Post on 17-Dec-2015
transcript
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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