Mathematics that the professor loved

徳山 豪
東北大学
Geometry that professors love

博士たちの愛する幾何Agenda• Great problems in mathematics• 23 problems proposed by Hilbert （1900）• Hilbertの23の問題(1900)– The list of grand challenges of modern mathematics• 現代数学をリードした数学プラン• How great mathematicians think–––

Equi-decomposability problem (分割同値問題）
Theorem of Bolyai and Gerwien
What Gauss thought….

• Wide scope is powerful tool for a scientist–

Geometric problem is often not solvable by only usinggeometry• Number theory and group theory to solve geometric problems– Same in any area of science and engineeringEqui-decomposability• Given two polygons P and P‟ , can we show thatthey have the same area?
• 二つの多角形PとP‟があるとき、それらが同一面積を持つことをどうやって示す？
• Equidecomposing polygons– Decompose P into a finite number of pieces, andassemble them to make P‟– Pを部分多角形に分割して、組み立てなおしてP‟を作れるか？

Theorem of Bolyai-Gerwien（１８３２）• Any pair of polygons with same area are
Equidecomposable(分割同値）• Example:

タングラムパズル

演習問題による証明
Solution by a series of exercises• Show that a unit area rectangle isequidecomposable to a unit square– If the side edge is rational, not difficult, but in general..
– 面積１の長方形は単位正方形と分割同値であることを示せ• Using the above fact, sbhow that any pair of equiarea polygons are equidecomposable.
• 上の事実を用いて、任意の2つの面積が等しい多角形が分割同値であることを示せ• Theorem of Bolyai and Gerwien– Farkas Boyai: Invented non-Euclidean geometry with hisson, Janos BolyaiEquideomposablity means• In 2dim, decomposition, scaling, translation,and rotation can create any “shape”• What is area of a figure? It is the size ofsquare equidecomposable to the figure.

• Gauss： Different in 3D space– 三角錐の体積の公式をどうやって小学生に教えるか？
How to teach the volume formula of tetrahedra
to kid?

The 3rd problem of Hilbert• Can we show that two prisms with thesame base area and height areequidecomposable?
– Solved in 1900: Earliest solution among 23problems• Max Dehn (22 years old）：– Unit volume simplex is not equidecomposable
to the unit cube
• 正4面体と立方体は分割同値ではない– This is number theory problem

Dehn‟s idea: １
Two values for an edge e of a polytope P
多面体Pの辺e について、2つの量を考える。
１： Length l(e)2: Angle θ(e) between two faces
３： Use the pair （ｌ(e), θ(e) mod 2π）

Idea of Dehn: ２We define equivalence that for a rational
number q （a,qθ) = (qa, θ）
2次元のイメージ図： 面積不変、分割も簡単

3次元の場合でも、辺の長さがｃ倍で角度1/cだと、辺の周りでの体積は不変に思える（これは動機なので、当面証明不要）

qθ
θ
aqa

Tensor product over the rational field Q： Almost
multiplication, but restricted (like filtration)a  q  qa   if q  Qa   a   a  (   )a    b    ( a  b)  0    a  0 0
注意： 物理や量子計算などでの通常のテンソル積では、
項 はベクトルで、Qの代わりに実数や複素数を用いる

Example：2  ( / 3)( 2 / 6)  2  ( 2 / 6)  0  01  2  3  ( 2 / 3)  0a   a  (   ) a    0

Dehn invariantf ( P)   l (e)   (e)e:edgesLemma： If we decompose P into two
polytopes P1 and P2
ｆ(P)= ｆ（P1)+f(P2)Theorem： If P and P‟areequidecomposable, ｆ（P)=f(P„)
つまり、Dehn不変量が異なる多面体は互いに分割同値でない

Dehn invariants of cube and simplex• Dehn invariant of any cube is 0–Why?

• The Dehn invariant of a simplex of unit-edge
length is:f ( P )   l ( e)   ( e )  6   ( e )e:edges 6  arccos(1 / 3)
There is not quotient number q such thatcos(qπ)=1/3

There is no quotient number q such thatcos(qπ)=1/n for odd ncos (k+1) x = 2 cos x cos kx – cos (k-1) x
Thus, if cos x = 1/n, cos k x = A(k)/ nk
where A(k) is coprime to n.
Thus, cos kx cannot be an integer for any k.
But, if q = a/b, cos 2bx= 1. A contradiction.

The implication of Dehn‟s theorem• Difference between 2-dim and 3-dim geometry– Volume is not really analogue of area

• We need analytic method to show the volume
formula of a tetrahedra.
• Invariant is fundamental for classification ofgeometric objects– Area, volume– We need more

• Algebraic method is fundamental for geometry– Dehn‟s surgery, Dehn‟s twist– Homology (Poincare‟s claim in 1900)More about it• If two polytopes P and Q have same volume and Dehn
invariant, can we decide whether P and Q areequidecomposable?
• Difficult looking：Isomorphism problem of graphs,
geometric objects, etc• After 65 years、Sydler solved it by using group theory,– Yes, P and Q are always equidecomposable.

• Algorithmic problems–––

Computation of Dehn invariant
Finding exact method for equi-decomposing
Minimum decomposition for 2-dim problem• Probably NP-hard, how to approximate??

Modern geometry• Invariance and transformation–Invariants and classification: Homology, Homotopy, etc• Poincare‟s claim: Homology classify geometric surfaces– Wrong!: “Exiotic sphere” exists.

• Poincare‟s conjecture(1904): Homotopy determines spheres.
– Perelman（2003) proved it• Thurston program, Morse theory– Graph isomorphism• Invariants： Order sequence、Neighbor degree set• We have not yet found good invariants.

• Invariants and information science– Geometric matching and geometric information retrieval• E.g., We can obtain geometric objects in a database that has the same
homology as a torus (donut)– We need invariants to represent more detail shape.
• This is not completely solved, and very challenging problem• Serious: We do not know whether our space is exotic or not!
• Dehn seminar: organized by Prof. Fujiwara