1. Hyperbolic Polyhedra: Volume and Scissors Congruence
Yana Mohanty
Hyperbolic Polyhedra: Volume and Scissors Congruence
Ph.D. Defense
Department of Mathematics
University of California, San Diego
June 6, 2002

2. Scissors Congruence
Example in 2-D:

3. Scissors Congruence 2 polytopes are scissors congruent if
you can cut one up into polygonal pieces
that can be reassembled to give the other.
Example in 3-D:
Equal
volume
Not equidecomposable!
[Max Dehn, 1900]

4. Key scissors congruence results
2 Dimensions
Euclidean: Equal area scissors congruence [Euclid]
Hyperbolic and spherical: Equal area scissors congruence [19th century]
3 Dimensions
Euclidean: Equal volume+same Dehn invariant scissors congruence [Dehn, 1900( );
Sydler, 1965 ( )]
Hyperbolic and spherical:Open conjecture

5. The Classical Dehn Invariant
P=polyhedron
E=edge
q(E)=dihedral angle at edge E (radians=# half revolutions)
l(E)=length of E
Idea: find a function on P invariant under slicing
where g(a+b)=g(a)+g(b) and g(p)=0
Mention the non-tensor way of seeing these formulae
Modern version: R
R/pZ R
R/pQ R
R/pQ

6. Regge symmetries This relation applies to side lengths and
where s=(b+b’+c+c’)/2. a b c a’ c’ b’
This relation applies to side lengths and
dihedral angles![Regge and Ponzano, 1968]
Involutive
6j-symbol invariant under these
Gives another tetrahedron
tral aa
…with the same volume and Dehn invariant!! [Justin Roberts, 1999]
Generate a family of 12 scissors congruent tetrahedra

7. H3: The Poincare and upper half-space models (obtained by inversion)
metric:
CONFORMAL
z=0
z>0
metric:
Inversion:

8. H3: The upper halfspace model (obtained by inversion)
metric:
z>0
z=0

9. Ideal tetrahedron in H3 (Poincare model)

10. Ideal tetrahedron in H3 (half-space model)B b B b g a g a C C A A
View from Above

11. Important facts about volumes of ideal hyperbolic tetrahedra
At any vertex
Opposite dihedral angles are equal a
“Isosceles” ideal tetrahedra are the basic building blocks

12. Isosceles ideal tetrahedron with apex angle a
L(a/2) A B

13. Reflections in H3 (half-space model) = Inversions in hemispheres
sphere,plane
line, circle
sphere,plane

14. Isosceles tetrahedra as basic building blocks:
Klein model picture
T(ABCD)+T(ABC )=
T(ABD )+
T(BCD ) +
T(ACD
View from Above B C A b a g D
2T(a,b,g)=2L(a)+2L(b)+2L(g) B D D’ C A
T(ABCD)~T(ABC )

15. Notation: L(b)= T(CBD’ )
An arbitrary ideal tetrahedron as a linear combination of isosceles ideal tetrahedra D’
T(ABCD)+T(ABC )=
T(ABD )+
T(BCD ) +
T(ACD
View from Above B C A b a g D 2g
Notation: L(b)= T(CBD’ ) B D C A
T(ABCD)~T(ABC )

16. Derivation of the volume of an isosceles ideal tetrahedron using integrals
metric:
Hemisphere
V(L(a)/2) D’ B a D 1 C A

17. Derivation of the volume of an isosceles ideal tetrahedron using integrals
V(L(a)/2)=L(a)/2
where
L(q) q

18. V(a,b,g)=L(a)+L(b)+L(g)
Volume of an arbitrary ideal tetrahedron in terms of the Lobachevsky function
V(a,b,g)=L(a)+L(b)+L(g)
where

19. What about non-ideal tetrahedra?
1 non-ideal point:
Step 1:
Extend edges to infinity

20. Step 2: view as a combo of ideal tetrahedra
“Twisted prism”=2 {a,b,c,p}=
{a,b,c,p}+{c’,b’,a’,p}=
{a,b,c,c’}+{a,a’,b’,c’}-{a,b,b’,c’} p A C B B’ c A’ a C’ b

21. Why bother with the twist?a’ a b c c’ C’ B’ A’ A B C
Prism=2{a,b,c,a’’,b’’,c’’}=
{a,b,c,c’}+{a,a’,b’,c’}-{a,b,b’,c’}
b’’
a’’
c’’

22. Claim: 3/4-ideal --> stump continuous deformation
2-D analogue:
Klein or hyperboloid model
hypoideal
ideal
hyperideal

23. Main idea of Leibon’s formula:
Take the idea of the (un)twisted prism to the extreme A B C A’ B’ C’ A’ B’ C’ A B C

24. Triangulation Octahedronb c d e f g h A B C A’ B’ C’
Octahedron
Remark: the chiseled away stuff is all linear in A,B,C,A’,B’,C’

25. Determining angles of the octahedron, cont’d
Half-space model
Linear constraints:
AB+BA+e=p
BC+CB+f=p
CD+DC+g=p
DA+AD+h=p
AB+AD=a
BA+BC=b
CB+CD=c
DC+DA=d AB BA BC CB CD DC DA AD e f g h a b c d
1-dim space of solutions

26. Determining angles of the octahedron, cont’db c d
This is can occur!
Choose a point in the 1-dim space of solutions
Solve for Z by ensuring holonomy condition:
Substitute sin(q)=(eiq-e-iq)/2
 quadratic in e2iArgZ

27. Which root is the correct one?
They both are!
V(T) if Z=Arg(- sqrt…)/2
-V(T) if Z=Arg(+ sqrt…)/2
Second root octahedron with angles that are p-angles of original octahedron

28. Roots correspond to “dual” octahedra
Warm-up: a c b b’ c’ a’
Half prism=({a,b,c,c’}+{a,b,b’,a’,c’})/2
{a,b,b’,a’,c}
Note: angles are p-each other b2 a2 c2 b1 a1 c1
a’1
b’2
a’2
b’1
c’2
c’1 b2 a2 c2 b1 a1 c1
a’1
b’2
a’2
b’1
c’2
c’1
Actual
case:
T=Average of
the 2 octahedra

29. Geometric interpretation of V(T)+ AB BA BC CB CD DC DA AD e f g h
AB’
BA’
BC’
CB’
CD’
DC’
DA’
AD’
p-e
p-f
p-g
p-h
By Dupont’s unique 2-divisibility result T= + AB BA BC CB CD DC DA AD
AB’
BA’
BC’
CB’
CD’
DC’
DA’
AD’

30. How to obtain T(s-A,B,s-C,s-A’,B’,s-C’) from
T(A,B,C,A’,B’,C’) (s=(A+C+A’+C’)/2) AB BA BC CB CD DC DA AD
AB’
BA’
BC’
CB’
CD’
DC’
DA’
AD’
T(A,B,C,A’,B’,C’)= + AB DC BC CB BA DA AD
AB’
DC’
BC’
CB’
CD’
BA’
DA’
AD’ +
T(s-A,B,s-C,s-A’,B’,s-C’)=

31. Future work Constructive example of unique 2-divisibility:
How do you make this with ideal tetrahedra without dividing by 2?
Construct the Regge scissors congruence in Euclidean space