arxiv:1310.3410v2 [math.gt] 29 nov 2013dedicated to professor sadayoshi kojima on the occasion of...

arX

iv1

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3410

v2 [

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29

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3

VERIFIED COMPUTATIONS FOR HYPERBOLIC 3-MANIFOLDS

NEIL HOFFMAN KAZUHIRO ICHIHARA MASAHIDE KASHIWAGI HIDETOSHI MASAISHINrsquoICHI OISHI AND AKITOSHI TAKAYASU

Dedicated to Professor Sadayoshi Kojima on the occasion of his 60th birthday

Abstract For a given cusped 3-manifoldM admitting an ideal triangulationwe describe a method to rigorously prove that either M or a filling of M admitsa complete hyperbolic structure via verified computer calculations Central toour method are an implementation of interval arithmetic and Krawczykrsquos TestThese techniques represent an improvement over existing algorithms as theyare faster while accounting for error accumulation in a more direct and userfriendly way

Contents

1 Introduction 2Acknowledgements 32 Gluing equations 321 Edge equations 422 Unfilled cusp equations 523 Filled cusp equations 724 Full set of equations 83 Krawczykrsquos Test 1031 Brief Review of Interval Analysis 1032 Machine Interval Arithmetic 1333 Automatic differentiation 1434 Complex arithmetic 1535 Example 164 Verification package 1741 The main algorithm 1742 Contents of our package hikmot 1843 How to use the package 185 Applications 1951 Census manifolds 1952 Exceptional surgeries on alternating knots 206 Advantages over existing methods 22Appendix A Verified computations for arg(z) 23A1 Arctangent 23A2 Evaluation of arg(z) 24References 25

2000 Mathematics Subject Classification Primary 57M50 Secondary 65G40Key words and phrases hyperbolic 3-manifold verified computation interval arithmetic

Krawczykrsquos Test

1

2 NHOFFMAN KICHIHARA MKASHIWAGI HMASAI SOISHI AND ATAKAYASU

1 Introduction

The study of 3-dimensional manifolds often abbreviated by 3-manifolds startswith the seminal papers by H Poincare in 1984ndash1904 In the last of these sixpapers he raised a question which became the famous Poincare Conjecture Thishas been one of the driving forces for (low-dimensional) topology and a greatamount of effort was spent in pursuit of a solution After nearly a century later in2002ndash2003 G Perelman finally reached to the end of the struggles by providing anaffirmative answer to the Geometrization Conjecture an extended version to thePoincare Conjecture in [28 29 30] See [23 25] as detailed references for example

The Geometrization Conjecture was raised by W Thurston who brought aboutdramatic changes to the study of 3-manifolds He introduced a geometric view inthe sense of Klein to the 3-manifold theory and in particular he incorporated anamazing application of non-Euclidean geometry namely hyperbolic geometry intothe study of 3-manifolds

As is well-known every closed 2-dimensional manifold ie closed surface admitsa geometric structure a complete Riemannian metric of constant sectional curva-ture Surprisingly Thurston predicted a similar situtation for the 3-dimensional case[35] That is every compact orientable 3-manifold can be canonically decomposedby cutting along essential 2-spheres and tori into the pieces each of which admitsa locally homogeneous geometric structure This is the Thurstonrsquos GeometrizationConjecture Actually he showed that there are exactly EIGHT geometries (iemodels of geometric structures) to be considered and essentially gave a proof forthe case of 3-manifolds containing decomposing tori

Among the eight geometries six have been well studied and are generally under-stood the Seifert fibered geometries The seventh is sol -geometry All manifoldsadmitting a sol geometric structure are torus bundles over the circle or n-fold quo-tients of torus bundles over the circle where n = 2 4 By most accounts the mostcommon and yet most interesting geometry occurs if a manifold M admits a hyper-bolic structure ie M admits a complete Riemannian metric of constant sectionalcurvature minus1 of finite volume See [22] for a good survey

In this paper we describe a computer-aided practical method to rigorously provethat a given 3-manifold admits a complete hyperbolic structure via verified calcu-lations This procedure is already implemented as a software and available at [11]

In the following we give a sketch of the contents with the organization of thepaper In the discussions that follow we will consider only 3-manifolds that areorientable compact with boundary consisting of the (possibly empty) disjoint unionof tori

Our program takes in as an initial input the combinatorial data of an ideal trian-gulation for the compact bounded case and a surgery description for the closed orpartially filled case From that data as Thurston first described we can find somealgebraic equations called a system of Gluing equations with complex variables Ifthis system of equations has a system of complex solutions with positive imaginaryparts then the given manifold admits a hyperbolic structure In the next sectionwe will recall a concise explanation for that mainly based on the well-known book[3] Note that this process has already been implemented as SnapPea by J WeeksUsing the kernel code by Weeks M Culler and N Dunfield implemented SnapPywhich is roughly speaking a SnapPea on python (For further background onSnapPy or the SnapPea kernel see the SnapPy documentation and [7]) By abusing

VERIFIED COMPUTATIONS FOR HYPERBOLIC 3-MANIFOLDS 3

notation in this paper when we say SnapPea we mean the Weeksrsquos kernel codeof SnapPea or the Weeksrsquos version of SnapPea By SnapPy we mean the actualprogram available at [7] SnapPea uses the Newtonrsquos method to solve the equa-tions and hence the solutions are just approximated ones These approximatedsolutions in principle would not prove the convergence of the Newtonrsquos methodThus although SnapPea is very practical it cannot give any rigorous mathematicalproof for a given manifold to be hyperbolic

To prove that a given system of equations actually have a desired solution weemploy interval arithmetic and Krawczykrsquos test both of which are explained in sect3We will give a brief review of how to obtain mathematically rigorous conclusionsfrom results of numerical computations

In sect4 the actual program hikmot and its implementation will be explainedWe further explain how to use it to rigorously prove the hyperbolicity of a giventriangulated manifold

The last two sections provide some of conclusions of our work open problemsand expected future work Also we will explain one application of our work to thestudy of exceptional surgeries on alternating knots which is a joint work of a partof the authors Ichihara and Masai This is followed by an appendix explaining theargument function

Acknowledgements

The authors would like to thank Mark Bell Marc Culler Nathan Dunfield CraigHodgson Sadayoshi Kojima and Bruno Martelli for a number of helpful conver-sations N H would like to thank Nihon University and the Tokyo Institute ofTechnology for graciously hosting him during a visit to Japan in the early stages ofthis project Finally for the remainder of this project N H has been supported bythe Max Planck Institute for Mathematics and Australian Research Council Discov-ery Grant DP130103694 K I was partially supported by Grant-in-Aid for YoungScientists (B) No 23740061 Ministry of Education Culture Sports Science andTechnology Japan The work of H M was supported by JSPS Research Fellowshipfor Young Scientists S O and A T were partialy supported by the CREST JSTproject tilted Establishment of Foundations of Verified Numerical Computationsfor Nonlinear Systems and Error-free Algorithms in Computational Engineering

Finally the authors wish to thank Tokyo Institute of Technology the Universityof Texas Math Department and Waseda University for providing computer supportfor this project

2 Gluing equations

This section establishes our much notation for the rest of our paper For themost part the notation is consistent the documentation for [7] to ease the readingof this paperrsquos companion computer code For further background on the hyper-bolic geometry used in this paper we refer the reader to [3] and [34] For somedetails on the various methods used by SnapPy or the SnapPea Kernel to constructthe necessary system of equations we refer the reader to [37] and of course thedocumentation for SnapPy [7]

As noted in the introduction we will assume our manifolds are orientable Wepoint out that our algorithm can verify the hyperbolicity of a non-orientable man-ifold Q by establishing the hyperbolicity of the orientable double cover of Q


Using the upper half space as our model for hyperbolic 3-space H3 we canidentify the ideal boundary partH3 with C cup infin Also we denote by H3 the unionH3cuppartH3 The group of orientation preserving isometries acting on H3 is identifiedwith PSL(2C) in the standard way

Refining our definition of the introduction a 3-manifold M is said to admit ahyperbolic structure if M sim= H3Γ where Γ sub PSL(2C) is discrete and the integralof the volume form for H3 over the fundamental domain for the quotient H3Γ isfinite For ease of notation when M admits a hyperbolic structure we will oftenjust consider it as the quotient H3Γ

An ideal tetrahedron T is a tetrahedron inH3 with all four vertices z1 z2 z3 z4 subpartH3 There is an isometry γ of H3 such that γ(T ) has vertices 0 1infin z suchthat infin 0 1 z and z1 z2 z3 z4 have the same cross ratio Under this definitionz is only well defined up to the choice of which ordered subset of points we send tothe ordered set 0 1infin However z is defined up to z zminus1

z and 11minusz Furthermore

we can label each edge in our ideal tetrahedron by z zminus1z and 1

1minusz such that the

argument of the complex number is the dihedral angle along that edge (see Figure1) By convention we will choose a complex argument function arg(w) such thatthe range is (minusπ π] for w isin C 0 Also we define log z = log |z| + i arg(z) Ifw is any complex parameter associated to the edge of an ideal tetrahedron witharg(w) lt 0 we say the corresponding ideal tetrahedron is negatively oriented andif arg(w) gt 0 we say the corresponding ideal tetrahedron is positively orientedUltimately the algorithm described by this paper certifies when there is a solutionto the Gluing equations such that all tetrahedral parameters are positively oriented

A truncated tetrahedron is constructed by removing a neighborhood of the idealpoint (see Figure 2) Technically a cusped manifold (eg a knot complement) istriangulated by ideal tetrahedra and compact manifold with non-empty toroidalboundary (eg a knot exterior) is triangulated by truncated tetrahedra Truncatedtetrahedra have the property that opposite dihedral angles are equal as well as theproperty that all angles on a triangular face sum to π (the product of the complexparameters is minus1)

We now begin the process of describing the gluing equations associated to atriangulation T of a hyperbolic 3-manifold M Ultimately there will be n+2k+ hequations where n is the number of tetrahedra in T k in the number of unfilledcusps of M and h is the number of filled cusps of M For convenience we will usethe parameter m throughout this section to denote the mth equation in the set ofn+2k+h equations we construct in the following paragraphs We will first assumethat h = 0 and then extend to the general case in sect23 In sect24 we give a completeset of n+ 2k + h equations before finally providing the set of n equations that aresolved by our computer algorithm

21 Edge equations Each edge em in a triangulation T of M is an equivalenceclass of edges in the set of n tetrahedra that make up T In this disjoint union oftetrahedra we fix a labeling of the edges in the jth tetrahedron by the complex

parameters zj zjminus1zj

or 11minuszj

such that the opposite dihedral angles get the same

label By Euler characteristic conditions there are n equivalence classes of edgesin T determined by the gluing identification We can record by ajm the numberof times an edge labeled by zj is a member of the equivalence class em Similarly

we can define bjm as the number of times an edge labeled byzjminus1zj

is part of the


equivalence class em and finally cjm as the number of times an edge labeled by1

1minuszjis part of the equivalence class em If T is a tessellation of H3 with em an edge

between 0 and infin then the product of all the complex parameters around em is 1and the sum of their arguments is 2π In fact for any edge em there is an isometryof H3 that sends the endpoints of em to 0 and infin Thus around the mth edge weget the equations

nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(1)

andnsum

j=1

ajm middot arg (zj) + bjm

(1

1minus zj

)middot arg+cjm middot arg

(zj minus 1

zj

)= 2πi(2)

Note for a fixed jsumn

ℓ=1 ajℓ = 2sumn

ℓ=1 bjℓ = 2 andsumn

ℓ=1 cjℓ = 2 and sinceeach of these coefficients is non-negative each ajm bjm or cjm is in 0 1 2

The notation above can also be presented in a more standard way using loga-rithms

nsum

j=1

ajm log (zj) + bjm log

(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 2πi(3)

22 Unfilled cusp equations Although this sets up a system of n equations withn unknowns this system does not specify a unique hyperbolic structure (see [34Theorem 56] for example) Instead we must add more equations correspondingto the k cusps of M The construction is as follows For the tth cusp of M thereis a peripheral torus Ct When viewed from a cusp any fundamental domainfor Ct can be identified with a quotient of the plane by two translations microt λtIntersecting this plane with the 2-skeleton of T results the plane tessellated bytriangles In the 1-skeleton of this tessellation microt λt lift to two oriented piece-wiselinear curves which we denote by γtm and γtl respectively (see Figure 3) Thefollowing conditions ensure that Ct is the quotient having a Euclidean structureX by a group of translations generated by microt and λt At each vertex v along thepath γtm or γtl we can take the product of dihedral angles to the right side ofthe oriented curve pv It is a consequence of [3 Lemma E68] that along a pathcontaining v vertices this condition is equivalent to the product of all complexparameters the pv is (minus1)v and so we build the mth equation corresonding tothe meridian microt as follows For the jth tetrahedron let (minus1)vajm be the numberof times the complex parameter zj is part of this product and let (minus1)vbjm be

the number of times the complex parameterzjminus1zj

shows up in pv Finally let

(minus1)vcjm be the number of times the complex parameter 11minuszj

corresponds to an

angle along this product This method yields the following 2k pairs of equations

nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(4)


Figure 1 An ideal tetrahedron with vertices at 01 infin and z

Figure 2 A truncated tetrahedron with labeled edges

and

nsum

j=1

ajm middot arg (zj) + bjm middot arg(

1

1minus zj

)+ cjm middot arg

(zj minus 1

zj

)= 0(5)

The m + 1st equation will follow the same recipe but we will replace the pathdetermined by microt with the path determined by λt

We note that SnapPea uses a slightly different method however using identities

such as zj middot 11minuszj

middot zjminus1zj

= minus1 the equations can be seen as equivalent Furthermore

in the definition above the coefficients ajm bjm and cjm would have the same sign

and have absolute value less than 4 but under the identity zj middot 11minuszj

middot zjminus1zj

= minus1 the

sign of the coefficients can change Just as above both conditions can be satisfied


Figure 3 The fundamental domain for Ci with an identifiedboundary In this example γtm is the path between the two dotsand the product at point p2 that contributes to the equation forγtm is (minus1)3w1ws where wi denotes the complex parameter ofan edge with endpoint at the cusp

ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)

Thus since M has n tetrahedra and k cusps (all unfilled) we have a system ofn+ 2k equations

23 Filled cusp equations A cusped 3-manifold M can be Dehn filled alonga cusp Ct

sim= T 2 times [0infin) if the set T 2 times (0infin) is removed and the boundaryT 2 times 0 is identified with the boundary of a solid torus These identifications canbe parametrized by the curve γ in π1(Ct) = 〈microjt λjt|[microjt λjt]〉 that gets identifiedwith the curve that bounds a disk in the solid torus If M is hyperbolic all but atmost finitely many choices of γ = pmicrojt + qλjt will yield a hyperbolic manifold byThurstonrsquos Hyperbolic Dehn Surgery Theorem [34] Furthermore the process canbe repeated if M is filled along multiple cusps

Denote by N a manifold that comes from filling h cusps of a manifold M Usingthe convention that M decomposes into n tetrahedra and has k cusps we willassume that the system of n+2k equations described above is already constructed

Denote by ΩM the set of gluing equations for M For each filled cusp in N there is of course a corresponding unfilled cusp of M and so assume that rth andr + 1st equations in ΩM correspond to the meridian and longitude (respectively)of this cusp in M For the equation set of N we replace these two equations by (asimplified form of) the following

nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)

and similarly for the argument functions


p middotnsum

j=1

(ajr middot arg (zj) + bjr middot arg

(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1

(ajr+1 middot arg (zj) + bjr+1 middot arg

(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π

We can also express these equations using logarithms

p

nsum

j=1

ajr log (zj) + bjr log

(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1

ajr+1 log (zj) + bjr+1 log

(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)

Again to show that a given triangulation of N corresponds to a hyperbolicstructure the solution set to these equations must be 0 dimensional and have atleast one solution corresponding to all positively oriented tetrahedra Howeverin the case where N has filled cusps the solution corresponds to an incompletehyperbolic structure We refer the reader to [3 sectE6-iv] for the details of how toextend this to a complete hyperbolic structure

24 Full set of equations In the paragraphs above there are two equivalentsets of equations The equations with polynomials and arguments we will callthe rectangular equations and the equations with logarithms we will call the logequations For a manifold N with n tetrahedra k unfilled cusps and h filled cuspsthen there are n+2k+h log equations needed to be solved and n+2k+ h pairs ofpolynomial and argument equations It is common to combine the features of bothof these equations in a often more convenient form by removing the third term ineach summand of the log equations (see [3 page 235] for example)

nsum

j=1

(ajm minus cjm) log (zj) + (minusbjm + cjm) log (1minus zj) + cjm middot πi = ǫm

n+2k+h

m=1

(10)

Here the constant term ǫm is 0 if the equation corresponds to an unfilled cusp and2πi otherwise Finally Equation 10 can be expressed as polynomial and argumentequations as follows

nprod

j=1

(zj)(ajmminuscjm) middot (1minus zj)

(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1

arg((zj)(ajmminuscjm)) + arg((1 minus zj)

(minusbjm+cjm)) = ǫm minusnsum

j=1

cjm middot πi

n+2k+h

m=1


where just as above we choose ǫm to be 2πi for equations corresponding to edgesor filled cusps and 0 in the case of an unfilled cusp This structure is unique by thefollowing result of Thurston It is essentially a direct consequence of [34 Corollary573] however we present as follows to better fit the discussion of this section

Theorem 21 (Thurston) If the gluing equations admit a solution each of whoseimaginary part is positive (positive solution in short) then M admits a completehyperbolic structure of finite volume Furthermore the structure is determined by thesolution and is unique up to homeomorphism and re-triangulation of the manifold

Also to ease the notation we define αjm = ajmminus cjm and βjm = minusbjm+ cjmFurthermore we can define (n+ 2k + h)times 2n matrix ΛM as follows

ΛM =

α11 middot middot middot αn1 β11 middot middot middot βn1

α1n+2k+h middot middot middot αnn+2k+h β1n+2k+h middot middot middot βnn+2k+h

(12)

The numbering of the equations in this section most closely describes the num-bering convention of SnapPea However in the computer argument that followswe start by changing the order of the equations such that for a manifold M withn tetrahedra k unfilled cusps and h filled cusps equations n such that the first nequations correspond to edge equations the next 2k equations correspond to un-filled cusps and the final h equations correspond to filled cusps Namely we can

make a new matrix ΛM such that the first k + h rows are selected from rows ofΛM corresponding to either filled cusps or meridian equations of unfilled cusps andthen choose a set of nminus (k+ h) equations such that the corresponding rows of ΛM

have maximal rank finally denote by S the total set of rows of selected to form

ΛM The following theorem of Neumann and Zagier [27] shows that if the gluing

equations admit a positive solution then ΛM has rank n (see also [26 Lemma 24]and [3 E6ii-iv])

Theorem 22 (Neumann amp Zagier [27]) If M and the associated triangulation

correspond to a hyperbolic structure ΛM has rank n Furthermore in this casesolutions to the following equations

nprod

j=1

(zj)αjm middot (1minus zj)

βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1

arg((zj)αjm) + arg((1minus zj)

βjm) = ǫm minusnsum

j=1

cjm middot πi

misinS

form a zero dimensional algebraic set

As it suffices to solve only the equations (both polynomial and argument) ofin the theorem above the remainder of this paper will be dedicated to verifyingthat a solution to the system of equations in (13) can be obtained from a suitableapproximation Also we have taken care to define the right hand side of (13)especially with regard to the right hand side of the argument equations We makethe following remark to justify this attention


Remark By [3 Lemma E63] there is no need to check the argument conditionfor any equation corresponding to an edge However this condition must be checkedfor any equation coming from a filled or unfilled cusp For example the completesolution to the equations for the census manifold m007 is also a solution to thepolynomial part of the rectangular equations for m007(3 1) but the argumentcondition of the equation corresponding to the filled cusp is not satisfied by thatsolution

3 Krawczykrsquos Test

In this section we present an algorithm which performs a mathematically rig-orous numerical existence test for solutions of gluing equations (11) In such analgorithm one must deal with all errors in numerical computation including trun-cation errors and rounding errors In usual numerical computations one uses thefloating-point arithmetic The floating-point arithmetic approximates the real andcomplex arithmetic It is well-known that it may occur a rounding error in everyfloating-point operation Thus it is necessary to discuss how to obtain mathemat-ically rigorous conclusions from results of numerical computations based on thefloating-point arithmetic For that purpose the idea of interval arithmetics is use-ful Historically the concept of interval arithmetic has been proposed in 1950rsquos[32 33 24] Since then a lot of work has been done in this area (see [19 24 31] forexample) In the following we first review how to implement the interval arithmeticby using the floating-point arithmetic In the end of this section a simple exampletreating the gluing equation is shown for demonstration

31 Brief Review of Interval Analysis Let us write an interval on the set ofreal numbers R as X = x isin R x le x le x x x isin R = [x x] We denote theset of such intervals by IR The midpoint of the interval mid(X) isin R is defined by

mid(X) =x+ x

2

The radius of the interval X is defined by

rad(X) =xminus x

2

Let us assume that a unary operation u R rarr R and a binary operation RtimesR rarrR are defined Most unary operations we consider here are standard functionsThen we can extend these operations to an unary operation u IR rarr IR anda binary operation IR times IR rarr IR including basic four arithmetic operations isin +minus middot by

(14) u(X) = u(x) x isin X

and

(15) X Y = x y x isin X y isin Y

respectively Here X Y isin IR Although in order to calculate four basic operations isin +minus middot which are called the interval arithmetics it seems to need infinitelymany computations only finitely many calculations of end points of intervals are


sufficient Namely the interval arithmetic can be executed by

X + Y =[x+ y x+ y

]

X minus Y =[xminus y xminus y

]

X middot Y =[minx middot y x middot y x middot y x middot ymaxx middot y x middot y x middot y x middot y

]

XY = X middot[1

y1

y

] (0 6isin Y )

for X = [x x] and Y = [y y]An interval vector is defined as m-tuple of interval entries satisfying

X = x isin Rm xi isin Xi for 1 le i le m

for Xi isin IR which is the Cartesian product of one-dimensional intervals The set ofinterval vectors is denoted by IRm The set of interval matrices IRmtimesm is definedsimilarly The midpoint and the radius of an interval vector or an interval matrixare also defined component-wisely For example letting A isin IRmtimesn be an intervalmatrix the mid(A) is defined by

mid(A) =

mid(a11) middot middot middot mid(a1n)

mid(am1) middot middot middot mid(amn)

isin R

mtimesn

for aij isin IRNow we shall discuss how to solve the system of gluing equations (11) via interval

arithmetic The system (11) can be rewritten as

nprod

j=1

zαjm

j (1minus zj)βjm = γm m = 1 middot middot middot n

Note that as discussed in sect24 αjm βjm γm isin Z Theorem 22 states that thesystem of equations (11) has at most n-independent equations To compute thematrix S in Theorem 22 we need to compute the ranks of matrices Note that ifwe compute the rank naively it involves a lot of calculations and hence it is notvery easy to rigorously compute the rank An efficient way to conjecture the rankof a matrix A is to use the singular value decomposition of A because the rank isequal to the number of non-zero singular values In actual numerical computationone can only have approximated singular values Thus we set a threshold δ gt 0which on the order of 10minus8 Then if all singular values are greater than δ thenwe can expect that a given matrix has full rank We can thus compute a candidateof S by choosing n minus k rows of ΛM one by one so that at each step the resultingmatrix has full rank In the event that our threshold was too big and we can notfind any candidate we randomize the triangulation of M with SnapPy and try tosolve another system of gluing equations

We assume here that we have successfully selected a candidate of S and hence

ΛM isin Rntimes2n in Theorem 22Setting α+

jm = maxαjm 0 αminusjm = minusminαjm 0 β+

jm = maxβjm 0 andβminusjm = minusminβjm 0 we can rewrite selected n equations corresponding to ΛM


as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm

j (1minus zj)βminus

jmγk = 0 k = 1 middot middot middot n

For z = (z1 middot middot middot zn) isin Cn the gluing equation (16) can be rewritten as g(z) = 0where g is a mapping from Cn onto itself Setting zj = x2jminus1 + ix2j (j = 1 middot middot middot n)we can rewrite (16) further as

(17) f(x) = 0

where f R2n rarr R2n is a differentiable nonlinear real mapping Let us put m = 2nBased on this preparation we now discuss how to use the interval arithmetic to

prove the existence and local uniqueness of a solution for (17) in the interval vectorX isin IRm Let c isin X be an approximation of a solution of f(x) = 0 We introducea simplified Newton mapping s Rm rarr Rm for the mapping f by

(18) s(x) = xminusRf(x)

where R isin Rmtimesm is a certain matrix Usually R is chosen to be an approximateinverse of f prime(c) If R is invertible f(x) = 0 and s(x) = x becomes equivalent Fromthe contraction mapping principle if we can show s to be a contraction mappingfrom X into itself say s is a contraction on X for short then we can prove theexistence and local uniqueness of solution for (17) on X provided that R is invert-ible To check whether s is a contraction on X usually Newton-Kantorovich typetheorem [14] is applied For the finite dimensional case so-called Krawczyk test[15 16 31] is often used since in many cases its implementation is straightforwardand easy compared with the direct application of Newton-Kantorovich type theo-rem For equation (17) we will show it is the case So-called Krawczykrsquos mappingK IRm rarr IRm a map proposed in [15] with respect to the gluing equation (17)is defined by

K(X) = cminusRf(c) + (I minusRf prime(X))(X minus c)(19)

where I isin Rmtimesm is a unit matrix We note that K(X) is a mean value formof s If we consider s(X) based on the naive interval extension which is justobtained by replacing the basic four arithmetic operation on reals with those ofintervals it follows that s(X) sup X so that we cannot expect s(X) sub int(X) Hereint(X) = x = (x1 middot middot middot xm) isin X xi lt xi lt xi (i = 1 middot middot middot m) On the contraryto this

K(X) sub int(X)

can be expected Indeed the following theorem holds which states a sufficientcondition that there exists a solution of (17) which is unique in X On the basis ofKrawczykrsquos mapping [15] Krawczykrsquos test is established by SM Rump [31]

Theorem 31 (Krawczykrsquos test) For a given interval X isin IRm let int(X) be theinterior of X If the condition

K(X) sub int(X)(20)

holds then there uniquely exists an exact solution xlowast of (17) in X Furthermore itis also shown that R and all matrices C isin f prime(X) including f prime(xlowast) are nonsingular

In the following we discuss how to implement Krawczykrsquos test using the floating-point arithmetic


32 Machine Interval Arithmetic While there exist many definitions of float-ing point systems in order to simplify the discussion we consider the floating-pointnumber system obeying IEEE 754 standard [13] Let F be a set of IEEE 754 doubleprecision floating-point numbers In IEEE 754 standard four rounding operationsfrom R to F are defined The rounding to the nearest fl R rarr F is one of themThis is defined through

|fl(x) minus x| = min|cminus x| c isin Ffor x isin R with |x| le max|a| a isin F As well as rounding to nearest IEEE754 standard defines the rounding towards minusinfin fldown R rarr F and the roundingtowards infin flup R rarr F by

fldown(x) = maxc isin F c le xand

flup(x) = minc isin F x le crespectively Although we omit the explanation IEEE 754 also defines roundingtowards zero It is known that almost all CPUs including Intel processors aredesigned to satisfy IEEE 754 standards

Let IF sub IR denote the set of intervals with floating-point end points [x x] x x isin F x le x We define a rounding IR rarr IF by

([x x]) = [fldown(x) flup(x)]

Using this the basic four arithmetic operations of intervalsXY isin IF IFtimesIF rarrIF are defined by

X Y = (X Y )

where isin +minus middot This is called a machine interval arithmetic The inter-val vectors and matrices with floating-point end points IFm IFmtimesm are also theCartesian product of one-dimensional intervals

Although we introduce this notation to stress the operations the computer isactually perfroming one of the advantages of interval arithmetic is that one canoverwrite a computerrsquos calls for the basic operations of arithmetic with the defini-tions above Similarly for the remainder of the paper we will suppress the notation in favor of the less cumbersome +minus middot

Since the gluing equation (17) is based on the basic four arithmetic operations wecan replace each arithmetic by its corresponding interval arithmetic Then by thisreplacement we can construct mappings F IFm rarr IFm and F prime IFm rarr IFmtimesm

for any X isin IFm satisfying

(21) F (X) sup f(x) forallx isin Xand

(22) F prime(X) sup f prime(x) forallx isin X respectively where f prime(x) is the derivative of f A map satisfying (21) (resp (22))is called an interval extension of f isin Rm rarr Rm (resp f prime isin Rm rarr Rmtimesm) Tocompute F prime(X) we use so-called automatic differentiation which we will explainin sect33

The extended Krawczyk mapping KF IFm rarr IFm is defined by

KF (X) = cminus RF (c) + (I minusRF prime(X))(X minus c)


Obviously KF (X) sup K(X) holds for any X isin IFm Thus if KF (X) sub int(X)then K(X) sub int(X) is satisfied Therefore if we can find computable intervalextension F of f which is the case of the gluing equation (17) as mentioned abovewe can have computable Krawczykrsquos test KF (X) sub int(X) Now we consider howto choose the candidate interval X isin IF

m on which s may be contractive Letc isin Fm be an approximate solution of (17) given by a certain numerical methodeg Newtonrsquos method via SnapPea For a given vector x = (x1 middot middot middot xm) isin Fm letus define the maximum norm of x by xinfin = max1leilem |xi| In our code namedhikmot we choose

X =

[c1 minus r c1 + r][c2 minus r c2 + r]

[cm minus r cm + r]

(23)

as a candidate interval where r = 2Rf(c)infin

33 Automatic differentiation Now we explain how to calculateKF (X) for aninterval vectorX isin IFm Most of the difficulty comes from the evaluation of F prime(X)Therefore we now explain how to calculate F prime(X) for X isin IFm For calculatingF prime(X) one may consider using symbolic computation However the computationcosts easily become too high to get useful results There is a reasonable alternativea notion called the automatic differentiation which enables us to calculate F prime(X)in a mathematically rigorous way Although there are several ways of implementingautomatic differentiation we here explain a method called bottom up automaticdifferentiation For its implementation we first prepare an automatic differentiationobject which is a pair of a data structure and set of operations among the objectsin that structure The data structure of an automatic differentiation object is

(d0 d1 d2 middot middot middot dm) isin IFm+1

We now define several operators on this data structure Let u R rarr R be aunary operation and U IR rarr IR be its interval extension We assume u is adifferentiable map We denote the derivative of u by uprime R rarr R Note that thederivative uprime we use here will be computed by hand and expressed in terms of thestandard operations Let U prime IR rarr IR denote an interval extension of uprime Thenwe define a map U(p) IFm+1 rarr IFm+1 which will be associate to u in the bottomup automatic differentiation as a map that maps p = (p0 p1 middot middot middot pm) isin IF

m+1 to

U(p) = (r0 r1 middot middot middot rm) isin IFm+1 where

r0 = U(p0)

ri = U prime(p0)pi (i = 1 middot middot middot m)

Next let b Rtimes R rarr R be a differentiable binary operation (x y) 7rarr b(x y) Theinterval extension of b is denoted by B IR times IR rarr IR We prepare by handspartial derivatives partxb R times R rarr R and partyb R times R rarr R of b with respect tox and y respectively For example Table 1 shows partxB(p0 q0) and partyB(p0 q0) forbasic four arithmetic operations Let partxB IRtimes IR rarr IR and partyB IRtimes IR rarr IR

denote the interval extensions of partxb and partyb respectively We now define a map

B IFm+1 times IF

m+1 rarr IFm+1 which will be associated to B in the automatic


differentiation as follows for p = (p0 p1 middot middot middot pm) q = (q0 q1 middot middot middot qm) isin IFm+1

we define B(p q) = (r0 r1 middot middot middot rm) by

r0 = B(p0 q0)

ri = partxB(p0 q0)pi + partyB(p0 q0)qi (i = 1 middot middot middot m)

Table 1 Partial derivative of basic four arithmetic operations

partxB(p0 q0) partyB(p0 q0)p+ q 1 1pminus q 1 minus1p middot q q0 p0pq 1q0 minusp0q

20

Based on the above discussion let us now consider a map f = (f1 middot middot middot fm)T

with fi Rm rarr R (i = 1 middot middot middot m) being differentiable functions of x = (x1 middot middot middot xm)

We assume that for the calculation of the value f(x) we have an algorithm whichconsists of differentiable unary operations and differentiable binary operations Nowwe will explain how to calculate F prime(X) for X = (X1 middot middot middot Xm) isin IFm Let Fi IRm rarr IR be the interval extension of fi For j = 1 middot middot middot m we denote partialderivatives of fi by partxj

fi Rm rarr R whose interval extensions are defined by

partxjFi IR

m rarr IR We define X isin IFmtimes(m+1) by

X =

X1

X2

Xm

=

X1 [1 1] [0 0] middot middot middot [0 0]X2 [0 0] [1 1] middot middot middot [0 0]

Xm [0 0] [0 0] middot middot middot [1 1]

for Xi isin IFm+1 Starting with X replacing each operation in the algorithm for

calculating the function f by operations of the bottom up automatic differentiationwe have an extension of F as

F (X) =

F1(X) partx1F1(X) partx2

F1(X) middot middot middot partxmF1(X)



Fm(X) partx1

Fm(X) partx2Fm(X) middot middot middot partxm

Fm(X)

isin IF

mtimes(m+1)

where F IFmtimes(m+1) rarr IFmtimes(m+1) A submatrix of F (X) consists of the secondcolumn to the last column of matrix is nothing but F prime(X)

34 Complex arithmetic We will now show how to compute F (X) using thecomplex mapping g in (16) For i = 1 middot middot middot n a data structure of complex object isdefined by

(24) Zi =(X2iminus1 X2i

)isin IF

(m+1)times2

for a pair of automatic differentiation objects X2iminus1 and X2i For given z =(zre zim) and w = (wre wim) zre zim wre wim isin IF

m+1 the basic four arithmetic


operations between z and w are defined by

z + w = (zre + wre zim + wim)

z minus w = (zre minus wre zim minus wim)

z middot w = (zrewre minus zimwim zrewim + zimwre)

z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)

Here these expressions +minus middot are defined through the operations of the automaticdifferentiation

Let us denote the relation (24) by

X = idCrarrR(Z)

or

Z = idRrarrC(X)

Replacing all operations for calculating the complex function g by operations ofthe bottom up automatic differentiation mixed with complex transformation wecan define an extension of g by G IFntimes((m+1)times2) rarr IFntimes((m+1)times2) Using GF (X) can be calculated as

F (X) = idCrarrR

(G(idRrarrC

(X)))

35 Example As an example we shall consider the rectangular equation for thecensus manifold 4 1(51) Assume that indices αjm βjm and γm are given by

αjm =

(5 92 2

) βjm =

(0 minus7minus1 minus1

) γm =

(minus11

)

Then the gluing equation (16) becomes

(25)

z51(1minus z1)

0z92(1 minus z2)minus7 = minus1

z21(1minus z1)minus1z22(1 minus z2)

minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z

22 minus (1minus z1)(1 minus z2) = 0

The equation (25) can be written as g(z) = 0 where g C2 rarr C2 is defined by

g1(z) = z51z92 + (1 minus z2)

7

g2(z) = z21z22 minus (1 minus z1)(1minus z2)

Assuming that an approximate solution1 of g(z) = 0 is given by

z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)

Our candidate interval X isin IF4 is chosen by

X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]

1 The decimal notation of approximate solution may include a rounding error when printing to

the screen The exact value of the approximate solution is described as ((0x109478e0b57659times2minus3 )+(0x17dfbed38abdaetimes2minus2)i (0x128cbf134a88detimes22 )+(0x1afeb2e24accfdtimes20)i) in the hexa-decimal In the rest of the paper we note that the decimal numbers might include some roundingerrors


The Krawczyk mapping of X is calculated as

KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]

Thus KF (X) sub int(X) holds Since we obtain the sufficient condition of thecomputable Krawczykrsquos test the existence of a unique solution of (25) in X isproved

4 Verification package

In this section we explain our package hikmot (v100) that verifies the hyperbol-icity of a given manifold M Our package is available at [11] We will first explainthe main algorithm and then give instructions on how to use our package

41 The main algorithm We here explain how hikmot proves the hyperbolicityof a given triangulated manifold We call a triangulation with positively orientedapproximated solutions by SnapPea a good triangulation In the parlance of Snap-Pea a good triangulation corresponds to an approximation with solution type lsquoalltetrahedra positively orientedrsquo First we assume that there is a good triangulationfor M A rough picture of our verification package is shown in Algorithm 1

Algorithm 1 Verify hyperbolicity of M

Require M has a good triangulationEnsure M admits hyperbolic metric of finite volumeApply Krawczykrsquos test to the approximated SnapPearsquos solution for (7)if The convergence has been verified and the imaginary parts of the solutions areall positive then

Check arguments condition (8)if The arguments condition is also satisfied thenreturn [True the set of intervals that contains rigorous solutions]

end if

end if

return [False a collection of empty sets]

We will now explain

bull how we apply Krawczykrsquos test andbull how to check argument condition (8)

in detail Note that we use the machine interval arithmetic for every computationGiven a good triangulation we first apply Newtonrsquos method five more times to

the approximated solution by SnapPea to get a more precise solution Then weapply Krawczykrsquos test (Theorem 31) to the Krawczyk mapping KF and the set ofintervals X explained in (19) and (23) respectively If the condition (20) that isKF (X) sub int(X) in Theorem 31 holds then we make a list L = [True X ] Thismeans that our verification of the convergence of Newtonrsquos method have succeededand X is a set of intervals each contains the rigorous shape of a tetrahedron IfKrawczykrsquos test fails we put L = [False E] where E is a collection of the emptysets


Next we will explain how to check argument condition (8) We first assume thatKrawczykrsquos test has succeeded ie L[0] = True We will use the set X = L[1] ofintervals First we note that the sum of arguments is in Z middot 2πi Hence to check theargument condition (8) we only need to ensure that the sum of arguments containsonly desired value We have prepared the function atan2 whose input is an intervalIC of complex numbers and output is an interval of real numbers that contains theset arg(z) | z isin IC Essentially we have used the theory of Taylor expansionSee appendix A for more detail about this atan2 Then by using our atan2 wecompute the sum of arguments and then check if the resulting interval containsonly our desired value Thus we can verify that our solution satisfies condition (8)

If L[0] = True and the argument condition (8) is ensured then we have a proofthat M is a hyperbolic 3-manifold of finite volume and our package hikmot returnsL = [True X ] Otherwise ie if L[0] = False or (8) is not satisfied then we returnLprime = [False E]

Note that even if our verification fails for a given triangulation it is still possiblethat the manifold M has other good triangulations whose solutions can be verifiedby our package In practice we

bull randomize triangulations of M in order to get another good triangulationand

bull try to verify the hyperbolicity by using Krawczykrsquos test

several times

Remark By using X = L[1] of the set of intervals for shapes of tetrahedra inprinciple we can compute other invariants with rigorous error control

42 Contents of our package hikmot Our package hikmot contains 3 foldersand 1 python files and READMEtxt First the folder ldquopython srcrdquo contains

bull hikmotpy the main file of our program The function verify hyperbolicityin hikmotpy is the one that proves the hyperbolicity of a given triangulatedmanifold see sect41

bull intervalpy complexpy and ftostrpy files that prepare interval arithmeticsee sect31

bull rankpy we use this file to guess the rank of S in Theorem 22 see alsosect31

bull manifoldpy this is file has been adapted from a file originially written byBruno Martelli [20]

The folder ldquocpp srcrdquo contains

bull the folder ldquokvrdquo that is a set of header files(autodifhpp hwroundhppmake-candidatehpp complexhpp interval-vectorhpp intervalhpp matrix-inversionhpp converthpp kraw-approxhpp rdoublehpp)

bull krawczykcc this implements Krawczykrsquos test see sect3

We will use setuppy to install our package see next subsection

43 How to use the package In this section we explain how to use our packageon Linux or Mac machine

We assume that SnapPy has been installed as a python module (see the docu-mentation of [7] for installation instructions) We also assume boost [4] has beeninstalled Our package depends on the OS CPU and compiler Therefore usersneed to compile the code on the machines which they will use


Installing command To install hikmot as a python module use the commandas a superuser

bull python setuppy install

This will automatically compile our code We would like to thank Nathan Dunfieldfor writing setuppy which eases the installation processTo use the code

(1) Install the code(2) Run python and import snappy and hikmot(3) The module hikmotverify hyperbolicity(M print data) is the verifier

(a) M should be a SnapPyrsquos manifold with solution type ldquoall tetrahedrapositively orientedrdquo

(b) If print data = True then it prints out several data

The function hikmotverify hyperbolicity returns a list L = [ldquoTruerdquo or ldquoFalserdquoX or E] where X (resp E) is a set intervals of tetrahedra shapes (resp emptysets) as explained in section 4Sample On a python interpreter we can use our codes as follows

gtgtgt import snappy

gtgtgt import hikmot

gtgtgt M = snappyManifold(4_1(51))

gtgtgt L = hikmotverify_hyperbolicity(M)

gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]

gtgtgt Mtetrahedra shapes(rect)

[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications

51 Census manifolds SnapPy has several censuses of manifolds (see the docu-mentation of [7] for the list of censuses) In this section we report a computer aidedverification of the hyperbolicity of manifolds in several censuses

Theorem 51 All the manifolds in OrientableCuspedCensus [5] are hyperbolic

Proof We use VerifyCuspedCensuspy (available at [11]) that applies our packageto every manifold in OrientableCuspedCensus Here we summarize the result

python VerifyCuspedCensuspyOut of 17661 manifolds in the closed census 17661 are hyperbolic and 0remain


So these manifolds remain []

On Mac OS X 1068 with Intel Core 2 Duo of speed 213 GHz it takes about 8minutes

We also verified the hyperbolicity of manifolds in Hodgson-Weeks closed census[10]

Theorem 52 All the manifolds in OrientableClosedCensus are hyperbolic

Proof For several manifolds in this SnapPy census it is difficult to find good trian-gulations The census contains 11031 manifolds and among them we can find goodtriangulations for 10989 manifolds either directly or by randomization of triangu-lations Then our package proves the hyperbolicity of those 10989 manifolds Forthe remaining of 42 manifolds we will use the list dehngz which came with theolder versions of SnapPea The list contains several surgery descriptions for eachmanifold in the census and often yields other good triangulations More specificallyby looking at surgery descriptions on the list dehngz and randomization among42 manifolds we get good triangulations and our package proves the hyperbolicityfor 38 of 42 with only [m007(31) m135(13) v3377(-12) v2808(-51)] remainingHere we are using SnapPearsquos notation For example m007(31) is a closed mani-fold obtained by filling 1-cusped manifld m007 with slope (31) For the remainingfour manifolds we apply Algorithm 2 to get good triangulations The authors wereinformed the main idea of Algorithm 2 by Craig Hodgson Note that in practicewe also randomize triangulations at each step of Algorithm 2 We can find a goodtriangulation by this code for [m135(13) v3377(-12) v2808(-51)] and our veri-fication package works for those triangulations and proves the hyperbolicity Form007(31) we need to take covering of degree 3 Since the first homology of m007is Z oplus Z3Z there is a covering N of degree 3 with 3 cusps Then by filling eachcusp of N by the slope (31) we have a degree 3 covering N prime of m007(31) Then byapplying Algorithm 2 to N prime we can get a good triangulation Our package provesthe hyperbolicity of N prime and hence m007(31) is hyperbolic

Remark The randomization function on SnapPea utilizes rand() function of clanguage The function rand() depends on compiler and machine For the proof ofTheorem 52 we used Mac machine with OS 1075 For the repeatability we pre-pared the set ClosedManifoldszip (available at [11]) of closed manifolds with goodtriangulations each corresponds to a manifold in OrientableClosedCensus Onecan check the hyperbolicity of those manifolds by using VerifyClosedCensuspyalso available at [11]

Finally hikmotverify hyperbolicity(Mprint data = False save data=True) gen-erates a file for M that gives all of the internal data used to compute Krawczykrsquostest (Theorem 31) which is the key step of our verification scheme Thus in prin-ciple one can just use the data from this file together with an independent schemethat rigorously does computations with this data to comfirm that Krawczykrsquos testyields a contraction mapping and hence verify the hyperbolicity of the manifoldM

52 Exceptional surgeries on alternating knots Two of the authors Ichi-hara and Masai applied the method developed in this paper to study exceptionalsurgeries on alternating knots in the 3-sphere S3


Algorithm 2 Find positive solutions by drilling out

Require M is a closed manifold with a surgery descriptionEnsure M has a good triangulationwhile We can find a short closed geodesic γ sub M do

Drill out γ to get M γTake filled triangulation N of M γFill the cusp of N by the slope (1 0)(By the above procedure we forget original surgery description and get newsurgery description)if N has positively oriented solution thenreturn [True N ]

end if

end while

return False

A knot is an embedded circle in S3 and it is represented by a diagram on theplane meaning that a projected image with under-over information at each doublepoint A knot is called alternating if it admits a diagram with alternatively arrangedover-crossings and under-crossings running along it

From a given knot in S3 by removing its tubular neighborhood and gluing solidtorus back one can obtain a ldquonewrdquo closed orientable 3-manifold Such an operationis called a Dehn surgery on the given knot The homeomorphism type of the 3-manifold so obtained is determined by the isotopy class of the loop bounding a diskin the attached solid torus which is called the surgery slope

Due to pioneering works by Thurston [34] all but at most finitely many Dehnsurgeries on a hyperbolic knot yield hyperbolic manifolds Here a knot is calledhyperbolic if its complement admits a complete Riemannian metric of constantsectional curvature minus1 of finite volume In view of this such an exceptional casethat is a Dehn surgery on a hyperbolic knot giving a non-hyperbolic manifold iscalled an exceptional surgery There are many of results about the classification ofexceptional surgeries on knots See [2] for a survey

In [12] Ichihara and Masai applied our package in this paper to the study ofexceptional surgeries on hyperbolic alternating knots and achieved a complete clas-sification

Here we include a rough sketch of this work In theory a result of [17] impliesthat there are only finitely many links (ie disjoint unions of knots) so that ifone could classify exceptional surgeries on those links completely then a completeclassification of exceptional surgeries on all hyperbolic alternating knots is obtained

Unfortunately the number of such links is in the millions Thus the first taskis to reduce the number of such links We now explain an outline of this stepThe links correspond to reduced alternating diagrams of alternating knots and arefiltered in terms of the complexity of the diagrams defined by Lackenby called thetwist number In the same paper Lackenby proved that the alternating knots withthe reduced alternating diagrams of twist number more than 8 have no exceptionalsurgeries Further the knots with the alternating diagrams of twist number atmost 5 are shown to be arborscent knots for which the classification of exceptionalsurgeries is almost known See [12] for more details Therefore our target is theknots with the alternating diagrams of twist number t satisfying 6 le t le 8 At this


point however the number of the corresponding links is more than 120000 Toreduce the number of the links further we applied some technique using essentiallaminations in the link exteriors based on the result obtained by Wu [38] Afterthat we have about remaining 30000 links

For each of the link we apply a computer-aided approach to get a classificationof exceptional surgeries on the link essentially developed in [20] (As noted beforethe actual code is available in the web page of B Martelli one of the authors of [20])Their method is based on the so-called 6-theorem obtained by Agol [1] and Lackenby[17] and seems very efficient in practice The key point is to compute the ldquolengthrdquoof the surgery slope on a horotorus in hyperbolic link compliments Actually thelength is more than 6 implies the corresponding surgery is not exceptional due to6-theorem However at the time of writing the procedure obtained in [20] mainlydepends upon the Moserrsquos work and utilizes floating point arithmetic Hence wemodified their code with using interval arithmetic and our package developed inthis paper

The final problem is about the computational time For each individual link ourprocedure have to be applied recursively since the links have several componentsFor example in the worst case for one link we have to apply our procedure to morethan 1400 cusped manifolds It then takes more than 2 hours in single standard per-sonal computer Therefore we used a super-computer called ldquoTSUBAMErdquo [21 36]set in Tokyo Institute of Technology Consequently after all computations whichtook about a day in computational time in TSUBAME we could do verify thatthere are no links among those we have targeted which have exceptional surgeries

6 Advantages over existing methods

The algorithm described in this paper is an improvement over existing methodsused in 3-manifold topology in three key ways the control over error the ease ofextending these computations to other topological invariants and the ability toimplement for large scale verifications For the purposes of this section we willcompare this method based on the Krawczyk test to an implementation of theKontorovich test popular in the study of 3-manifold topology [26] and finally anexact arithmetic algorithm

The code associated to [26] is available via the website in the citation In thatspecific implementation the eigenvalues of a conjugate transpose matrix are com-puted via solving a characteristic polynomial The eigenvalues need to be computedto high precision because they are used to bound the size of a neighborhood of theapproximate solution that contains an exact solution

However an undesirable amount of precision loss can occur during this com-putaiton of eigenvalues It is well known that given ntimes n matrices A = aij andAǫ = aij such that |aij minus aij | le ǫ for all pairs i j the minimum of differencethe eigenvalues for A and Aǫ can be at least n

radicǫ To be more explicit choose A

to the matrix with ones along the main and submain diagonal and let Aǫ be thematrix with ones along the main and submain diagonal and ǫ in the (n 1)-entryIn this case the eigenvalues of A are all 1 and the eigenvalues of Aǫ are of the form1minus ζn n

radicǫ where ζn is an n-th root of unity

Despite this potential for precision loss in [26] as described above it is doubtfulthat a small complexity 3-manifold exploits such an error in order to a false cer-tificate of hyperbolicity via that computation More concretely we point out that


every manifold verified in [26] has also been verified by the methods of this paper(including the large links of Leininger [18] which have 32 and 44 tetrahedra)

The second feature of the interval arithmetic technique is that the computerkeeps track of the accumulated error throughout the computation Therefore thesame techniques of computation allow for further rigorously computed invariantscoming from the solutions to the gluing equations In principle any computationthat Snappea preforms can be made rigorous using the techniques of this paper

Finally a third method exists for rigorous computer verification of hyperbolicitynamely snap which uses an exact arithmetic algorithm (see [8 6]) Snap is able toverify the exact hyperbolic structure by representing algebraic numbers exactly andalgebraically solve the polynomial equations of a manifold (13) With this datathe user can compute arithmetic data related to hyperbolic 3-manifolds (which ourmethods cannot) However the exact arithmetic methods are designed principallyfor the computations of this arithmetic data and so the methods of this paperare more efficient to rigorously verify hyperbolicity For instance using the snapsoftware would be less practical than our methods for a large scale verificationinvolving large sets of manifolds with relatively high numbers of tetrahedra such asthe computation preformed in [12] In addition snap does not report the error onits ldquonon-arithmeticrdquo computations such as volume or parabolic length Thereforea separate argument addressing such error would have to be derived one is able toclaim rigorous computations using data from snap

Appendix A Verified computations for arg(z)

In this appendix we describe how to rigorously compute arg(z) for z = x+iy isin C

using the floating-point arithmetic The complex argument of z is defined by

arg(z) = atan2(y x)

Here atan2(y x) is one of commonly used mathematical functions in programminglanguages including C C++ JAVA etc and defined by Table 2 In this table the

Table 2 Definition of atan2(y x)

Conditions for x and y atan2(y x)

y le x y gt minusx arctan(yx)

y gt x y gt minusx π2 minus arctan(xy)

y gt x y le minusx y ge 0 π + arctan(yx)

y gt x y le minusx y lt 0 minusπ + arctan(yx)

y le x y le minusx minusπ2 minus arctan(xy)

arctan function is assumed to have the range (minusπ2

π2 ) Here we first show how to

rigorously evaluate an interval extension of arctan function Then based on thiswe present a rigorous method of evaluating atan2 function

A1 Arctangent First we show how to construct an interval extension of arctanfunction It is seen from Table 2 that the evaluation of arctan(x) for x isin F reduces

to that for arctan function on the interval [minus(radic2 minus 1)

radic2 minus 1] On the interval


x |x| leradic2minus1 an interval inclusion of arctan(x) is obtained from the Maclaurin

expansion with remainder term

(26) arctan(x) isin xminus 1

3x3 +

1

5x5 minus 1

7x7 + middot middot middot+ 1

n[minus1 1]xn

with n being chosen as ∣∣∣∣1

n[minus1 1]xn

∣∣∣∣ le 2minus53

Calculating the right-hand side of (26) with interval arithmetic we define atan point F rarr IF by

atan point(x) = xminus 1

3x3 +

1

5x5 minus 1


n[minus1 1]xn

For I = [a b] isin IF we define I = a and I = b respectively Then an intervalextension Arctan IF rarr IF is given by

Arctan(X) =[atan point(X) atan point(X)

]isin IF

A2 Evaluation of arg(z) Next we show how to construct an interval extensionof atan2 function whose range is assumed to be (minusπ π] For x y isin F let us denoteIyx = [fldown(yx) flup(yx)] and Ixy = [fldown(xy) flup(xy)] Furthermore[π2 ] and [π] denote the interval inclusion of π

2 and π respectively An intervalinclusion of atan2(y x) which is denoted by atan2 point(y x) can be calculatedthrough Arctan function defined above Namely Table 3 shows a realization

Table 3 Realization of atan2 point(y x)

Conditions for x and y atan2 point(y x)

y le x y gt minusx Arctan(Iyx)

y gt x y gt minusx [π2 ]minusArctan(Ixy)

y gt x y le minusx y ge 0 [π] + Arctan(Iyx)

y gt x y le minusx y lt 0 minus[π] + Arctan(Iyx)

y le x y le minusx minus[π2 ]minusArctan(Ixy)

For Iy Ix isin IF Table 4 shows a realization of Atan2 IFtimes IF rarr IF which is akind of interval extension of atan2 function It is easily seen from Table 4 Atan2function is an interval extension of atan2 function except the case of Ix 6ni 0 Iy ni 0

and Ix lt 0 In this case a natural interval extension of atan2 function becomesthe union of the following two intervals

[minus[π] atan2 point(Iy Ix)

]and

[atan2 point(Iy Ix) [π]

]

Since this makes the algorithm multivalued we modify the algorithm to return asingle interval for such Iy and Ix

Atan2(Iy Ix) =[atan2 point(Iy Ix) 2[π] + atan2 point(Iy Ix)

]

It is just a change of expression so that this modification does not cause any con-fusion


Table 4 Atan2(Iy Ix)

Conditions for Ix and Iy Atan2(Iy Ix)

Ix ni 0 Iy ni 0[minus[π] [π]

]

Ix ni 0 Iy 6ni 0 Iy gt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

Ix ni 0 Iy 6ni 0 Iy lt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

Ix 6ni 0 Iy ni 0 Ix gt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

Ix 6ni 0 Iy ni 0 Ix lt 0[atan2 point(Iy Ix) 2[π] + atan2 point(Iy Ix)

]

Ix 6ni 0 Iy 6ni 0 Ix gt 0 Iy gt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

Ix 6ni 0 Iy 6ni 0 Ix gt 0 Iy lt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

Ix 6ni 0 Iy 6ni 0 Ix lt 0 Iy gt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

Ix 6ni 0 Iy 6ni 0 Ix lt 0 Iy lt 0[atan2 point(Iy Ix) atan2 point(Iy Ix)

]

References

[1] I Agol Bounds on exceptional Dehn filling Geom Topol 4 (2000) 431ndash449[2] S Boyer Dehn surgery on knots in Handbook of geometric topology 165ndash218 North-Holland

Amsterdam[3] R Benedetti and C Petronio Lectures on hyperbolic geometry Universitext Springer-Verlag

Berlin 1992[4] Boost C++ libraries httpwwwboostorg[5] P Callahan M Hildebrand and J Weeks A census of cusped hyperbolic 3-manifolds Math

Comp 68 (1999) no 225 321ndash332[6] D Coulson and O Goodman and C Hodgson and W Neumann Computing arithmetic

invariants of 3-manifolds Experiment Math 9 (2000) no 1 127ndash152[7] M Culler N M Dunfield and J R Weeks SnapPy a computer program for studying the

geometry and topology of 3-manifolds httpsnappycomputoporg[8] O Goodman Snapbased on the SnapPea kernel of Jeff Weeks and the Pari software

httpsourceforgenetprojectssnap-pari[9] M Hildebrand and J Weeks A computer generated census of cusped hyperbolic 3-manifolds

Computers and mathematics (Cambridge MA 1989) 53ndash59 Springer New York 1989[10] C Hodgson and J Weeks Symmetries isometries and length spectra of closed hyperbolic

three-manifolds Experiment Math 3 (1994) 261ndash274[11] N Hoffman K Ichihara M Kashiwagi H Masai S Oishi and A Takayasu hikmot

httpwwwoishiinfowasedaacjp~takayasuhikmot[12] K Ichihara and H Masai Exceptional surgeries on alternating knots in preparation[13] IEEE 754 ANSIIEEE 754-2008 IEEE Standard for Floating-Point Arithmetic New York

2008

[14] LV Kantorovich and GP Akilov Functional analysis in normed spaces Pergamon Press1964

[15] R Krawczyk Newton-Algorithmen zur Bestimmung von Nullstellen mit FehlerschrankenComputing 4 (1969) 187ndash201


[16] R Krawczyk Fehlerabschatzung reeller Eigenwerte und Eigenvektoren von Matrizen Com-puting 4 (1969) 281ndash293

[17] M Lackenby Word hyperbolic Dehn surgery Invent Math 140 (2000) no 2 243ndash282[18] C Leininger Small curvature surfaces in hyperbolic 3-manifolds J of Knot Theory and Its

Ramifications 15 (2006) no 3 379ndash411[19] S Markov and K Okumura1999 rdquoThe contribution of T Sunaga to interval analysis and

reliable computingrdquo in Developments in Reliable Computing Kluwer (1999) 167ndash188[20] B Martelli C Petronio and F Roukema Exceptional Dehn surgery on the minimally twisted

five-chain link preprint arXiv11090903[21] S Matsuoka T Endo N Maruyama H Sato S Takizawa The Total Picture of TSUB-

AME20rdquo The TSUBAME E-Science Journal Tokyo Tech GSIC Vol 1 pp 16-18 Sep2010

[22] J Milnor Hyperbolic geometry the first 150 years Bull Amer Math Soc (NS) 6 (1982)no 1 9ndash24

[23] J Milnor The Poincare conjecture The millennium prize problems 71ndash83 Clay Math InstCambridge MA 2006

[24] RE Moore (1966) Interval Analysis Prentice-Hall Englewood Cliffs 1966[25] JW Morgan Recent progress on the Poincare conjecture and the classification of 3-

manifolds Bull Amer Math Soc (NS) 42 (2005) no 1 57ndash78

[26] H Moser Proving a manifold to be hyperbolic once it has been approximated to be so Al-gebraic amp Geometric Topology 9 (2009) 103ndash133 Code associated to the paper is availablehere httpwwwmathcolumbiaedu~moser

[27] W Neumann and D Zagier Volumes of hyperbolic three-manifolds Topology 24 (1985)no 3 307ndash332

[28] G Perelman The entropy formula for the Ricci flow and its geometric applications preprintavailable at arxivemathDG0211159

[29] G Perelman Ricci flow with surgery on three-manifolds preprint available atarxivemathDG0303109

[30] G Perelman Finite extinction time for the solutions to the Ricci flow on certain three-

manifolds preprint available at arxivmathDG0307245[31] SM Rump Verification methods Rigorous results using floating-point arithmetic Acta

Numerica 19 (2010) 287ndash449[32] T Sunaga Geometry of numerals Masterrsquos thesis University of Tokyo 1956[33] T Sunaga Theory of an interval algebra and its application to numerical analysis RAAG

Memoirs 2 (1958) 29ndash46[34] WP Thurston The geometry and topology of 3ndashmanifolds Lecture notes Princeton Uni-

versity 1978[35] WP Thurston Three-dimensional manifolds Kleinian groups and hyperbolic geometry Bull

Amer Math Soc (NS) 6 (1982) no 3 357ndash381[36] Tokyo Institute of Technology The Global Scientific Information and Computing Center

TSUBAME httptsubamegsictitechacjpen[37] J Weeks rdquoComputation of Hyperbolic Structures in Knot Theoryrdquo in Handbook of Knot

Theory Elsevier Science (2005) 461-480[38] Y-Q Wu Persistently laminar branched surfaces Comm Anal Geom 20 (2012) no 2

397ndash434


Department of Mathematics and Statistics University of Melbourne Victoria 3010

Australia

E-mail address nhoffmanmsunimelbeduau

Department of Mathematics College of Humanities and Sciences Nihon University

3-25-40 Sakurajosui Setagaya-ku Tokyo 156-8550 Japan

E-mail address ichiharamathchsnihon-uacjp

Department of Applied Mathematics Waseda University 3-4-1 Okubo Shinjuku Tokyo

169-8555 Japan

E-mail address kashiwasedajp

Department of Mathematical and Computing Sciences Tokyo Institute of Technol-

ogy O-okayama Meguro-ku Tokyo 152-8552 Japan

E-mail address masai9istitechacjp


169-8555 Japan

E-mail address oishiwasedajp


169-8555 Japan

E-mail address takitoshiaoniwasedajp

1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations

22 Unfilled cusp equations

23 Filled cusp equations

24 Full set of equations

3 Krawczyks Test

31 Brief Review of Interval Analysis

32 Machine Interval Arithmetic

33 Automatic differentiation

34 Complex arithmetic

35 Example


41 The main algorithm

42 Contents of our package hikmot

43 How to use the package

5 Applications

51 Census manifolds

52 Exceptional surgeries on alternating knots



A1 Arctangent

A2 Evaluation of arg(z)

References


1 Introduction

The study of 3-dimensional manifolds often abbreviated by 3-manifolds startswith the seminal papers by H Poincare in 1984ndash1904 In the last of these sixpapers he raised a question which became the famous Poincare Conjecture Thishas been one of the driving forces for (low-dimensional) topology and a greatamount of effort was spent in pursuit of a solution After nearly a century later in2002ndash2003 G Perelman finally reached to the end of the struggles by providing anaffirmative answer to the Geometrization Conjecture an extended version to thePoincare Conjecture in [28 29 30] See [23 25] as detailed references for example

The Geometrization Conjecture was raised by W Thurston who brought aboutdramatic changes to the study of 3-manifolds He introduced a geometric view inthe sense of Klein to the 3-manifold theory and in particular he incorporated anamazing application of non-Euclidean geometry namely hyperbolic geometry intothe study of 3-manifolds

As is well-known every closed 2-dimensional manifold ie closed surface admitsa geometric structure a complete Riemannian metric of constant sectional curva-ture Surprisingly Thurston predicted a similar situtation for the 3-dimensional case[35] That is every compact orientable 3-manifold can be canonically decomposedby cutting along essential 2-spheres and tori into the pieces each of which admitsa locally homogeneous geometric structure This is the Thurstonrsquos GeometrizationConjecture Actually he showed that there are exactly EIGHT geometries (iemodels of geometric structures) to be considered and essentially gave a proof forthe case of 3-manifolds containing decomposing tori

Among the eight geometries six have been well studied and are generally under-stood the Seifert fibered geometries The seventh is sol -geometry All manifoldsadmitting a sol geometric structure are torus bundles over the circle or n-fold quo-tients of torus bundles over the circle where n = 2 4 By most accounts the mostcommon and yet most interesting geometry occurs if a manifold M admits a hyper-bolic structure ie M admits a complete Riemannian metric of constant sectionalcurvature minus1 of finite volume See [22] for a good survey

In this paper we describe a computer-aided practical method to rigorously provethat a given 3-manifold admits a complete hyperbolic structure via verified calcu-lations This procedure is already implemented as a software and available at [11]

In the following we give a sketch of the contents with the organization of thepaper In the discussions that follow we will consider only 3-manifolds that areorientable compact with boundary consisting of the (possibly empty) disjoint unionof tori

Our program takes in as an initial input the combinatorial data of an ideal trian-gulation for the compact bounded case and a surgery description for the closed orpartially filled case From that data as Thurston first described we can find somealgebraic equations called a system of Gluing equations with complex variables Ifthis system of equations has a system of complex solutions with positive imaginaryparts then the given manifold admits a hyperbolic structure In the next sectionwe will recall a concise explanation for that mainly based on the well-known book[3] Note that this process has already been implemented as SnapPea by J WeeksUsing the kernel code by Weeks M Culler and N Dunfield implemented SnapPywhich is roughly speaking a SnapPea on python (For further background onSnapPy or the SnapPea kernel see the SnapPy documentation and [7]) By abusing






Acknowledgements



2 Gluing equations















or 11minuszj




is part of the





nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(1)

andnsum

j=1


(1

1minus zj


(zj minus 1

zj

)= 2πi(2)


ℓ=1 ajℓ = 2sumn




nsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 2πi(3)





corresponds to an


nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(4)




and

nsum

j=1


1

1minus zj

)+ cjm middot arg

(zj minus 1

zj

)= 0(5)




middot zjminus1zj




middot zjminus1zj

= minus1 the




ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)






nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)



p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References






Acknowledgements



2 Gluing equations















or 11minuszj




is part of the





nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(1)

andnsum

j=1


(1

1minus zj


(zj minus 1

zj

)= 2πi(2)


ℓ=1 ajℓ = 2sumn




nsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 2πi(3)





corresponds to an


nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(4)




and

nsum

j=1


1

1minus zj

)+ cjm middot arg

(zj minus 1

zj

)= 0(5)




middot zjminus1zj




middot zjminus1zj

= minus1 the




ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)






nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)



p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References













or 11minuszj




is part of the





nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(1)

andnsum

j=1


(1

1minus zj


(zj minus 1

zj

)= 2πi(2)


ℓ=1 ajℓ = 2sumn




nsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 2πi(3)





corresponds to an


nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(4)




and

nsum

j=1


1

1minus zj

)+ cjm middot arg

(zj minus 1

zj

)= 0(5)




middot zjminus1zj




middot zjminus1zj

= minus1 the




ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)






nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)



p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References





nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(1)

andnsum

j=1


(1

1minus zj


(zj minus 1

zj

)= 2πi(2)


ℓ=1 ajℓ = 2sumn




nsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 2πi(3)





corresponds to an


nprod

j=1

(zj)ajm

(1

1minus zj

)bjm (zj minus 1

zj

)cjm

= 1(4)




and

nsum

j=1


1

1minus zj

)+ cjm middot arg

(zj minus 1

zj

)= 0(5)




middot zjminus1zj




middot zjminus1zj

= minus1 the




ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)






nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)



p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References




and

nsum

j=1


1

1minus zj

)+ cjm middot arg

(zj minus 1

zj

)= 0(5)




middot zjminus1zj




middot zjminus1zj

= minus1 the




ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)






nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)



p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References



ifnsum

j=1


(1

1minus zj

)+ cjm log

(zj minus 1

zj

)= 0 + 0πi(6)






nprod

j=1

((zj)

ajr

(1

1minus zj

)bjr (zj minus 1

zj

)cjr)p

middot((zj)

ajr+1

(1

1minus zj

)bjr+1(zj minus 1

zj

)cjr+1

)q

= 1(7)



p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References


p middotnsum

j=1


(1

1minus zj

)+ cjr middot arg

(zj minus 1

zj

))(8)

+ q middotnsum

j=1


(1

1minus zj

)+ cjr+1 middot arg

(zj minus 1

zj

))= 2π


p

nsum

j=1


(1

1minus zj

)+ cjr log

(zj minus 1

zj

)+

q

nsum

j=1


(1

1minus zj

)+ cjr+1 log

(zj minus 1

zj

)= 0 + 2πi(9)



nsum

j=1


n+2k+h

m=1

(10)


nprod

j=1


(minusbjm+cjm) =

nprod

j=1

(minus1)cjm

n+2k+h

m=1

and

(11)

nsum

j=1



j=1

cjm middot πi

n+2k+h

m=1





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References





ΛM =



(12)








nprod

j=1


βjm = γm =

nprod

j=1

(minus1)cjm

misinS

and(13)

nsum

j=1



j=1

cjm middot πi

misinS








mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References






mid(X) =x+ x

2


rad(X) =xminus x

2



and





X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References



X + Y =[x+ y x+ y

]


]


]

XY = X middot[1

y1

y

] (0 6isin Y )




mid(A) =



isin R

mtimesn



nprod

j=1

zαjm








as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References


as

(16)

nprod

j=1

zα+

jm

j (1minus zj)β+

jm minusnprod

j=1

zαminus

jm




(17) f(x) = 0







K(X) sub int(X)



K(X) sub int(X)(20)











X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References









X Y = (X Y )












X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References




X =


[cm minus r cm + r]

(23)





m+1 to


r0 = U(p0)




B IFm+1 times IF





r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References




r0 = B(p0 q0)




20





partxjFi IR


X =

X1

X2

Xm

=





F (X) =





Fm(X) partx1


Fm(X)

isin IF

mtimes(m+1)




)isin IF

(m+1)times2








z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References






z

w=

(zrewre + zimwim

w2re + w2

im

zimwre minus zrewim

w2re + w2

im

)



X = idCrarrR(Z)

or

Z = idRrarrC(X)


F (X) = idCrarrR

(G(idRrarrC

(X)))


αjm =

(5 92 2

) βjm =


) γm =

(minus11

)


(25)

z51(1minus z1)



minus1 = 1lArrrArr

z51z

92 + (1minus z2)

7 = 0z21z



g1(z) = z51z92 + (1 minus z2)

7



z =

(01295310113154524+ 03730313363875791i46374476446382840+ 16871823157824217i

)


X =

[01295310113154227 01295310113154820][03730313363875496 03730313363876089][46374476446382538 46374476446383142][16871823157823886 16871823157824481]





KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References



KF (X) =

[01295310113154520 01295310113154527][03730313363875788 03730313363875796][46374476446382680 46374476446382999][16871823157824033 16871823157824335]








end if

end if


We will now explain










several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References







several times

























gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References













gtgtgt L[0]

True

gtgtgtL[1][([012953101131545199012953101131545273])+([037303133638757879037303133638757963])i

([4637447644638267946374476446383])+([1687182315782403216871823157824335])i]


[(012953101131545247+03730313363875793j)

(4637447644638281+16871823157824253j)]

5 Applications


















end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References














end if

end while

return False


























arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References




















arg(z) = atan2(y x)



















3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References





3x3 +

1

5x5 minus 1


n[minus1 1]xn


n[minus1 1]xn




3x3 +

1

5x5 minus 1


n[minus1 1]xn



]isin IF













]and


]



]






]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References





]


]


]


]


]


]


]


]


]

References











2008


























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References
























397ndash434



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References



Australia






169-8555 Japan






169-8555 Japan



169-8555 Japan


1 Introduction

Acknowledgements

2 Gluing equations

21 Edge equations




3 Krawczyks Test





35 Example





5 Applications

51 Census manifolds




A1 Arctangent


References

arxiv:1310.3410v2 [math.gt] 29 nov 2013dedicated to professor sadayoshi kojima on the occasion of...

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