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The question remains whether there is a competitive quan- Our construction is to concatenate a sequence {τa j }1≤ j≤m of tum algorithm for any natural question in 3-manifold topol- these mapping cylinders together with a solid torus at each ogy. This is a natural thing to look for, since for instance it is end, as in Figure 1. We also assume that a1 > 1. The tetra- known that unknottedness is the complexity class NP ∩ coNP hedron number t is thus O(m). If the solid tori σ1 and σ2 are [4, 8, 10]. (See the Complexity Zoo [19] for a survey of com- positioned suitably, then the result is M =∼ L(n,k), where n and putational complexity classes.) While NP ∩ coNP is thought k are given as a finite continued fraction: to neither contain nor be contained in quantum polynomial time BQP, some key problems (such as discrete logarithm) n 1 = am + . k a 1 are known to be in both of them. Note that Aharonov, Jones, m−1 + . .. and Landau [1] give an algorithm to approximate the Jones + 1 a1 polynomial of a knot at a principal root of unity; this algo- rithm also has a version for 3-manifolds [3]. However, the If we let n j/k j be the jth partial evaluation, then we can also approximation is exponentially poor; any fair approximation express the calculation with the recurrence that could be useful for is #P-hard [9]. k j = n j−1 n j = a jn j−1 + k j = a jn j−1 + n j−2.

The answer n/k determines the monodromy numbers {a j} ACKNOWLEDGMENTS since the continued fraction is unique under the constraint a1 > 1. Since the integers {n j} increase, we obtain the in- The author would like to thank an anonymous MathOver- equality flow user for help with part of the calculation. n j < (a j + 1)n j−1. If we choose the sequence of monodromynumbers at random, 2. LARGE LENS SPACES WITH SMALL TRIANGULATIONS we obtain the probabilistic relation

Ex[log(n j)] < Ex[log(a j + 1)] + Ex[log(n j)]. In this section, we will construct lens spaces M =∼ L(n,k) where n is much larger than the number of tetrahedra t, and Also, k has many possible values. The manifolds that we con- log(2)+ log(3)+ ···+ log(6) struct are easy to identify given their specific triangulations. Ex[log(a j + 1)] = However, the triangulations can then be obfuscated with lo- 5 cal moves (e.g., Newman-Pachner bistellar moves). Proposi- < log(3.73). tion 2.1 makes both Theorem 1.1 and the Lackenby-Schleimer result look more interesting. For the latter question, it is easy By the law of large numbers, most monodromy sequences produce n < 3.73m. On the other hand, there are 4 · 5m−1 se- to compute whether H1(M) =∼ Z/n is cyclic. If it is, and if n is polynomially bounded in t, then Schleimer’s prior result quences of length m, so by the pigeonhole principle, some [16] gives an algorithm in NP to compute whether the abelian value of n must see exponentially many values of k. Any such cover M˜ is homeomorphic to S3, which then implies that M is value of n must also be exponentially large. In any case, for a lens space. every choice of numbers {a j}, {n j} grows at least as fast as the Fibonacci numbers, which also implies that n is exponen- Proposition 2.1. There exists a family of triangulated lens tially large. spaces {M =∼ L(n,k)} with t = t(n,k) tetrahedra, such that n is exponential in t and there are exponentially many choices for k for each fixed n. 3. REIDEMEISTER TORSION

Proof. Our construction is equivalent to a well-known con- We review Reidemeister torsion [18] and its value for lens struction of lens spaces using on a chain of un- spaces. knots [15, Ex. 9H13]. Suppose that We choose a fixed triangulation σ of the torus T = S1 × 1 ∂ S , and we choose two solid tori X1,X2 with ∂X1,∂X2 = T , C∗ = {Ck −→ Ck−1}0≤k≤m and with triangulations σ1,σ2 that extend σ. We can describe an element of the mapping class group of T by an element is a finite, acyclic chain complex over a field F. (Reidemeister of GL(2,Z) that describe its action on the homology group torsion is well defined for a free complex over any commuta- H1(T ). For each 1 ≤ a ≤ 5, we choose a fixed triangulation tive ring, but it is easier to discuss algorithms in the field case.) τa of a torus bundle over an interval, T ⋊ I, that connects the Suppose in addition that each term Ck has a distinguished ba- triangulation σ of T to itself using the monodromy matrix sis. Since C∗ is acyclic and finite, it is isomorphic to a direct sum of complexes of the form 0 1 Fa = . =∼ 1 a 0 −→ F −→ F −→ 0. 3

σ1 τa1 τa2 τa3 ··· τam σ2

1 1 Figure 1. A lens space L(n,k) as solid tori with triangulations σ1 and σ2, connected by twisted bundles (S × S ) ⋊ I with triangulations τa j .

Define an adapted basis A∗ for C∗ to be one induced by such Then to compute the Reidemeister torsion of M, we let F = a decomposition. In other words, if αk ∈ Ak is a basis vector, Q(ζn), where ζn is an abstract primitive nth root of unity, i.e., then either ∂αk = 0, or ∂αk ∈ Ak−1 is another basis vector. an abstract root of the nth cyclotomic polynomial. We also 1 Then the Reidemeister torsion of C∗ is choose a cellular cocycle ω ∈ C (M;Z/n) such that [ω] gen- erates H1(M;Z/n). We use ω to define a twisted coefficient def −1 (−1)m ∆(C∗) = (detA0)(detA1) (detA2)···(detAm) , system Q(ζ)ω on M, and we let where each A j is also interpreted as the change-of-basis ma- def R = C (M;Q(ζn)ω ) trix from the distinguished basis to the adapted basis. The ∗ ∗ following two facts are standard: to define the Reidemeister torsion ∆(R∗) of M. A change in the choice of the generator [ω] can change ∆(R∗) by a Galois 1. Every adapted basis yields the same value of ∆(C∗). automorphisms of Q(ζn). After fixing [ω], a change in the 2. Let C∗ be the chain complex of a finite CW complex σ with choice of its representative ω can change ∆(R∗) by a factor of c PL attaching maps, possibly with twisted coefficients, ζn for some residue c ∈ Z/n. Otherwise ∆(R∗) is a topological and using the cells of Φ as its distinguished basis. Then invariant of M, provided that the cells of M are ordered and the Reidemeister torsion ∆(C∗) is invariant under refine- oriented so that ∆(Q∗) > 0. ment of Φ. In particular, if M = L(n,k), then The second fact essentially says that Reidemeister torsion is c a b ∆(R∗)= ζn (1 − ζn )(1 − ζn ), (2) a PL topological invariant. We have to be careful because the sign of ∆(C∗) depends on the ordering and orientation of the where cells of Φ, and ambiguities in the local coefficient system can a = k±1 ∈ Z/n. also make ∆(C∗) multivalued. b Let M be a closed, oriented rational homology 3-sphere with a triangulation, or more generally a cellulation which This answer is easy to calculate using the standard cellulation may support a combinatorial local system. We first calculate of L(n,k) with one cell in each dimension, as follows. For its untwisted Reidemeister torsion with coefficients in F = Q. a convenient choice of twisted coefficients, this CW complex Using the orientation, we can canonically augment the chain yields complex C (M;Q) at both ends to obtain the acyclic complex ∗ k 1−ζn 0 1−ζn 0 −→ Q(ζn) −→ Q(ζn) −→ Q(ζn) −→ Q(ζn) −→ 0. 0 −→ Q −→ C3(M;Q) −→ C2(M;Q) Q∗ = . (1)  −→ C1(M;Q) −→ C0(M;Q) −→ Q −→ 0 Thus,

k Then it is standard that ∆(R∗) = (1 − ζn)(1 − ζn ).

∆(Q∗)= ±|H1(M;Z)|. The formula (2) is the same as this one, except generalized to let ∆(R∗) change with a change in the choice of ω. The The sign is not a topologicalinvariant, because the j-simplices exponents a and b are also ambiguous, as follows. First, the of M are unordered and unoriented, so they only provide value of the torsion (2) does not determine the global sign of Cj(M;Q) with an unordered, unsigned basis. We choose an a and b, only their relative sign, since ordering and an orientation of the cells such that ∆(Q∗) > 0. c a b a+b+c −a −b We can then use the same ordering and orientation for a let ζn (1 − ζn )(1 − ζn )= ζn (1 − ζn )(1 − ζn ). Reidemeister torsion calculation on M with twisted coeffi- cients. The formula is also symmetric in a and b, so we cannot dis- Suppose further that tinguish k from 1/k. This stands to reason because

H1(M)= H1(M;Z) =∼ Z/n. L(n,k) =∼ L(n,1/k). 4

4. PROOF OF THEOREM 1.1 Note that the constants a, b, and c depend only on ω and not on the exponent ℓ. To prove Theorem 1.1, we begin with two basic results in If the cell complex Φ has g edges, then it also has g 2-cells, numerical algorithms. and we can write the complex R∗(ζ) as

Theorem 4.1 (Edmonds [2]). The determinant detM of a ∂3 ∂2 ∂1 0 −→ C −→ Cg −→ Cg −→ C −→ 0. square matrix M defined over Q(i), the field of complex num- bers with rational real and imaginary parts, can be computed The complex R∗(ζ) is acyclic, so ∂3 is injective while ∂1 is in deterministic polynomial time in the bit complexity of M. surjective. We can now make an adapted basis as follows: Edmonds states his result over an integral domain with suit- 1. We use the canonical basis vector 1 ∈ C in degree 3 able arithmetic algorithms; the context of the paper suggests of the chain complex R∗(ζ), and its image under ∂3 in integer matrices. However, his construction works just as well degree 2. using exact arithmetic in the field Q(i). He defines a varia- tion of Gaussian elimination such that every number that ever 2. We choose a non-zero entry of the vector ∂3. Ifwe appears is a minor of the original matrix M. As a result, all choose the jth entry (∂3) j, then we can omit the jth numbers that arise in the calculation have polynomial bit com- canonical basis vector of Cg in degree 2. We also use plexity. the image of these g − 1 vectors under ∂2 in degree 1. Remark. There are many ways to prove Theorem 4.1 and we 3. We choose a non-zero entry of the dual vector ∂1. If do not know the best attribution. The hard part of the result we choose the kth entry, then we include the kth basis is to bound the bit complexity of intermediate expressions, g vector of C in degree 1 and its image under ∂1, which rather than just the number of arithmetic operations. is simply the scalar value (∂1)k. Theorem 4.1 is related to the problem of calculating the ( j,k) Smith normal form of a matrix. Let ∂2 denote the matrix of ∂2 omitting the jth column and the kth row. Then we can express the Reidemeister torsion of Theorem 4.2 (Kannan-Bachem [5]). The Smith normal form R (ζ) as of a square or rectangular matrix M defined over Z, together ∗ with left and right multipliers, can be computed in determin- (∂3) j(∂1) ∆(R (ζ)) = k . istic polynomial time in the bit complexity of the M. ∗ ( j,k) det∂2 Let M be an oriented rational homology 3-sphere described by a triangulation Θ. As a first step which will be important To compute ∆(R∗(ζ)) over C, the most important question later, we can simplify Θ to a cellulation Φ with one vertex is how many digits of precision we need throughout the cal- and by removing enough triangles until all of the tetrahedra culation for an accurate final answer. merge into a single 3-cell, and dually by collapsing edges that Lemma 4.3. Suppose ζ = exp(2πiℓ/n) ∈ C and that R (ζ) connect two distinct vertices until only one vertex is left. Us- ∗ is the chain complex of M with its local system Cω . Suppose ing either Θ or Φ, we can calculate the cellular chain complex that we want to calculate z ∈ Q(i) such that C∗(M;Z) in polynomial time. We can use Theorem 4.1 to cal- −α culate the torsion ∆(Q∗) of the augmentation Q∗ in equation ∆(R∗(ζ)) = z + O(n ) (1); in particular to determine whether C∗(M;Z) has a positive or negative basis. We can assume a positive basis. ( j,k) for some constant α. Then it suffices to calculate det∂2 We can iteratively use Theorem 4.2 to calculate a change by estimating its entries with d digits of precision, where d is of basis of the chain complex C∗(M;Z) to put every differ- polynomial in log(n), g, and α. Moreover, the determinant ential ∂k into Smith normal form. This also puts the dual can be calculated in polynomial time. complex C∗(M;Z) into Smith normal form. Using Smith nor- a b mal form, if H1(M;Z) =∼ Z/n, then we can calculate a cocycle Proof. Both (∂3) j and (∂1)k are of the form ζ − ζ for some 1 1 1 ω ∈ C (M;Z/n) that generates H (M;Z/n), and we can ex- constants a and b, so each of these factors of order Ω( n ). press ω in the original basis of C∗(M;Z/n). ( j,k) −α−2 Thuswe need to estimate det∂2 to a precisionof O(n ). After calculating ω, we can form the chain complex R∗ de- ( j,k) Each entry ∂2 is O(g), and therefore each (g − 1) × (g − 1) scribed in Section 3. However, we will want to generalize 2g the calculation, and instead of computing torsion over the ab- minor of the same matrix is O(g ) since the determinant ex- pansion has g! = O(gg) terms and each term is O(gg). So stract field Q(ζn) which may have exponential dimension over 2 Q, we will compute it over the complex numbers C. To this it suffices to estimate each of the O(g ) terms to precision ℓ −α−2 −2g−2 ( j,k) end, let ζn = exp(2πi/n), and let ζ = ζn for certain exponents O(n g ) in order for det∂2 (if it is then computed ℓ ∈ (Z/n)∗. Note that ℓ need not be a prime residue, only non- exactly) to have the desired accuracy. Moreover, each term is zero, so ζ may have some lower order m|n with m > 1. Then O(g), which requires O(log(g)) digits to the left of each dec- we can form the chain complex R∗(ζ), and its torsion has the imal point. Thus the total number of digits need to express same form as in equation (2): each entry is c a b −α−2 −2g−2 ∆(R∗(ζ)) = ζ (1 − ζ )(1 − ζ ). (3) d = O(log(g)+ log(n g )), 5

c which is polynomial in log(n), α, and g. We can then apply we can learn the sum g+(ζn) and the product h(ζn) of ζn and a+b+c a+c Theorem 4.1 to exactly compute the determinant with these ζn ; and the sum g−(ζn) and the product h(ζn) of ζn b+c approximate entries. and ζn . We can thus solve quadratic equations to obtain all four of these numbers. We can then learn the values of an ℓ To complete the proof of Theorem 1.1, recall that ζ = ζn . unordered pair {ζ sa,ζ sb}, where s = ±1, by taking ratios. Recall from equation (3) that the Reidemeister torsion is n n

def c a b If we calculate the torsion values (4) using floating point f−(ζ) = ∆(R∗(ζ)) = ζ (1 − ζ )(1 − ζ ). arithmetic over C, we obtain floating point approximations to We want to calculate several values of f to obtained simplified sparse sums:

2 sa 2πisa sb 2πisb def f−(ζ ) c a b ζ = exp( ) ζ = exp( ). f+(ζ) = = ζ (1 + ζ )(1 + ζ ) n n f−(ζ) def f (ζ)+ f (ζ) g (ζ) = + − = ζ c + ζ a+b+c + 2 def f (ζ) − f (ζ) We can then numerically calculate logarithms to obtain the ar- g (ζ) = + − = ζ a+c + ζ b+c − 2 guments 2πsa/n and 2πsb/n. If at this point we know z and 2 2 arg(z) to O(log(n)) digits of precision, we can calculate the def g+(ζ) − g+(ζ ) h(ζ) = = ζ a+b+2c. residues sa,sb ∈ Z/n by rounding their computed values to 2 the nearest integer. We can then take their ratio in the ring Z/n At this pointwe assume that n > 4, which we can do since oth- to obtain k±1. Working backwards, we it suffices to compute erwise we can compute the Reidemeister torsion of M directly each values of f+, g±, and h with O(log(n)) digits of preci- over the field Q(ζn). Using the three evaluations sion. We can use Lemma 4.3 to specify the precision at the beginning of the calculation in order to have enough precision 2 4 ∆(R∗(ζn)) ∆(R∗(ζn )) ∆(R∗(ζn )), (4) at this last stage.

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