}(z) + k(1 − k)A_{1}(z) + zk^{2}A_{∞} (z)]}{(1 − kz + k^{2})(1 − k)} × \frac{[(1 − \overline{z}\frac{x}{k})(1 − \frac{x}{k})A_{0}(\overline{z}) + \frac{x}{k}(1 − \frac{x}{k})A_{1}(\overline{z}) + \overline{z}\frac{x^{2}}{k^{2}}A_{∞} (\overline{z})]dk}{(1 − \frac{x}{k}\overline{z}+ \frac{x^{2}}{k^{2}})(1 − \frac{x}{k})k}

= \frac{1}{2πi} ∮_{|k|=\sqrt{x}} \frac{[(1 − zk)(1 − k)A_{0}(z) + k(1 − k)A_{1}(z) + zk^{2}A_{∞} (z)] [(1 − \overline{z}\frac{x}{k})(1 − \frac{x}{k})A_{0}(\overline{z}) + \frac{x}{k}(1 − \frac{x}{k})A_{1}(\overline{z}) + \overline{z}\frac{x^{2}}{k^{2}}A_{∞} (\overline{z})] k^{2}dk}{(1 − kz + k^{2})(1 − k)(k^{2}− xk\overline{z}+ x^{2})(k − x)} ,

where ‘|k| = \sqrt{x}’ means the sphere in ℂ with origin zero and radius \sqrt{x}, positively oriented. As x → 0, the relevant zeros of the denominator are x, x · \overline{k_{0}} and x/\overline{k_{0}}, where k = k_{0} is one zero of k^{2}− zk + 1, so that

z = k_{0}+ \frac{1}{k_{0}}.  (4)

To express ourselves more briefly, set

Φ(k,x,z) := [(1 − zk)(1 − k)A_{0}(z) + k(1 − k)A_{1}(z) + zk^{2}A_{∞} (z)][(k − \overline{z}x)(k − x)A_{0}(\overline{z}) + x(k − x)A_{1}(\overline{z}) + \overline{z}x^{2}A_{∞} (\overline{z})].

The denominators of the residues are (in the order of appearance)

(1 − xz + x^{2})(1 − x)x^{2}(2 − \overline{z}),

(1 − x\overline{k_{0}}z + x^{2}\overline{k_{0}}^{2})(1 − x\overline{k_{0}})(−x + x\overline{k_{0}})(x\overline{k_{0}}− \frac{x}{\overline{k_{0}}}),

( 1 − \frac{x}{\overline{k_{0}}}z + \frac{x^{2}}{\overline{k_{0}}^{2}}) ( 1 − \frac{x}{\overline{k_{0}}}) ( −x + \frac{x}{\overline{k_{0}}}) ( \frac{x}{\overline{k_{0}}} − x\overline{k_{0}}) .

We may assume that k_{0} ≠ 1 (else z = 2, which is clearly not of the kind we want). Then, regarded as functions in x, they have the following zeros (with multiplicities)

\begin{matrix}\overline{k_{0}}, & \frac{1}{\overline{k_{0}}}, & 1, & 0, & 0; \\ _{\frac{1}{\overline{k_{0}}^{2}}}, & 1, & \frac{1}{\overline{k_{0}}}, & 0, & 0; \\ \overline{k_{0}}^{2}, & 1, & \overline{k_{0}}, & 0, & 0. \end{matrix}

If k_{0} is an n-th unity root we would have z = 2cos2πl/n. Now assume, that k_{0} is not an n-th unity root. Then, because \overline{k_{0}}^{±2} appear as zeros in only one of the denominators above, we must have that they divide the corresponding numerators. Therefore, x − \overline{k_{0}}^{−2} divides the numerator of the first residue, and x − \overline{k_{0}}^{2} divides the numerator of the second residue for any choice of A_{0,1,∞}. These numerators are Φ(x\overline{k_{0}},x,z) and Φ(x/\overline{k_{0}},x,z), respectively.

Therefore, Φ(\overline{k_{0}},\overline{k_{0}}^{2},z) = Φ(\frac{1}{\overline{k_{0}}},_{\frac{1}{\overline{k_{0}}^{2}}},z) = 0, for the given choice of z. Set again A_{∞} = A_{1} = 1 and A_{0} = 2/z − 1. The

equality Φ(\overline{k_{0}},\overline{k_{0}}^{2},z) = 0 yields

ẑ^{2}k_{0}+ ẑ(k_{0}^{2}− 1) + (2 − 2k_{0}) = 0,  (5)

where ‘ẑ’ means that one of both identities with ẑ replaced by z or \overline{z} is to be satisfied. But because of (4), the substitution k_{0} → 1/k_{0} does not change z (and \overline{z}), and hence from (5) we obtain under this substitution

ẑ^{2}k_{0}+ ẑ(1 − k_{0}^{2}) + (2k_{0}^{2}− 2k_{0}) = 0,

with ‘ẑ’ meaning the same choice between z and \overline{z}as in (5).

Taking difference with (5), we find k_{0} = ±1, hence z = ±2, or z = −1, which are evaluations where the picture is clear by [BLM].

Hence, assume that z = 2cos2πk/n. As in the previous proof, we get

|Q(2cos2πk/n)| = | \frac{1}{cos2πk/n}− 1|^{dimH_{1}(D_{L},ℤ_{n})}.

Again with the trefoil we find for n > 3 this implies

| − 3 + 4c + 8c^{2}| = 1,

with c as before. But −3+4c+8c^{2} is real, so −3+4c+8c^{2} = ±1. In former case we get where we did in the previous proof, and in latter case we obtain c = ^{\frac{−1±\sqrt{5}}{4}} , giving the case n = 5 of Jones. □

Acknowledgement. I wish to thank to P. Traczyk and W. B. R. Lickorish for interesting discussions and their helpful remarks on unknotting numbers. I also wish to thank to MPI Bonn for the stimulating working atmosphere.

References 5

References

[Al] J. W. Alexander, Topological invariants of knots and links, Trans. Amer. Math. Soc. 30 (1928), 275–306.

[BLM] R. D. Brandt, W. B. R. Lickorish and K. Millett, A polynomial invariant for unoriented knots and links, Inv. Math. 74 (1986), 563–573.

[Co] J. H. Conway, On enumeration of knots and links, in “Computational Problems in abstract algebra” (J. Leech, ed.), 329–358, Pergamon Press, 1969.

[Ho] C. F. Ho, A polynomial invariant for knots and links – preliminary report, Abstracts Amer. Math. Soc. 6 (1985), 300.

[H] P. Freyd, J. Hoste, W. B. R. Lickorish, K. Millett, A. Ocneanu and D. Yetter, A new polynomial invariant of knots and links, Bull. Amer. Math. Soc. 12 (1985), 239–246.

[HT] J. Hoste and M. Thistlethwaite, KnotScape, a knot polynomial calculation program, available at

      http://www.math.utk.edu/~morwen.

[HTW]  ”  ,  ”  and J. Weeks, The first 1,701,936 knots, Math. Intell. 20 (4) (1998), 33–48.

[J] V. F. R. Jones, A polynomial invariant of knots and links via von Neumann algebras, Bull. Amer. Math. Soc. 12 (1985), 103–111.

[J2] ” , On a certain value of the Kauffman polynomial, Comm. Math. Phys. 125 (1989), 459–467.

[Ka] L. Kauffman, An invariant of regular isotopy, Trans. Amer. Math. Soc. 318 (1990), 417–471.

[Kw] A. Kawauchi, A survey of Knot Theory, Birkhäuser, Basel-Boston-Berlin, 1996.

[L] W. B. R. Lickorish, The panorama of polynomials of knots, links and skeins, in “Braids”, Santa Cruz, 1986 (J. S. Birman and A. L. Libgober, eds.), Contemp. Math. 78, 399–414.

[LM] ” and K. C. Millett, Some evaluations of link polynomials, Comment. Math. Helv. 61 (1986), 349–359.

[Pr] J. Przytycki, t_{k} moves on links, in “Braids”, Santa Cruz, 1986 (J. S. Birman and A. L. Libgober, eds.), Contemp. Math. 78, 615–656.

[Ro] D. Rolfsen, Knots and links, Publish or Perish, 1976.

[Tr] P. Traczyk, A criterion for signed unknotting number, preprint.

[We] H. Wendt, Die Gordische Auflösung von Knoten, Math. Z. 42 (1937), 680–696.