Meromorphic CFT’s - Nikheft58/Presentations/Freiburg_Merom...This procedure can be applied to all...

Meromorphic CFT’sBert Schellekens

DEF-CFT Seminar, Freiburg, 5 July 2019

P(τ, τ) = ∑i

Mij χi(τ)χj(τ)

χi(−1τ ) = Sij χi(τ)

χi(τ + 1) = ehi− c24 χi(τ) ≡ Tij χj(τ)

Partition functions

χi = Tr eiτ(L0−c/24) (Extended) Virasoro algebra character

Mij ≥ 0; M00 = 1

S2 = (ST )3 = 1Modular group:

Single character CFT’s*Positivity of character coefficients: S = 1

T3 = 1; T = e−ic/24 → c = 8n

n = 1 : E8,1

n = 2 : E8,1 ⊗ E8,1; D16,1, (χ0 + χs)

Xp,k denotes affine Xp at level k

[Jam, Jb

n] = if abcJcm+n + kmδabδm+n,0

(*) Meromorphic CFT (P. Goddard, 1989)

Lowest � cases:n

Free BosonsA large subset of meromorphic CFTs originate from free bosons with momenta quantised on an even self-dual lattice

P(τ) = η(τ)−8n ∑λ∈Λ

eiπτλ2

Λ is an ! -dimensional even self-dual lattice: � The ! -expansion of ! counts the number vectors of given length

8n Λ = Λ*, λ2 ∈ 2ℤ for all λ ∈ Λq P(τ)

η(τ)8nP(τ) = 1 + 𝒩2 q + 𝒩4 q2 + …

N2ℓ = Number of vectors of length2 2ℓ

η(τ) = eπiτ12

∏n=1

(1 − e2nπiτ)Dedekind � -function:η

Even self-dual latticesThese have been enumerated completely for � .n = 1,2 and 3

! Root lattice of ! ! Root lattice of ! Root lattice of ! with spinor conjugacy class, ! ! Niemeier lattices: 23 lattices with roots ( � ) 1 without roots (the Leech lattice)

n = 1 E8

n = 2 E8 ⊕ E8

D16 Λ0 + Λs

n = 3𝒩2 ≠ 0

W. Lerche ci a!., Lattices and strings 115

Table 10The 23 Niemeier lattices that are Lie algebra lattices. squarebrackets indicate cyclic

permutation

Lie algebra conjugacy class generators

(s)D16E4 (s, 0)E~ (0,0,0)A24 (5)D~2 (s, v), (u,s)A,7E, (3, 1)D10E~ (s,1,0), (c, 0, 1)A,,D, (2, s)D~ ([s, v, v])A~, (1,5)

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

even permutations of (0, s, u, c)A~ ([1,1,41)A~D~ (1,1,s,v),(1,7,v,s)A (1,[2,1,6])A~D4 (2,[0,2,4],0),(3,3,0,0,s),(3,0,3,0,v),(3,0,0,3,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

itself. It consists of at least two elements, the identity and the reflection through the origin. Denote thetotal number of elements qf this group for a lattice A as g(A). Then the following identity can be proved[100, 109].

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

where B21 are the Bernoulli numbers. This formula is called the Minkowski—Siegel mass formula. The

sum is over all lattices in 8n dimensions, and adds up to a number that is less than half the total numberof lattices in this dimension. Thus the right-hand side, multiplied by 2, is a lower limit of the number oflattices. Calculating it for the lowest dimensions, one finds

dimension lower limit actual number8 2.8x i0~ 116 4.9 x 1018 224 15.8xi0

15 2432 8.0x107

Niemeier lattices (not including Leech lattice)

The Leech lattice.

(D1)24 with “glue vectors” or conjugacy class generators

Simple CurrentsConway and Sloane introduced the notion of “glue vectors”, combinations ofconjugacy classes that “glue” components of the lattice together.

Conjugacy classes are the decomposition of the weight lattice modulo the root lattice.For example ! has four classes denoted ! .DN (0), (v), (s), (c)

Simple currents can be viewed as a generalisation of glue vectors to general CFTs (in particular higher level affine Lie algebras).

They are defined as primaries ! with the simplest possible fusion rules with every other primary:

J × ϕi = ϕj

If such a � has integer conformal weight, there exists a correspondingextension of the chiral algebra.

A.N. Schellekens and S. Yankielowicz, Nucl. Phys. B327 (1989) 673-703 (See also K. Intriligator, Nucl. Phys. B332 (1990) 541-565, “Bonus Symmetry”)

g(Λ)−1 = (8k)−1B4k

4k−1

∏j=1

(4j)−1B2j

Siegel Mass FormulaThe sum over the inverse of the orders of the lattice automorphism group is a computable number

Minimal automorphism group is Maximal automorphism group is probably the Weyl group of D8k

[this is true for . In particular, has a larger automorphism group than ]k ≥ 3 D8k (E8)k

Even self-dual lattice of dimension !8k

A.N. Schellekens, “Big numbers in string theory” arXiv:1601.02462

A Mass Formula for Unimodular Lattices with No Roots, O.D. King (2001)

(13218 lines)

Beyond free bosonsOrbifold using X → − X

Applied to Leech lattice: all 24 spin-1 fields � are projected out.One obtains a new orbifold CFT with four primaries.

E, O, σE, σO h = 0, 1, 2, 32with

Fusion rules E2 = E; O2 = E; σ2O = E; σ2

Extend by O ⟶ return to Leech latticeExtend by σE ⟶ Monster Module (Frenkel, Lepowsky, Meurman (1986))

(see also Dixon, Ginsparg, Harvey (1988))

E: evenO: odd

The fields � and � are simple currents of integer spin.We can extend the chiral algebra with them.

This procedure can be applied to all Niemeier lattice CFT’s. One obtains an orbifold theory with

𝒩 − 12

spin-1 currents. In 15 cases (including the monster module) this turnsout to yield a new meromorphic CFT that cannot be constructed out of free bosons. [Dolan, Goddard, Montague (1990)]

(Application to the ! and ! theories yields nothing new)n=1 n=2

The boson string map*Bosonic strings:

([free boson CFT with c = 26] + ghosts, [free boson CFT with c = 26] + ghosts)

Xμ, bL, cL, bR, cR, μ = 1,…26

Heterotic strings:

([free boson CFT with c = 26] + ghosts, [free super CFT with c = 10] + ghosts)Xμ, XI, ψμ, bL, cL, bR, cR, βR, γR; μ = 1,10; I = 11,…26

Both are based on a choice of a modular invariant partition function

(*) Casher, Englert, Nicolai, Taormina (1985) Englert, Nicolai, Schellekens (1986) Lerche, Lüst, Schellekens (1986)

Yα =(det D1/2)α

(det D3/2)α

The boson string mapMulti-loop heterotic string partition function

ψμ βR, γRXIXμ, b, c

Ω : Period matrixα : spin structure

Genus 1 : Y3 = − Y4 = − Y2 (GSO)

The boson string map

Xα(Ω) = (det D1/2(Ω))4α

∑β (det D1/2(Ω))8βDefine:

Xα and Yα have the same modular transformations at any genus

P16;16 ≡ ∑α

P16α (det D1/2)8

α ∑β

(det D1/2)8β

We have constructed a modular invariant boson string with partition function

The boson string map(E8 ⊕ E8)L × (fermionic)R ↔ (E8 ⊕ E8)L × (E8 ⊕ E8)R

(D16(0 + s))L × (fermionic)R ↔ (D16(0 + s))L × (E8 ⊕ E8)R

(Pα)L × (fermionicα)R ↔ ∑α

(Pα)L × [(D8)α ⊕ E8]R

Now we can strip off (E8)R and replace (D8)R by (D8)L

(Pα)L × (fermionicα)R ↔ ∑α

(Pα)L × ((D8)α)L

Consistent Heterotic 10D string ↔ c = 24 Meromorphic CFT(with a D8 factor)

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

E8 × E8 superstring

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

D16 superstring; D8 × E8 tachyonic stringE8 × E8 superstring

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

O(16) × O(16) tachyon−free string

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

D16 tachyonic string

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

D12 × D4 tachyonic string

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

D2 × (E7)2 tachyonic string

permutation

A,1D,E6 (is, 1)(1,[0, 1,2])

A~D6 (2,4,0),(5,0,s),(0,5,c)

(s,s,s,s,s,s),(0,[0, u,c,c,uI)(1, [0, 1, 4, 4, 1])

A~ (3,[2,0,0,1,0,1,1])A~’ (2,[1,1,2,1,1,1,2,2,2,1,2])

A~4 (1,lO,0,0,0,0,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1,1,ll)

B 4n—1‘V ‘ — ~ fT 2]~g(A) 8n I 4j

15 2432 8.0x107

D2 × (E7)2 tachyonic stringD1 × A15 tachyonic string

References:

Supersymmetric Heterotic Strings: Gross, Harvey, Martinec, Rohm (1985)

O(16) x O(16) tachyon-free string: Dixon, Harvey (1986) Alvarez-Gaumé, Ginsparg, Moore, Vafa (1986) Tachyonic Strings: Seiberg, Witten (1986) Dixon, Harvey (1986) Kawai, Lewellen, Tye (1986)

Relation to Niemeier lattices: Lerche, Lüst, Schellekens (1986)

But Kawai et. al. found another (tachyonic) solution, with a Lie Algebra E8,2 This implies the existence of an additional meromorphic CFT, with an algebra

D9 × E8,2

So now we know 24+15+1=40 meromorphic � CFT’sc = 24

Another example:

Algebra F4,6

χ0000 + χ0004 + χ0030 + χ11002

+ χ0003 + χ0006 + χ0021 + χ20102

+ 2 χ0101 + χ10122

+ 2 χ0102 + χ20002

(A.N. Schellekens and S. Yankielowicz, Nucl. Phys. B327 (1989))

Fusion rules isomorphic to !

Spectrum opposite (mod 1) Central charges add up to 24.Hence one can build a meromorphic CFT with algebra

F4,6 × A2,2

(A.N. Schellekens and S. Yankielowicz, “Curiosities at c=24”, Phys. Lett. B226 (1989))

This is nr. 41

Motivations

Check completeness of the set of � stringsFind relatives of the Monster ModuleCuriosity

The Elliptic Genus

AnomaliesClassical symmetries that are not symmetries of the quantum theory

kµ( ) = 0 ?

Gauge vector boson or Graviton

Fermion

Anomalies from the index theorem

Field strength two-form

Curvature two-form

Index (γaDa) = ∫M[ A(R)Ch(F)]Vol

F ≡ 12 Fμν dxμ ∧ dxν

Rαβ ≡ 1

2 Rαβδγ dxδ ∧ dxγ

A(R) Dirac Genus Ch(F) Chern Character

Atiyah-Singer index theorem for a Weyl fermion

xa : Skew eigenvalues of R

Ch(F) = Tr eiF/2π

A(R) = ∏a

xa/2sinh (xa/2)

A(R) = 1 +1

(4π)2

(4π)4 [ 1288 (Tr R2)2 +

Tr R2]+

1(4π)6 [ 1

10368 (Tr R2)3 +1

4320Tr R2 Tr R4 +

Tr R6]⋯

xa : skew eigenvalue of R

Compute the anomaly polynomial by summing the contributions of all the particles in the loop.

In " dimensions, take the " volume-form from that polynomial (this is of order ! in ! and !

Use the “method of descent” on this term to compute the actual anomaly.

But to check anomaly cancellation, it is sufficient to check that this term vanishes.

2N 2N + 2N + 2 F R

Green-Schwarz (1984)

If a group can be found that satisfies:

Tr F6 = 148 Tr F2 Tr F4 − 1

14400 (Tr F2)3

Anomaly = I3/2(R) − I1/2(R) + I1/2(R, F)

∝ −115

Tr F6 +124

Tr R2 Tr F4 +18

Tr R2 Tr R4

+132 (Tr R2)3 −

Tr F2 Tr R4 −1

192Tr F2 (Tr R2)2

Anomaly ∝ (Tr R2 −130

Tr R2) × X8(R, F)

There are two solutions two these conditions:

� (Green and Schwarz, 1984) � (Thierry-Mieg, 1984)

With fermions in the adjoint representation

SO(32)E8 × E8

The anomalies still don’t cancel, but now they can be cancelled by adding extra terms to the action.

SO(32)E8 × E8

Additional particle

SO(32)E8 × E8

Additional interactions

Additional particle

Why do these miracles occur?All these field theories originated from string theories.All can be described as field theory limits of closed string theories:Heterotic strings (Gross, Harvey, Martinec, Rohm (1984))

Loop graphs of closed string theories have a remarkable property: Modular Invariance.This singles out the gauge groups SO(32) and E8 × E8 in 10 dimensions.

Level Matching

Massless particles

Level Matching

Massive excited states

Level Matching

Massive excited states

Level Matching

Unphysical

Level Matching

Unphysical Tachyon

Chiral Sector

HL HR (P,P)

Chiral Massless particles

Now we can write down an anomaly generating function for the entire chiral sector, a “character valued partition function”

The anomaly generating function

A(q, F, R) = A(R)∑k

qkCh(F)kCh(R)k

Only the coefficient of ! and the terms of order 6 in ! and ! are relevant for the anomaly

q0 F R

dimk → A(R)Ch(F)kCh(R)k

Modular transformationThe anomaly generating function must be modular invariant for ! .

On the other hand, it can be written in terms of characters of affine Lie algebras. It is known how such characters transform for ! or ! .Consider, for example, the θ-functions (related to ! characters)

F = R = 0

F ≠ 0 R ≠ 0SO(N )

Modular Weight

The phases ! cancel in the final assembly, because the result is modular invariant.The overall weight is also determined by modular invariance.

θi ( Fcτ + d

,aτ + bcτ + d ) = ∑

Sij cτ + d eiπF2c/(cτ+d) θj(F, τ)

Modular transformation

A ( aτ + bcτ + d

cτ + d,

Rcτ + d )

= exp [ ic32π3(cτ + d) (Tr F2 − Tr R2)](cτ + d)− (D − 2)

2 A(τ, F, R)

The character valued partition function transforms as follows

The anomaly

This is a meromorphic modular function of weight 2

fanomaly ( aτ + bcτ + d ) = (cτ + d)2 fanomaly(τ)

Expand A(q, F, R) to order (D + 2)/2 in F and R

One obtains a coefficient function fT(τ)for each combination T(F, R) of traces of F and R .

If we ignore the phases involving ! , the coefficient function of the anomaly transforms as

TrF2 − TrR2

The transformation properties of these functions follow from those of A(q, F, R)

Theorem: Any meromorphic function of weight 2 can be written as

Therefore there is no term q0.

Hence there is no anomaly if ! .Therefore the anomaly is proportional to !

TrF2 = TrR2

TrF2 − TrR2

f(τ) =ddτ

Related mathematical work:

Published:

Eisenstein functions

G2k(τ) = ∑(m,n)∈ℤ2\(0,0)

1(m + nτ)2k E2k(τ) =

G2k(τ)2ζ(2k)

E2k ( aτ + bcτ + d ) = (cτ + d)2kE2k(τ) for k ≥ 2

E2 ( aτ + bcτ + d ) = (cτ + d)2E2(τ) −

c(cτ + d)

! is a holomorphic modular formE2k, k > 2,

Eisenstein functions

E2(q) = 1 − 24∞

∑n=1

1 − qn

E4(q) = 1 + 240∞

∑n=1

1 − qn

E6(q) = 1 − 504∞

∑n=1

1 − qn

(q = e2πiτ)q expansions

All E2k, k ≥ 4 are polynomials of E4 and E6 (for example E8 = (E4)2)

Hence ℰ4n is a polynomial in E4 and E6 of weight 4n

Meromorphic Partition Functions

ℰ4 = αE4

ℰ8 = α (E4)2

ℰ12 = α (E4)2 + β (E6)2

ℰ16 = α (E4)4 + β (E6)2 E4⋯

Define P8n = Tr eiτH =ℰ4n

Taking out the η-function removes the pole at q=0and the phase of the modular transformation τ→τ+1This implies that ℰ8n is a modular form .(assuming P(τ) has no poles in the upper half−plane)

The absolute modular invariant

Its higher coefficients are equal to sums of dimensions of the monster group. This is a the largest “sporadic” group, a discrete group withelements.

808017424794512875886459904961710757005754368000000000

P(q) = j(q) + constant

j(q) =1q

+ 744 + 196884 q + 21493760 q2 + 864299970 q3…

For ! the partition function has one free parameter,and is usually written as

n = 3 (c = 24)

The N spin-1 excitations must form an “affine Lie algebra” or Kac-Moody algebra:

[Jam, Jb

n] = if abcJcm+n + kmδabδm+n,0

P(q) =1q

+ 𝒩 + 196884 q + …

The number 196884 must be built out of dimensions of representations of that algebra.

But we will need more information than that.

The partition function of a general � meromorphic CFT isc = 24

� : Dual coxeter number (depends on algebra)gℓ

This function can be expressed in terms of basic modular functions

P ( aτ + bcτ + d

cτ + d ) = exp ( −ic8π(cτ + d)

ℱ2) P(τ, F )

So we will consider the “character valued partition function” or “elliptic “genus”

ℱ2 = ∑ℓ

gℓTradj F2

P(q, F ) = Tr e F ⋅ J 0qL0−c/24

This transform in the following way under modular transformations

First we modify the function to take into account the modular anomaly

E2 ( aτ + bcτ + d ) = (cτ + d)2E2(τ) −

c(cτ + d)

P(τ, F ) = e− 148 E2(τ)ℱ2η8n(τ) P(τ, F )

Define

P ( aτ + bcτ + d

cτ + d ) = (cτ + d)4nP(τ, F )

Then, because

we get

Now � can be expanded in terms of holomorphic modular forms. P(τ)

P (q, F1, …, FL) = e148 E2(q)ℱ2(η(q))−8n

∑m=0

ℰi4n+m(q)𝒯m

Reassembling the original partition function we get

where � is some combination of traces of total order � .For example: � has � .

𝒯mi m

Tr(F1)4Tr(F2)2Tr(F3)6 m = 12

The coefficient functions � are modular forms of weight � .ℰi4n+m(q) 4n + m

Now we make use of the fact that we already know the ground state representation, associated with the coefficient of � in the expansion.This representation is trivial, so all its higher traces vanish.

In the absence of the “anomalous” pre-factor, this would imply that all � for � must contain a factor � .

ℰi4n+m(q) m > 0 q

P(τ, F) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

η24ℰi4n+m−12(τ)𝒯m

Taking this into account we get

Expanding to the first excited level and second order in �F

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

P(τ, F ) = exp ( 148

E2(τ)ℱ2) η(τ)−8n

× ℰ04n(τ) + (E4(τ))n [cosh ( 1

48E4(τ)ℱ2) − 1]

−(E4(τ))n−3/2 E6(τ)[sinh ( 148

E4(τ)ℱ2)]+

∑m=2

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

� vanishes for �ℰ4n+m−12 m = 2, n = 1,2,3ℰ4n+m−12(q) = αℓ + 𝒪(q)…

∑ℓ

(dimℓ +12

Tradj F2ℓ) = 𝒩 + (15 −

n +𝒩48 ) ℱ2 + ∑

αℓ Tr F2ℓ

∑ℓ

dimℓ = 𝒩

(30 −313

n +𝒩24 ) kℓ

gℓ+ αℓ = 1

This implies:

ctot = ∑ℓ

kℓ dimℓ

kℓ + gℓ

= 24𝒩

248(3 − n) + 𝒩

kℓ= 30 −

n +𝒩24n = 1,2,3

Total affine Lie algebra central charge

= 8 (n = 1, 𝒩 = 248)= 16 (n = 2, 𝒩 = 496)= 24 (n = 3, any𝒩)

Independent of � ! ℓ

If there is any spin-1 current, they saturate the central charge

Lattices: Venkov (1980)

(30 −313

n +𝒩24 ) x + αℓ = 0

For � ’s (with an arbitrary normalisation � )U(1) x

𝒩 = 248n − 720

Only possible with 24 � ’s and �U(1) n = 3

n = 1,2,3

(Leech lattice)

The Lie groups

From the quadratic traces:

This determines the total Virasoro central charge of the “gauge” part: � .

Hence either there are no gauge symmetries at all, or they saturate the full central charge.

In the latter case, there are 221 solutions.

c = 24

𝒩 − 1

Trace IdentitiesThe vacuum is a singlet; all its representation traces vanish.

At the first excited level we encounter only adjoints, whose traces are now known.

Given the vacuum and the first excited level, all traces of order 0, 2, 4, 6, 8, 10 and 14 are now known at all levels.

This is sufficient information to determine all solutions.

The 221 candidates are reduced to 69 that satisfy the trace identitiesat level 2. In particular, we know how 196884 is built out of combinations of Lie algebra representations.

Modular InvarianceWe could go on. Knowing the state decomposition at level 2 fixes even more trace identities. But it also fixes the affine character combinations completely for all 69 candidates.

Therefore at this stage it is more convenient to check modular invariance for those 69 candidates. This guarantees that the trace identities are satisfied to any order.

All 69 candidates survive this check. This is not really surprising, because the trace identities are a very strong constraint already.

The list• One CFT without any Lie-algebra.

(“The Monster Module”)• One U(1)24 lattice (“the Leech Lattice”)• 23 Niemeier lattices with non-abelian groups• 14 Z2 Orbifolds of Niemeier lattices.

(Goddard, Olive, Montague, 1990)

• One case related to the � string theory: �

• One already known “curious” case: � .

• 30 new cases

E8,2 E8,2B8,1

F4,8 × A2,2

71 in total

− 31 −

No. N Spin-1 algebra Glue Orbits Ref.

0 0 — [10]

1 24 U(1)24 (0) [41]

2 36 (A1,4)12 1[1; (0; )10] see text [12]

3 36 D4,12A2,6 (0, 1) + (s, 0) + (v, 0) (0000 + 0006 + 0060 + 0066 + 0400 + 3033, 00)

+(0204 + 0240 + 0300 + 0244 + 1411 + 2122, 03)

+(0044 + 0600 + 1213 + 1231 + 1233 + 2022, 11)

+(0004 + 0040 + 0048 + 0320 + 0302 + 0324

+1033 + 1035 + 1053 + 3 × 2222, 22)

4 36 C4,10 1 0000 + 0024 + 0040 + 0044 + 00,10,0 + 0260 + 0321

+0323 + 0500 + 0800 + 1051 + 1430 + 1431 + 2 × 2222

+2242 + 3031 + 4140

5 48 (A1,2)16 11[11; (00; )6] see text [12]

+1010[1010; (0000; )2]

+(1000)4

6 48 (A2,3)6 1[1; (0; )4] (00)6 + {(11; )4(00; )2} + (01)5(12) + (10)5(21) + 6 × (1, 1)6

7 48 (A3,4)3A1,2 [1; 0; 0]1 ((000)3 + (012)3, 0) + ({002; 010; 111}, 1)+ 4 × ((111)3, 1)

+([000; 020; 020], 2)+ ([012; 020; 020], 2)

8 48 A5,6C2,3A1,2 (1, 0, 1) + (0, 1, 1) (00000 + 02020, 00, 0) + (00003 + 00211, 30, 1)

+(00200 + 02020, 20, 2)+ (00130 + 03100, 11, 1)

+(00022, 01, 0)+ (00030, 00, 2) + (01102, 10, 1)

+(01121, 20, 0) + (01210, 01, 2) + 2 × (11111, 11, 1)

9 48 (A4,5)2 (1, 0) + (0, 1) (0000, 0000) + (0102, 0102) + 4 × (1111, 1111)

+[0021; 0110] + [1111; 0013]

10 48 D5,8A1,2 (s, 0) (00000 + 00222 + 03011 + 10111, 0)+

(00113 + 00131 + 2 × 11111, 1)+

(00044 + 00200 + 01022 + 01211, 2)

11 48 A6,7 1 000000 + 001301 + 103100 + 002030

+010122 + 3 × 111111

12 60 (C2,2)6 1[1; (0; )5] (00)6 + [00; (20; )5] + [00; (01; )5] + [11; (10; )5] [12]

+{(01; )3(20; )3} + [(00; )2[(01; )2; (20; )2]]

13 60 D4,4(A2,2)4 (0, 1, 1, 1, 0) (0000, (00)4) + (0100, (11)4)

+(0, 2, 1, 0, 1) +(0200, [11; (00; )3]) + (1011, [00; (11; )3])

+(v, 0, 0, 0, 0) +(0002, 00, 11, 11, 00)+ (0002, 11, 00, 00, 11)

+(s, 0, 0, 0, 0) +(0020, 00, 00, 11, 11)+ (0020, 11, 11, 00, 00)

+(0022, 00, 11, 00, 11)+ (0022, 11, 00, 11, 00)

The number 71808017424794512875886459904961710757005754368000000000

71 is the largest prime factor in the order of the monster group...

246 ⋅ 320 ⋅ 59 ⋅ 76 ⋅ 112 ⋅ 133 ⋅ 17 ⋅ 19 ⋅ 23 ⋅ 29 ⋅ 31 ⋅ 41 ⋅ 47 ⋅ 59 ⋅ 71

The number 71N

∑i=0

i2 = M2

Has (only?) the following solutions (checked for � ):

� � �

The vector � in the Lorentzian lattice � can be used to characterise the Leech lattice.

N < 3000000000

N = 0; M = 0N = 1; M = 1N = 24; M = 70

(0,1,2,3,…,24; 70) Γ25,1

Gerald Höhn, arXiv:1708.05990

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