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M A T H E M A T I K O N Florian Pop UPenn Open House Feb 2017

FROM MYTHOS TO LOGOS: Construction Problems of the Ancient Greeks: The Questions a) Squaring the circle: = b) Trisecting the angle c) Doubling the cube Delian Problem ( 420 BC?) d) Constructing regular n- gones Triangle, Square, Pentagon, Hexagon, Octogon...... even the Heptagon(!)...

Construction Problems of the Ancient Greeks: Axiomatic One allows exclusively Compass & Straightedge! Given: P 0, P 1 fixed points in the plane, P 0 P 1 = 1. For finite sets M = {P 0, P 1,..., P n } of points in the plane, do: Step 1: Draw all the lines P i P j, and the circles C(P k, P i P j ), where P i, P j, P k M. - Let P M,1 be the set of intersection points from Step 1. N.B., M P M,1. - Let Z M,1 be the set of lengths P i P j with P j, P i P 1, e.g., 1 Z M,1. [Exercise : Estimate the cardinality of P M,1... ] Step 2: Proceed inductively/recursively to construct P M,2, Z M,2 ; P M,3, Z M,3 ; etc.... Define: - P M := n P M,n the set of M-constructible points. - Z M := n Z M,n the set of M-constructible numbers. - P := M P M the set of (absolutely) constructible points. - Z := M Z M the set of (absolutely) constructible numbers. [Comment: The points P M, P = P 0, P 1 are called auxiliary, etc.... ] THM. The domains of numbers Z, Z M are closed w.r.t addition, multiplication, und taking square roots.

Construction Problems of Ancient Greeks: Reformulation /K K M Reformulation of the Construction Problems: a) Is π a constructible number? b) Is 3 2 a constructible number? c) Is cos(α/3) M-constructible, if cos(α) is so (in general)? d) Is cos(2π/n) constructible for all n? The ancient Greeks could only do d), for n = 3, 4, 5, 6, 8, 12, 15,.... Nowadays we know all the Answers: NO, NO, NO, NO (!) Q : What is that we know, and the ancient Greeks did not? Answer: Notation & algebraic /analytic Structure! Coordinates (Descartes) & Complex plain (Gauß): - C = {z = a + bi a, b R}, z P(a, b) - P 0 := P(0, 0) z = 0 und P 1 := P(1, 0) z = 1. - Intersection points P P become complex numbers z = a + bi C... THM. The following hold: 1) K K M := {z C z P P} are subfields of C. 2) K is the closure of Q under taking square roots.

Construction Problems of the Ancient Greeks: The Answers a) Lindemann (1882): π is transcendental, hence π K b) Field theory + Galois theory: 3 2 K c) In general: cos(α) K cos(α/3) K. d) Gauß (1796, 1801): cos(2π/n) K iff n = 2 k p 1... p l with p i Fermat Fermat prime numbers are of the Form F n := 2 2n + 1, n 0. Example: F 0 := 2 20 + 1 = 3, F 1 := 2 22 + 1 = 5, F 2 := 2 22 + 1 = 17, F 3 := 2 23 + 1 = 257, F 4 := 2 24 + 1 = 65537,... Conjecture (?): All the numbers F n are prime numbers. Euler: Unfortunately, F 5 := 2 25 + 1 = 4294967297 = 641 6700417,... Proofs: Fact. u K Mipo u has degree a power of 2. To b): Mipo 3 2 = t3 2 over Q [WHY], hence 3 2 K. To c): u := 2 cos( α 3 ) u3 6u 2 cos(α) = 0, etc. [Exercise: For which values of u := cos(α) does it hold: cos(α/3) K {u}?] To d): By Gauß, ζ := cos(2π/n) + i sin(2π/n) has Mipo ζ = Φ n (t) [WHY]. n = 2 k p e 1+1 1... p e r+1 r deg(φ n ) = 2 k 1 p e 1 1... pe r(p 1 1)... (p r 1). Hence ζ K iff e 1 = = e r = 0, & p i 1 = 2 m i for some m i, etc.

MODERN POINT OF VIEW: Arithmetic in finitely generated fields Recall: M = {P 0, P 1,..., P n } = { 0, 1,..., z n } C finite subset - P M, Z M constructed inductively from M = {P 0, P 1,... }. - K M = Q(z 1,..., z n ) C is a finitely generated subfield - K M C is the closure of K M under square roots. - Z = { z, z K M } = K M are the M-constructible numbers A map T : P M P M is called algebraic symmetry, if the map T : K M K M is compatible with +, [Comment: One can characterize the algebraic symmetries in purely geometric way... ] The symmetry group Π M of K M is the set of all algebraic symmetries T : K M K M, which fix all points of M. Fact. K M K M is a Galois extension with Gal(K M K M ) = Π M. Examples. 1) The conjugation z := a + bi a bi =: σ(z) is an algebraic symmetry, which fixes R point wise. Further, one has: σ Π M M R. 2) Let M = { 0, 1}, hence K M = K. Then (i, 2) (±i, ± 2) can be prolonged to algebraic symmetries of K.

Quadratic / Pfister forms & Transcendence degree Pfister Forms are higher dimensional variants of q(x) = x 2... - n = 1: q a (x 1, x 2 ) := x 2 1 + ax2 2 - n = 2: q a,b (x 1, x 2, x 3, x 4 ) := x 2 1 + ax2 2 + bx2 3 + abx2 4 - n > 2: q a1,...,a n (x 1,..., x 2 n ) has 2 n Variables x 1,..., x 2 n. Fact. The images of q a, q a,b, q a,b,c,... are subgroups of K M. Quadratic forms q = i,j a ijx i X j generalize the above... Example: q = n i=1 X 2 i is a Pfister form iff n = 2 m [WHY]. THM A. Let i M. Then the transcendence degree d = td(k M ) is the smallest d such that n = d + 2 satisfies: q a1,...,a n (K 2n M ) = K M, for all a 1,..., a n K M. Hence the transcendence degree d = td(k M ) is in the arithmetic of K M uniformly encoded. (Big) Open Questions: - What is the Pythagoras number of K M? - What is the u-invariant of K M? - Do the quadratic/pfister forms distinguish K M from K M?

Symmetries and Isomorphism Type: Decoding K M from Π M Question: 1) To what extent are the properties of M reflected in the group Π M, and how does that take place? 2) Which properties of the field K M are encoded in the group Π M, and how does that take place? THM B. There exists a concrete group theoretical recipe which recovers the field K M functorially from the group Π M. In particular, the arithmetic of K M, e.g., tr.deg(k M ), can be recovered from the group Π M. (Big) Open Question: How does encode Π M the arithmetic of K M? Proofs: To THM A... Milnor Conjecture: proven (Vojevodsky, Rost,... ) To THM B... Pro-2 Anabelian Geometry... [Comment: These new insights would have been impossible without Ideas and Work of... ]

Alexander Grothendieck... Alexander Grothendieck March 28, 1928 - Nov 13, 2014