Volume of a doubly truncated hyperbolic tetrahedron

Transcription

1 Volume of a doubly truncated hyperbolic tetrahedron Alexander Kolpakov, Jun Murakami To cite this version: Alexander Kolpakov, Jun Murakami. Volume of a doubly truncated hyperbolic tetrahedron. Aequationes Mathematicae, Springer Verlag, 01, pp < /s y>. <hal v3> HAL Id: hal Submitted on 6 Dec 01 (v3), last revised Aug 017 (v5) HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 Volume of a doubly truncated hyperbolic tetrahedron Alexander Kolpakov Jun Murakami Abstract The present paper regards the volume function of a doubly truncated hyperbolic tetrahedron. Starting from the previous results of J. Murakami, U. Yano and A. Ushijima, we have developed a unified approach to expressing the volume in different geometric cases by dilogarithm functions and to treat properly the many analytic strata of the latter. Finally, several numeric examples are given. Key words: hyperbolic tetrahedron, Gram matrix, volume formula. 1 Introduction The real vector space R 1,n of dimension n + 1 with the Lorentzian inner product x,y = x 0 y 0 +x 1 y 1 + +x n y n, where x = (x 0,x 1,...,x n ) and y = (y 0,y 1,...,y n ), is called an (n+1)-dimensional Lorentzian space E 1,n. Consider the two-fold hyperboloid H = {x E 1,n x,x = 1} and its upper sheet H + = {x E 1,n x,x = 1,x 0 > 0}. The restriction of the quadratic form induced by the Lorentzian inner product, to the tangent spaceto H + ispositivedefinite, andsoitgivesariemannianmetricon H +. The space H + equipped with this metric is called the hyperboloid model of the n-dimensional hyperbolic space and denoted by H n. The hyperbolic distance d(x,y) between two points x and y with respect to this metric is given by the formula coshd = x,y. Consider the cone K = {x E 1,n x,x = 0} and its upper half K + = {x E 1,n x,x = 0,x 0 > 0}. A ray in K + emanating from the origin corresponds to a point on the ideal boundary of H n. The set of such rays forms a sphere at infinity S n 1 = H n. Thus, each ray in K + becomes an ideal point of H n = H n H n. supported by the Schweizerischer Nationalfonds SNF no /1 supported by Grant-in-Aid for Scientific Research no

3 Let p denote the radial projection of E 1,n \{x E 1,n x 0 = 0} onto the affine hyperplanep n 1 = {x E1,n x 0 = 1} along a ray emanating from the origin o. Theprojection pis adiffeomorphismof H n onto theopenn-dimensionalunit ball B n in P n 1 centred at (1,0,0,...,0) which defines a projective model of H n. The affine hyperplane P n 1 includes not only Bn and its set-theoretic boundary B n in P n 1, which is canonically identified with Sn 1, but also the exterior of the compactified projective model B n = B n B n = H n S n 1. Let ExtB n denote the exterior of B n in P n. Thus p could be naturally extended to a map from E 1,n \o onto an n-dimensional real projective space P n = P n 1 Pn, where Pn is the set of straight lines in the affine hyperplane {x E 1,n x 0 = 0} passing through the origin. Consider the one-fold hyperboloid H = {x E 1,n x,x = 1}. Given some point u in H define in E 1,n the half-space R u = {x E 1,n x,u 0} and the hyperplane P u = {x E 1,n x,u = 0} = R u. Denote by Γ u (respectively Π u ) the intersection of R u (respectively P u ) with B n. Then Π u is a geodesic hyperplane in H n, and the correspondence between the points in H and the half-space Γ u in H n is bijective. Call the vector u normal to the hyperplane P u (or Π u ). Let v be a point in ExtB n. Then p 1 (v) H consists of two points. Let ṽ denote one of them, so we may define the polar hyperplane Πṽ to v, independent on the choice of ṽ p 1 (v) H. Now we descend to dimension n = 3. Let T be a tetrahedron in P 3 1, that is a convex hull of four points {v i } 4 i=1 P3 1. We say a vertex v {v i} 4 i=1 to be proper if v B 3, to be ideal if v B 3 and to be ultra-ideal if v ExtB 3. Letv beanultra-ideal vertex oft. Wecall a truncation of v theoperationof removingthepyramidwithapexv andbaseπṽ T. Ageneralised hyperbolic tetrahedron T is a polyhedron, possibly non-compact, of finite volume in the hyperbolic space obtained from a certain tetrahedron by polar truncation of its ultra-ideal vertices. In case when only two vertices are truncated, we call such a generalised tetrahedron doubly truncated. Depending on the dihedral angles, the polar hyperplanes may or may not intersect. In Fig. 1 a tetrahedron T with two truncated vertices is depicted. The corresponding polar planes do not intersect. We shall call this kind of generalised tetrahedron mildly truncated. In Fig. the case when the polar planes intersect is shown. Here the tetrahedron T is truncated down to a prism, hence we shall call it prism truncated. In this paper we study the volume function of a doubly truncated hyperbolic tetrahedron. The question arises first in the paper by R. Kellerhals [4], where a doubly truncated hyperbolic orthoscheme is considered. The whole evolution of an orthoscheme, starting from a mildly truncated one down to

4 a Lambert cube has been investigated. The case of a general hyperbolic tetrahedron was considered in numerous papers [1, 3, 8]. The case of a mildly truncated tetrahedron is due to J. MurakamiandA.Ushijima[9,13]. Thepaper[13]isthestartingpointwherethe question about intense truncations of a hyperbolic tetrahedron was posed. Thus, the case of a prism truncated tetrahedron remains unattended. As we shall see later, it brings some essential difficulties. First, the structure of the volume formula should change, as first observed in [4]. Second, the branching properties of the volume function come into sight. This phenomenon was first observed for tetrahedra in the spherical space and is usually related to the use of the dilogarithm function or its analogues, see [5, 9, 10]. For the rest of the paper, a mildly (doubly) truncated tetrahedron is given in Fig. 1. Its dihedral angles are θ k and its corresponding edge lengths are l k, k = 1,6. Figure 1: Mildly (doubly) truncated tetrahedron A prism truncated tetrahedron is given in Fig.. The dashed edge connects the ultra-ideal vertices and corresponds to the edge l 4 in the previous case. The dihedral angles remain the same, except that the altitude l of the prism replaces the dihedral angle θ 4. The altitude carries the dihedral angle µ and is orthogonal to the bases because of the truncation. Right dihedral angles in Fig. 1 are not indicated by symbols. The other dihedral angles and for the given integers n 1, m 0, the notation k = n,n+m means that k {n,n+1,...,n+m}. 3

5 the corresponding edge length are called essential. Figure : Prism (intensely) truncated tetrahedron T and its geometric parameters Acknowledgement. The first named author is grateful to Waseda University, Tokyo for hospitality during his stay in December, 011. Part of this work was performed at the Institut Henri Poincaré, Paris in February- March, 01. Erratum. This is a published version of the paper. For formulas (4), (14) and (17) please see Erratum. Preliminaries The following propositions reveal a relationship between the Lorentzian inner product of two vectors and the mutual position of the respective hyperplanes. Proposition 1 Let u and v be two non-collinear points in H. Then the following holds: i) The hyperplanes Π u and Π v intersect if and only if u,v < 1. The dihedral angle θ between them measured in Γ u Γ v is given by the formula cosθ = u,v. 4

6 ii) The hyperplanes Π u and Π v do not intersect in B n if and only if u,v > 1. They intersect in ExtB n and admit a common perpendicular inside B n of length d given by the formula coshd = ± u,v. iii) The hyperplanes Π u and Π v intersect on the ideal boundary B n only, if and only if u,v = 1. In case ii) we say the hyperplanes Π u and Π v to be ultra-parallel and in case iii) to be parallel. Proposition Let u be a point in B n and let Π v be a geodesic hyperplane whose normal vector v H is such that u,v < 0. Then the length d of the perpendicular dropped from the point u onto the hyperplane Π v is given by the formula sinhd = u,v. Let T be a generalised tetrahedron in H 3 with outward Lorentzian normals n i, i = 1,4, to its faces and vertex vectors v i, i = 1,4, as depicted in Fig.. Let G denote the Gram matrix for the normals G = n i,n j 4 i,j=1. The conditions under which G describes a generalised mildly truncated tetrahedron are given by [6] and [13]. For a prism truncated tetrahedron, its existence and geometry are determined by the vectors ñ i, i = 1,6, where ñ i = n i for i = 1,4 and ñ 5 = v 1, ñ 6 = v. Hence the above matrix G does not have necessarily a Lorentzian signature. However, it will suffice for our purpose. In case T is prism truncated, as Fig. shows, we obtain G = 1 cosθ 1 cosθ cosθ 6 cosθ 1 1 cosθ 3 cosθ 5 cosθ cosθ 3 1 coshl cosθ 6 cosθ 5 coshl 1 in accordance with Proposition 1. We will define the edge length matrix G of T by G = 1 cosµ isinhl isinhl 6 cosµ 1 isinhl 3 isinhl 5 isinhl isinhl 6 1 coshl 1 isinhl 6 isinhl 5 coshl 1 1, (1) () in order to obtain a Hermitian analogue of the usual edge length matrix for a mildly truncated tetrahedron (for the latter, see [13]). we chose the minus sign if Γ u Γ v =, otherwise we choose the plus sign. 5

7 By [6] or [13], we have that g ij = c ij cii cjj, (3) wherec ij arethecorresponding(i,j) cofactors ofthematrixg. Thecomplex conjugation gij = g ji, i,j = 1,4, corresponds to a choice of the analytic strata for the square root function. Proposition 3 ([6, 13]) The vertex v i, i = 1,4, of T is proper, ideal, or ultra-ideal provided that c ii > 0, c ii = 0, or c ii < 0, respectively. Hence, Propositions 1- imply that the matrices G and G agree concerning the relationship of the geometric parameters of T. The matrix G has complex entries since the minors c ii, i = 1,4, in formula (3) may have different signs by Proposition 3. To perform our computations later on in a more efficient way, we shall introduce the parameters a k, k = 1,6, associated with the edges of the tetrahedron T. If T is a prism truncated tetrahedron, then we set a k := e iθ k, k {1,,3,5,6}, a 4 := e l and then G = 1 a 1+1/a 1 a +1/a a 6+1/a 6 a 1+1/a 1 1 a 3+1/a 3 a 5+1/a 5 a +1/a a 3+1/a 3 1 a 4+1/a 4 a 6+1/a 6 a 5+1/a 5 a 4+1/a 4 1. (4) The meaning of the parameters becomes clear if one observes the picture of a mildly truncated tetrahedron T (see Fig. ). In case two vertices v 1 and v of T become ultra-ideal and the corresponding polar hyperplanes intersect (see [13, Sections -3] for more basic details), the edge v 1 v becomes dual in a sense to the altitude l of the resulting prism. Since in case of a mildly truncated tetrahedron we have a k = e iθ k according to [13], the altitude l for now corresponds still to the parameter a 4, in order to keep consistent notation. 3 Volume formula Let U = U (a 1,a,a 3,a 4,a 5,a 6,z) denote the function U = Li (z)+li (a 1 a a 4 a 5 z)+li (a 1 a 3 a 4 a 6 z)+li (a a 3 a 5 a 6 z) (5) Li ( a 1 a a 3 z) Li ( a 1 a 5 a 6 z) Li ( a a 4 a 6 z) Li ( a 3 a 4 a 5 z) 6

8 depending on seven complex variables a k, k = 1,6 and z, where Li ( ) is the dilogarithm function. Let z and z + be two solutions to the equation e z U z = 1 in the variable z. According to [8], these are z = q 1 q1 4q 0q and z + = q 1 + q 1 4q 0 q, (6) q q where q 0 = 1+a 1 a a 3 +a 1 a 5 a 6 +a a 4 a 6 +a 3 a 4 a 5 +a 1 a a 4 a 5 +a 1 a 3 a 4 a 6 +a a 3 a 5 a 6, (( q 1 = a 1 a a 3 a 4 a 5 a 6 a 1 1 )( a 4 1 ) )+ (a )(a 1a 5 1a5 a 1 a 4 )) + (a 3 )(a 1a3 6 1a6, (7) q = a 1 a a 3 a 4 a 5 a 6 (a 1 a 4 +a a 5 +a 3 a 6 +a 1 a a 6 +a 1 a 3 a 5 +a a 3 a 4 + a 4 a 5 a 6 +a 1 a a 3 a 4 a 5 a 6 ). Given a function f(x,y,...,z), let f(x,y,...,z) z=z z=z + denote the difference f(x,y,...,z ) f(x,y,...,z + ). Now we define the following function V = V (a 1,a,a 3,a 4,a 5,a 6,z) by means of the equality V = i ( U (a 1,a,a 3,a 4,a 5,a 6,z) z U ) z=z 4 z logz. (8) z=z + Let W = W (a 1,a,a 3,a 4,a 5,a 6,z) denote the function below, that will correct possible branching of V resulting from the use of (di-)logarithms: W = 6 k=1 ( V a k i ) V a k 4 loge 4ia k a k loga k. (9) Given a generalised hyperbolic tetrahedron as in Fig., truncated down to a quadrilateral prism with essential dihedral angles θ k, k {1,,3,5,6}, altitude l and dihedral angle µ along it, we set a k = e iθ k, k {1,,3,5,6}, a 4 = e l, as above. Then the following theorem holds. Theorem 1 Let T be a generalised hyperbolic tetrahedron as given in Fig.. Its volume equals ( VolT = R V + W µl ). 7

9 Note. In the statement above, the altitude length ( is) l = Rloga 4 and the V corresponding dihedral angle equals µ = R a 4 a 4 modπ. 3.1 Preceding lemmas. Before giving a proof to Theorem 1, we need several auxiliary statements concerning the branching of the volume function. Lemma 1 The function W has a.e. vanishing derivatives W a k, k = 1,6. Proof. By computing the derivative of (9) with respect to each a k, k = 1,6, outside of its branching points and by making use of the identities d dz logz = 1/z, e z e w = e z+w for all z,w C. Since the branching points of a finite amount of log( ) and Li ( ) functions form a discrete set in C, the lemma follows. Lemma The function R( V + W ) does not branch with respect to the variables a k, k = 1,6, and z. Proof. Let us consider a possible branching of the function defined by formula (5). Let U comprise only principal strata of the dilogarithm and let U correspond to another ones. Then we have U z=z± = U z=z± +πik ± 0 log(z ±)+πik ± 1 log(a 1a a 4 a 5 z ± ) +πik ± log(a 1a 3 a 4 a 6 z ± )+πik ± 3 log(a a 3 a 5 a 6 z ± ) +πik ± 4 log( a 1a a 3 z ± )+πik ± 5 log( a 1a 5 a 6 z ± ) (10) +πik ± 6 log( a a 4 a 6 z ± )+πik ± 7 log( a 3a 4 a 5 z ± ) +4π k ± 8, with some k j Z, j = 0,8. From the above formula, it follows that U z ± U 7 z logz ± = z ± z=z± z logz ± +πi k j ± logz ±. (11) z=z± Then, according to formulas (8), (10) (11), the following expression holds for the corresponding analytic strata of the function V : V = V π 6 j=1 j=0 m j loga j + iπ m 7, (1) where m j Z, j = 1,7 and we have used the formula log(uv) = logu + logv +πik, k Z. Hence, according to (1), we compute a j V a j = a j V a j πm j, 8

10 for each j = 1,6. The latter implies that V a j i a j 4 log V e 4ia j a j V = a j i a j 4 log V e 4ia j a j πm j, for j = 1,6, since we choose the principal stratum of the logarithm function log( ). Thus, by formula (9), V + W = V + W + π 6 j=1 m j loga j iπ m 7 π with m j Z, j = 1,7. The proof is completed. 6 m j loga j j=1 = V + W iπ m 7, 3. Proof of Theorem 1. The scheme of our proof is the following: first we show that θ k VolT = l k, k {1,,3,5,6}, µ VolT = l and second we apply the generalised Schläfli formula [7, Equation 1] to show that the volume function and the one from Theorem 1 coincide up to a constant. Finally, the remaining constant is determined. Now, let us prove the three statements below: (i) θ 1 VolT = l 1, (ii) θ k VolT = l k for k {,3,5,6}, (iii) µ VolT = l. Note that in case (ii) it suffices to show θ VolT = l. The statement for another k {3,5,6} is completely analogous. Let us show that the equality in case (i) holds. First, we compute ( U z U ) ( θ 1 z logz = ia 1 U z U ) a 1 z logz = ( U = ia 1 logz ( z U )) a 1 a 1 z Upon the substitution z := z ±, we see that ( ) U z ± a 1 z (a 1,a,a 3,a 4,a 5,a 6,z ± ) = 0, 9

11 by taking the respective derivative on both sides of the identity e z ± U z (a 1,a,a 3,a 4,a 5,a 6,z ± ) = 1, c.f. the definition of z ± and formula (6). Finally, we get V = a 1 U z=z θ 1 4 a 1 = 1 z=z + 4 log φ(z )ψ(z + ) φ(z + )ψ(z ) + iπ k, for a certain k Z, where the functions φ( ) and ψ( ) are φ(z) = (1+a 1 a a 3 z)(1+a 1 a 5 a 6 z), ψ(z) = (1 a 1 a a 4 a 5 z)(1 a 1 a 3 a 4 a 6 z). The real part of the above expression is R V = 1 θ 1 4 log φ(z )ψ(z + ) φ(z + )ψ(z ). (13) Let us set = detg and δ = detg. We shall show that the expression ( E = φ(z )ψ(z + ) c 34 δ a ) ( 1 1/a 1 φ(z + )ψ(z ) c 34 +δ a ) 1 1/a 1 (14) is identically zero for all a k C, k = 1,6, which it actually depends on. In order to perform the computation, the following formulas are used: z = ˆq 1 4δ ˆq, z + = ˆq 1 +4δ ˆq, where ˆq l = q l / 6 k=1 a k for l = 1, (c.f. formulas (6) (7)). Note that one may consider the expression E as a rational function of independent variables a k, k = 1,6, and δ, making the computation easier to perform by a software routine [15]. First, we have 1 E = 4a 1Y δ a 1 ˆq 3 a 1, where Y = Y (a 1,a,a 3,a 4,a 5,a 6 ) is a certain technical term explained in Appendix. Since a 1 = δ δ a 1, the above expression gives us E = 8a 1 δ Y. (15) a 1 a 1 ˆq 3 10

12 Second, we obtain E δ = 8a 1 Y. (16) Finally, we recall that δ is a function of a k, k = 1,6, and the total derivative of E with respect to a 1 is ( ) E a δ:=δ(a1,a,a 3,a 4,a 5,a 6 ) = E + E δ = 0, 1 a 1 δ a 1 according to equalities (15) (16). ) An analogous computation shows that a k (E δ:=δ(a1,a,a 3,a 4,a 5,a 6 ) ˆq 3 = 0 for all k = 1,6. Then by setting a k = 1, k = 1,6, we get = 16 and so δ = 8i. In this case E = 0. Thus the equality E 0 holds for all a k C, k = 1,6. Together with (13) it gives R V = 1 θ 1 4 log On the other hand, by formula (3), we have c 34 +δ a 1 a 1 1 c 34 δ a 1 a 1 1 g 34 = coshl 1 = c 34 c33 c44.. (17) Here, both c 33 and c 44 are positive by Proposition 3, since the vertices v 3 and v 4 are proper. The formula above leads to the following equation e l 1 +g 34 el 1 +1 = 0, the solution to which is determined by 1 1 e l 1 = c 34 +δ a 1 a c 34 δ a 1 a 1 1, (18) in analogy to [13, Equation 5.3]. Thus, equalities (17) (18) imply R V θ 1 = 1 4 log e l 1 = l 1. Together with Lemma 1, this implies that claim (i) is satisfied. As already mentioned, in case (ii) it suffices to prove θ VolT = l. The statement for another k {3,5,6} is analogous. 11

13 By formula (3) we have that g 4 = isinhl = c 4 c c44. Since the vertex v is ultra-ideal and the vertex v 4 is proper, by Proposition 3, c < 0 and c 44 > 0. Thus, The formula above implies sinhl = e l = c 4 +c 4 c 4 c c 44. c 4 c c 44 +c 4 c c 44 c c 44 (19) By applying Jacobi s theorem [11, Théorème.5.] to the Gram matrix G, we have ( c 4 c a a 1 ) c 44 =. (0) Combining (19) (0) together, it follows that 1 e l = c 4 +δ a a c 4 δ a a 1. (1) By analogy with (13), we get the formula R V = 1 θ 4 log φ(z )ψ(z + ) φ(z + )ψ(z ), () where φ(z) = (1+a 1 a a 3 z)(1+a a 4 a 6 z), ψ(z) = (1 a 1 a a 4 a 5 z)(1 a a 3 a 5 a 6 z). Similar to case (i), the following relation holds: 1 φ(z )ψ(z + ) φ(z + )ψ(z ) = c 4 δ a a c 4 +δ a a 1 Then formulas (1) () yield R V = 1 θ 4 log 1 e l = l. 1.

14 The first equality of case (ii) now follows. Carrying out an analogous computation for R V θ k, k = 3,5, we obtain R V = l 3 θ 3 Thus, all equalities of case (ii) hold. As before, by formula (3), we obtain that and R V θ 5 = l 5. g 1 = cosµ = c 1 c11 c. (3) Since both vertices v 1 and v are ultra-ideal, by Proposition 3, the cofactors c 11 and c are negative. Then (3) implies the equation e iµ + c 1e iµ c11 c +1 = 0. Without loss of generality, the solution we choose is e iµ = c 1 + c 1 c 11c c11 c. (4) By squaring (4) and by applying Jacobi s theorem to the corresponding cofactors of the matrix G, the formula below follows: 1 4 e iµ = c 1 δ a 4 a c 1 +δ a 4 a 1 4. (5) On the other hand, a direct computation shows that R V l = i ( ) 4 log φ(z )ψ(z + ), (6) φ(z + )ψ(z ) where φ(z) = (1+a 3 a 4 a 5 z)(1+a a 4 a 6 z), ψ(z) = (1 a 1 a a 4 a 5 z)(1 a 1 a 3 a 4 a 6 z). Similar to cases (i) and (ii), we have the relation 1 4 φ(z )ψ(z + ) φ(z + )ψ(z ) = c 1 δ a4 a c 1 +δ a 4 a ,

15 which together with (5) (6) yields V l = i 4 logeiµ = µ π k, k Z. Since, 0 µ π, we choose k = 0 and so V l = µ. The latter formula implies the equality of case (iii). Now, by the generalised Schläfli formula [7], dvolt = 1 k {1,,3,5,6} l k dθ k l dµ. Together with the equalities of cases (i)-(iii) it yields ( VolT = R V + W µl ) + C, (7) where C R is a constant. Finally, we prove that C = 0, and the theorem follows. Passing to the limit θ k π, k = 1,6, the generalised hyperbolic tetrahedron T shrinks to a point, since geometrically it tends to a Euclidean prism. Thus, we have µ π and l 0. By setting the limiting values above, we obtain that a k = i, k {1,,3,5,6}, a 4 = 1. Then z = z + = 1 by equality (6), and hence V = 0. Since the dilogarithm function does not branch at ±1, ±i, we have W = 0. Since T shrinks to a point, VolT 0, that implies C = 0 by means of (7) and the proof is completed. 14

16 (θ 1,θ,θ 3,θ 5,θ 6 ) (l,µ) Volume Reference ( π, π 3, π 3, π, π ) (0,0) [] ( π, π 3, π 3, π, π ) ( , π 4 ) [] ( π, π 3, π 4, π, π ) (0,0) [] ( π, π 3, π 4, π, π ) ( , π 4 ) [] ( π, π 3, π 5, π, π ) (0,0) [] ( π, π 3, π 5, π, π ) ( , π 4 ) [] ( π,0, π,0, π ) (0,0) [1] ( π, π 3, π, π 3, π ) ( , π 3 ) [1] ( π, π 4, π, π 4, π ) ( , π 4 ) [1] ( π, π 5, π, π 5, π ) ( , π 5 ) [1] ( π, π 6, π, π 6, π ) ( , π 6 ) [1] ( π, π 10, π, π 10, π ) ( , π 10 ) [1] this value is misprinted in [1, Equation 4.11] Table 1: Some numerically computed volume values 4 Numeric computations In Table 1 we have collected several volumes of prism truncated tetrahedron computed by using Theorem 1 and before in [, 1]. The altitude of a prism truncated tetrahedron is computed from its dihedral angles, as the remark after Theorem 1 states. All numeric computations are carried out using the software routine Mathematica [15]. Appendix The term Y from Section 3. Let us first recall of the expressions q k, k = 0,1,, that are polynomials in the variables a k, k = 1,6 defined by formula (7) and the expressions ˆq l, l = 1,, defined in the proof of Theorem 1 by ˆq l = q l / 6 k=1 a k, l = 1,. Then, the following lemma holds concerning the technical term Y mentioned 15

17 above, that actually equals Y = a 1 a a 3 a 4a 5 +a 1a a 3a 4a 5 +a 3 a 3a 4a 5 +a 1 a a 3 3a 4a 5 +a 1 a a 4 a 5 +a 1a a 3 a 4 a 5 +a 3 a 3 a 4 a 5 +a 1 a a 3a 4 a 5 +a 1a a 3 a 4a 6 +a 1 a a 3a 4a 6 +a 1 a 3 a 3 a 4 a 6 +a a3 3 a 4 a 6 +a 1 a a 4a 5 a 6 +a 1 a a 3 a 4 a 5 a 6 +a 3 1 a a 3 a 4 a 5 a 6 +a 1 a 3 a 3a 4 a 5 a 6 +a 1 a 3 a 4a 5 a 6 +a a 3 a 4a 5 a 6 a 1 a a 3 a 4a 5 a 6 +a 4 1 a a 3 a 4a 5 a 6 +a 1 a a 3 3a 4 a 5 a 6 +a 1 a a 3 a 3 4a 5 a 6 a 3 1a a 3 a 3 4a 5 a 6 +a a 3a 3 4a 5 a 6 a 1a a 3a 3 4a 5 a 6 +a a 3 a 5a 6 a 1a a 3 a 5a 6 +a 1 a a 3a 5a 6 a 3 1a a 3a 5a 6 +a 1 a a 4a 5a 6 +a 1a 3 a 4a 5a 6 +a a 3 a 4a 5a 6 a 1a a 3 a 4a 5a 6 +a 4 1a a 3 a 4a 5a 6 +a 1 a a 3a 4a 5a 6 +a 3 1 a a 3 a 4 a 5 a 6 +a 1 a 3 a 3 a 4 a 5 a 6 +a 1 a a3 3 a 4 a 5 a 6 +a a 4a 3 5 a 6 +a 1 a a 3 a 4 a 3 5 a 6 +a 1 a 3 a 3a 4 a 3 5 a 6 +a 1 a a 3 a 4a 3 5 a 6 +a 1 a a 3 a 4 a 6 +a 1a 3 a 4a 6 +a 1a a 3 a 4a 6 +a a 3 3 a 4a 6 +a 1a a 3a 5 a 6 a3 1 a a 3a 5 a 6 +a a 3 a 5a 6 a 1 a a 3 a 5a 6 +a 1a a 4a 5 a 6 +a 1 a 3 a 4a 5 a 6 +a 1 a a 3 a 4a 5 a 6 +a 3 1a a 3 a 4a 5 a 6 +a a 3a 4a 5 a 6 a 1a a 3a 4a 5 a 6 +a 4 1a a 3a 4a 5 a 6 +a 1a 3 a 3a 4a 5 a 6 +a 1 a a 3 3a 4a 5 a 6 +a 1 a a 4 a 5a 6 +a a 3 a 4 a 5a 6 a 1a a 3 a 4 a 5a 6 +a 4 1a a 3 a 4 a 5a 6 +a 1a 3 a 3 a 4 a 5a 6 +a 1 a 3a 4 a 5a 6 +a 1 a a 3 a 4a 5 a 6 +a3 1 a a 3 a 4a 5 a 6 +a 1 a a 3 3 a 4a 5 a 6 +a a 3 a 3 4 a 5 a 6 a 1 a a 3 a 3 4 a 5 a 6 +a 1 a a 3 a3 4 a 5 a 6 a3 1 a a 3 a3 4 a 5 a 6 +a a 4 a3 5 a 6 +a 1a 3 a 4 a3 5 a 6 +a 1a a 3a 4 a3 5 a 6 +a 1a a 3a 4a 3 5a 6 +a 1 a a 3 a 4 a 5 a 3 6 +a 3a 4 a 5 a 3 6 +a 1a a 3a 4 a 5 a 3 6 +a 1 a a 3 3a 4 a 5 a 3 6 +a 1 a a 4a 5a 3 6 +a 3 a 4a 5a 3 6 +a 1a a 3 a 4a 5a 3 6 +a 1 a a 3a 4a 5a 3 6. Lemma 3 The above expression Y is not a polynomial neither in the variables q 0, q 1, q, nor in the variables q 0, q 1, ˆq. Proof. By setting a 1 = a 6 := 0, we get q 0 = 1+a 3 a 4 a 5, q 1 = q = 0 and Y a1 =a 6 :=0 = a 3 a 3 a 4 a 5 (a 3 a 4 +a 5 ) = a 3 (q 0 a1 =a 6 :=0 1)(a 3 a 4 +a 5 ). As well, we have ˆq = a (a 3 a 4 +a 5 ) and Y a1 =a 6 :=0 = a (q 0 a1 =a 6 :=0 1) ˆq a1 =a 6 :=0. The former equality proves that Y is not a polynomial in the variables q 0, q 1, q. The latter shows that Y is not a polynomial in q 0, q 1, ˆq either. Erratum This erratum indicates several typos present in the paper entitled Volume of a doubly truncated hyperbolic tetrahedron by A. Kolpakov and J. Murakami. The first named author bears the responsibility for these typos 16

18 appeared in print. He would like to thank Prof. Atakan T. Yakut, who drew his attention to that fact. 1. The matrix G from formula () should read as G = 1 cosµ isinhl 5 isinhl 3 cosµ 1 isinhl 6 isinhl isinhl 5 isinhl 6 1 coshl 1 isinhl 3 isinhl coshl In formula (14), the function E should read as ( E = φ(z )ψ(z + ) c 34 +δ a ) ( 1 1/a 1 φ(z + )ψ(z ) c 34 δ a ) 1 1/a Formula (17), consequently, should have the form R V = 1 θ 1 4 log c 34 δ a 1 a 1 1 c 34 +δ a 1 a 1 1. All of the typos listed above do not affect the result of the paper. References [1] Y. Cho, H. Kim On the volume formula for hyperbolic tetrahedra, Discreet Comput. Geom. (3), (1999). [] D.A. Derevnin, A.C. Kim The Coxeter prisms in H 3 in Recent advances in group theory and low-dimensional topology (Heldermann, Lemgo, 003), pp [3] D.A. Derevnin, A.D. Mednykh A formula for the volume of a hyperbolic tetrahedron, Russ. Math. Surv., 60 (), (005). [4] R. Kellerhals On the volume of hyperbolic polyhedra, Math. Ann. 85 (4), (1989). [5] A. Kolpakov, A. Mednykh, M. Pashkevich Volume formula for a Z -symmetric spherical tetrahedron through its edge lengths, Arkiv för Matematik, to appear; arxiv: [6] A.D. Mednykh, M.G. Pashkevich Elementary formulas for a hyperbolic tetrahedron, Siberian Math. J. 47 (4), (006). 17

19 [7] J. Milnor, The Schläfli differential equality in Collected Papers. I. Geometry (Publish or Perish, Houston, TX, 1994), pp [8] J. Murakami, M. Yano On the volume of hyperbolic and spherical tetrahedron, Comm. Annal. Geom. 13 (), (005). [9] J. Murakami, A. Ushijima A volume formula for hyperbolic tetrahedra in terms of edge lengths, J. Geom. 83 (1-), (005); arxiv:math/ [10] J. Murakami, The volume formulas for a spherical tetrahedron, Proc. Amer. Math. Soc. 140, (01); arxiv: [11] V. Prasolov Problèmes et théorèmes d algèbre linéaire, Enseignement des mathématiques (Cassini, Paris, 008), pp [1] J. Szirmai Horoball packings for the Lambert-cube tilings in the hyperbolic 3-space, Beitr. Algebra Geom. 44 (1), (005). [13] A. Ushijima A volume formula for generalised hyperbolic tetrahedra in Mathematics and Its Applications 581 (Springer, Berlin, 006), pp [14] E.B. Vinberg, Geometry II: Spaces of constant curvature, Encycl. Math. Sci. 9, (1993). [15] Wolfram Research, Mathematica 8, a computational software routine; Alexander Kolpakov Department of Mathematics University of Fribourg chemin du Musée 3 CH-1700 Fribourg, Switzerland kolpakov.alexander(at)gmail.com Jun Murakami Department of Mathematics Faculty of Science and Engineering Waseda University Okubo Shinjuku-ku Tokyo, Japan murakami(at)waseda.jp 18