BBIPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS
COLIN ADAMS
Abstract.
We utilize ideal bipyramids to obtain new upper bounds on volume for hyperboliclink complements in terms of the combinatorics of their projections. Introduction
Knots fall into three disjoint categories: torus knots, satellite knots (including compositeknots) and hyperbolic knots. A knot is hyperbolic if its complement S − K carries a hyperbolicmetric. Such a metric is uniquely determined and hence, the hyperbolic volume of its complementbecomes an invariant that can be used to distinguish it from other knots.Given a projection P of a knot, we will say it is reduced if there are no Type I or TypeII Reidemeister moves that would individually lower the number of crossings, or a circle in theprojection plane that intersects the knot in only one crossing, with two strands to either side of it.In a reduced projection, define a maximal bigon chain to be a sequence of bigonal complementaryregions touching end-to-end that is as long as possible. The crossing length of such a bigon chain isthe number of crossings it contains. Note that a crossing that does not lie on a bigon is consideredto be a maximal bigon chain of crossing length 1. The number of such maximal bigon chains in aprojection P is called the twist number of the projection, denoted t ( P ). We further let t i ( P ) be thenumber of maximal bigon chains of crossing length i and g i ( P ) be the number of maximal bigonchains of crossing number at least i . We typically assume that our diagrams are twist reduced,which mean flypes have been applied to minimize the number of distinct bigon chains.In a variety of papers, there have been attempts to determine bounds on hyperbolic volumefrom projections of knots and links. In [1] it was proved that with the exception of the figure-eightknot, a hyperbolic knot satisfies vol( S − K ) ≤ (4 c − v tet , where c is the crossing number and v tet is the volume of an ideal regular tetrahedron, approximately 1 . L has a reduced alternating projection P , then vol( S − L ) ≤ v tet ( t ( P ) − L is a hyperbolic alternating linkin a reduced alternating projection P , then vol( S − L ) ≤ (4 t ( P )+6 t ( P )+8 t ( P )+10 g ( P ) − a ) v tet where a = 10 when g ≥ a = 7 when g = 0 but t ≥ a = 6 otherwise. We refer to this asthe DT bound on volume.There is another approach to bounding volume that follows from a construction of D. Thurston.One places an octahedron at each crossing with its top vertex on the bottom of the overstrand andits bottom vertex on the top of the understrand as in Figure 1. Date : August 24, 2018. a r X i v : . [ m a t h . G T ] N ov COLIN ADAMS
163 7 10 42 5 8
Figure 1.
Placing an octahedra between each crossing.Then, as per the edge labelings in Figure 1, one pulls two of the opposite remaining vertices upto meet above the crossing at a point denoted U , thereby identifying two edges on the octahedron,and pulls the remaining two vertices down to meet below the crossing at a point denoted D , againidentifying two edges of the octahedron.Then one can glue together the faces on the octahedra atthe various crossings in order to fill the complement of the knot. If the two vertices at U and D are included, we have a decomposition of the knot complement into octahedra with some finitevertices and some ideal vertices. This construction has been used to attempt to prove Kaeshaev’sVolume Conjecture for various categories of knots. See [17] and [23] for instance. However, onealso obtains an upper bound for the hyperbolic volume of the knot or link complement since anoctahedron in hyperbolic 3-space has volume at most the volume of an ideal regular octahedron,which is v oct ≈ . S − L ) ≤ cv oct . This bound was improved in [5] by pulling thefinite vertices to ideal vertices. Theorem 1.1.
Let L be a hyperbolic knot or link with c ≥ crossings. Then vol ( S − L ) ≤ ( c − v oct + 4 v tet . We call this the octahedral bound.In this paper, we utilize Thurston’s octahedral construction to generate several other construc-tions in terms of bipyramids, one based on bigon chains and one based on faces of the projection.Both give improved upper bounds on volume. For instance, the first construction, which is de-scribed in Section 3, yields the following improvement on the DT bound.
Theorem 3.1.
Let L be a hyperbolic alternating link with five or more crossings in a reduced twistreduced alternating diagram that has at least three twist regions and that is not the Borromeanrings. Then vol( S − L ) < t v oct + t (6 v tet ) + t (7 . . . . ) + t (9 . . . . ) + g (10 v tet ) − a , whereif g = 0, then a = 15 . g = 0 but t ≥
1, then a = 11 v tet , if g = 0 but t ≥
1, then a = 10 . g = 0 but t ≥
1, then, a = 10 . g ≥
1, then a = 12 . IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 3
The second construction, which is described in Section 4, yields a means of obtaining a boundon volume in terms of the faces of the projection of any hyperbolic link. This face-centeredbipyramid bound is called the FCB bound. In the cases of the 4 and 6 , it yields the exactvolume. For small volume knots, it appears to yield the smallest upper bound on volume of thevarious bounds discussed here. In Section 5, we compare the different bounds.Note that various authors, in addition to considering upper bounds either in general or forspecific classes of knots and links, have also determined lower bounds. See for instance [11], [13]and [15].Additional applications of the bipyramid construction appear in [7], where it is used to obtainvolume density results for 2-bridge links and [8], where it is generalized and used to obtain volumedensity results for links in S x I where S is a surface. Acknowledgement
Thanks to Aaron Calderon, Alexander Kastner, Xinyi Jiang, Gregory Kehne, and NathanielMayer, for very helpful discussions and Mia Smith for the same and particularly for pointing outimprovements to Theorem 2.2. Thanks also to NSF Grant DMS-1347804, and Williams College,which fund the SMALL REU program at Williams College.2.
Ideal Bypyramids
An ideal n -bypyramid B n appears as in Figure 2. We call such an ideal hyperbolic bipyramid regular if if can be constructed by gluing together n ideal tetrahedra along a central edge, eachisometric to the ideal tetrahedron T n = T (2 π/n, ( n − pi n , ( n − π n ), where each tetrahedron intersectsthe central edge with an edge of diherdral angle 2 π/n . Notice that all of the vertical edges of theresulting bipyramid have dihedral angle ( n − πn , as do the edges going to the central vertex, whilethe edges connecting vertices that are neither the central vertex nor the vertex at ∞ have dihedralangle 2 π/n . The volume of T n is given by:vol( T n ) = (cid:90) π/n − ln (sin θ ) dθ + 2 (cid:90) π ( n − / n − ln (sin θ ) dθ Theorem 2.1.
The maximal volume for an ideal n -bipyramid is obtained uniquely by the regular n -bipyramid.Proof. An n -bipyramid can always be decomposed into n ideal tetrahedra, meeting at the centraledge. Let them be denoted T , T , . . . T n , where T i has angles α i at the central edge and then β i and γ i , reading clockwise. Then α + α + · · · + α n = 2 π . Let f be the volume of the n -bipyramid.Then f = n (cid:88) i =1 (cid:32)(cid:90) α i − ln (2 sin θ ) dθ + (cid:90) β i − ln (2 sin θ ) dθ + (cid:90) γ i − ln (2 sin θ ) dθ (cid:33) . We use Lagrange Multipliers to maximize f subject to the following constraints:For i = 1 , . . . , n : g i = α i + β i + γ i − π = 0Additionally, g n +1 = α + α + · · · + α n − π = 0 COLIN ADAMS
Figure 2. An n -bipyramid for n = 6.Then we generate the following equations:For i = 1 , . . . , n : − ln(2 sin α i ) = λ i + µ, − ln(2 sin β i ) = λ i , − ln(2 sin γ i ) = λ i From the second and third equations, we see that β i = γ i for all i . From the first and secondequations, we have that for i = 1 , . . . , n : µ = ln(2 sin β i ) − ln(2 sin α i )Thus, sin β sin α = sin β sin α = · · · = sin β n sin α n Since, β i = π − α i , this implies that α = α = · · · = α n . Hence, α i = πn and β i = γ i = n − n π ,yielding a regular n -bipyramid. (cid:3) Volumes for the n -bipyramids for certain values of n are provided in Table 1.Note that with the exception of n = 6, the volume of the corresponding n -bipyramid is alwaysless than the volume that would be obtained by taking the sum of the volumes of n regulartetrahedra. It is this volume saving that we take advantage of.As n increases, the volumes of the n -bipyramids increase without bound. We show that thevolumes grow logarithmically. Theorem 2.2. vol ( B n ) < π ln( n/ for n ≥ and vol ( B n ) grows asymptotically like π ln( n/ .Proof. We show the second fact first.
IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 5 n Volume2 03 2.02984 3.66385 4.98676 6.08967 7.03258 7.85499 8.583610 9.237511 9.830412 10.372513 10.871914 11.334720 13.5668100 23.67091,000 38.13821,000,000 81.54091,000,000,000 124.944
Table 1.
Volumes of n -bypyramids.lim n →∞ vol( B n )ln( n/
2) = lim n →∞ n vol( T n )ln( n/
2) = lim n →∞ vol( T n )(ln( n/ /n (2.1) = lim n →∞ − ln (cid:0) (cid:0) πn (cid:1)(cid:1) − πn + − (cid:16) (cid:16) ( n − π n (cid:17)(cid:17) πn − ln( n/ n (2.2) = lim n →∞ π ln (cid:0) (cid:0) πn (cid:1)(cid:1) − π ln (cid:16) (cid:16) ( n − π n (cid:17)(cid:17) − ln( n/
2) = lim n →∞ π ln (cid:0) (cid:0) πn (cid:1)(cid:1) − ln( n/ n →∞ − πn πn (2 π ) n − n = 2 π (2.4)Hence, the volumes of the B n grow asymptotically like f ( n ) = 2 π ln( n/ B n ) < π ln( n/ T n ) < π ln( n/ n . Let f ( n ) = vol( T n )and let g ( n ) = 2 π ln( n/ n . Since these two functions agree asymptotically and since the result doeshold for n = 3 and 4, if it were the case that f ( a ) > g ( a ) for some a , then it would have to be truethat f (cid:48) ( b ) < g (cid:48) ( b ) for some b > a . So we will prove that f (cid:48) ( x ) > g (cid:48) ( x ) for all x ≥ f (cid:48) ( n ) = 2 πn ln sin (cid:0) πn (cid:1) sin (cid:16) ( n − π n (cid:17) = 2 πn ln (cid:32) sin (cid:0) πn (cid:1) cos (cid:0) πn (cid:1) (cid:33) = 2 πn ln (cid:16) (cid:16) πn (cid:17)(cid:17) g (cid:48) ( n ) = 2 π (cid:18) − ln( n/ n (cid:19) COLIN ADAMS
Then if f (cid:48) ( b ) < g (cid:48) ( b ), we have ln (cid:16) (cid:16) πb (cid:17)(cid:17) < − ln b/ b sin (cid:16) πb (cid:17) < e However, at b = 4, b sin (cid:0) πb (cid:1) ≈ . > e and b sin (cid:0) πb (cid:1) is an increasing function approaching2 π , so this inequality never holds for b ≥ (cid:3) Note that the proof can be altered to show the slightly stronger result that vol( B n ) < π ln( n/ . Conjoining Octahedra
In this section we consider sequences of octahedra in the Thurston construction that lie at thecrossings of a maximal bigon chain. Consider two octahedra at the crossings of a bigon as in Figure3. They are glued together along four faces on each, however, we will focus on the two shadedfaces on each. These are glued together as squares with a diagonal and no rotation. Hence wecan glue the two octahedra together along this pair of faces to obtain a 6-bipyramid. If there is asubsequent bigon in the maximal bigon chain, then it can be glued onto the 6-bipyramid to createan 8-bipyramid. Thus, the octahedra that make up a maximal bigon chain of crossing length n can be consolidated into a single (2 n + 2)-bipyramid. Thus, if a reduced projection P has twistnumber t ( P ), then the complement can be decomposed into t ( P ) bipyramids, each with two idealvertices and the rest divided into two equivalence classes of finite vertices. Figure 3.
Octahedra at the crossings of a bigon.Note that in the case of a maximal bigon chain of length n , if the Thurston constructionis applied, there will be n octahedra, and the contribution to the volume from that sequence ofcrossings will be on the order of n (3 . . . . ). The volume associated with the corresponding(2 n + 2)-bipyramid will be substantially less. For instance, for n = 499, the octahedra produce avolume bound of 1828.2, whereas the corresponding1000-bipyramid yields a volume bound of 38.1. IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 7
On the other hand, in the Agol-Thurston method, we can augment each bigon chain with anadditional link component, and obtain a bound from the bigon chain of just 10 v tet . Thus, in orderto minimize the volume bound, it makes sense to use a (2 n + 2)-bipyramid when its volume isless than 10 v tet and to augment with an additional component when its volume is greater than10 v tet . Hence, by considering the table of volumes of n -bipyramids, we see that we should use usethe (2 n + 2)-bipyramid for maximal bigon chains of crossing length n ≤ n >
4. However, to do so, we need to know that thetwo constructions can be fit together.
Theorem 3.1.
Let L be a hyperbolic alternating link with five or more crossings in a reduced twistreduced alternating diagram that has at least three twist regions and that is not the Borromeanrings. Then vol ( S − L ) < t v oct + t (6 v tet ) + t (7 . . . . ) + t (9 . . . . ) + g (10 v tet ) − a , whereif g = 0 , then a = 15 . , if g = 0 but t ≥ , then a = 11 v tet , if g = 0 but t ≥ , then a = 10 . , if g = 0 but t ≥ , then, a = 10 . and if g ≥ , then a = 12 . .Proof. We begin by inserting one octahedron per crossing, as in the original construction of DylanThurston. For each maximal bigon chain of crossing length n less than 5, we consolidate thecorresponding octahedra at each crossing into a single (2 n + 2)-bipyramid, as described previously.For any maximal bigon chain that has crossing length 5 or greater, we add a trivial componentthat wraps once around the bigon chain, which we call a vertical component. The resulting link L (cid:48) is hyperbolic by work in [2], and its volume is greater than the original link complement bywork of W. Thurston. From [3], it follows that we can cut the link complement open along thetwice-punctured disk bounded by the new component, and twist to change the number of crossingsin the chain from whatever it was to 2, and the resulting link L (cid:48)(cid:48) will have complement, possiblyhomeomorphic to the complement of L (cid:48) and possibly not, with the same volume as the complementof L (cid:48) .We now consider the pairs of octahedra corresponding to the two crossings that were createdby the augmentation process. For each such pair, we first glue them together along two facesas described above to obtain a 6-bipyramid (see Figure 6(a) and (b)). We then drill a verticalcomponent out of the complement as in Figure 6(a) by drilling it out of this 6-bipyramid, as inFigure 6(b). By adding edges to the faces of the 6-bipyramid and collapsing the new drilled arcsdown to vertices, we obtain an ideal polyhedron as in Figure 6(c). This decomposes into fourtetrahedra and one 6-bipyramid, as in Figure 6(d). Hence, its contribution to the total volume isat most 10 v tet .Note that the drilling avoids the faces of the 6-bipyramid that are glued to the faces of otherbipyramids corresponding to other bigon chains, so gluing faces does yield a manifold homeomor-phic to the complement of L (cid:48)(cid:48) .Finally, we can collapse the finite vertices U and D to the cusp, by choosing edges from eachof them to one of the cusps and shrinking the edges away.In the case that g = 0, so all bigon chains have crossing length 1, We first assume that theprojection is not the alternating projection of the Borromean rings. Then there exists an edge inthe projection that bounds two regions, at least one of which has more than three edges. As inthe proof of Theorem 5.1 in [5], we collapse a vertical edge reaching from this strand to U andanother from this strand to D, as in edges 1 and 2 in Figure 4. This flattens the two octahedra atthe crossings on the ends of this strand.The collapse of the edges labelled 1 and 2 also causes the edges labelled 3 and 4 to collapse.This collapses each of the at least three octahedra corresponding to additional crossings on theseregions to two tetrahedra each. It is also true that the edges labelled 1 and 2 can be slid out past COLIN ADAMS
123 4 12
Figure 4.
Collapsing the edges labelled 1 and 2 causes the edges labelled 3 and4 to collapse, eliminating the volume contribution of two octahedra and shrinkingthe volume contribution of another five octahedra.the two crossings so that the octahedron to the left and the octahedron to the right each containone of the edges that is collapsed. Because there are no bigons, all of these octahedra are distinct.Thus, the total volume drop is 7 v oct − v tet = 15 . . . . .When g = 0, but t ≥
0, we have a 6-bipyramid corresponding to a bigon chain B of crossinglength 2. The edges around its equator fall into four equivalence classes, two of size two and twoof size one, each edge class of which connects the vertex D to the vertex U. We collapse one ofthe edges with an equivalence class of two to collapse the 6-bipyramid to a 4-bipyramid. Thisedge appears as the central edge of an adjacent rgion inthe projection plane and it identifies U toD.We also collapse an edge from U to an ideal vertex. In Figure6, we can realize these collapsesby collapsing edge 7 and edge 5. This forces edge 2 to collapse as well and in fact collapses theentire face bounded by these three edges, depicted in grey at the top of Figure 5. This ultimatelyentirely flattens the original 6-bipyramid into two triangular faces with no volume. We also seea volume drop from any bigon chain bipyramid with a crossing in region A that has its centraledge collapsed. Because the diagram is twist reduced and has at least three bigon chains, one ofthe two adjacent regions to B has at least two additional crossings on it. We choose this to be A .The least volume drop comes from when there are exactly two additional crossings and they bothoccur in another bigon. In this case, appearing as in the projection of the link 6 we collapse anadditional 6-bipyramid to a 4-bipyramid. There are also two collapsed edge that slide out past theoriginal bigon region B . Their smallest impact occurs when the other adjacent region to B hasonly one crossing. Then these two collapsed edges turn its octahedron into a single tetrahedron.Thus, the total volume drop is at least 2vol( B ) + vol( B ) − (vol( B ) + v tet ) = 11 v tet .When g = 0, but t ≥
0, we have an 8-bipyramid corresponding to a bigon chain of crossinglength 3. The edges around its equator fall into four equivalence classes, two of size three and twoof size one. We collapse one of the edges corresponding to an adjacent region A with an equivalenceclass of three and we collapse an edge from U to an ideal vertex, which causes two faces on itsboundary to collapse as in the middle of Figure 5. This takes the 8-bipyramid to one tetrahedronwith an additional flat face. We also see a volume drop from any bigon chain bipyramid witha crossing in region A . Again we can assume there is more than one additional crossing in theregion. The least volume drop comes from having two crossings on this region at the end of two IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 9
Figure 5.
Collapsing U and D to a cusp.bigon chains of length 3. However, then two bigon chains of length 3 are adjacent. We do betterby taking a bigon chain of length 3 adjacent to a bigon chain of length 2. Then the 6-bipyramidcorresponding to the bigon chain of length 2 collapses to a 4-bipyramid. There is also an edgethat slides out of this region with its minimal decrease on volume taking another 8-bipyramid to a7-bipyramid. This corresponds to the case of the 8 knot projection for instance. Thus, the totalvolume drop is at least 2vol( B ) + vol( B ) − v tet − vol( B ) − vol( B ) = 10 . . . . .When g = 0, but t ≥
0, we have a 10-bipyramid corresponding to a bigon chain of crossinglength 4. The edges around its equator fall into four equivalence classes, two of size four and twoof size one. We first collapse one of the edges with an equivalence class of four. This edge is thecentral edge perpendicular to an adjacent region A in the projection plane connecting U to D. Wealso collapse an edge from U to an ideal vertex, causing two faces on its boundary to collapse as inthe bottom of Figure 5.This collapses the 10-bipyramid to two tetrahedra and a flat face. We alsosee a volume drop from any bigon chain bipyramid with a crossing on the region A . Once againwe can assume an adjacent region A has two or more additional creossings on it. The least volumedrop come from a bigon having both crossings on the region, in which case the corresponding6-bipyramid collapses to two tetrahedra. The extra collapsed edge can slide out of the region andthe minimal volume drop from it occurs whenit takes a 10-bipyramid to a 9-bipyramid. Thus, thetotal volume drop is at least 2vol( B ) + vol( B ) − v tet − vol( B ) − vol( B ) ≈ . g >
0, we take the 6-bipyramid and its four associated tetrahedra as in Figure 6(c)and (d) corresponding to one of the vertical components that were added, and we collapse the edgeg created at the top of edge 8 and the edge h created at the bottom of edge 8. There are 6 edgeson the 6-bipyramid and the four associated tetrahedra in the equivalence class of each of theseedges, and their collapse flattens all four tetrahedra and the entire 6-bipyramid. It also forces thecollapse of edges 7 and 10. Since edges 7 and 10 are central edges to two regions adjacent to thenew vertical component, their collapse will impact any bipyramids that touch these regions. Theminimal volume drop comes from the case of having just one additional crossing on the first region,which is the end of a bigon chain of crossing length 4. Then the collapse of the central edge ofthis region will collapse a 10-bipyramid to a 9-bipyramid and drop volume by 0.6539. The other adjacent region must have more than one other crossing in it as otherwise the projection would notbe twist reduced. Considering the various options, one finds that the smallest volume drop occurswhen there are two bigon chains of crossing length 4 ending on this region. Then the collapse ofthe central edge of this region results in collapsing two 10-bipyramids to 9-bipyramids, and yieldsan additional volume drop of 2(0.6539). Thus, the total volume drop is at least 12.1111. (cid:3)
Note that the excluded cases of the Borromean rings and of all reduced twist reduced alter-nating diagrams with just two twist regions, which all come from Dehn filling the Borromean rings,must have volume bounded above by 2 v oct .As in [10], this upper bound on volume can be translated into a result in terms of coefficientsof the colored Jones polynomial. Let J K ( n ) = ± ( a n q k n − b n q k n − + c n q k n − ) + · · · ± ( γ n q k n − r n +2 − β n q k n − r n +1 + α n q k n − r n )be the colored Jones polynomial of K , where a n and α n are positive. Theorem 3.2. If L is a hyperbolic alternating link with at least three twist sequences in a twistreduced diagram other than the Borromean rings, then vol ( S − L ) < (10 v tet − v oct )(( c + γ ) − ( c + γ )) − (10 v tet − v oct )( b + β ) − a , where a = 10 . if b + β (cid:54) = ( c − c ) + ( γ − γ ) and a = 15 . otherwise. Face-Centered Bipyramids
We introduce a second method for decomposing a link complement into polyhedra. Given areduced projection of a knot or link, thought of as a projection on a sphere, let b i be the numberof complementary regions of the projection with exactly i edges. By Euler characteristic, it mustbe the case that: 2 b + b = 8 + b + 2 b + 3 b + . . . Note that the number of regions with four edges is not restricted by this formula.There will be one n -bipyramid corresponding to each complementary face with n edges inthe projection.To show this, we begin with the octahedral construction, placing one octahedron ateach crossing with top vertex on the underside of the overcrossing and bottom vertex on the topof the undercrossing. Now we cut each octahedron into four tetrahedra along the central verticaledge, as in Figure 7.Once we do so, we can glue the resulting tetrahedra together along an edge perpendicular toand at the center of each complementary face. The result is a decomposition of the link complementinto bipyramids, each with n ideal vertices and two finite vertices. Note that if a face is a bigon,the resulting bipyramid is a 2-bipyramid, which has no volume. These “flat” bipyramids can beglued to the adjacent bipyramids resulting in no volume contribution but rather just a change inthe gluing of the faces involved.Since any n -bipyramid has volume no more than the ideal regular bipyramid, we see immedi-ately that vol( S − L ) ≤ r (cid:88) i =1 b i vol( B n )We have not yet taken advantage of the fact that the two vertices U and D are finite vertices.But before we drill or collapse to reduce volume, it should be noted that with those finite vertices inplace, in the case of a knot, the collection of faces on the bottom halves of the bipyramids, which all IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 11 g hh a)b) c)d) g gg gg gh h h Figure 6.
Drilling a vertical component out of two paired octahedra.share the vertex denoted D together generate the singular punctured disk that appears in [14] andsubsequently [4], where it was used to obtain bounds on cusp volumes. This immersed punctureddisk is obtained by coning from the knot to a point beneath the projection plane, the role of whichis played by D . Similarly, the top faces of the bipyramids form a second singular punctured diskwhich is obtained by coning from the knot to a point above the projection plane corresponding tothe point U . Put the other way around, one can take any knot (or link) in an alternating projection Figure 7.
Cutting up octahedra and reassembling them into bipyramids.and cone to a point above and below the projection. Then cutting the complement open alongthese singular disks yields the decomposition into face-centered bipyramids.We also note that the ideal equatorial n -gons for each bipyramid glue together to form theunion of the two checkerboard surfaces for the projection.We now consider means to eliminate the finite vertices and reduce volume. We can choose thetwo bipyramids of largest n that are not adjacent to each other, meaning the faces of the projectionthat generate them do not share an edge. We then drill out an additional link component C thatis obtained by removing the union of the two vertical central edges for the pair of bipyramids. Ifthe result is a hyperbolic link, then the volume of the complement of our original link will be lessthan the volume of the complement of this link, which is at most the sum of the volumes of theideal regular n -bipyramids corresponding to the complementary regions other than the two we havedrilled through. The drilling collapses their central edges to ideal points, and thereby eliminatestheir volume contribution.In the case that the original link is alternating in a reduced alternating diagram, the resultinglink is a generalized augmented alternating link and by Theorem 2.1 of [3], it is hyperbolic.A second operation that we can do to lower the volume bound other than drilling is collapsing.In collapsing, we choose an edge running from the cusp to D and an edge running from the cuspto U . Then we collapse both of these edges, pulling the finite vertices to the cusp. To have themaximal effect on volume, we choose our edges to be along a strand of the original link complementthat bounds two faces with as many edges as possible. See Figure 4.The collapse of the edges labelled 1 and 2 causes the edges labelled 3 and 4 to collapse aswell. Those two edges are central to the two bipyramids corresponding to the two faces to eitherside, so their volume is eliminated. It is also true that the edges labelled 1 and 2 can be slid outpast the two crossings so that the two bipyramids to the left and the two bipyramids to the righteach contain one of the edges that is collapsed. As in Figure 8, for n ≥
3, such a collapse on an n -bipyramid results in a collection of n − n − − bipyramid. This collection of tetrahedras can therefore have no more volume than theresulting ( n − n ≥
12, the resulting ( n − n − ≤ n ≤
11, the n − n − IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 13 volumes of all of the bipyramids, except for the two adjacent ones that we have collapsed and thefour others that have each been reduced from a contribution of vol( B n ) to a contribution of either( n − v tet or vol( B n − ). Thus, we have the following theorem. Figure 8.
Collapsing the thickened edge on an n -bipyramid yields n − n − Theorem 4.1. If L is a hyperbolic alternating link with reduced alternating projection with b i facesof i edges, then vol ( S − L ) ≤ (cid:88) b i vol ( B i ) − a where a = max { r, s } , with r the maximum value that can be attained when r = vol ( B c ) + vol ( B d ) for c and d the numbers of edges for a pair of nonadjacent faces, and s the maximum value thatcan be attained when s = vol ( B c ) + vol ( B d ) + (cid:88) j =1 vol ( B e j ) − (cid:88) j =1 V e j where c and d are the numbers of edges in a pair of adjacent faces F c and F d , and { e j } are theedge numbers for the four faces that each share a crossing with both of F c and F d . When e j ≥ , V e j = vol ( B e j − ) . When ≤ e j ≤ , V e j = ( n − v tet . Example 4.2.
We consider the figure-eight knot. Its standard projection consists of comple-mentary regions with four triangles and two bigons. Choosing an edge between two triangles, wecollapse away the two 3-bipyramids to either side of the edge. At each end of the edge, there is alsoa 2-bipyramid and a 3-bipyramid. Collapsing an edge on the 2-bipyramid does nothing since italready contributes zero volume. Collapsing an edge on the 3-bipyramid yields a singe tetrahedron.So we know that the complement of the figure-eight knot has a volume of at most 2 v tet . This is infact the actual volume of the complement of the figure-eight knot, demonstrating that there is atleast one knot for which this upper bound on volume is exact.Note that in the case of an alternating hyperbolic link, between drilling and collapsing offace-centered bipyramids, we know that we can eliminate the volume contribution of any twobipyramids, whether they are adjacent or not.Of course, we would like to do better. We would like to be able to drill out additional linkcomponents to eliminate other face-centered bipyramids of high n .If we have two faces that share a bigon chain, and we would like to remove the bigon chainby drilling out a vertical component, as in the Agol-Thurston construction, we use the method depicted in Figure 6. That is, we replace the bigon chain by a bigon with two crossings, we inserttwo octahedra at the crossings, we glue them together and drill out the vertical component fromthem, and then decompose them into four tetrahedra and one 6-bipyramid. Then, as describedpreviously, we take all other crossings and fill them with octahedra, and then decompose the otheroctahedra into four tetrahedra. We then glue the resulting tetrahedra together along the verticaledges through the centers of all faces. When it comes to the two crossings along our bigon, weglue the four individual tetrahedra in to the bipyramids corresponding to the two faces adjacentto the bigon. The resulting n for each n -bipyramid that results has n -value corresponding to thelink diagram without the extra vertical component but with the inserted bigon.If we begin with a bigon chain of crossing length k , then by drilling the corresponding verticalcomponent, we switch two bipyramids corresponding to the adjacent faces from r and s to r − k + 2and s − k + 2. However, we have also replaced the chain of 2-bipyramids given by the chain ofbigons, all of which had zero volume, by a single 6-bipyramid. So it only makes sense to drill whenvol( B r ) + vol( B s ) > vol( B r − k +2 ) + vol( B s − k +2 ) + 6 . Flyping.
Given a particular reduced alternating projection of a hyperbolic link, we can useface-centered bipyramids to obtain an upper bound on the volume of the link complement. Onewould like to know that the bound obtained is as low as possible.
Theorem 4.3.
The smallest upper bound on volume obtained from a reduced alternating projectionof a hyperbolic link L via face-centered bipyramids occurs in a twist reduced diagram.Proof. We show that if a diagram of L is not twist-reduced, the sum of the volumes of the corre-sponding face-centered bipyramids can be reduced. Suppose P is a reduced alternating diagramthat is not twist-reduced. Then there must be a portion of the diagram with crossings to eitherside of a tangle as in Figure 9. T T Figure 9.
Flyping to twist reduce lowers the volume bound.If there are bigons to the outer side of both of these crossings, then we can flype to move allbut one of the crossings to one side of the tangle, without changing the numbers of n -bipyramidsfor all n . However, when we are down to only one crossing remaining to one side, then when weflype to twist-reduce and move it across, we replace an a -bipyramid and a b -bipyramid with asingle ( a + b − B a ) + vol( B b ) > vol( B a + b − ) forall a, b ≥
3. Assume that b ≥ a . Suppose first that a ≥ . Note that the tetrahedron T a , withangles πa , ( a − π a and ( a − π a has volume a decreasing function in a for a ≥
6. Thenvol( B a ) + vol( B b ) = a vol( T a ) + b vol( T b )(4.1) ≥ ( a + b )vol( T b )(4.2) ≥ ( a + b − T a + b − ) = vol( B a + b − )(4.3) IPYRAMIDS AND BOUNDS ON VOLUMES OF HYPERBOLIC LINKS 15
In the case that a = 3, 4 and 5, one can check that vol( B a ) > ( a − . ... ).Hence, we havevol( B a ) + vol( B b ) ≥ ( a − . ... ) + b vol( T b )(4.4) ≥ ( a + b − T b )(4.5) ≥ ( a + b − T a + b − ) = vol( B a + b − )(4.6) (cid:3) Comparisons of the upper bounds
In this section, we compare the bounds we have discussed. Note first that the octahedralupper bound is always better than the tetrahedral upper bound for c ≥
6. Similarly, the BCBbound always improves the DT bound, which itself improved the AT bound.For low crossing number, the FCB (face-centered bipyramid) bound always seems to do betterthan the BCB (bigon chain bipyramid) bound. However, if one chooses a link such as the oneappearing in Figure 10, but with n hexagonal faces, then we can think of the volume contributionto the BCB bound of each bigonal chain of length 3 weighted against the volume contribution ofits two adjacent hexagonal faces tom the FCB bound. We add in the volume contribution of eachsingleton crossing to the BCB bound and weight it against the volume contribution of the fouradjacent faces to the FCB bound. Therefore, the contribution to the FCB bound of a hexagonalface is vol( B ) = 6 . B ) / B ) / . n rather than n . So for large enough n , the BCB bound will always be lower than the FCBbound.However, at least for small crossing number, the FCB bound appears more effective, so wewill focus on it. In calculating a volume bound using the FCB bound, one can either drill twonon-adjacent bipyramids or one can collapse as described in the discussion of the algorithm forthe bound. In the case of certain knots and links, drilling can do better than collapsing, as occurswith 9 for instance. The same holds true for any (3 , , . . . , , , . . . , and 5 , this also holds for the twist knots, wherein all cases, drilling yields a bound of 4 v tet , and the sequence of volumes are known to approach v oct = 3 . . . . from below.In the case of 7 , drilling and collapsing in the FCB bound yield the same value. But moregenerally, at least for low crossing number, collapsing yields a better bound than does drilling. Inthe two cases of 4 and 6 , the FCB bound obtained by collapsing yields the actual volume, namely2 v tet and 4 v tet respectively. For hyperbolic knots of seven or fewer crossings, the FCB bound iswithin 4.1% of the actual volume for 7 , within 10.6% for the additional knots 5 , , , , and7 . Excluding the twist knots 6 and 7 , all the hyperbolic knots of seven or fewer crossings haveFCB bound within 19% of the actual volume.In the case of 5 , both the FCB bound and the octahedral bound yield the same bound of4 v tet , whereas the actual volume is v oct . But in general, the FCB bound beats the octahedralbound. Figure 10.
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