The mathematical universe is infinitely large. Most of what mathematicians do is look for regular, repeating, infinite patterns. Patterns provide ways that the universe is smooth and comprehensible: the laws of nature in one part of the mathematical universe are the same as in all other parts.

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EXCEPTIONAL : from Egyptian fractions to heterotic strings

Theo Johnson-Freyd

Canada/USA Mathcamp Colloquium, 23 July 2019 map.gsfc.nasa.gov/media/121238/ The mathematical universe is infinitely large. Most of what mathematicians do is look for regular, repeating, infinite patterns. Patterns provide ways that the universe is smooth and comprehensible: the laws of nature in one part of the mathematical universe are the same as in all other parts.

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How do exceptional objects arise? Imagine plucking two violin strings that are almost, but not quite, in tune. You will here a throbbing noise, called a beat.

↵ ↵ + cos(↵t) + cos(t) = 2 cos t cos t 2 2 regular patterns ✓ ◆ ✓ ◆ throbbing “beat” average pitch | {z } At low volumes, the beat is just| another{z repeating} | pattern.{z } But at very high volumes, sound waves can scatter o↵ other sound waves. The “beat” becomes a sonic boom, an exceptional object. Waves map.gsfc.nasa.gov/media/121238/ passing by the exceptional object get di↵racted and irregularized. But some patterns have exceptions. Exceptional objects can be heard far away, with their strange, Here be dragons irregular calls. Exceptional objects are entertaining and enticing, like a Siren to Odysseus.

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My goal for this talk is to explain the first exceptional object.

Five short chapters: From fractions to finite groups From fractions to finite groups Species of finite groups: solvable and simple How to analyze a finite group I: spin representations How to analyze a finite group II: diagrams and lattices Applications of the E8 lattice 7 8

What are all solutions to 1 1 1 + + > 1, a b c 2, 3, 4,... ? a b c   2 { } One infinite family:

(a, b, c)=(2, 2, c), c 2, 3, 4,... . “Type D” 2 { } Three exceptional solutions:

(2, 3, 3), (2, 3, 4), (2, 3, 5). “Type E” wikimedia.org/wiki/File:Rhind_Mathematical_Papyrus.jpg The lettering is from the 20th century. For later convenience, the Its genesis begins in ancient Egypt. The Egyptians were interested Type D solutions are called “Dn” and the Type E solutions are in sums of simple fractions. What are all solutions to called “E ”, where n = a + b + c 2=b + c. E.g. the solution n 1 1 1 (2, 3, 4) is called “E7.” + + > 1, a b c 2, 3, 4,... ? a b c   2 { }

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What are the rotational symmetries of a regular solid (a, b, c)? The Greeks understood these solutions. Given any solution I There is a symmetry ↵ of order a = 2 which rotates the solid (a, b, c), take some regular c-gons, and attach them so that b of around an edge. Which edge? Pick one arbitrarily. them meet at each vertex, and of course a =2facesmeetateach There is a symmetry of order b which rotates the solid edge. Together with their duals (b c), you will form all the I $ around a vertex. Which vertex? One of the ends of the edge (possibly degenerate) regular three-dimensional solids. that you picked. I There is a symmetry of order c which rotates the solid around a face. Which face? One of the sides of the edge that you picked. Exercise: ↵ = 1. It turns out that every symmetry of the solid is a composition of

D8 E6 E7 E8 these three symmetries. Furthermore, it turns out that the relation (shrink squares to edges) in the Exercise generates all relations between these symmetries. wikimedia.org/wiki/File:Hexagonal_Prism_BC.svg I.e. the full group of rotational symmetries is wikimedia.org/wiki/File:Tetrahedron.svg wikimedia.org/wiki/File:Hexahedron.svg wikimedia.org/wiki/File:Dodecahedron.svg G = Aut(solid) = ↵, , ↵a = b = c = ↵ =1. h | i

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So the Egyptian puzzle led us to discover some finite groups. Species of finite groups

A group is an abstract collection of “symmetries,” which is closed under composition. By “abstract,” I mean that the elements don’t have to be symmetries of anything — what matters is the structural relations between the di↵erent elements of the group. 13 14

Solvable groups are built out Simple groups, on the other of thin abelian layers. The layers are hand, have no layers. They are rigid, only loosely attached to each other. crystalline, and highly structured.

It is impossible to classify all solvable You could try to mix solvable and groups. The problem is that there simple groups together, by including are a lot of ways to add a layer to the simple layers into your solvable jello. top of a solvable group, but the exact wikimedia.org/wiki/File: Rainbow-Jello-Cut-2004-Jul-30.jpg It is near-impossible to layer wikimedia.org/wiki/File: number depends on how all the other Pyrite_60608.jpg layers are attached to each other. a simple group on top of a solvable group: it sinks towards the bottom. Solvable groups are flexible, almost liquid. But if you push too much they shear apart. They feel like blocks of jello. There are a few ways to layer a solvable group on top of a simple group, like jello resting on a crystalline foundation.

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The “D” groups G = Aut(D-type solid) are dihedral. They are solvable with only two layers. layers are minimally attached top layer bottom layer Simplicity is exceptional among finite groups: if you choose a finite Aut(Dn solid) ⇠= Cn 2 o C2 ( ) group at random, with 100% probability you will choose a solvable cyclic groups The “E” groups are: group. But the exceptional solutions to the Egyptian fraction problem led us to discover a simple group! 2 Aut(E6 solid) ⇠= C2 o C3 ( ) 2 Aut(E7 solid) ⇠= C2 o C3 o C2 ( ) Aut(E8 solid) ⇠= Alt(5) ( )

Aut(E8 solid) is the smallest simple group. It has 60 elements.

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To completely understand a group, you should understand its representations. A representation of a group is a way of assigning matrices to group elements, preserving the group law.

The group SO(3) of three-dimensional rotations can (almost!) be represented by 2 2 complex matrices: ⇥ 10 0 cos ✓ sin ✓ How to analyze a finite group I 0 cos ✓ sin ✓ 2 2 x-axis rotation 0 1 7! ✓ ✓ 0 sin ✓ cos ✓ sin 2 cos 2 ! @ A cos ✓ 0 sin ✓ cos ✓ ı˙ sin ✓ 010 2 2 y-axis rotation 0 1 7! ✓ ✓ sin ✓ 0 cos ✓ ı˙ sin 2 cos 2 ! @ A cos ✓ sin ✓ 0 eı˙✓/2 0 sin ✓ cos ✓ 0 z-axis rotation 0 1 7! ı˙✓/2 001 0 e ! @ A formula for arbitrary rotation 19 20

It is only an “almost” representation because when ✓ =2⇡, you For each group G = Aut(solid), write G˜ for the set of elements in 10 get 0 1 . What you can actually represent is a double cover of Spin(3) corresponding to rotations in G.ThenG˜ is a double cover SO(3), called “Spin(3)”. Every element in SO(3) corresponds to a ˜ of G: each element of G corresponds to two elements in G. pair of elements in Spin(3). In Spin(3), 360 rotation is not trivial, but 720 rotation is trivial. Example: The D4 solid is very degenerate. If you expand the degenerate You can realize Spin(3) by taking an object, and attaching some faces, you get a rectangular prism, with stretchy ribbons to it, with the other ends of the ribbons attached to the walls of the room. (You may use as many ribbons as you G(D )=Aut(D solid) = C C . 4 4 ⇠ 2 ⇥ 2 want.) If you rotate the object by 360, the ribbons will twist up. Its double cover is the group But if you rotate by 720, you can untwist the ribbons! www.k6-geometric-shapes.com/ rectangular-prisms.html G˜(D ) = Q = 1, i, j, k , Watch this in action: 4 ⇠ 8 {± ± ± ± } https://commons.wikimedia.org/w/index.php?title=File: i 2 = j2 = k2 = ijk = 1. Belt_Trick.ogv

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Draw a graph whose vertices are the nontrivial indecomposable representations of G˜. Draw an edge between V and W if you get a standard 2 2 block when you decompose V W . ⇥ ⌦ Remarkably, you will get a Y-shape graph. If you started with 1 1 1 a + b + c > 1, then the arms of the Y have lengths a, b,andc. How to analyze a finite group II c ··· ··· a ··· b

To decompose arepresentation,findabasismakingallmatrices block-diagonal. The tensor product of representations is:

AB ab Aa Ba Ab Bb = Ca Da Cb Db . CD⌦ cd Ac Bc Ad Bd ✓ ◆ ✓ ◆ ✓ Cc Dc Cd Dd ◆

23 24 n We can picture this as a basis ~↵ 1,...,~↵ n of R ,suchthat~↵ i and These are almost all the graphs such that the Cartan matrix { } ~↵ j are orthogonal if vertex i is not connected to vertex j,butthey 2 (identity matrix) (adjacency matrix for ) should be at 120 if they are connected. Since the dot product is ⇥ ~↵ ~↵ = ~↵ ~↵ cos(angle), is positive definite. i · j | i || j | and cos(120 )= 1 , and since I don’t like fractions, I will decide To get the complete set, you must allow the degenerate case a = 1. 2 that each ~↵ has length ~↵ = p2. Take all vectors that are They are called “Type A.” The finite group in this case is cyclic. i | i | The “A” solutions are an infinite, regular, unexceptional family. integer sums of the ~↵ s. That set of vectors is the lattice of the given ADE type.

Example: The A lattice is the triangular lattice in 2. An = ··· , Dn = ··· , 2 R n n 2

En = ··· , n =6, 7, 8. n 3 ~↵ 2 ~↵ 1

AmatrixA is positive definite if ij vi Aij vj > 0 for all nonzero vectors ~v. P 25 26

The distance between any two vectors in an ADE lattice is p2. Indeed, the distance is always peven number. A lattice with this property is called even.

Fundamental fact: The density of an ADE lattice, i.e. the number of atoms per unit volume, is 1 Applications of the E8 lattice Density = . # one-dimensional representations of G˜ { } This number is < 1ifG solvable. The number of one-dimensional representations of a solvable group is equal to the size of its bottom layer.

But it is exactly 1 when G is simple. And that happens for exactly one example: the E8 solid (= ).

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Application 1 (Error correcting codes): The E8 lattice determines The E8 lattice is a crystalline arrangement (and can be built from) Hamming’s in eight dimensions such error-correcting code Ham(8, 4). This that there is one atom is a way of transmitting four bits in an eight-bit channel. The 24 codewords are: voyager.jpl.nasa.gov per unit volume, but Error-correcting codes are any two atoms are at used extensively. An ex- 00000000 11111111 least p2-distance apart. ceptional error-correcting 11110000 00001111 code, called the Golay code,washowNASA Each atom is adjacent 11001100 00110011 communicated with the to 240 neighbours. 11000011 00111100 Voyager probes. The Go- lay code is the “beat” that The picture on the 10101010 01010101 left is the projection of leads to Monstrous Moon- 10100101 01011010 shine. For more details, these 240 neighbours to take my class this week. a two-dimensional plane. 10011001 01100110 wikimedia.org/wiki/File:E8Petrie.svg 10010110 01101001

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Application 2 ((1+1)-dimensional quantum matter): Remarkably, this does not force the system to undergo a phase Start with a (very cold) system of eight fermions 1,..., 8 transition (e.g. from liquid to gas). Even more remarkably, the moving freely on a 1-dimensional wire, with no interactions. Slowly interacting system is trivial.ThisisbecauseG =Aut(E8 solid) is a (“adiabatically”) turn on a quartic potential energy, determined by simple group; if you started with a di↵erent group, the interacting the Hamming code: system would look like the bottom layer of G˜.

A system of eight (1+1)-dimensional fermions represents the same PE = 1 2 3 4 + 1 2 5 6 + 1 2 7 8 + 1 3 5 7 phase of matter as the trivial system. This can be seen in the

1 3 6 8 1 4 5 8 1 4 6 7 2 3 5 8 laboratory. + + 2 3 6 7 2 4 5 7 2 4 6 8 3 4 5 6 + 3 4 7 8 + 5 6 7 8 It leads to myriad examples of 8-fold periodicity in mathematics. For example, the “homotopy groups” of many important Each term is a (nontrivial) Hamming-code word. The signs are due topological objects exhibit an approximate 8-fold periodicity. It is to the fermionic nature of the particles. only approximate because it is caused by an exceptional object. 31 32

Application 4 (): Any n-dimensional lattice L determines an Application 3 ((2+1)-dimensional n-dimensional torus T = Rn/L. Consider a quantum matter): (quantum) string propagating in T .Locally, A similar construction in two spatial any solution to the equations of motion dimensions is responsible for the factorizes into a “chiral” piece and an “thermal quantum Hall e↵ect.” “antichiral” piece. Usually this factorization It can be used to construct wikimedia.org/wiki/File: Weierstrass_elliptic_ cannot be done globally. It only works interesting “topological phases function_P.png www.nature.com/articles/ globally if L is an even lattice of density = 1. of matter.” These are important d41586-018-05913-4 for quantum computing. When the factorization works, the chiral piece is a “chiral string theory.” The E8 chiral string theory is important for the construction of “heterotic strings.”

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What are the symmetries of the E8 string theory? There are classical symmetries: A is an infinite group which is also a smooth manifold. 8 translating along T = R /(E8 lattice). Just like for finite groups, most are solvable (layers of jello) and a There are also quantum symmetries that few are simple (crystals). Remarkably, the classification of simple convert a state into a superposition of Lie groups is the same as the classification of solutions to 1 1 1 other states (including states with a + b + c > 1, together with some decoration called “folding.” nontrivial topology). There are four infinite families:

wikimedia.org/wiki/File: A , D , C (folded A ), B (folded D ), It turns out that there is one dimension E8Petrie.svg n n n n n n of quantum symmetries for each of the 240 neighbouring atoms to a given atom and five exceptions: in the E8 lattice. E6, E7, E8, G2 (folded E6), F4 (folded E7). Together with the classical symmetries, this produces a E8 is the most exceptional. That exceptional Egyptian “beat” 248-dimensional group, called the “E8 Lie group.” It is an 1 + 1 + 1 > 1 can be heard very far away! exceptional group, with exceptional properties. For instance, its 2 3 5 smallest representation is itself!

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Spin(3) is isomorphic to SU(2), which is by definition the (infinite) group consisting of the following complex matrices:

µ 2 2 , ,µ C, + µ =1. µ¯ ¯ 2 | | | | ✓ ◆ Exercise: Check that SU(2) is closed under matrix multiplication.

Consider the function f : SU(2) SO(3) which takes ! Re(2 + µ2)Im(2 µ2)Im(2µ) 1 1 1 µ f : Im(2 + µ2)Re(2 µ2)Re(2µ) . + + > 1 µ¯ ¯ 7! 0 1 ✓ ◆ Im(2µ¯) Re(2µ¯) 2 µ 2 2 3 5 | | | | @ A Exercise: Show that f (AB)=f (A)f (B)foranytwomatrices Thank you! A, B SU(2), i.e. show that f is a homomorphism. return2 to main slideshow