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Yana Mohanty

Hyperbolic Polyhedra: Volume and Scissors Congruence

Ph.D. DefenseDepartment of Mathematics

University of California, San DiegoJune 6, 2002

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Scissors Congruence

Example in 2-D:

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Scissors Congruence

2 polytopes are scissors congruent ifyou can cut one up into polygonal pieces that can be reassembled to give the other.

Example in 3-D:

Equalvolume

Not equidecomposable![Max Dehn, 1900]

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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

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The Classical Dehn Invariant

ZR RZ R RQ

Z

P=polyhedron E=edge(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()=0,))(()()( Pofedgesall

EgEPf

Pofedgesall

EEP )()(:)( Modern version:

R RQQ

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Regge symmetries

),',,',',,()',',',,,( cscsabsbsacbacba

where s=(b+b’+c+c’)/2.

•Involutive

•6j-symbol invariant under these

•Gives another tetrahedron

•…with the same volume and Dehn invariant!! [Justin Roberts, 1999]

•Generate a family of 12 scissors congruent tetrahedra

a

b

c

a’ c’

b’

•This relation applies to side lengths and dihedral angles![Regge and Ponzano, 1968]

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H3: The Poincare and upper half-space models (obtained by inversion)

z=0

z>0

metric:

2

2222

z

dzdydxds

d

Rd

2

Inversion:

metric:

2222

2222

)](1[4

zyx

dzdydxds

;1222 zyx

CONFORMAL

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H3: The upper halfspace model

(obtained by inversion)

metric:

2

2222

z

dzdydxds

z=0

z>0

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Ideal tetrahedron in H3 (Poincare model)

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Ideal tetrahedron in H3 (half-space model)

A

B

C

B

CAView from Above

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Important facts about volumes of ideal hyperbolic tetrahedra

• At any vertex

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1ianglesdihedral

• Opposite dihedral angles are equal

• “Isosceles” ideal tetrahedra are the basic building blocks

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Isosceles ideal tetrahedronwith apex angle

A B

C

L()

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Reflections in H3 (half-space model)

= Inversions in hemispheres

sphere,planesphere,plane

line, circle line, circle

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T(ABCD)~T(ABC)

A

B

C

Isosceles tetrahedra as basic building blocks:Klein model picture

D’

T(ABCD)+T(ABC)=

T(ABD)+ T(BCD)+T(ACD)

View from Above

B

CA

D

2T(=2L()+2L()+2L()

D

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An arbitrary ideal tetrahedron as a linear combination of isosceles ideal tetrahedra

T(ABCD)~T(ABC)

A

B

C

D

T(ABCD)+T(ABC)=

T(ABD)+ T(BCD)+T(ACD)

View from Above

B

CA

D

Notation: L()= T(CBD’ )

D’

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Derivation of the volume of an isosceles ideal tetrahedron using integrals

Hemisphere

,1 22 yxz 0z

A

B

C

D

D’

cos

0

tan

0 1

322x

x

y yxzz

dxdydz

metric:

2

2222

z

dzdydxds

1

V(L()/2)

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Derivation of the volume of an isosceles ideal tetrahedron using integrals

where

V(L()/2)=()/2

-3 -2 -1 1 2 3

-0.4

-0.2

0.2

0.4

duu

0

sin2log)(

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Volume of an arbitrary ideal tetrahedron in terms of the

Lobachevsky function

V()=

where duu

0

sin2log)(

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What about non-ideal tetrahedra?

1 non-ideal point:

Step 1:Extend edges to infinity

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Step 2: view as a combo of ideal tetrahedra

b’

a’

“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’}

ab

c

c’

C’

B’

A’

A’

B’

C’

p

AB

C

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Why bother with the twist?

Prism=2{a,b,c,a’’,b’’,c’’}={a,b,c,c’}+{a,a’,b’,c’}-{a,b,b’,c’}

b’a’

a

b

c

c’

C’

B’

A’

A’

B’

C’

A B C

b’’

a’’c’’

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Claim:3/4-ideal --> stump

continuousdeformation

2-D analogue:Klein or hyperboloid model

hypoideal ideal hyperideal

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Main idea of Leibon’s formula:

Take the idea of the (un)twisted prism to the extreme

A’

B’

C’

AB

C

A

BC

A’

B’C’

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Triangulation

A

BC

A’

B’C’

Octahedron

a

b

c

d

e

f

gh

Remark: the chiseled away stuff is all linear in A,B,C,A’,B’,C’

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Determining angles of the octahedron, cont’d

Half-space model

Linear constraints:

AB+BA+e=BC+CB+f=CD+DC+g=DA+AD+h=

AB+AD=aBA+BC=bCB+CD=cDC+DA=d

AB

BA

BCCB

CD

DC

DAAD

e

f

g

h

a

b

c

d 1-dim space of solutions

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Determining angles of the octahedron, cont’d

e

f

g

h

a

b

c

d

This is can occur!

ZAB

ZBA

ZBC ZCB

ZCD

ZDC

ZDA

ZAD

),,,,,,,( ADDADCCDCBBCBAABChoose a point in the 1-dim space of solutions

Solve for Z by ensuring holonomy condition:

1)sin()sin()sin()sin(

)sin()sin()sin()sin(

ZADZDCZCBZBA

ZDAZCDZBCZAB

Substitute sin()=(ei-e-i)/2 quadratic in e2iArgZ

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Which root is the correct one?

They both are!

)',',',,,(

)()()()()()()()(

CBACBAoffnslinear

ZADZDAZDCZCDZCBZBCZBAZAB

V(T) if Z=Arg(- sqrt…)/2

-V(T) if Z=Arg(+ sqrt…)/2

)]()()()()()()()([

)()()()()()()()()(2

ZADZDAZDCZCDZCBZBCZBAZAB

ZADZDAZDCZCDZCBZBCZBAZABTV

Second root octahedron with angles that are -angles of original octahedron

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b2

a2c2

b1

a1

c1a’1

b’2

a’2

b’1

c’2

c’1

Roots correspond to “dual” octahedra

Half prism=({a,b,c,c’}+{a,b,b’,a’,c’})/2

Actual case:

Note: angles are -each other

T=Average of the 2 octahedra

b2

a2c2

b1

a1

c1a’1

b’2

a’2

b’1

c’2

c’1

Warm-up:

a

c

b

b’c’

a’

a

c

b

b’c’

a’

{a,b,b’,a’,c}

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Geometric interpretation of V(T)

)]()()()()()()()([

)()()()()()()()()(2

DAADCDDCBCCBABBA

ADDADCCDCBBCBAAB

ZADZDAZDCZCDZCBZBCZBAZAB

ZADZDAZDCZCDZCBZBCZBAZABTV

2 T=

+

AB

BABC

CBCD

DC

DAAD

e

f

g

h

AB’

BA’BC’CB’

CD’

DC’

DA’ AD’

-e

-f

-g

-h

By Dupont’s unique 2-divisibility result

T=+

AB

BA

BC CB

CDDC

DAAD

AB’

BA’

BC’ CB’

CD’DC’

DA’AD’

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How to obtain T(s-A,B,s-C,s-A’,B’,s-C’) fromT(A,B,C,A’,B’,C’) (s=(A+C+A’+C’)/2)

T(A,B,C,A’,B’,C’)= +AB

BA

BC CB

CDDC

DAAD

AB’

BA’

BC’ CB’

CD’DC’

DA’AD’

T(s-A,B,s-C,s-A’,B’,s-C’)= AB

DC

BC CB

BA

DC

DAAD

AB’DC’

BC’CB’

CD’BA’

DA’AD’

+

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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