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2 General formulas
In hyperbolic geometry, we have a good chance to get a concrete value of the volume function if we can transform our problem into either a suitable coordinate system or a model of the space, respectively. In this section, we give volume-integrals with respect to some important system of coordinates. In our computation we also use the parameter k (used by J.Bolyai for the express the curvature of the hyperbolic space).
2.1 Coordinate system based on paracycles (horocycles)
P p
T t
Figure 1: coordinate system based on paracycles.
Consider a parasphere of dimension n 1 and its pencil of rays of parallel lines. The last − coordinate axis let be one of these rays, the origin will be the intersection of this line by the parasphere. The further axes are pairwise orthogonal paracycles. The coordinates of P are T (ξ ,ξ , ,ξn) , where the last coordinate is the distance of P and the parasphere, while the 1 2 · · · further coordinates are the coordinates of the orthogonal projection T with respect to the Carte- sian coordinate system giving by the mentioned paracycles. In Rn we can correspond to P a point p (see on Fig.1) with ordinary Cartesian coordinates:
T T ξn ξn ξn (x1,x2, ,xn) = e− k ξ1, e− k ξ2, , e− k ξn 1,ξn . · · · · · · − By definition let the volume of a Jordan measurable set D be
v(D) := vn dx dxn, 1 · · · DZ⋆
⋆ where D is the image of D by the above mapping and vn is a constant which we will choose later. Our first formula on the volume is:
(n 1) ξn v(D)= vn e− − k dξ dξn, 1 · · · DZ depending on the coordinates of the points of D, with the given system of coordinates. Let now the domain D = [0, a1] [0, an 1] be a sector of parallel segments of length an based on a ×···× −
2 coordinate-brick of the corresponding parasphere, then
a1 an n 1 (n 1) ξn kvn − (n 1) an 0 v(D)= vn e− − k dξn dξ = ai e− − k + e = · · · · · · 1 n 1 − i Z0 Z0 − Y=1 h i n 1 kvn − (n 1) an = ai[1 e− − k ]. n 1 − i − Y=1 kvn If an tends to infinity and ai = 1 for i = 1 (n 1), then the volume is equal to n 1 . Note · · · − − that J.Bolyai and N.I.Lobachevski used the value vn = 1 so in their calculations the value of the volume is independent from the dimension but depends on the constant k which determine the measure of the curvature of the space. We follows them we will determine the constant vn such that for every fixed k the value of the measure of a thin layer divided by its height tends to the value of the measure of the limit figure of lower dimension. Now the limit:
n 1 (n 1) an n 1 v(D) kvn − [1 e− − k ] − lim = ai lim − = vn ai, an a n 1 an a →∞ n i →∞ n i − Y=1 Y=1 n 1 − is equal to vn 1 ai showing that 1 = v1 = v2 = . . . = vn = . . .. − i=1 Thus vn = 1 asQ it used by earlier. On the other hand if for a fixed n the number k tends to infinity the volume of a body tends to the euclidean volume of the corresponding euclidean body. In every dimension n we also have a k for which the corresponding hyperbolic n-space contains a natural body with unit volume, if k is equal to n 1 then the volume of the paraspheric sector − based on a unit cube of volume 1 is also 1. So with respect to paracycle coordinate system our volume function by definition is
(n 1) ξn v(D)= e− − k dξ dξn. 1 · · · DZ
2.2 Volume in the Poincare half-space model
x , n P
exn
x n-1
x1
Figure 2: Coordinate system in the half-space model.
In the Poincare half-space model we consider a Cartesian coordinate system via Fig. 2. The first n 1 axes lie in the bounding hyperplane the last one is perpendicular to it. If the hyperbolic − T coordinates of a point P (with respect to a paracycle coordinate system) is (ξ ,ξ , ,ξn) , 1 2 · · · correspond to P the point P ′ with coordinates:
T T ξn (x1,x2, ,xn) = ξ1,ξ2, ,ξn 1, e k . · · · · · · − 3 The Jacobian of this substitution is k showing that xn 1 v(D)= k n dx1 dxn, xn · · · DZ′ where D means the image D via the mapping ξ x. ′ →
2.3 Hyperbolic orthogonal coordinate system
Put an orthogonal system of axes to paracycle coordinate system such that, the new half-axes let the tangent half-lines at the origin of the old one. (We can see it on Fig.3.) To determine the new coordinates of the point P we project P orthogonally to the hyperplane spanned by the axes x1,x2, ,xn 2,xn. The getting point is Pn 1. Then we project orthogonally Pn 1 onto · · · − − − th the (n 2)-space spanned by x1,x2, ,xn 3,xn. The new point is Pn 2. Now the (n 1) − · · · − th − − coordinate is the distance of P and Pn 1, the (n 2) is the distance of Pn 1 and Pn 2 and th− − − − so on ... In the last step we get the n coordinate which is the distance of the point P1 from the origin O. Since the connection between the distance 2d of two points of a paracycle and the
x n-1
P
xn-1 Pn-2 x n xn xn-2
Pn-1 x1
Figure 3: Coordinate system based on orthogonal axes. arclength of the connecting paracycle arc 2s is d s = k sinh k thus the distance z of the respective halving points can be calculated as: d z = k ln cosh . k Now elementary calculation shows that the connection between the coordinates with respect to the two system of coordinates is:
ξn xn 1 ξn 1 = e k k sinh − − k x − ξn +ln cosh n 1 xn 2 ξn 2 = e k k k sinh − − k . . x − x ξn +ln cosh n 1 + +ln cosh 2 x1 ξ = e k k k k sinh 1 ··· k xn 1 x2 x1 xn = ξn + k ln cosh − + + k ln cosh + k ln cosh . k · · · k k
T n T From this we get a new one, corresponding point (ξ ,ξ , ,ξn) H to point (u , un) 1 2 · · · ∈ 1 · · · ∈ Rn as in our first calculation. The corresponding system of equation is:
4 x2 xn 1 x1 u = k cosh cosh − sinh 1 k · · · k k . . xn 1 un 1 = k sinh − − k x1 xn 1 un = xn k ln cosh k ln cosh − − k −···− k The Jacobian of this transformation is n x1 x2 2 xn 1 1 cosh cosh cosh − − , k k · · · k and we get our third formula on the volume:
n 1 2 xn 1 − x2 x1 v(D)= cosh − cosh cosh dx dxn. k · · · k k 1 · · · DZ Here we use hyperbolic orthogonal coordinates.
2.4 Coordinate system based on spherical hyperbolic coordinates
x n-1
P
rn xn-1 O x x n r 1 n-1 , P
x1
Figure 4: Spherical coordinates
From the hyperbolic orthogonal coordinates x ,x , ,xn we can get the spherical coordi- 1 2 · · · nates of a point P . Let the distance of P = Pn from the origin is rn and denote by φi the angle th between the i coordinate axis and the segment OPi+1 for i = n 1,n 2, 1. ( Here Pn 1 − − · · · − is the orthogonal projection of P into the coordinate subspace of the axes x1,xn 2,xn, and so − on...) We have xn 1 rn sinh − = sinh cos φn 1 k k − by the hyperbolic theorem of Sin. For general i, we get that
xn 1 xn i+1 xn i rn cosh − cosh − sinh − = sinh sin φn 1 sin φn i+1 cos φn i, k · · · k k k − · · · − − and by the Pythagorean theorem we can get a last equation:
xn 1 x2 x1 rn cosh − cosh sinh = sinh sin φn 1 sin φ2 cos φ1. k · · · k k k − · · · Straightforward computation shows that:
n 1 n 1 rn − n 2 v(D)= k − sinh sin − φn 1 sin φ2dφ1 dφn 1drn. k − · · · · · · − DZ
5 2.5 Volume in the projective model
At the origin of the projective (Kayley-Clein) model we consider a Cartesian system of coor- dinates. Regarding the factor k we assume that the radius of the sphere is k. The considered system is an orthogonal coordinate system both of the embedding Euclidean and the modeled hyperbolic space.
xn Xn r k th _n k =Rn
P O Rn x2 X2
x1 X1
Figure 5: Coordinates in the Cayley-Klein model
We connect the hyperbolic spherical coordinates and the euclidean spherical ones by the system of equations:
1 Rn rn = k tanh− , φi = θi, i = 1, ,n 1. k · · · − Since the Jacobian is 1 2 , 1 Rn − k and n 1 n 1 − r n 1 1 1+ Rn − Rn sinh n − = sinh ln k = k k 2 Rn 2 1 k !! 1 Rn − − k the volume is: q
n 1 Rn− n 2 v(D)= n+1 sin − θn 1 sin θ2dθ1 dθn 1dRn. R 2 − · · · · · · − DZ 1 n − k q Transforming it into the usual Cartesian coordinates (X , , Xn) the new formula is: 1 · · · 1 v(D)= n+1 dX1 dXn 1dXn. n 2 2 · · · − DZ 1 Xi − k i=1 P 3 The three dimensional case
3.1 Formulas of J. Bolyai
In this section k is a constant giving the curvature of the hyperbolic space and the value of our constant vn is 1. The following formulas can be found in [2], [1] and [11]. Most of it can be easily determined using the results of the previous section. An important exception is the volume of the orthosceme, we will give a new formula for it in section 3.3.
6 Equidistant body: Volume of the body determined by a disk of area p and the segments orthogonal to it with edge lengthes t, respectively:
1 2q 2q 1 v = pk e k e− k + pq. 8 − 2 Paraspherical sector: Volume of the sector of parallel half-lines intersecting orthogonally an paraspherical basic domain of area p: 1 pk. 2 Sphere: Volume of the sphere of radious x:
− 1 3 2x 2x 2 3 2x 2 πk (e k e k ) 2πk x = πk sinh 2πk x. 2 − − k − Barrel: Volume of the set of those points of H3, which distances from a fixed segment AB of length p is not greater then q:
1 2 q q 2 πk p(e k e− k ) . 4 − Orthosceme: Volume of a special tetrahedron. Two edges a and b are orthogonal to each other and a third one c (skew with respect to a) is orthogonal to the plane of a and b.
z c y
g x b a a b
Figure 6: Orthosceme
The dihedral angle at a is α, the angle opposite to b of the triangle with edges a and b is β and the angle opposite to the edge c in the triangle with edges c and z is γ, respectively (see Fig.6). J.Bolyai gave two formulas:
c tan γ z sinh z v = dz, 2 tan β cosh2 z cosh2 z 2 1 2 1 Z0 cos α − cos γ − q and
α 1 sinh a cos φ v = a + dφ 2 2 2 2 − 2 2 2 cosh a cos φ+√tanh b+sinh a cos φ Z0 2 tanh b + sinh a cos φ ln 2 2 cosh a cos φ √tanh b+sinh a cos2 φ − p Asymptotic orthosceme: Volume of the orthosceme with ideal vertex (which is the common endpoint of the edges a, x and z):
7 c sin 2α z v = dz 4 cosh2 z cos2 α Z0 − and
a 1 cos φ v = ln dφ 2 cos2 φ tanh2 b Z0 − p Circular cone: Volume of a cone with a basic circle of radious b and with half-angle β at its apex.
b sinh2 y v = π dy. cosh2 y Z cosh y 2 1 0 cos β − q Asymptotic circular cone: The apex B of a circular cone tends to an ideal point on its axis of rotation.
v = π ln cosh b.
3.2 Formulas of N.I.Lobachevsky
The formulas of this subsection can be found in [6] or [8]. Barrel-wedge: Barrel-wedge is a sector of a barrel intersected from it by two meridian-plane through its axis of rotation. Let T be the area of a meridian-intersection and p be the length of its parallel circular arcs. Then we have
1 v = pT. 2 Orthosceme: Let the non-rectangular dihedral angles of an orthosceme be α, β and γ, respec- tively. There admitted at the edges a, z and c, respectively. (See in Fig.6 ). Introduce the parameter δ by the equalities:
tanh δ := tanh a tan α = tanh c tan γ, and the Milnor form of the Lobachevsky-function (see in [7]):
x Λ(x)= ln 2sin ζ dζ, − | | Z0 respectively. Then the volume v of the orthosceme is 1 π π π Λ(α + δ) Λ(α δ) Λ β + δ + Λ β δ + Λ(γ + δ) Λ(γ δ) + 2Λ δ . 4 − − − 2 − 2 − − − − 2 − h i
8 3.3 Once more again on the volume of the orthosceme
As an application of our general formulas we determine the volume of the orthosceme as the function of its edge-lengthes a, b and c. We note that there are formulas to transform the dihedral angles into the edge-lengthes. By the notation of the previous section these are:
1 sin(α + δ) 1 sin(γ + δ) 1 sin( π β + δ) a = ln , c = ln , z = ln 2 − . 2 sin(α δ) 2 sin(γ δ) 2 sin( π β δ) − − 2 − − It is clear that there is no simple way to get a new volume formula using these ones.
3.3.1 The 3-dimensional case
We now follow another way for computation, we will determine the integral
a φ(x) ψ(x,y) v(D)= (cosh z)2(cosh y)dydzdy = (cosh z)2(cosh y)dydzdx, DZ Z0 Z0 Z0 getting it from hyperbolic orthogonal coordinates using the parameter value k = 1. The functions φ(x) and ψ(x,y) can be determined as follows. Consider the orthosceme on Fig.7. In the
z=x2
P
y=x1 , Q c , c zmax P O 2 y , Q a b b
P 1
x=x3
Figure 7: Orthosceme and orthogonal coordinates rectangular triangle OP P the tangent of the angle P OP ∠ is: △ 2 1 2 1 tanh b tanh y tan P OP ∠ = = 1 . 2 1 sinh a sinh x So tanh b tanh y = sinh x, max sinh a
9 hence 1 tanh b 0 y φ(x) = tanh− sinh x =: λ. ≤ ≤ sinh a Consider now the triangle P P P . The line O(x,y, 0) intersects that point Q for which P Q = △ 1 2 3 | 1 | b′, and let denote the point of the segment P P1 above Q be Q′. Thus we get the equality tanh c tanh c′ = sinh b′. sinh b Take into consideration again the equality tanh y tanh b′ = sinh a, sinh x and using the hyperbolic Pythagorean theorem, from the triangle OQQ′ we get that △ 1 2 2 sinh cosh− (cosh x cosh y) tanh c cosh x cosh y 1 tanh zmax = tanh c′ = sinh b′ − = sinh cosh 1(cosh a cosh b ) sinh b 2 2 − ′ pcosh a cosh b′ 1 − p tanh c sinh2 y + sinh2 x cosh2 y tanh c 1+ sinh2 x coth2 y = sinh b′ = sinh y = sinh b 2 2 2 sinh b 2 2 psinh b′ + sinh a cosh b′ p1+ sinh a coth b′ tanh c p = sinh y. p sinh b Hence the assumption
1 tanh c 0 z ψ(x,y) = tanh− sinh y =: ν ≤ ≤ sinh b holds if we fixed the first two variables. Thus the required volume is:
a λ ν v = (cosh z)2(cosh y)dzdydx = Z0 Z0 Z0
a λ 1 1 ν = z + (sinh 2z) (cosh y)dydx. 2 2 0 Z0 Z0 Using now the equality 1 sinh b + tanh c sinh y ν = ln , 2 sinh b tanh c sinh y − we get that a λ 1 sinh b + tanh c sinh y v = ln cosh ydy+ 4 sinh b tanh c sinh y Z0 Z0 − λ sinh b + tanh c sinh y + sinh ln cosh ydy dx . sinh b tanh c sinh y Z0 − ex e−x To determine the second integral we use that sinh x = −2 . Now
λ sinh b + tanh c sinh y sinh ln cosh ydy = sinh b tanh c sinh y Z0 −
10 λ 1 sinh b + tanh c sinh y sinh b tanh c sinh y = − cosh ydy = 2 sinh b tanh c sinh y − sinh b + tanh c sinh y Z0 − λ λ sinh y cosh y sinh 2y = 2 sinh b tanh c 2 dy = 2 sinh b tanh c tanh c dy = c b sinh y 2 c b cosh 2y + b Z0 tanh − sinh Z0 tanh − sinh sinh sinh b sinh b tanh c tanh c λ = ln 2 cosh 2y + = −tanh c tanh c − sinh b sinh b 0 sinh b sinh b tanh c tanh c sinh b sinh b = ln 2 cosh 2λ + + ln 2 . −tanh c tanh c − sinh b sinh b tanh c tanh c From the definition of λ we can calculate cosh 2λ and get:
1 sinh a + tanh b sinh x sinh a tanh b sinh x cosh 2λ = + − , 2 sinh a tanh b sinh x sinh a + tanh b sinh x − thus the value of the second integral (denoted by II) is:
sinh b tanh2 c sinh2 x II := ln 1 . −tanh c − cosh2 b(sinh2 a tanh2 b sinh2 x) − The first part can be integrated as follows:
λ sinh b + tanh c sinh y sinh b + tanh c sinh y λ ln cosh ydy = ln sinh y sinh b tanh c sinh y ( sinh b tanh c sinh y 0 − Z0 − −
λ tanh c cosh y[(sinh b tanh c sinh y) + (sinh b + tanh c sinh y)] − sinh ydy = − sinh2 b tanh2 c sinh2 y Z0 − λ sinh b + tanh c sinh λ 2 tanh c sinh b cosh y sinh y = sinh λ ln dy = sinh b tanh c sinh λ − sinh2 b tanh2 c cosh2 y + tanh2 c − Z0 − sinh b + tanh c sinh λ sinh b λ = sinh λ ln + ln(sinh2 b tanh2 c cosh2 y + tanh2 c) = sinh b tanh c sinh λ tanh c − 0 − sinh b + tanh c sinh λ sinh b = sinh λ ln + ln(sinh2 b tanh2 c sinh2 λ) ln(sinh2 b) . sinh b tanh c sinh λ tanh c − − − Since tanh2 b sinh2 x sinh2 λ = , sinh2 a tanh2 b sinh2 x − the first integral is:
sinh b + tanh c sinh λ sinh b tanh2 c sinh2 x sinh λ ln + ln 1 = sinh b tanh c sinh λ tanh c − cosh2 b(sinh2 a tanh2 b sinh2 x) − − sinh b + tanh c sinh λ = sinh λ ln II. sinh b tanh c sinh λ − − The sum of the two parts is: sinh b + tanh c sinh λ sinh λ ln . sinh b tanh c sinh λ − 11 1 tanh b From λ = tanh− sinh a sinh x follows 2 2 2 1 tanh λ sinh a tanh λ sinh a + tanh λ sinh a + tanh b x = sinh− = ln sinh b tanh b p and we get that
b 1 tanh λ sinh a sinh b + tanh c sinh λ v = ln dλ, 4 tanh2 b cosh2 λ + sinh2 a sinh2 λ sinh b tanh c sinh λ Z0 − proving the theorem:p
Theorem 1 Let the edges of an orthosceme be a, b, c, respectively where a b and (a, b) c. If ⊥ ⊥ k = 1 then the volume of it is:
b 1 tanh λ sinh a sinh b + tanh c sinh λ v = ln dλ. 4 tanh2 b cosh2 λ + sinh2 a sinh2 λ sinh b tanh c sinh λ Z0 − p Corollary: This formula can be simplified in the case of asymptotic orthoscemes. If the edge- length a tends to infinity, the function tanh λ sinh a tends to 1 showing that √tanh2 b cosh2 λ+sinh2 a sinh2 λ cosh λ the volume of the orthosceme with one ideal vertex is b 1 1 sinh b + tanh c sinh λ v = ln dλ. 4 cosh λ sinh b tanh c sinh λ Z0 − If now the length of the edge c also grows to infinity, then this formula simplified into:
b 1 1 sinh b + sinh λ v = ln dλ, 4 cosh λ sinh b sinh λ Z0 − which is the volume of an orthosceme with two ideal vertices. If now we reflect it in the face containing the edges b and c then we get a tetrahedron with three ideal vertices. If than we reflect the getting tetrahedron in the face containing the edges b and a we get another one with four ideal vertices. The volume of it is b 1 sinh b + sinh λ v = ln dλ. cosh λ sinh b sinh λ Z0 − Of this tetrahedron there are two edges (a and c) which are skew and orthogonal to each other (its common normal transversal is b). Since the reflection in the line of b is a symmetry of this ideal tetrahedron, we can see that there are two types of its dihedral angles, two opposite (at the edges a and c) are equal to each other, ( say A); and the other four ones are also equal to each other ( say B). Then we have A + 2B = π, and its volume by Milnor’s formula is equal to π v′ = Λ(π 2B) + 2Λ(B) = Λ(2B) + 2Λ(B) = 4Λ(B) + 2Λ B + . − 2 (We used that the Lobachevsky function is odd, periodic of period π, and satisfies the identity π Λ(2B) = 2Λ(B) + 2Λ(B + 2 ).) We have the following connection between the two integrals:
π b B+ 2 B 1 sinh b + sinh λ 0= ln dλ + 2 ln 2sin ζ dζ + 4 ln 2sin ζ dζ. cosh λ sinh b sinh λ | | | | Z0 − Z0 Z0
12 Remark: If we substitute to this formula the first-order terms of the Taylor series of the functions in the integrand, respectively, we get that
b 1 tanh λ sinh a sinh b + tanh c sinh λ v = ln dλ = 4 tanh2 b cosh2 λ + sinh2 a sinh2 λ sinh b tanh c sinh λ Z0 − p b b 1 λa cλ ac λ2 abc = dλ = dλ = . 2 √b2 + a2λ2 b 2b2 √1 6 Z0 Z0 This shows that our formula for little values gives back the euclidean one.
3.3.2 The case of dimension two
The following calculation shows that area of a rectangular triangle also can be get with our method. a φ(x) a tanh b sinh x (cosh y)dydx = sinh a dx = 2 Z Z Z 1 tanh b sinh x 0 0 0 − sinh a q (tanh b)2 1 sinh a 1 = dt 2 tanh b sinh a 2 sinh a 2 Z0 1+ 1 t t2 tanh b − − tanh b r At this point we can use the following result on antiderivative: if y = ax2 + bx + c and a< 0 ′ 1 1 1 y then dt = sin − . Thus the required area is: √y √ a − √b2 4ac − − R 2 2 2 2 tanh b 1 1 (2 sinh a)t sinh a + tanh b 1 1 2 2 1 sin− − = sin− (2 sin β sinh a cos 2β) sin− ( cos 2β) . 2 sinh2 a + tanh2 b 2 − − − 0 sinh b sinh a sinh2 a+tanh2 b Using now the equalities sin β = , sin α = we get that 2 2 = cos(2β), sinh c sinh c −sinh a+tanh b 2 2 − (2 sinh a) tanh b 2 2 sinh2 b 2 sinh2 b sinh2 a sinh2 b+sinh2 a 2 2 = 2sin β sinh a = 2 2 sinh a and 1 = 2 + 2 . Now our sinh a+tanh b sinh c sinh c sinh c formula simplified into the form
2 2 1 1 sinh b + sinh a 2 2 π π sin− (2 2 + sin β cos β) + ( 2β) = π α + β + , 2 − sinh2 c − 2 − − 2 as we stated.
3.3.3 The case of dimension n
The following lemma plays an important role in the n-dimensional case.
Lemma 1 We have two k-dimensional affine subspaces Hk and Hk′ , respectively for which they intersection have dimension k 1. Assume that the points P Hk, P H and P Hk H − ∈ ′ ∈ k′ ′′ ∈ ∩ k′ hold the relations P P H and P P Hk H , respectively. Then the angle ′⊥ k′ ′ ′′⊥ ∩ k′
1 tanh(P P ′) α = tan− , sinh P ′P ′′ is independent from the position of P in Hk.
13 Proof: Let P and Q be two arbitrary points of Hk. Then by the theorem of three perpendiculars we also have that P P Hk H and Q Q Hk H implying that P Q orthogonal to the ′′⊥ ∩ k′ ′ ′′⊥ ∩ k′ ′′ ′′ lines P P ′, P ′P ′′,QQ′ and Q′Q′′, respectively. Thus the angles P P ′P ′′∠, QQ′Q′′∠ are equals. This immediately implies the statement.
Let now our orthosceme is the convex hull of the vertices 0 = A0, A1,...,An situated into a hyperbolic orthogonal coordinate system as we did it in the three dimensional case. T T T More precisely, the coordinates of the vertices are (0,..., 0) , (0,..., 0, an) , (a1, 0,..., 0, an) , T T (a1, a2, 0 ..., 0, an) . . . (a1, a2, . . . , an) , respectively. Introduce the function giving the up- per boundary of the successive integrals. These are φ0 = an, φ1 : xn x1 for a point T 7→ T (x , 0,..., 0,xn) of the edge conv (OA ), φ : (xn,x ) x on the points (x ,x , 0 ...,xn) of 1 2 2 1 7→ 2 1 2 the triangle conv (OA2A3). In general φk : (x1,...,xk 1, 0,..., 0,xn) xk if the correspond- T − 7→ ing point (x1,x2 ...xk 1,xk, 0,..., 0,xn) is on the k 1-face conv (OA2A3 Ak+1) and so − − · · · on... From Lemma 1 we get that
tanh φ (x ,x ,...,x ) tanh a k+1 n 1 k = k+1 , sinh xk sinh ak implying that 1 tanh ak+1 φ (x ,x ,...,x ) = tan− sinh x . k+1 n 1 k sinh a k k In the case of k = 1, the volume can be determined by the following n-times integral: v(O) = tanh −1 tanh a1 −1 an−1 tan sinh xn tan sinh sinh xn−1 an sinh an an−2
0 0 · · · 0 R R R n 1 n 2 (cosh − xn 1)(cosh − xn 2) (cosh x1)dxn 1 dx1dxn. − − · · · − · · ·
3.4 Further results on concrete volumes
Here we mention some important results from the last few decades, which are calculate hyperbolic volumes. To investigate it we can see immediately that our formula on orthosceme has very simple building up. In this section the great latin letters mean the measures of the dihedral angles at the edges denoted by the corresponding small ones.
Theorem 2 (J. Milnor [7]) Opposite dihedral angles of ideal tetrahedron are equal to each other, A + B + C = π and its volume v is
v = Λ(A) + Λ(B) + Λ(C),
x where Λ(x)= ln 2sin ζ dζ. − 0 | | R Theorem 3 (R.Kellerhaus [5]) Let R be a Lambert cube with essential angles wk, 0 wk π ≤ ≤ 2 , k = 0, 1, 2. Then the volume V (R) of R is given by
2 1 π V (R)= (Λ(wi + θ) Λ(wi θ)) Λ(2θ) + 2Λ θ 4 − − − 2 − ( 0 ) X with 2 2 2 1 cosh V1 sin w0 sin w2 π 0 <θ = tan− − . p cos w0 cos w2 ≤ 2
14 Theorem 4 (Y. Mohanty [9]) Let O be an ideal symmetric octahedron with all vertices on the infinity. Then C = π A, D = π B, F = π E and the volume of O is: − − − π + A + B + E π A B + E π + A B E v = 2 Λ + Λ − − + Λ − − + 2 2 2 π A + B E +Λ − − . 2 Theorem 5 (D. Derevin and A.Mednykh [4]) The volume of the hyperbolic tetrahedron T = T (A,B,C,D,E,F) is equal to
z2 1 cos A+B+C+z cos A+E+F +z cos B+D+F +z cos C+D+E+z V ol(T )= − log 2 2 2 2 dz, 4 A+B+D+E+z A+C+D+F +z B+C+E+F +z z sin 2 sin 2 sin 2 sin 2 zZ1 where z1 and z2 are the roots of the integrand, given by
1 k2 1 k4 1 k2 1 k4 z1 = tan− tan− , z2 = tan− + tan− , k1 − k3 k1 k3 with k = (cos S + cos(A + D) + cos(B + E) + cos(C + F ) + cos(D + E + F )+ 1 − cos(D + B + C) + + cos(A + E + C) + cos(A + B + F )),
k2 = sin S + sin(A + D) + sin(B + E) + sin(C + F ) + sin(D + E + F )+ sin(D + B + C) + sin(A + E + C) + sin(A + B + F ),
k3 = 2(sin A sin D + sin B sin E + sin C sin F ),
k = k2 + k2 k2, 4 1 2 − 3 and S = A + B + C + D + E + F . q
The above theorem implies the theorem of Murakami and Yano:
Theorem 6 (J.Murakami, M.Yano [10]) The volume of the simplex T is 1 v = (U(z , T ) U(z , T )) , 2ℑ 1 − 2 where 1 U(z, T )= (l(z)+ l(A + B + D + E + z)+ l(A + C + D + F + z)+ l(B + C + E ++F + z) 2 − l(π + A + B + C + z) l(π + A + E + F + z) l(π + B + D + F + z) l(π + C + D + E + z), − − − − iz and l(z)=Li2(e ) by the Dilogarithm function
x log(1 t) Li (z)= − dt. 2 − t Z0
Here I mention the theorem of Y.Cho and H. Kim described the volume of a tetrahedron using Lobachevsky function, too. It is also very complicated formula the reader can find it in [3].
15 References
[1] Bonola, R. Non-euclidean geometry Dover Publication, 1955.
[2] Bolyai, J. Appendix in Tentamen written by F.Bolyai, Marosv´as´arhely, 1832
[3] Cho, Y., Kim, H., On the Volume Formula for Hyperbolic Tetrahedra Discrete Comput. Geom. 22 1999, 347366.
[4] Derevnin, D. A., Mednykh, A.D., A formula for the volume of a hyperbolic tetrahedon Uspekhi Mat. Nauk 60/2 2005, 159-160.
[5] Kellerhaus, R., On the volume of hyperbolic polyhedra Math. Ann. 285,(1989) 541-569.
[6] Lobachevsky, N.I. Zwei Geometrische Abhandlungen B.G.Teubner, Leipzig and Berlin, 1898, (reprinted by Johnson Reprint Corp., New York and London, 1972)
[7] J.W.Milnor, Hyperbolic geometry: The first 150 years. Bull. Amer. Math. Soc. (N.S.) 6/1 (1982), 9-24.
[8] Moln´ar E.,Lobachevsky and the non-euclidean geometry. (in hungarian) in Bolyai eml´ekk¨onyv Bolyai J´anos sz¨ulet´es´enek 200. ´evfordu- l´oj´ara 221-241, Vince Kiad´o 2004.
[9] Mohanty, Y., The Regge symmetry is a scissors congruence in hyperbolic space Algebraic and Geometric Topology 3 (2003) 131.
[10] Murakami, J., Yano, M., On the volume of Hyperbolic and Spherical tetrahe- dron,Communications in analysis and geometry 13/2, (2005) 379-400.
[11] T.Weszely, The mathematical works of J.Bolyai (in hungarian) Kriterion, 1981.
Akos´ G.Horv´ath, Department of Geometry Budapest University of Technology and Economics 1521 Budapest, Hungary e-mail: [email protected]
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