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# Scissors Congruence The Birth of

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```									Scissors Congruence: The Birth of Hyperbolic Volume
Gregory Leibon Department of Mathematics Dartmouth College

Scissors Congruence: The Birth of Hyperbolic Volume – p.1/70

The Ideal Tetrahedron
Here we see the oriented convex hull of four ideal points, an ideal tetrahedron.

8

8

8

8

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Ideal Tetrahedron
The boundary at inﬁnity is the Riemann sphere with hyperbolic isometries corresponding to conformal mappings. Hence we label the points...

p s

q

r

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Ideal Tetrahedron
..and compute the cross ratio z. This cross ration parameterizes these labeled oriented ideal tetrahedra.

p

0
s

q

1

z=[p,q;r,s]

r

8

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Ideal Tetrahedron
It is easy to see that this cross ration depends really only on a choice of orientation and a choice of a pair of opposite edges. Hence, the complex coordinate parameterize the space of ideal oriented tetrahedra with a speciﬁed pair of opposite edges.

z
1/(1−z) (z−1)/z (z−1)/z 1/(1−z)

z

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Another Big Free Groups

Let <C> denote the free Abelian group generated by all complex numbers, i.e. all ideal tetrahedra.

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Key Relations

Two understand the need relations, take a pair of ideal tetrahedra and...

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Key Relations

and...

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Key Relations

and...

Scissors Congruence: The Birth of Hyperbolic Volume – p.9/70

Key Relations

and...

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Key Relations

and...

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Key Relations

and...

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Key Relations

glue them together.

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Key Relations

Now "ﬁrepole" this pair and...

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Key Relations

and...

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Key Relations

and...

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Key Relations

and...

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Key Relations

and...

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Key Relations

and...

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Key Relations

and...

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Key Relations

we can re-express this pair as three ideal tetrahedra. This is called a 2-3 move.

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Key Relations

In terms of the z coordinates we have

[z] + [w]

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Key Relations

equals

z − zw w − zw [zw] + + 1 − zw 1 − zw

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The Relations

Let T the subgroup of < C > generated by all elements in the form w − zw z − zw − , [z] + [w] − [zw] − 1 − zw 1 − zw where z and w are complex numbers, together with all elements in the form [z] + [¯]. z

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The Dupont and Sah Theorem

Wonderfully enough these are all the relations we need.

Theorem:

(Dupont, Sah) ∼ < C >, Sis(H ) = T
n

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The Proof
Recall Sis∞ (H n ) ∼ Sis(H n ). A key step in the proof is showing we can express a ﬁnite = tetrahedron using ideal tetrahedra.

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From Finite to Inﬁnite
Let us make a ﬁnite vertex inﬁnite. First extend an edge to inﬁnity.

8

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From Finite to Inﬁnite
Then form the red tetrahedra, with an ideal vertex.

+ + −
8

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From Finite to Inﬁnite
and note...

+ + −

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From Finite to Inﬁnite
and note...

+ + −

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From Finite to Inﬁnite
and note...

+

−

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From Finite to Inﬁnite
and note...

− +

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From Finite to Inﬁnite
and note...

−

+

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From Finite to Inﬁnite
Hence, we have expressed the ﬁnite tetrahedron using two ideal tetrahedra each with only 3 ﬁnite vertices.

− 8

+ 8

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From Finite to Inﬁnite

One can continue this till one is using only ideal tetrahedra. The hard step is removing the ﬁnal vertex. The best known method to do this is due to Yana Mohanty (2003).

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Getting a Grip on Volume

At this point, we see that understanding hyperbolic volume can be reduced to understanding the volume of an ideal tetrahedron. To this it useful to take a close look at the ideal tetrahedron’s angles.

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Ideal Tetrahedron’s Angles

Given any ideal polyhedron, at each ideal vertex we see this. The red sphere is a horosphere.

A B C

8

Scissors Congruence: The Birth of Hyperbolic Volume – p.37/70

Euclidean Angles

Sending the ideal vertex to the point at inﬁnity in the upper-half space model, we ﬁnd that the angles at an ideal vertex are Euclidean.

A B

8 C

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Ideal Tetrahedron’s Angles

We view our tetrahedron in the upper-half space model.

8
z 0 1

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Ideal Tetrahedron’s Angles

Looking down from inﬁnity we see.

z A+B+C= π C

0

A

B 1

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Ideal Tetrahedron’s Clinants
It is best not to think in terms of the dihedral angles but rather the dihedral clinants. Namely e2Iθ is the clinant associated to the angle θ.

z c abc=1

0

a

b 1

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Ideal Tetrahedron’s Clinants
The compactiﬁcation of the space of ideal tetrahedra is all clinants triples (a, b, c) such that abc = 1, "blown up" at (1, 1, 1). To see this, note that the z coordinate equals 1−a . 1−¯ b

z c abc=1

0

a

b 1

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Decomposing Ideal Tetrahedron

8
c a P b

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Decomposing Ideal Tetrahedron

and double it.

c a P b

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Decomposing Ideal Tetrahedron

Firepole this doubled ideal tetrahedron.

c a b

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Decomposing Ideal Tetrahedron

Then we have our 2-3 move which....

c a b

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Decomposing Ideal Tetrahedron

Then we have our 2-3 move which....

c a b

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Decomposing Ideal Tetrahedron

Then we have our 2-3 move which....

c a b

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Decomposing Ideal Tetrahedron

Then we have our 2-3 move which....

c a b

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Decomposing Ideal Tetrahedron

Then we have our 2-3 move which....

c a b

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Decomposing Ideal Tetrahedron

allows to view our double tetrahedron as three ideal tetrahedra.

c a b

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Decomposing Ideal Tetrahedron

From inﬁnity we see these three ideal tetrahedra are very special ideal tetrahedra, the isosceles ideal tetrahedron.

−1/b −1/a

b −1/b −1/c

2 2

a c2

−1/a

−1/c

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The Isosceles Ideal Tetrahedron
Let us denote this isosceles ideal tetrahedron as II(a). We have just proved IT (a, b, c) = II(a) + II(b) + II(c). So we have reduced ﬁnding the volume of an ideal tetrahedron to ﬁnding the volume of an isosceles ideal tetrahedron.

−1/a −1/a a2 −1/a −1/a a
2

=II(a)

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The Isosceles Ideal Tetrahedron
Equally important is that the z coordinate of an Isosceles ideal tetrahedron II(a) i s a itself, and a z coordinate corresponds to an isosceles ideal tetrahedron if and only if it is unit sized.

−1/a −1/a a2 −1/a −1/a a
2

=II(a)

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A Tetrahedron’s Root
Theorem:(Dupont,
n

Sah)
n

[z ] = n
k=1

[e

ik2π n

z]

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In particular

Corollary:(Kubert)
n

V ol(z ) = n
k=1

n

V ol(e

ik2π n

z)

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Milnor’s Theorem
Theorem:(Milnor)

A continuous function f : S1 → R

that satisﬁes f (z) = f (¯) z and f (z ) = n must be equal c (Li2 (z)). Li2 (ζ) is the Euler dilogarithm Li2 (ζ) = Z
ζ 0 n n X

f (e

ik2π n

z)

k=1

log(1 − s) ds. s

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The Birth of Volume

After normalizing, we have a formula due to Lobachevski, 2V ol(IT (a, b, c)) = (a) + (b) + (c).

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The Milnor Conjecture

Let M = spanQ {[e
i2πp q

]}

and view the volume as a map, V ol, from M to RQ. Conjecture: ker(V ol) is the Q span of elements in the from [e
i2πp q

n

]−n
k=1

[e

ik2π n

e

i2πp nq

]

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The Milnor conjecture

In words: all rational relations are consequences of the Kubert identities.

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Dehn Invariant
Recall, Sis(H n ) ≡ <C> . Let us extend the Dehn invariant to <C> . If we have an ideal T T points cut off with a horoball, we may use the cut off lengths to deﬁne Dehn(P ) = X l(e) ⊗ θ(e).

e∈P

A B C

8

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Dehn Invariant
s Notice this is well deﬁned since if you use a different horosphere, then the difference of our two candidate Dehn Invariants is x⊗ X θ = x ⊗ nπ = 0.

θ∈∞

x x

8
x

A B

C

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Dehn Example?

There is no know explicit "Dehn counter example" in H 3 ! Below we have graphed V ol(II(e2Iθ )), with respect to θ.

0

π

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Dehn Example?

We’d like (and expect) that every such ϕ(p/q) is irrational, and hence provides a "Dehn counter example". But not one is known to be! We even have...

(p/q)π φ

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Dehn Example?

Theorem:(Dupont,Sah

) If Dehn(ϕ(1/N )) = 0

for any 1/N ∈ (0, 1/6), then the Milnor conjecture is false.

(p/q)π φ

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Dehn Kernel

Denote the kernel of Dehn restricted to
<C> T

as D(C).

Notice: Dehn Sufﬁciency is equivalent to (V ol, Dehn) being injective. In other words that (V ol, Dehn) has trivial kernel, or even more simply that V ol is 1-1 when restricted D(C).

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Countability Conjectures

Conjecture: V ol is 1-1 when restricted D(C). Conjecture: D(C) is countable. Conjecture: dimQ (D(C)) > 1.

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Evidence
<C> T

Theorem

(Suslin)

has the unique

division property. Theorem: (Dupont, Sah) V ol(D(C)) is countable.

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Suslin’s Theorem
The rectangle on top and the triangle below are both the middle polygon divided by 2. Hence they are scissors congruent.

+

+

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Evidence

The unique division property say that for every [P ] there exist a class (1/n)[P ] ∈ Sis(H n ) and that if n[Q] = n[R] then [Q] = [R]. Notice that
n

[z ] = n
k=1

n

[e

ik2π n

z]

is a candidate for division. Suslin showed this candidate obeys the 2 − 3 relation.

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