Siberian Mathematical Journal, Vol. 45, No. 5, pp , 2004 Original Russian Text c 2004 Derevnin D. A., Mednykh A. D. and Pashkevich M. G.

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1 Siberian Mathematical Jornal Vol. 45 No. 5 pp Original Rssian Text c 2004 Derevnin D. A. Mednykh A. D. and Pashkevich M. G. On the volme of symmetric tetrahedron 1 D.A. Derevnin A.D. Mednykh M.G. Pashkevich Abstract An elementary formla is obtained for the volme of symmetric tetrahedron in hyperbolic and spherical spaces. Mathematical Sbject Classification: 30F40 51M10 57M50 Key words: hyperbolic tetrahedron spherical tetrahedron volme formla Gram matrix 1 Introdction The calclation of volme of polyhedron is very old and difficlt problem. A few years ago it was shown by I.H. Sabitov [Sb] that the volme of Eclidean polyhedron is a root of algebraic eqation whose coefficient are fnction combinatorial type and lengths of polyhedra. In hyperbolic and spherical spaces the sitation is march more complicated. Since Lobachevsky and Schläfli see [L] and [Sh] respectively) the volme formla for biorthogonal tetrahedron orthoscheme) is know. The volme of the Lambert cbe and some other polyhedron were calclated by R. Kellerhals [K] D. A. Derevnin A. D. Mednykh [DM] A. D. Mednykh J. Parker A. Y. Vesnin [MPV] and other. The volme of reglar polyhedron was obtained by G. Martin [M]. The volme of ideal hyperbolic polyhedron in many important particlar cases was fond by E. Vinberg [V]. The volme formla for arbitrary hyperbolic and spherical tetrahedron for a long time was nknown. Some attempt to obtain sch a formla contains in W Yi Hsiang [H]. Jst recently simple volme formla for tetrahedron was obtained by Y. Cho H. Kim [ChK] and J. Mrakami U. Yano [MY]. Easy proof for this formla which covers also the volme of trncated tetrahedron can be find in A. Ushijima [U]. The aim of this paper is to find an elementary formla for volme of symmetric tetrahedron both in hyperbolic and spherical spaces. 1 Partially spported by the Rssian Fondation for Basic Research Grant ) INTAS Grant ) and the State Maintenance Program for the Leading Scientific Schools of the Rssian Federation Grant SS ). 1

2 2 Preliminary reslts Denote by X n the Eclidean hyperbolic or spherical n-space. Let compact tetrahedron T A B C D E F ) X 3 have vertices v 1 v 2 v 3 v 4 and dihedral angles A B C D E F with edge lengths l A l B l C l D l E l F respectively see Fig. 1). Figre 1: The tetrahedron T We will call tetrahedron T symmetric if A D B E C F. Or calclation of volme of tetrahedron can be based on the following Schläfli formla see for instance [Sh] [Hd] [K]). Theorem 1 The Schläfli volme formla). Let compact simplex S X n n 2) have vertices P 1... P n+1 and dihedral angles α jk S j S k ) 1 j < k n + 1 of order n 1 formed by the faces S j S k of S with apex S jk : S j S k. Then the differential of the volme fnction V n on the set of all simplices in X n can be represented by KdV n S) 1 n 1 where K is the crvatre of X n. n+1 j k1 j<k V n 2 S jk )dα jk V o S jk ) : 1) 2

3 In the present paper we set K 1 for hyperbolic space and K 1 for spherical space. The Schläfli formla for hyperbolic and spherical 3-spaces can be redced to KdV 1 4 l jk dα jk 2 j k1 j<k where l jk are the lengths of the correspondent edges of S. 3 The volme of hyperbolic tetrahedron Let T be a hyperbolic tetrahedron. Denote by G cos α ij ij cos A cos B cos F cos A 1 cos C cos E cos B cos C 1 cos D cos F cos E cos D 1 the Gram matrix of T and by H c ij ij1234 the associated with G matrix form by c ij 1) i+j M ij where M ij is i j) th minor of G. Following argments by A. Ushijima [U] we obtain Proposition 1. Let T be a proper hyperbolic tetrahedron. Then i) det G < 0 ii) c ii > 0 i sin A c33 c 44 iii) 3.1) sinh l A det G As an immediate conseqence of this proposition we have the following reslt Proposition 2. Let T be a proper hyperbolic tetrahedron. Then sin A sin D sinh l A sinh l D sin B sin E sinh l B sinh l E sin C sin F sinh l C sinh l F P 3.2) where P c 11 c 22 c 33 c 44 and det G From now on we sppose that the tetrahedron T is symmetric. straightforward calclation we obtain c 11 c 22 c 33 c 44 γ where By direct γ 1 cos 2 A cos 2 B cos 2 C 2 cos A cos B cos C. 3.3) 3

4 And also det G 1 a + b + c)1 + a b + c)1 + a + b c) 1 + a + b + c) 3.4) where a cos A b cos B c cos C. Ptting this calclations into Proposition 2 we have Proposition 3 The Sine Rle). Let T be a symmetric hyperbolic tetrahedron. Then sin A sin B sin C 3.5) sinh l A sinh l B sinh l C where γ and γ are defined by 3.3) and 3.4) respectively. We note the following sefl identity cos A + cos B cos C)cos B + cos A cos C)cos C + cos B cos A). 3.6) The following lemma can be obtained by elementary calclations Lemma 1. Let t is defined by eqality t 2 4a + bc)b + ac)c + ab) 1 a + b + c)1 + a b + c)1 + a + b c) 1 + a + b + c) 3.7) where a cos A b cos B c cos C and A B C are the dihedral angles of a symmetric hyperbolic tetrahedron T. Then arcsin a t + arcsin b t + arcsin c t arcsin 1 t. 3.8) Proof. Notice first that from 3.6) follows t > 1. Show that t defined by eqality 3.7) satisfies to eqality 3.8). By the basic formla we transform 3.8) to a arcsin 1 b2 t t + b 2 t Hence 3.8) is eqivalent to arcsinx ± y) arcsinx 1 y 2 ± y 1 x 2 ) 1 a2 t 2 ) arcsin 1 t 1 c2 t c 1 1 ). 2 t t 2 a t 2 b 2 + b t 2 a 2 t 2 c 2 c t ) 4

5 From the other side the straightforward calclation shows that 3.7) implies ) 2 a1 a 2 + b 2 + c 2 ) + 2bc t 2 a 2 ) 2 b1 b 2 + a 2 + c 2 ) + 2ac t 2 b 2 ) 2 c1 c 2 + a 2 + b 2 ) + 2ab t 2 c 2 ) 2 1 a 2 b 2 c 2 2abc t 2 1. By 3.1) ii) we have 1 a 2 b 2 c 2 2abc c 11 > 0 and it is not difficlt to see that t2 a 2 a1 a2 + b 2 + c 2 ) + 2bc t2 b 2 b1 b2 + a 2 + c 2 ) + 2ac 3.10) t2 c 2 c1 c2 + a 2 + b 2 ) + 2ab t2 1 1 a2 b 2 c 2 2abc. Sbstitting 3.10) into 3.9) we have the identity. Let T be a symmetric hyperbolic tetrahedron. Denote by V V A B C) the hyperbolic volme of T. Since A D B E C F l A l D l B l E l C l F by Theorem 1 we have dv l A da l B db l C dc. 3.11) Hence V A l V A B l V B C l C. 3.12) We note that if A B C arccos 1 then T is going to reglar Eclidean tetrahedron. In this case 0 and. By the Sine Rle l A l B l C 0 and 3 conseqently V 0. So we have V arccos 1 3 arccos 1 3 arccos 1 ) ) 3 Now we are able to prove the following 5

6 Theorem 2. Let T be a symmetric hyperbolic tetrahedron whose dihedral angles corresponding to pairs of opposite edge are A B C. The hyperbolic volme of T is given by the formla V where arcsin arcsin cos A ν arcsin 1 ) dν ν2 + 1 ν cos B ν arcsin cos C ν ) 1 a 2 b 2 c 2 2abc 1 a + b + c)1 + a b + c)1 + a + b c) 1 + a + b + c) Proof. We set F A B C ν) arcsin and Ṽ A B C) a cos A b cos B c cos C. a ν arcsin b ν arcsin c ν2 + 1 arcsin 1 ν2 + 1 F A B C ν)dν. To prove the theorem is sfficient to show that Ṽ satisfies 3.12) with initial data 3.13). By the Leibnitz Rle we have A F A B C ) A + F A B C ν) dν. 3.15) A For t in 3.6) by Lemma 1 we have F A B C ) 0. Hence taking into accont that F A B C ) sin A A and by the Sine Rle 2 + sin 2 A l A arcsinh sin A we obtain A F A B C ν) sin A dν A ν ν 2 + sin 2 A dν l A. 3.16) The eqalities B l B C l C 3.17) can be obtained by the similar way. Let A B C arccos 1 then and 3 the relation Ṽ arccos 1 3 arccos 1 3 arccos 1 3 ) ) follows from the convergence of integral F A B C ν)dν. 6

7 Sbstitting ν tan t in the above proposition we have Theorem 3. Let T be a symmetric hyperbolic tetrahedron whose dihedral angles corresponding to pairs of opposite edge are A B C. Then the hyperbolic volme of T is given by the formla 2 π 2 θ dt arcsincos A cos t)+arcsincos B cos t)+arcsincos C cos t) arcsincos t)) sin 2t where θ [0 π ] is defined by 2 tan 2 θ 1 a 2 b 2 c 2 2abc 1 a + b + c)1 + a b + c)1 + a + b c) 1 + a + b + c) 3.19) a cos A b cos B c cos C. The obtained reslt nmerically coincides with a reslt obtained early by Y. Cho H. Kim [ChK] and can be expressed in term of the Lobachevsky fnction x Λx) ln 2 sin ξ dξ. 0 4 The volme of spherical tetrahedron Let T be a spherical tetrahedron with Gram matrix G cos α ij ij cos A cos B cos F cos A 1 cos C cos E cos B cos C 1 cos D cos F cos E cos D 1 and associated matrix H c ij ij1234. The next proposition essentially follows from [L]. Proposition 4. Let T be a spherical tetrahedron. Then i) det G > 0 ii) c ii > 0 i sin A c33 c 44 iii) 4.20) sin l A det G 7

8 Proof. Conditions i) and ii) follows from existence of spherical tetrahedron whose Gram matrix is G see [L] [V]). To prove iii) we sed the following assertion de to Jacobi [P] Theorem p.12). Lemma 2 Jacobi). Let A a ij ij1...n be a matrix and det A is determinant of A. Denote by C c ij ij1...n the matrix formed by elements c ij 1) i+j det A ij where A ij is n 1) n 1) minor obtained by removing i th line and j th colmn of the matrix A. Then for any k 1 k n 1 we have det c ij ij1...k k 1 det a ij ijk+1...n 4.21) By applying Lemma 2 to matrices G and H for k 2 we obtain 1 cos 2 A det G c 33 c 44 c 2 34 ). Since cos l A c 34 c33 c 44 the relation iii) follows compare A. Ushijima [U]). As a conseqence of Proposition 4 similar to hyperbolic case we have Proposition 5. Let T be a spherical tetrahedron. Then sin A sin D sin l A sin l D sin B sin E sin l B sin l E sin C sin F sin l C sin l F P det G 4.22) where P c 11 c 22 c 33 c 44. Now we apply the obtained reslt to symmetric tetrahedron. Proposition 6 The Sine Rle). Let T T A B C A B C) be a symmetric spherical tetrahedron. Then where v γ sin A sin B sin C v 4.23) sin l A sin l B sin l C γ c 11 c 22 c 33 c 44 1 cos 2 A cos 2 B cos 2 C 2 cos A cos B cos C det G 1 a + b + c)1 + a b + c)1 + a + b c)1 a b c) and a cos A b cos B c cos C. We note also that v 2 1 4a + bc)b + ac)c + ab). 4.24) The following lemma can be obtained by the same argments as Lemma 2 8

9 Lemma 3. Let p is defined by eqality p 2 4a + bc)b + ac)c + ab) 1 a + b + c)1 + a b + c)1 + a + b c)1 a b c). where a cos A b cos B c cos C and A B C are the dihedral angles of a symmetric spherical tetrahedron T. Then arcsinh a p + arcsinh b p + arcsinh c p arcsinh 1 p. Denote by V V A B C) the spherical volme of tetrahedron T A B C A B C). Then by Theorem 1 The Schläfli volme formla) we have dv l A da + l B db + l C dc Hence V A l V A B l V B C l C. 4.25) As in hyperbolic case we note that T collapsed to a point as 0 or V. In particlar for A B C arccos 1 we obtain 3 V arccos 1 3 arccos 1 3 arccos 1 3 ) ) Theorem 4. Let T T A B C A B C) be a symmetric spherical tetrahedron whose dihedral angled corresponding to pairs of opposite edges are A B C. Then the spherical volme of T is given by the formla where v V v arcsinh cos A cos B + arcsinh ν2 1 ν ) arcsinh cos C ν2 1 arcsinh 1 ) dν ν2 1 ν 1 a 2 b 2 c 2 2abc 1 a + b + c)1 + a b + c)1 + a + b c)1 a b c) Proof. We set Ṽ A B C) a cos A b cos B c cos C. v ˆF A B C ν)dν where ˆF A B C ν) arcsinh cos A cos B cos C +arcsinh +arcsinh ν2 1 ν2 1 ν2 1 arcsinh 1 ν2 1. 9

10 We have to show that Ṽ satisfy 4.25) and 4.26). Then we have Ṽ A B C) V A B C) By the Leibnitz Rle A ˆF A B C v) v A v ˆF A B C ν) dν. A By Lemma 3 for p 2 v 2 1 we have ˆF A B C v) 0. Since ˆF A B C ν) sin A A ν ν 2 sin 2 A and by the Sine Rle l A arcsin sin A v we obtain A The eqalities v ˆF A B C ν) dν A B l B v sin A ν ν 2 + sin 2 A dν l A. 4.28) C l C can be obtained by the similar way. In the case A B C arccos 1 we have v 3. The relation Ṽ arccos 1 arccos 1 arccos 1 ) 0 follows from the convergence of the integral ˆF A B C ν)dν. v Corollary 1. Let T T A B C A B C) be a symmetric spherical tetrahedron. Sppose that π A π B and π C are sides of a right angled spherical triangle that is one of the three conditions cos A+cos B cos C 0 cos B+cos A cos C 0 or cos C + cos B cos A 0 is satisfied. Then the spherical volme of T is eqal to A 2 + B 2 + C 2 2 π2 4. Proof. Since the spherical space S 3 is tessellated by sixteen copies of tetrahedron T T π π π π π π ) we have V π π π ) 1 V ols3 ) π2. Hence by Theorem 8 4 we get V π 2 π 2 π 2 ) 1 dν arcsinh ν2 1 ν π ) 1 Sppose that A B C satisfying the condition of the theorem. Then by 4.24) v Hence v 1. 10

11 Lemma 4. Proof. Indeed IA) I A) 1 1 arcsinh cos A ν2 1 sin A dν 1 + cos2 A ν ν 2 1 ν 2 1 dν ν A2 2 π2 8 0 A π 2. 1 sin A ν2 sin 2 A and verified by 4.29) I0) π2 A2. Hence IA) π dν ν A By Theorem 4 we obtain V A B C) IA)+IB)+IC) I0) A2 +B 2 +C 2 2 π 2. 4 Sbstitting ν coth t in the statement of Theorem 4 we obtain Theorem 5. Let T T A B C A B C) be a symmetric spherical tetrahedron whose dihedral angles corresponding to pairs of opposite edges are A B C. Then the spherical volme of T is given by the formla 2 τ 0 dt arcsinh cos A sinh t)+arcsinh cos B sinh t)+arcsinh cos C sinh t) t) sinh 2t where τ is a positive nmber defined by coth 2 τ References 1 a 2 b 2 c 2 2abc 1 a + b + c)1 + a b + c)1 + a + b c)1 a b c) a cos A b cos B c cos C. 4.30) [Sb] I. H. Sabitov The volme of polyhedron as a fnction of the lengths of its edges. Rssian) Vestnik Moskov. Univ. Ser. I Mat. Mekh.) : English translation in: Moskow Univ. Math. Bll. 516) [M] G J. Martin The volme of reglar tetrahedra and sphere packings in hyperbolic 3 space Math. Chronicle [P] V. V. Prasolov Problems and theorems in linear algebra Translations of mathematical monographs 134 Amer. Math. Soc. Providence RI pp. 11

12 [MPV] A. D. Mednykh J. Parker A. Y. Vesnin On Hyperbolic Polyhedra Arising as Convex Cores of Qasi-Fchsian Pnctred Tors Grops Seol National University 2002 RIM-GARC Preprint Series pp. [ChK] Y. Cho H. Kim On the volme formla for hyperbolic tetrahedra Discrete and Comptational Geometry p [DM] D. A. Derevnin A. D. Mednykh On the volme of spherical Lambert cbe arxiv: math.mg/ pp. [K] R. Kellerhals On the volme of hyperbolic polyhedra Math.Ann p [H] W Yi Hsiang On infinitesimal symmetrization and volme formla for spherical or hyperbolic tetrahedrons Qart.J.Math Oxford Second Series p [MY] J. Mrakami and M. Yano On the volme of hyperbolic tetrahedron Preprint 2001 available at [U] A. Ushijima A volme formla for generalized hyperbolic tetrahedra Preprint 2002 available at [Sh] L. Schläfli Theorie der vielfachen Kontinität In: Gesammelte mathematishe Abhandlngen Basel: Birkhäser [Lb] N. I. Lobatschefskij Imaginäre Geometrie nd ihre Anwendng af einige Integrale Detsche Übersetzng von H.Liebmann Leipzig: Tebner [Hd] C. D. Hodgson Schläfli revisited: Variation of volme in constant crvatre spaces Preprint. [L] F. Lo On a problem of Fenchel Geometriae Dedicata p [V] E. B. Vinberg Geometry II New York: Springer-Verlag Athors addresses: A.D. Mednykh Sobolev Institte of Mathematics Novosibirsk State University Novosibirsk Rssia mednykh@math.nsc.r 12

13 D.A. Derevnin Novosibirsk State University Novosibirsk Rssia M.G. Pashkevich Novosibirsk State University Novosibirsk Rssia 13