Wednesday, November 6, 2019
Levels 126 & 252
I. Non-moonshine functions. There are no moonshine functions with uppercase index (in Atlas notation) for level >119. However, surprisingly we can still find trinomial identities for level 144 (some a consequence of level 72) and as high as level 252 (a consequence of level 126). For the latter, define the two pairs of eta quotients, a(τ)=(d27d29d3d14d18d21),b(τ)=(d21d263d2d3d21d126)c(τ)=(d214d218d6d7d9d42),d(τ)=(d22d2126d1d6d42d63)Given one moonshine function of level 21, namely j21B(τ)=(d1d3d7d21) and define, e(τ)=j21B(τ)j21B(3τ)=(d1d63d7d9) then ratios of the pairs are simply, ab=e(2τ)e2(τ)cd=e(τ)e2(2τ) They obey,a(τ)−2=b(τ)c(τ)−1=d(τ) a(12+τ)−2=b(12+τ)c(12+τ)−1=d(12+τ) which is the pair of trinomial identities each for level 126 and level 252, respectively, and the latter seems to be the highest known.
Levels 42 & 84
I. Moonshine functions. The four functions for level 42,
j42A(τ)=(√j42B+1√j42B)2j42B(τ)=(d1d6d14d21d2d3d7d42)2j42C(τ)=3√j14B(3τ)=(d3d21d6d42)j42D(τ)=(d2d6d7d21d1d3d14d42) and the three functions for level 84,
j84A(τ)=√j42A(2τ)j84B(τ)=√j42B(2τ)j84C(τ)=3√j28B(3τ)=(d26d242d3d12d21d84)
II. Non-moonshine functions. Define the two pairs of eta quotients, a(τ)=(d21d214d2d3d7d42),b(τ)=(d22d27d1d6d14d21)c(τ)=(d23d242d1d6d14d21),d(τ)=(d26d221d2d3d7d42) Interestingly, these satisfy a(τ)b(τ)c(τ)d(τ)=1 And in terms of the moonshine functions,
ab=j42Bj42Ddc=j42Bj42D They obey, a(τ)+3=b(τ)c(τ)+1=d(τ) a(12+τ)+3=b(12+τ)c(12+τ)+1=d(12+τ) which is the pair of trinomial identities each for level 42 and level 84, respectively.
j42A(τ)=(√j42B+1√j42B)2j42B(τ)=(d1d6d14d21d2d3d7d42)2j42C(τ)=3√j14B(3τ)=(d3d21d6d42)j42D(τ)=(d2d6d7d21d1d3d14d42) and the three functions for level 84,
j84A(τ)=√j42A(2τ)j84B(τ)=√j42B(2τ)j84C(τ)=3√j28B(3τ)=(d26d242d3d12d21d84)
II. Non-moonshine functions. Define the two pairs of eta quotients, a(τ)=(d21d214d2d3d7d42),b(τ)=(d22d27d1d6d14d21)c(τ)=(d23d242d1d6d14d21),d(τ)=(d26d221d2d3d7d42) Interestingly, these satisfy a(τ)b(τ)c(τ)d(τ)=1 And in terms of the moonshine functions,
ab=j42Bj42Ddc=j42Bj42D They obey, a(τ)+3=b(τ)c(τ)+1=d(τ) a(12+τ)+3=b(12+τ)c(12+τ)+1=d(12+τ) which is the pair of trinomial identities each for level 42 and level 84, respectively.
Levels 14 & 28
For certain even levels divisible by 7, the situation is now different. We can still find an eta quotient a such that a+m=b is also an eta quotient, but only for one rational m.
I. Moonshine functions: Define a(τ)=(d1d7d4d28),b(τ)=(d32d314d1d24d7d228) then a+2=b or, (d1d7d4d28)+2=(d32d314d1d24d7d228) for the unique m=2. This is the single trinomial identity of level 28. First, we have the 3 moonshine functions of level 14,
j14A(τ)=(√j14C+1√j14C)2j14B(τ)=(d1d7d2d14)3=b+4b−4j14C(τ)=(d2d7d1d14)4 and the 4 moonshine functions of level 28,
j28A(τ)=√j14A(2τ)=j28D(τ)+1j28D(τ)j28B(τ)=(d22d214d1d4d7d28)3=a+4a+4j28C(τ)=d1d7d4d28=aj28D(τ)=√j14C(2τ)=(d4d14d2d28)2 Note that j14B(12+τ)=−j28B(τ).
I. Moonshine functions: Define a(τ)=(d1d7d4d28),b(τ)=(d32d314d1d24d7d228) then a+2=b or, (d1d7d4d28)+2=(d32d314d1d24d7d228) for the unique m=2. This is the single trinomial identity of level 28. First, we have the 3 moonshine functions of level 14,
j14A(τ)=(√j14C+1√j14C)2j14B(τ)=(d1d7d2d14)3=b+4b−4j14C(τ)=(d2d7d1d14)4 and the 4 moonshine functions of level 28,
j28A(τ)=√j14A(2τ)=j28D(τ)+1j28D(τ)j28B(τ)=(d22d214d1d4d7d28)3=a+4a+4j28C(τ)=d1d7d4d28=aj28D(τ)=√j14C(2τ)=(d4d14d2d28)2 Note that j14B(12+τ)=−j28B(τ).
Thursday, October 31, 2019
Level 72
These are not monster functions (with the exception of the first). But I'm including them since this seems to be the highest level such that the requirement of the triple relation works, namely that a,a+m,a+n are all eta quotients. For convenience, first define the eta quotient, fp=dpd2p=η(pτ)η(2pτ). Let,
a(τ)=(f24f26f236f2f412f18)b(τ)=(f23f4f36f1f9f212)f26c(τ)=(f1f4f26f9f36f2f23f212f18)f26 such that a+1=b,a−1=c. Then define, j12F(τ)=f46=(d6d12)4=bcak72A(τ)=f46+4f46k72B(τ)=(f1f4f26f9f36f2f23f212f18)2=acbk72C(τ)=(f23f4f36f1f9f212)2=acbk72D(τ)=(f24f26f236f2f412f18)=a Then we have an analogous linear relation,
k72A+2k72D=k72B+k72C+j12F+2 Curiously, notice that the form of the last three k72 appear in other functions,
1j36A(τ)=(d1d4d26d9d36d2d23d212d18)21k36A(τ)=(d23d4d36d1d9d212)21√j18A(2τ)=(d24d26d236d2d412d18)
a(τ)=(f24f26f236f2f412f18)b(τ)=(f23f4f36f1f9f212)f26c(τ)=(f1f4f26f9f36f2f23f212f18)f26 such that a+1=b,a−1=c. Then define, j12F(τ)=f46=(d6d12)4=bcak72A(τ)=f46+4f46k72B(τ)=(f1f4f26f9f36f2f23f212f18)2=acbk72C(τ)=(f23f4f36f1f9f212)2=acbk72D(τ)=(f24f26f236f2f412f18)=a Then we have an analogous linear relation,
k72A+2k72D=k72B+k72C+j12F+2 Curiously, notice that the form of the last three k72 appear in other functions,
1j36A(τ)=(d1d4d26d9d36d2d23d212d18)21k36A(τ)=(d23d4d36d1d9d212)21√j18A(2τ)=(d24d26d236d2d412d18)
Level 60
j60A(τ)=j60E(τ)+1j60E(τ)=√j30B(2τ)j60B(τ)=(d2d6d10d30)2d1d3d4d5d12d15d20d60j60C(τ)=ab=1a(d6d10)3+1=ab2(d2d30)3−1j60D(τ)=(d1d12d15d20d3d4d5d60)=(d2d6d10d30d3d4d5d60)−1j60E(τ)=d4d6d20d30d2d10d12d60=√j30D(2τ)j60F(τ)=(d12d30d6d60) where
a=d2d3d5d12d20d30b=d1d4d6d10d15d60
There are no linear relations between these functions.
a=d2d3d5d12d20d30b=d1d4d6d10d15d60
There are no linear relations between these functions.
Level 40
j40A(τ)=√j20B(τ)=4√j10A(2τ)j40B(τ)=√j20A(2τ)=(d24d220d2d8d10d40)2=j40C(τ)+1j40C(τ)j40C(τ)=√j20F(2τ)=(d8d10d2d40)
There is no linear relation between these functions. j40A is one of the few that involves a 4th root.
There is no linear relation between these functions. j40A is one of the few that involves a 4th root.
Level 36, part 1
j36A(τ)=(d2d23d212d18d1d4d26d9d36)2=j36B(τ)+3j36B(τ)+3j36B(τ)=(d1d4d18d2d9d36)j36C(τ)=(d26d212d2d4d18d36)=√j18B(2τ)j36D(τ)=(d4d9d1d36)
There is no linear relation between these functions.
There is no linear relation between these functions.
Level 30
j30A(τ)=(d1d6d10d15d2d3d5d30)3j30B(τ)=j30D(τ)+1j30D(τ)+2j30C(τ)=(d1d3d5d15d2d6d10d30)j30D(τ)=(d2d3d10d15d1d5d6d30)2j30E(τ)=(d6d15d3d30)2j30F(τ)=(d3d5d6d10d1d2d15d30)j30G(τ)=(d21d6d10d215d22d3d5d230)
These functions obey the triple relations. Let a=j30G,m=1,n=2, then,
j30F=a+mna+(m+n)j30C=a+m+m(m−n)a+m−(2m−n)j30A=a+n+n(m−n)a+n+(m−2n) as well as,
j30B=j30F+(m−n)2j30F+3=j30C+n2j30C+5=j30A+m2j30A+7 which implies the linear relation between 5 monster functions,
j30B+2j30G=j30A+j30C+j30F+3 This is the highest level with linear relations between 5 monster functions.
These functions obey the triple relations. Let a=j30G,m=1,n=2, then,
j30F=a+mna+(m+n)j30C=a+m+m(m−n)a+m−(2m−n)j30A=a+n+n(m−n)a+n+(m−2n) as well as,
j30B=j30F+(m−n)2j30F+3=j30C+n2j30C+5=j30A+m2j30A+7 which implies the linear relation between 5 monster functions,
j30B+2j30G=j30A+j30C+j30F+3 This is the highest level with linear relations between 5 monster functions.
Level 24, part 1
Level 24
j24A(τ)=(d24d212d2d6d8d24)3=√j12A(2τ)=j24D(τ)+4j24D(τ)=j24H(τ)+1j24H(τ)j24B(τ)=(d2d4d6d12d1d3d8d24)2j24C(τ)=(d21d6d28d12d2d23d4d224)j24D(τ)=(d2d6d8d24)=√j12E(2τ)j24E(τ)=(d212d6d24)4=√j12D(2τ)j24F(τ)=(d8d12d4d24)3=√j12F(2τ)j24G(τ)=(d4d8d12d24)=√j12G(2τ)j24H(τ)=(d6d8d2d24)2=√j12H(2τ)j24I(τ)=(d2d23d28d12d21d4d6d224)j24J(τ)=(d12d24)2=√j12J(2τ) The "important" functions (that aren't just square roots) are j24C and j24I. First define the auxiliary non-monster functions, U=(d1d6d8d12d2d3d4d24)4V=(d22d3d8d212d1d24d26d24)4 I. Let a1=j24C,m=1,n=3, then,
j24B=a1+mna1+(m+n)j12G=a1+m+m(m−n)a1+m−(2m−n)U=a1+n+n(m−n)a1+n+(m−2n) as well as, j12C=j24B+(m−n)2j24B−2=j12G+n2j12G=U+m2U+4
II. Let a2=j24I,m=1,n=−1, then,
j24B=a2+mna2+(m+n)j12F=a2+m+m(m−n)a2+m−(2m−n)V=a2+n+n(m−n)a2+n+(m−2n) as well as, j12C=j24B+(m−n)2j24B−2=j12F+n2j12F=V+m2V−4 These two families imply the two linear relations, j12C+2j24C=j24B+j12G+U−2j12C+2j24I=j24B+j12F+V−2
However, there is one linear relation using only monster functions,
j12C−j12F−j12G+j12E−2j12I=2(j24B−j24C−j24I−3) Note that,
j12C=j12F+1j12F=j12G+9j12Gj12E=j12I−3j12I−2
j24A(τ)=(d24d212d2d6d8d24)3=√j12A(2τ)=j24D(τ)+4j24D(τ)=j24H(τ)+1j24H(τ)j24B(τ)=(d2d4d6d12d1d3d8d24)2j24C(τ)=(d21d6d28d12d2d23d4d224)j24D(τ)=(d2d6d8d24)=√j12E(2τ)j24E(τ)=(d212d6d24)4=√j12D(2τ)j24F(τ)=(d8d12d4d24)3=√j12F(2τ)j24G(τ)=(d4d8d12d24)=√j12G(2τ)j24H(τ)=(d6d8d2d24)2=√j12H(2τ)j24I(τ)=(d2d23d28d12d21d4d6d224)j24J(τ)=(d12d24)2=√j12J(2τ) The "important" functions (that aren't just square roots) are j24C and j24I. First define the auxiliary non-monster functions, U=(d1d6d8d12d2d3d4d24)4V=(d22d3d8d212d1d24d26d24)4 I. Let a1=j24C,m=1,n=3, then,
j24B=a1+mna1+(m+n)j12G=a1+m+m(m−n)a1+m−(2m−n)U=a1+n+n(m−n)a1+n+(m−2n) as well as, j12C=j24B+(m−n)2j24B−2=j12G+n2j12G=U+m2U+4
II. Let a2=j24I,m=1,n=−1, then,
j24B=a2+mna2+(m+n)j12F=a2+m+m(m−n)a2+m−(2m−n)V=a2+n+n(m−n)a2+n+(m−2n) as well as, j12C=j24B+(m−n)2j24B−2=j12F+n2j12F=V+m2V−4 These two families imply the two linear relations, j12C+2j24C=j24B+j12G+U−2j12C+2j24I=j24B+j12F+V−2
However, there is one linear relation using only monster functions,
j12C−j12F−j12G+j12E−2j12I=2(j24B−j24C−j24I−3) Note that,
j12C=j12F+1j12F=j12G+9j12Gj12E=j12I−3j12I−2
Level 20
I. Monster functions: Let,
a(τ)=(d1d4d10d2d5d20)2,b(τ)=d1d4d1010d22d55d520,c(τ)=d82d31d34d5d20
such that a+1=b,a+5=c. This level 20 triple can be derived from a level 10 triple. Then define,
j20A(τ)=(d22d210d1d4d5d20)4=bca=(√j20F+1√j20F)2j20B(τ)=√j10A(2τ)=j20D(τ)+4j20D(τ)=j20E(τ)−1j20E(τ)j20C(τ)=(d1d4d10d2d5d20)2=aj20D(τ)=(d2d10d4d20)2=√j10B(2τ)j20E(τ)=(d4d10d2d20)3=√j10D(2τ)j20F(τ)=(d4d5d1d20)2
Examples: j10A(12√−19/10)=762,j20B(14√−19/10)=76
II. Non-monster functions. Also define,
u(τ)=(d42d5d20d1d4d410)2=acbv(τ)=(d1d4d210d22d5d20)6=abc
III. Relations. Like j30B, the function j20A is expressible by others in 4 ways. There is no linear relation between purely monster functions of level 20.
a(τ)=(d1d4d10d2d5d20)2,b(τ)=d1d4d1010d22d55d520,c(τ)=d82d31d34d5d20
such that a+1=b,a+5=c. This level 20 triple can be derived from a level 10 triple. Then define,
j20A(τ)=(d22d210d1d4d5d20)4=bca=(√j20F+1√j20F)2j20B(τ)=√j10A(2τ)=j20D(τ)+4j20D(τ)=j20E(τ)−1j20E(τ)j20C(τ)=(d1d4d10d2d5d20)2=aj20D(τ)=(d2d10d4d20)2=√j10B(2τ)j20E(τ)=(d4d10d2d20)3=√j10D(2τ)j20F(τ)=(d4d5d1d20)2
Examples: j10A(12√−19/10)=762,j20B(14√−19/10)=76
II. Non-monster functions. Also define,
u(τ)=(d42d5d20d1d4d410)2=acbv(τ)=(d1d4d210d22d5d20)6=abc
III. Relations. Like j30B, the function j20A is expressible by others in 4 ways. There is no linear relation between purely monster functions of level 20.
Level 18
This level is also special since we can find two a(τ) namely, j18D and j18C. Using S1 of the Intro, these two explain 2×4=8 of the 14 trinomial identities of level 18 found by Somos.
I. First triple. Define,
a(τ)=(d22d9d1d218),b(τ)=(d6d39d3d318),c(τ)=(d21d6d9d2d3d218) then,
(d22d9d1d218)−1=(d6d39d3d318)(d22d9d1d218)−3=(d21d6d9d2d3d218) so a−1=b,a−3=c, and we get the moonshine functions,
j18A(τ)=(d1d2d9d18)=acbj18B(τ)=(d23d26d1d2d9d18)2j18C(τ)=(d1d46d9d22d23d218)2=bca+1j18D(τ)=(d22d9d1d218)=aj18E(τ)=(d2d9d1d18)3=abc Naturally, these also satisfy S2 and S3. We also have
j18B=j18A+32j18A+3=j18E+1j18E−1 and the linear relation between these 5 moonshine functions,
j18B+2j18D=j18A+j18C+j18E+4
II. Second triple. Define,
a(τ)=(d1d46d9d22d23d218)2,b(τ)=(d31d26d39d32d23d318),c(τ)=(d63d1d2d26d9d18) then, (d1d46d9d22d23d218)2−1=(d31d26d39d32d23d318)(d1d46d9d22d23d218)2+3=(d63d1d2d26d9d18)so a−1=b,a+3=c, and we get,
j6d=j6F+16j6Fj18B(τ)=(d23d26d1d2d9d18)2=acbj18C(τ)=(d1d46d9d22d23d218)2=aβ(τ)=(d1d26d9d2d23d18)6=abcj6F(τ)=(d3d6)8=bca where j6d and β are non-monster functions with expansions (A007263) and (A128512), respectively. Naturally, these also satisfy S1,2,3. And by adding these additional relations, we get a second linear equation for level 18 of similar form to the first,
j6d+2j18C=j18B+j6F+β−2
I. First triple. Define,
a(τ)=(d22d9d1d218),b(τ)=(d6d39d3d318),c(τ)=(d21d6d9d2d3d218) then,
(d22d9d1d218)−1=(d6d39d3d318)(d22d9d1d218)−3=(d21d6d9d2d3d218) so a−1=b,a−3=c, and we get the moonshine functions,
j18A(τ)=(d1d2d9d18)=acbj18B(τ)=(d23d26d1d2d9d18)2j18C(τ)=(d1d46d9d22d23d218)2=bca+1j18D(τ)=(d22d9d1d218)=aj18E(τ)=(d2d9d1d18)3=abc Naturally, these also satisfy S2 and S3. We also have
j18B=j18A+32j18A+3=j18E+1j18E−1 and the linear relation between these 5 moonshine functions,
j18B+2j18D=j18A+j18C+j18E+4
II. Second triple. Define,
a(τ)=(d1d46d9d22d23d218)2,b(τ)=(d31d26d39d32d23d318),c(τ)=(d63d1d2d26d9d18) then, (d1d46d9d22d23d218)2−1=(d31d26d39d32d23d318)(d1d46d9d22d23d218)2+3=(d63d1d2d26d9d18)so a−1=b,a+3=c, and we get,
j6d=j6F+16j6Fj18B(τ)=(d23d26d1d2d9d18)2=acbj18C(τ)=(d1d46d9d22d23d218)2=aβ(τ)=(d1d26d9d2d23d18)6=abcj6F(τ)=(d3d6)8=bca where j6d and β are non-monster functions with expansions (A007263) and (A128512), respectively. Naturally, these also satisfy S1,2,3. And by adding these additional relations, we get a second linear equation for level 18 of similar form to the first,
j6d+2j18C=j18B+j6F+β−2
Level 16
I. Moonshine functions: Define,a(τ)=(d38d4d216)2,b(τ)=(d21d8d2d216),c(τ)=(d52d8d21d24d216) then, (d38d4d216)2−2=(d21d8d2d216)(d38d4d216)2+2=(d52d8d21d24d216)so a−2=b,a+2=c. This a,b,c also belong to the class where a(τ)=−a(12+τ) and b(τ)=−c(12+τ). And the three level 16 moonshine functions,j16A(τ)=(d4d8d2d16)4=√j8A(2τ)j16B(τ)=(d38d4d216)2=aj16C(τ)=(d32d38d21d24d216)2=acb
II. Relations: Using S1, this a,b,c explains 4 of the 6 trinomial identities of level 16 (prefixed with t16) by Somos. They also obey,
(bca+16abc)+2a=bca+acb+abc In general, (bca+(b−c)2abc)+2a=bca+acb+abc where b−c=4. Or in terms of named functions of level 8 and 16, j8B+2j16B=j8D+j16C+j16C′where, similar to level 8, j16C′(τ)=−j16C(12+τ)=(d21d38d32d216)2 The last 2 of the 6 trinomial identities involves the pair j16C and j16C′, hence is just the same function using different arguments, j16C(τ)−2=(d32d38d21d24d216)2−2=d104d21d32d38d216j16C(12+τ)−2=−(d21d38d32d216)2−2=−d21d124d92d38d216
III. Inter-level quadratic relations: We also have,
j4C=j8E+16j8Ej4D=j8E−16j8E j8D=j16B−4j16Bj8E=j16B+4j16B Adding each pair leaves functions with the same argument,
j4C+j4D=2j8Ej8D+j8E=2j16B which are 2 of the 9 linear relations found by Conway, Norton, and Atkins.
II. Relations: Using S1, this a,b,c explains 4 of the 6 trinomial identities of level 16 (prefixed with t16) by Somos. They also obey,
(bca+16abc)+2a=bca+acb+abc In general, (bca+(b−c)2abc)+2a=bca+acb+abc where b−c=4. Or in terms of named functions of level 8 and 16, j8B+2j16B=j8D+j16C+j16C′where, similar to level 8, j16C′(τ)=−j16C(12+τ)=(d21d38d32d216)2 The last 2 of the 6 trinomial identities involves the pair j16C and j16C′, hence is just the same function using different arguments, j16C(τ)−2=(d32d38d21d24d216)2−2=d104d21d32d38d216j16C(12+τ)−2=−(d21d38d32d216)2−2=−d21d124d92d38d216
III. Inter-level quadratic relations: We also have,
j4C=j8E+16j8Ej4D=j8E−16j8E j8D=j16B−4j16Bj8E=j16B+4j16B Adding each pair leaves functions with the same argument,
j4C+j4D=2j8Ej8D+j8E=2j16B which are 2 of the 9 linear relations found by Conway, Norton, and Atkins.
Level 12, part 1
Level 12 is very special because, given an eta quotient a, we can find a+mi as an eta quotient for four different integer values mi. However, that also makes the situation more complicated than in other levels. Let,
a(τ)=(d24d6d2d212)2,b(τ)=d33d4d1d312,c(τ)=d31d4d26d22d3d312 such that a+1=b,a−3=c. And the monster functions, j12A(τ)=(d22d26d1d3d4d12)6j12B(τ)=(d1d4d6d2d3d12)4=acbj12C(τ)=√j6A(2τ)=j12F(τ)+1j12F(τ)=j12G(τ)+9j12G(τ)j12D(τ)=(d26d3d12)8=3√j4A(3τ)j12E(τ)=(d1d3d4d12)2=bcaj12F(τ)=(d4d6d2d12)6=√j6B(2τ)j12G(τ)=(d2d4d6d12)2=√j6D(2τ)j12H(τ)=(d3d4d1d12)4=abcj12I(τ)=(d24d6d2d212)2=aj12J(τ)=(d6d12)4=√j6F(2τ)
Example: j12C(12√−17/6)=198
Let a=j12I,m=1,n=−3, then,
j12E=a−3a−2j12B=a+1+4a+1−5j12H=a−3+12a−3+7 as well as,
j12A=j12E+42j12E+8=j12B+32j12B+10=j12H+1j12H+2 which implies the linear relation between 5 monster functions,
j12A+2j12I=j12B+j12E+j12H+8 The functions j12I and j12B can be used for level 12, while versions of j12E and j12H will be useful in level 24.
a(τ)=(d24d6d2d212)2,b(τ)=d33d4d1d312,c(τ)=d31d4d26d22d3d312 such that a+1=b,a−3=c. And the monster functions, j12A(τ)=(d22d26d1d3d4d12)6j12B(τ)=(d1d4d6d2d3d12)4=acbj12C(τ)=√j6A(2τ)=j12F(τ)+1j12F(τ)=j12G(τ)+9j12G(τ)j12D(τ)=(d26d3d12)8=3√j4A(3τ)j12E(τ)=(d1d3d4d12)2=bcaj12F(τ)=(d4d6d2d12)6=√j6B(2τ)j12G(τ)=(d2d4d6d12)2=√j6D(2τ)j12H(τ)=(d3d4d1d12)4=abcj12I(τ)=(d24d6d2d212)2=aj12J(τ)=(d6d12)4=√j6F(2τ)
Example: j12C(12√−17/6)=198
Let a=j12I,m=1,n=−3, then,
j12E=a−3a−2j12B=a+1+4a+1−5j12H=a−3+12a−3+7 as well as,
j12A=j12E+42j12E+8=j12B+32j12B+10=j12H+1j12H+2 which implies the linear relation between 5 monster functions,
j12A+2j12I=j12B+j12E+j12H+8 The functions j12I and j12B can be used for level 12, while versions of j12E and j12H will be useful in level 24.
Level 10
I. Moonshine functions: Define, a(τ)=(d22d5d1d210)2,b(τ)=(d2d55d1d510),c(τ)=(d31d5d2d310) then (d22d5d1d210)2−1=(d2d55d1d510)(d22d5d1d210)2−5=(d31d5d2d310) so a−1=b,a−5=c and we get the moonshine functions j10A(τ)=j10B(τ)+42j10B(τ)+8j10B(τ)=(d1d5d2d10)4=bcaj10C(τ)=(d1d2d5d10)2=acbj10D(τ)=(d2d5d1d10)6=abcj10E(τ)=(d22d5d1d210)2=a Example: j10A(√−19/10)=762
II. Relations: Let a,b,c as defined above and m=−1,n=−5, then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let a=j10E then, j10B=a+5a−6j10C=a−1−4a−1−3j10D=a−5+20a−5+9 as well as S3, j10A=j10B+42j10B+8=j10C+52j10C+6=j10D+1j10D−2 which implies the linear relation between 5 moonshine functions,
j10A+2j10E=j10B+j10C+j10D+8
III. Special property. As mentioned in the Intro, there are special triples {a′,b′,c′} of level N=6,10,12,18,30 that, using a common formula, can generate another triple {a′,b′,c′} of level 2N. For N=10,
a′(τ)=−a(12+τ)=(d1d4d10d2d5d20)2=j20C(τ)b′(τ)=−b(12+τ)=(d1d4d1010d22d55d520)c′(τ)=−c(12+τ)=(d82d31d34d5d20) such that a′+1=b′,a′+5=c′.
II. Relations: Let a,b,c as defined above and m=−1,n=−5, then the system S1 in the introduction explains all four trinomial identities of level 10 (denoted t10) found by Somos. For S2, let a=j10E then, j10B=a+5a−6j10C=a−1−4a−1−3j10D=a−5+20a−5+9 as well as S3, j10A=j10B+42j10B+8=j10C+52j10C+6=j10D+1j10D−2 which implies the linear relation between 5 moonshine functions,
j10A+2j10E=j10B+j10C+j10D+8
III. Special property. As mentioned in the Intro, there are special triples {a′,b′,c′} of level N=6,10,12,18,30 that, using a common formula, can generate another triple {a′,b′,c′} of level 2N. For N=10,
a′(τ)=−a(12+τ)=(d1d4d10d2d5d20)2=j20C(τ)b′(τ)=−b(12+τ)=(d1d4d1010d22d55d520)c′(τ)=−c(12+τ)=(d82d31d34d5d20) such that a′+1=b′,a′+5=c′.
Level 8
I. Moonshine functions: Define, a(τ)=(d34d2d28)4,b(τ)=(d21d4d2d28)2,c(τ)=(d52d21d4d28)2 then, (d34d2d28)4−4=(d21d4d2d28)2(d34d2d28)4+4=(d52d21d4d28)2so a-4 = b,\, a+4 = c. This a,b,c also belong to the class where a(\tau) = -a\big(\tfrac12+\tau\big) and b(\tau) = -c\big(\tfrac12+\tau\big). We then have the six level 8 moonshine functions,
\qquad\begin{align} j_{8A}(\tau) &= \left(\frac{d_2\, d_4}{d_1\, d_8}\right)^{8}\;=\;\frac{ac}b\\ j_{8B}(\tau) &= \left(\frac{d_4^2}{d_2\, d_8}\right)^{12}=\sqrt{j_{4A}(2\tau)}\\ j_{8C}(\tau) &= \sqrt{j_{4B}(\tau)}\,=\,\sqrt[4]{j_{2A}(2\tau)}\\ &= \sqrt{j_{8A}(\tau)}-\frac{4}{\sqrt{j_{8A}(\tau)}}\\ j_{8D}(\tau) &= \left(\frac{d_2}{d_8}\right)^{4}\\ \color{red}{j_{8E}}(\tau) &= \left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4}\;=\; a\\ j_{8F}(\tau) &= \left(\frac{d_4}{d_8}\right)^{6} \end{align} Example: \quad j_{8C}\big(\tfrac14\sqrt{-58}\big) = 396
II. Relations: Using S1, this a,b,c explains all of the four trinomial identities of level 8 (prefixed with t8) by Somos. They also obey the relation,
\left(\frac{bc}{a}+64\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad In general, \left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad where b-c=8. Or in terms of named functions of level 4 and 8,
\quad\quad j_{4B}\;+\;2\color{red}{j_{8E}} = j_{4D}+j_{8A}+j_{8A'} where, j_{8A'}(\tau) = -j_{8A}\big(\tfrac12+\tau\big) = \left(\frac{d_1\, d_4^2}{d_2^2\, d_8}\right)^{8}
\qquad\begin{align} j_{8A}(\tau) &= \left(\frac{d_2\, d_4}{d_1\, d_8}\right)^{8}\;=\;\frac{ac}b\\ j_{8B}(\tau) &= \left(\frac{d_4^2}{d_2\, d_8}\right)^{12}=\sqrt{j_{4A}(2\tau)}\\ j_{8C}(\tau) &= \sqrt{j_{4B}(\tau)}\,=\,\sqrt[4]{j_{2A}(2\tau)}\\ &= \sqrt{j_{8A}(\tau)}-\frac{4}{\sqrt{j_{8A}(\tau)}}\\ j_{8D}(\tau) &= \left(\frac{d_2}{d_8}\right)^{4}\\ \color{red}{j_{8E}}(\tau) &= \left(\frac{d_4^3}{d_2\, d_8^2}\right)^{4}\;=\; a\\ j_{8F}(\tau) &= \left(\frac{d_4}{d_8}\right)^{6} \end{align} Example: \quad j_{8C}\big(\tfrac14\sqrt{-58}\big) = 396
II. Relations: Using S1, this a,b,c explains all of the four trinomial identities of level 8 (prefixed with t8) by Somos. They also obey the relation,
\left(\frac{bc}{a}+64\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad In general, \left(\frac{bc}{a}+(b-c)^2\frac{a}{bc}\right)+2a = \frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}\qquad where b-c=8. Or in terms of named functions of level 4 and 8,
\quad\quad j_{4B}\;+\;2\color{red}{j_{8E}} = j_{4D}+j_{8A}+j_{8A'} where, j_{8A'}(\tau) = -j_{8A}\big(\tfrac12+\tau\big) = \left(\frac{d_1\, d_4^2}{d_2^2\, d_8}\right)^{8}
Level 6
I. Moonshine functions: Let, a(\tau)=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b(\tau)= \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c(\tau)=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right) then, \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right) So m = -1,\,n=-9, and we get the moonshine functions, \begin{align}
j_{6A}(\tau) &= j_{6D}(\tau)+\frac{9^2}{j_{6D}(\tau)}+18\\
j_{6B}(\tau) &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} \\
j_{6C}(\tau) &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} \\
j_{6D}(\tau) &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} \\
j_{6E}(\tau) &= \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4 = a\\
j_{6F}(\tau) &=\;\left(\frac{d_3}{d_6}\right)^{8}\\
\end{align} Example: \; j_{6A}\left(\sqrt{-17/6}\right) = 198^2
II. Relations: Let a,b,c as defined above and m=-1,\,n=-9, then the system S1 given in the Introduction,
\begin{align} a+m &= b\\ a+n &= c\\ a(m-n)+bn &=cm\\ b-c &= (m-n)\end{align} explains all four trinomial identities of level 6 (with prefix t6) found by Somos. For S2, let a=\color{red}{j_{6E}} then,
\begin{align} j_{6B}(\tau) &= a-9+\frac{72}{a-9}+17\\ j_{6C}(\tau) &= a+\frac{9}{a}-10\\ j_{6D}(\tau) &= a-1-\frac{8}{a-1}-7\\ \end{align} as well as S3, j_{6A} = j_{6B}+\frac{1}{j_{6B}}+2 = j_{6C}+\frac{8^2}{j_{6C}}+20 = j_{6D}+\frac{9^2}{j_{6D}}+18 which implies the linear relation between 5 moonshine functions,
j_{6A}+2\color{red}{j_{6E}} = j_{6B}+j_{6C}+j_{6D}+20These are general phenomena but, in levels 6, 10, 12, 18, 30, the five functions are ALL moonshine.
III. Special property. As mentioned in the Intro, there are special triples {a',b',c'} of level N = 6, 10, 12, 18, 30 that using a common formula can generate another triple {a',b',c'} of level 2N. For N = 6,
\begin{align} a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6}{d_2\,d_3\,d_{12}}\right)^4 = j_{12B}(\tau)\\ b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6^6}{d_2^2\,d_3^3\,d_{12}^3}\right)^3\\ c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{14}}{d_1^5\,d_3\,d_4^5\,d_6^2\,d_{12}}\right) \end{align} such that a'+1=b',\; a'+9=c'.
II. Relations: Let a,b,c as defined above and m=-1,\,n=-9, then the system S1 given in the Introduction,
\begin{align} a+m &= b\\ a+n &= c\\ a(m-n)+bn &=cm\\ b-c &= (m-n)\end{align} explains all four trinomial identities of level 6 (with prefix t6) found by Somos. For S2, let a=\color{red}{j_{6E}} then,
\begin{align} j_{6B}(\tau) &= a-9+\frac{72}{a-9}+17\\ j_{6C}(\tau) &= a+\frac{9}{a}-10\\ j_{6D}(\tau) &= a-1-\frac{8}{a-1}-7\\ \end{align} as well as S3, j_{6A} = j_{6B}+\frac{1}{j_{6B}}+2 = j_{6C}+\frac{8^2}{j_{6C}}+20 = j_{6D}+\frac{9^2}{j_{6D}}+18 which implies the linear relation between 5 moonshine functions,
j_{6A}+2\color{red}{j_{6E}} = j_{6B}+j_{6C}+j_{6D}+20These are general phenomena but, in levels 6, 10, 12, 18, 30, the five functions are ALL moonshine.
III. Special property. As mentioned in the Intro, there are special triples {a',b',c'} of level N = 6, 10, 12, 18, 30 that using a common formula can generate another triple {a',b',c'} of level 2N. For N = 6,
\begin{align} a'(\tau) &= -a(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6}{d_2\,d_3\,d_{12}}\right)^4 = j_{12B}(\tau)\\ b'(\tau) &= -b(\tfrac12+\tau) = \left(\frac{d_1\,d_4\,d_6^6}{d_2^2\,d_3^3\,d_{12}^3}\right)^3\\ c'(\tau) &= -c(\tfrac12+\tau) = \left(\frac{d_2^{14}}{d_1^5\,d_3\,d_4^5\,d_6^2\,d_{12}}\right) \end{align} such that a'+1=b',\; a'+9=c'.
Level 4
I. Moonshine function Define, a(\tau) = \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8,\quad b(\tau) = \left(\frac{d_1}{d_4}\right)^8+8 then \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8=\left(\frac{d_1}{d_4}\right)^8+8 Equivalently, using the reciprocal of a(\tau), \left(\frac{\sqrt2\,d_1\,d_4^2}{d_2^3}\right)^8+\left(\frac{d_1^2\,d_4}{d_2^3}\right)^8 = 1 which are just versions of the single trinomial identity for level 4. We also have the four level 4 moonshine functions, \qquad\qquad\begin{align}
j_{4A}(\tau) &= \left(\left(\frac{d_1}{d_4}\right)^4+16\left(\frac{d_4}{d_1}\right)^4\right)^2=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24}\\
j_{4B}(\tau) &=\left(\left(\frac{d_2\,d_4}{d_{1}\;d_8}\right)^4-4\left(\frac{d_{1}\;d_8}{d_2\,d_4}\right)^4\right)^2 = \sqrt{j_{2A}(2\tau)} \\
&=j_{4D}(\tau)+\frac{2^6}{j_{4D}(\tau)} \\
\color{red}{j_{4C}}(\tau) &= \left(\frac{d_1}{d_4}\right)^8+8\; = \;b(\tau)\\
j_{4D}(\tau) &= \left(\frac{d_2}{d_4}\right)^{12}
\end{align} Note that the function j_{4C}(\tau) = \frac{16}{\lambda(2\tau)}-8=\left(\frac{d_1}{d_4}\right)^8+8 with modular lambda function \lambda(\tau) is a normalized Hauptmodul.
II. Comparisons: Notice that,j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8,\quad j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2The last two will play analogous roles for levels 8 and 16. Also, j_{4A} is remarkably a 24th power. Versions appear in levels that are 4m divisors of 24, j_{4A}=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24},\quad j_{8B}=\left(\frac{d_4^2}{d_2\,d_8}\right)^{12},\quad j_{12D}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^{8},\quad j_{24E}=\left(\frac{d_{12}^2}{d_6\,d_{24}}\right)^{4} Similarly, versions of j_{2B} appear in levels that are 2m divisors of 24, j_{2B}(\tau) = \left(\frac{d_1}{d_2}\right)^{24},\quad j_{4D}(\tau) = \left(\frac{d_2}{d_4}\right)^{12},\quad j_{6F}(\tau) = \left(\frac{d_3}{d_6}\right)^{8},\\ j_{8F}(\tau) = \left(\frac{d_4}{d_8}\right)^{6},\quad j_{12J}(\tau) = \left(\frac{d_6}{d_{12}}\right)^{4},\quad j_{24J}(\tau) = \left(\frac{d_{12}}{d_{24}}\right)^{2}
II. Comparisons: Notice that,j_{4C} =\left(\frac{d_2^3}{d_1\,d_4^2}\right)^8-8,\quad j_{8E} =\left(\frac{d_4^3}{d_2\,d_8^2}\right)^4,\quad j_{16B} =\left(\frac{d_8^3}{d_4\,d_{16}^2}\right)^2The last two will play analogous roles for levels 8 and 16. Also, j_{4A} is remarkably a 24th power. Versions appear in levels that are 4m divisors of 24, j_{4A}=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24},\quad j_{8B}=\left(\frac{d_4^2}{d_2\,d_8}\right)^{12},\quad j_{12D}=\left(\frac{d_6^2}{d_3\,d_{12}}\right)^{8},\quad j_{24E}=\left(\frac{d_{12}^2}{d_6\,d_{24}}\right)^{4} Similarly, versions of j_{2B} appear in levels that are 2m divisors of 24, j_{2B}(\tau) = \left(\frac{d_1}{d_2}\right)^{24},\quad j_{4D}(\tau) = \left(\frac{d_2}{d_4}\right)^{12},\quad j_{6F}(\tau) = \left(\frac{d_3}{d_6}\right)^{8},\\ j_{8F}(\tau) = \left(\frac{d_4}{d_8}\right)^{6},\quad j_{12J}(\tau) = \left(\frac{d_6}{d_{12}}\right)^{4},\quad j_{24J}(\tau) = \left(\frac{d_{12}}{d_{24}}\right)^{2}
Levels 1, 2, 3
Recall our notation for the Dedekind eta function, \color{blue}{d_m = \eta(m\tau)} where we suppress the argument \tau for ease of notation. Then we have the following moonshine functions:
Level 1
\begin{align} j_{1A}(\tau) &= \left(\left(\frac{d_1}{d_2}\right)^8+2^8\left(\frac{d_2}{d_1}\right)^{16}\right)^3 \end{align} This is the well-known j-function j(\tau). In this form, it is easily seen to be a cube. Its most famous value is j_{1A}\big(\tfrac{1+\sqrt{-163}}{2}\big)=-640320^3 .
Level 2
\begin{align} j_{2A}(\tau) &=\left(\left(\frac{d_1}{d_2}\right)^{12}+2^6\left(\frac{d_2}{d_1}\right)^{12}\right)^2\\ j_{2B}(\tau) &= \left(\frac{d_1}{d_2}\right)^{24} \end{align} Alternatively, j_{2A}(\tau) =\left(\left(\frac{d_1\,d_2}{d_{1/2}\;d_4}\right)^4-4\left(\frac{d_{1/2}\;d_4}{d_1\,d_2}\right)^4\right)^4 In this form, j_{2A} is seen to be a 4th power. The function doesn't have a name, but is notable for the value j_{2A}(\tfrac12\sqrt{-58})=396^4 in Ramanujan's well-known pi formula. Note that,
j_{4B}(\tau) = \sqrt{j_{2A}(2\tau)}\\j_{8C}(\tau) = \sqrt[4]{j_{2A}(2\tau)}
Level 3
\begin{align} j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\ j_{3B}(\tau) &= \left(\frac{d_1}{d_3}\right)^{12}\\ j_{3C}(\tau) &= \sqrt[3]{j_{1A}(3\tau)} \end{align} Alternatively, j_{3A}(\tau) = \left(\frac{d_1^2}{d_3^2}+\frac{9\,d_9^3}{d_1\,d_3^2}\right)^6 In this form, j_{3A} is seen to be also a cube. An example is j_{3A}\big(\tfrac{1+\sqrt{-89/3}}{2}\big)=-300^3 which can also be used in a Ramanujan-type pi formula. Notice that j_{3C} is a cube root.
Level 1
\begin{align} j_{1A}(\tau) &= \left(\left(\frac{d_1}{d_2}\right)^8+2^8\left(\frac{d_2}{d_1}\right)^{16}\right)^3 \end{align} This is the well-known j-function j(\tau). In this form, it is easily seen to be a cube. Its most famous value is j_{1A}\big(\tfrac{1+\sqrt{-163}}{2}\big)=-640320^3 .
Level 2
\begin{align} j_{2A}(\tau) &=\left(\left(\frac{d_1}{d_2}\right)^{12}+2^6\left(\frac{d_2}{d_1}\right)^{12}\right)^2\\ j_{2B}(\tau) &= \left(\frac{d_1}{d_2}\right)^{24} \end{align} Alternatively, j_{2A}(\tau) =\left(\left(\frac{d_1\,d_2}{d_{1/2}\;d_4}\right)^4-4\left(\frac{d_{1/2}\;d_4}{d_1\,d_2}\right)^4\right)^4 In this form, j_{2A} is seen to be a 4th power. The function doesn't have a name, but is notable for the value j_{2A}(\tfrac12\sqrt{-58})=396^4 in Ramanujan's well-known pi formula. Note that,
j_{4B}(\tau) = \sqrt{j_{2A}(2\tau)}\\j_{8C}(\tau) = \sqrt[4]{j_{2A}(2\tau)}
Level 3
\begin{align} j_{3A}(\tau) &= \left(\left(\frac{d_1}{d_3}\right)^6+3^3\left(\frac{d_3}{d_1}\right)^6\right)^2\\ j_{3B}(\tau) &= \left(\frac{d_1}{d_3}\right)^{12}\\ j_{3C}(\tau) &= \sqrt[3]{j_{1A}(3\tau)} \end{align} Alternatively, j_{3A}(\tau) = \left(\frac{d_1^2}{d_3^2}+\frac{9\,d_9^3}{d_1\,d_3^2}\right)^6 In this form, j_{3A} is seen to be also a cube. An example is j_{3A}\big(\tfrac{1+\sqrt{-89/3}}{2}\big)=-300^3 which can also be used in a Ramanujan-type pi formula. Notice that j_{3C} is a cube root.
Introduction
I. Moonshine functions: Dedekind eta quotients are very interesting, especially the class called moonshine functions (terminology after McKay, Ronan, Carnahan, et al). Conway and Norton found in 1979 that these functions span a linear space of 172-9 = \color{red}{163} dimensions which McKay, in an email, remarked was a "delicious coincidence". There is growing evidence that each of these 163 dimensions contain a Ramanujan-type pi formula family, hence my interest in these functions.
II. Algebraic identities: First, some basics which will be useful later. Let a+m = b,\; a+n = c for some rational m,n. Then the four trinomial identities of S1 are satisfied,
\begin{align} a+m &= b\\ a+n &= c\\ a(m-n)+bn &=cm\\ b-c &= (m-n)\end{align} These are responsible for quadruples of some trinomial identities found by Somos, as well as all for levels 6, 8, 10 (which only have four trinomial identities per level and prefixed by t6, t8, t10, respectively). The identities of S2 are also satisfied, \begin{align} x_1 &= \frac{bc}{a} = a+\frac{mn}{a}+(m+n)\\ x_2 &= \frac{ac}{b} = b+\frac{m(m-n)}{b}+(-2m+n)\\ x_3 &= \frac{ab}{c} = c-\frac{n(m-n)}{c}+(m-2n)\end{align}This implies S3, x_0 = x_1 + \frac{(m-n)^2}{x_1} = x_2 + \frac{n^2}{x_2} + 2m= x_3 + \frac{m^2}{x_3} + 2n and the linear relation S4, (x_0 + r)+2a = x_1+x_2+x_3+r where r may be chosen such that x_0 +r has nice properties, such as being a square. (Note that if m = -n, then one can just set r = 0 as x_0 is already a square, a special case that will be relevant in certain levels.) These identities are valid for any a,b,c but become exceedingly interesting when all three a,b,c are eta quotients.
III. Example: Given the Dedekind eta function, \color{blue}{d_m = \eta(m\tau)} where we suppress the argument \tau for ease of notation. Let,
a=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b = \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right) then, \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right) So m = -1,\,n=-9, and m-n = 8, and from S_2 we get, \begin{align} x_1 &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} = a+\frac{9}{a}-10\\ x_2 &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} = b-\frac{8}{b}-7\\ x_3 &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} = c+\frac{72}{c}+17\end{align} where a,b,c can be expressed in terms of each other. To make x_2 + \frac{9^2}{x_2} -2 a square, add r=20 and we have the linear relation,
\left(\sqrt{x_2} + \frac{9}{\sqrt{x_2}}\right)^2+2a = x_1+x_2+x_3+20 This is equivalent to one of the 9 linear dependencies in Conway and Norton’s 1979 paper, Monstrous Moonshine. For levels N = 6, 12, 10, 18, 30, those dependencies are, \begin{align} j_{6A}+2\color{red}{j_{6E}} &= j_{6B}+j_{6C}+j_{6D}+20\\ j_{10A}+2\color{red}{j_{10E}} &= j_{10B}+j_{10C}+j_{10D}+8\\ j_{12A}+2\color{red}{j_{12I}}&= j_{12A}+j_{12B}+j_{12E}+8\\ j_{18B}+2\color{red}{j_{18D}} &= j_{18A}+j_{18C}+j_{18E}+4\\ j_{30B}+2\color{red}{j_{30G}}&= j_{30A}+j_{30C}+j_{30F}+5 \end{align} Later we shall see this generalizes (but not completely) to other levels.
IV. Special property: The relevant triples {a,b,c} of level N = 6, 10, 12, 18, 30 connected to the linear dependencies above are rather special since, using a common formula, they can generate another triple {a',b',c'} of level 2N namely,
\begin{align} a'(\tau) &= \sqrt{\frac{a(2\tau)\,c(\tau)}{b(\tau)}}\\ b'(\tau) &= \frac{b(2\tau)}{b(\tau)}\\ c'(\tau) &= \frac{a(\tau)\,c(2\tau)}{a'(\tau)\,b(\tau)} \end{align} where a+m = b,\quad a+n = c\\a'-m = b',\quad a'-n = c'So the highest it can generate is a level 30\times2=60 identity. However, for particular triples of level N = 10, 12, 18, one can also generate a level 4N,
\begin{align} a'(\tau) &= \sqrt{\frac{a(4\tau)}{a(2\tau)}}\\ b'(\tau) &= a'(\tau)\,\sqrt[4]{\frac{b^2(2\tau)\,c(\tau)}{a(4\tau)\,b(\tau)}}\\ c'(\tau) &= a'(\tau)\,\sqrt[4]{\frac{b(\tau)\,c^2(2\tau)}{a(4\tau)\,c(\tau)}}\end{align} where a'\pm1 = b',\quad a'\mp1 = c' and the \pm sign chosen appropriately. Thus, it can generate {a',b',c'} with level 18\times4=72 which is the highest I've observed. We will explicitly give these in the other posts.
II. Algebraic identities: First, some basics which will be useful later. Let a+m = b,\; a+n = c for some rational m,n. Then the four trinomial identities of S1 are satisfied,
\begin{align} a+m &= b\\ a+n &= c\\ a(m-n)+bn &=cm\\ b-c &= (m-n)\end{align} These are responsible for quadruples of some trinomial identities found by Somos, as well as all for levels 6, 8, 10 (which only have four trinomial identities per level and prefixed by t6, t8, t10, respectively). The identities of S2 are also satisfied, \begin{align} x_1 &= \frac{bc}{a} = a+\frac{mn}{a}+(m+n)\\ x_2 &= \frac{ac}{b} = b+\frac{m(m-n)}{b}+(-2m+n)\\ x_3 &= \frac{ab}{c} = c-\frac{n(m-n)}{c}+(m-2n)\end{align}This implies S3, x_0 = x_1 + \frac{(m-n)^2}{x_1} = x_2 + \frac{n^2}{x_2} + 2m= x_3 + \frac{m^2}{x_3} + 2n and the linear relation S4, (x_0 + r)+2a = x_1+x_2+x_3+r where r may be chosen such that x_0 +r has nice properties, such as being a square. (Note that if m = -n, then one can just set r = 0 as x_0 is already a square, a special case that will be relevant in certain levels.) These identities are valid for any a,b,c but become exceedingly interesting when all three a,b,c are eta quotients.
III. Example: Given the Dedekind eta function, \color{blue}{d_m = \eta(m\tau)} where we suppress the argument \tau for ease of notation. Let,
a=\left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4,\quad b = \left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3,\quad c=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right) then, \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-1=\left(\frac{d_2\,d_3^3}{d_1\,d_6^3}\right)^3\\ \left(\frac{d_2^2\,d_3}{d_1\,d_6^2}\right)^4-9=\left(\frac{d_1^5\,d_3}{d_2\,d_6^5}\right) So m = -1,\,n=-9, and m-n = 8, and from S_2 we get, \begin{align} x_1 &= \left(\frac{d_1\,d_3}{d_2\,d_6}\right)^6 = \frac{bc}{a} = a+\frac{9}{a}-10\\ x_2 &= \left(\frac{d_1\,d_2}{d_3\,d_6}\right)^4 = \frac{ac}{b} = b-\frac{8}{b}-7\\ x_3 &= \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12} = \frac{ab}{c} = c+\frac{72}{c}+17\end{align} where a,b,c can be expressed in terms of each other. To make x_2 + \frac{9^2}{x_2} -2 a square, add r=20 and we have the linear relation,
\left(\sqrt{x_2} + \frac{9}{\sqrt{x_2}}\right)^2+2a = x_1+x_2+x_3+20 This is equivalent to one of the 9 linear dependencies in Conway and Norton’s 1979 paper, Monstrous Moonshine. For levels N = 6, 12, 10, 18, 30, those dependencies are, \begin{align} j_{6A}+2\color{red}{j_{6E}} &= j_{6B}+j_{6C}+j_{6D}+20\\ j_{10A}+2\color{red}{j_{10E}} &= j_{10B}+j_{10C}+j_{10D}+8\\ j_{12A}+2\color{red}{j_{12I}}&= j_{12A}+j_{12B}+j_{12E}+8\\ j_{18B}+2\color{red}{j_{18D}} &= j_{18A}+j_{18C}+j_{18E}+4\\ j_{30B}+2\color{red}{j_{30G}}&= j_{30A}+j_{30C}+j_{30F}+5 \end{align} Later we shall see this generalizes (but not completely) to other levels.
IV. Special property: The relevant triples {a,b,c} of level N = 6, 10, 12, 18, 30 connected to the linear dependencies above are rather special since, using a common formula, they can generate another triple {a',b',c'} of level 2N namely,
\begin{align} a'(\tau) &= \sqrt{\frac{a(2\tau)\,c(\tau)}{b(\tau)}}\\ b'(\tau) &= \frac{b(2\tau)}{b(\tau)}\\ c'(\tau) &= \frac{a(\tau)\,c(2\tau)}{a'(\tau)\,b(\tau)} \end{align} where a+m = b,\quad a+n = c\\a'-m = b',\quad a'-n = c'So the highest it can generate is a level 30\times2=60 identity. However, for particular triples of level N = 10, 12, 18, one can also generate a level 4N,
\begin{align} a'(\tau) &= \sqrt{\frac{a(4\tau)}{a(2\tau)}}\\ b'(\tau) &= a'(\tau)\,\sqrt[4]{\frac{b^2(2\tau)\,c(\tau)}{a(4\tau)\,b(\tau)}}\\ c'(\tau) &= a'(\tau)\,\sqrt[4]{\frac{b(\tau)\,c^2(2\tau)}{a(4\tau)\,c(\tau)}}\end{align} where a'\pm1 = b',\quad a'\mp1 = c' and the \pm sign chosen appropriately. Thus, it can generate {a',b',c'} with level 18\times4=72 which is the highest I've observed. We will explicitly give these in the other posts.
Sunday, April 7, 2019
Entry 39 Using the Jacobi and Borwein theta functions
Let \color{red}{q = e^{2\pi i\tau}}. Given \tau = \sqrt{-n} or \tau = \frac{1+\sqrt{-n}}{2}, then the solutions \alpha, \beta,\gamma to the following equations,
\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-\alpha\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,\alpha\big)}\,i=\color{red}{\sqrt{4}\,\tau}\frac{_2F_1\big(\tfrac13,\tfrac23;1;\,1-\beta\big)}{_2F_1\big(\tfrac13,\tfrac23;1;\,\beta\big)}\,i=\color{red}{\sqrt{3}\,\tau}\frac{_2F_1\big(\tfrac14,\tfrac34;1;\,1-\gamma\big)}{_2F_1\big(\tfrac14,\tfrac34;1;\,\gamma\big)}\,i=\color{red}{\sqrt{2}\,\tau} are given by, \;\alpha =\frac{16}{u_1^8+16} = \left(\frac{\sqrt2 }{u_2}\right)^8 = \left(\frac{\vartheta_2(q)}{\vartheta_3(q)}\right)^4\beta = \frac{27}{v_1^{12}+27} = \left(\frac{3}{v_2^3+3}\right)^3 = \left(\frac{c(q)}{a(q)}\right)^3\gamma = \frac{64}{w_1^{24}+64} = \left(\frac{8}{w_2^8+8}\right)^2 = \left(\frac{f(q)}{d(q)}\right)^2 where, \quad u_1 = \frac{\eta(\tau)}{\eta(4\tau)},\quad u_2 = \frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}v_1 = \frac{\eta(\tau)}{\eta(3\tau)},\quad v_2 = \frac{\eta(\tau/3)}{\eta(3\tau)}w_1 = \frac{\eta(\tau)}{\eta(2\tau)},\quad w_2 = \frac{\eta(\tau/2)}{\eta(2\tau)} and where the functions of \color{red}{q = e^{2\pi i\tau}} are the Jacobi and Borwein theta functions discussed in Entry 38. Also, \alpha= \lambda(2\tau) with the modular lambda function \lambda(\tau).
\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-\alpha\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,\alpha\big)}\,i=\color{red}{\sqrt{4}\,\tau}\frac{_2F_1\big(\tfrac13,\tfrac23;1;\,1-\beta\big)}{_2F_1\big(\tfrac13,\tfrac23;1;\,\beta\big)}\,i=\color{red}{\sqrt{3}\,\tau}\frac{_2F_1\big(\tfrac14,\tfrac34;1;\,1-\gamma\big)}{_2F_1\big(\tfrac14,\tfrac34;1;\,\gamma\big)}\,i=\color{red}{\sqrt{2}\,\tau} are given by, \;\alpha =\frac{16}{u_1^8+16} = \left(\frac{\sqrt2 }{u_2}\right)^8 = \left(\frac{\vartheta_2(q)}{\vartheta_3(q)}\right)^4\beta = \frac{27}{v_1^{12}+27} = \left(\frac{3}{v_2^3+3}\right)^3 = \left(\frac{c(q)}{a(q)}\right)^3\gamma = \frac{64}{w_1^{24}+64} = \left(\frac{8}{w_2^8+8}\right)^2 = \left(\frac{f(q)}{d(q)}\right)^2 where, \quad u_1 = \frac{\eta(\tau)}{\eta(4\tau)},\quad u_2 = \frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}v_1 = \frac{\eta(\tau)}{\eta(3\tau)},\quad v_2 = \frac{\eta(\tau/3)}{\eta(3\tau)}w_1 = \frac{\eta(\tau)}{\eta(2\tau)},\quad w_2 = \frac{\eta(\tau/2)}{\eta(2\tau)} and where the functions of \color{red}{q = e^{2\pi i\tau}} are the Jacobi and Borwein theta functions discussed in Entry 38. Also, \alpha= \lambda(2\tau) with the modular lambda function \lambda(\tau).
Wednesday, March 27, 2019
Entry 38 The Jacobi, Borwein, and derived Jacobi theta functions
For consistency, let q be the nome's square q=e^{2\pi i \tau}\,throughout.
I. The null Jacobi theta functions (with z=0) are,
\qquad\qquad\quad \vartheta_3(q) = \sum_{m=-\infty}^{\infty}q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}+\frac{4\,\eta^2(16\tau)}{\eta(8\tau)}=\frac{\eta^5(2\tau)}{\eta^2(\tau)\eta^2(4\tau)} \vartheta_4(q) = \sum_{m=-\infty}^{\infty}(-1)^n\,q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)} \vartheta_2(q) = \sum_{m=-\infty}^{\infty}q^{(n+1/2)^2} = \frac{2\,\eta^2(4\tau)}{\eta(2\tau)}
II. The Borwein cubic theta functions are,
a(q)=\sum_{m,n=-\infty}^{\infty}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)} +\frac{9\,\eta^3(9\tau)}{\eta(3\tau)} b(q)=\sum_{m,n=-\infty}^{\infty}\zeta^{m-n}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)}\quad\quad \quad c(q)=\sum_{m,n=-\infty}^{\infty}q^{(m+1/3)^2+(m+1/3)(n+1/3)+(n+1/3)^2} = \frac{3\,\eta^3(3\tau)}{\eta(\tau)}
where \zeta =e^{2\pi i/3}.
\qquad\qquad\quad \vartheta_3(q) = \sum_{m=-\infty}^{\infty}q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}+\frac{4\,\eta^2(16\tau)}{\eta(8\tau)}=\frac{\eta^5(2\tau)}{\eta^2(\tau)\eta^2(4\tau)} \vartheta_4(q) = \sum_{m=-\infty}^{\infty}(-1)^n\,q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)} \vartheta_2(q) = \sum_{m=-\infty}^{\infty}q^{(n+1/2)^2} = \frac{2\,\eta^2(4\tau)}{\eta(2\tau)}
II. The Borwein cubic theta functions are,
a(q)=\sum_{m,n=-\infty}^{\infty}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)} +\frac{9\,\eta^3(9\tau)}{\eta(3\tau)} b(q)=\sum_{m,n=-\infty}^{\infty}\zeta^{m-n}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)}\quad\quad \quad c(q)=\sum_{m,n=-\infty}^{\infty}q^{(m+1/3)^2+(m+1/3)(n+1/3)+(n+1/3)^2} = \frac{3\,\eta^3(3\tau)}{\eta(\tau)}
where \zeta =e^{2\pi i/3}.
III. The derived Jacobi theta functions are,
\qquad\qquad d(q)= \vartheta_4^4(q)+2\,\vartheta_2^4(q) = \frac{\eta^8(\tau)}{\eta^4(2\tau)} +\frac{32\,\eta^8(4\tau)}{\eta^4(2\tau)} \quad e(q)= \vartheta_4^4(q)= \frac{\eta^8(\tau)}{\eta^4(2\tau)} \quad f(q)= \tfrac12 \vartheta_2^4(q^{1/2})= \frac{8\,\eta^8(2\tau)}{\eta^4(\tau)} These obey the beautiful relations, \vartheta_3^4(q)=\vartheta_4^4(q)+\vartheta_2^4(q) a^3(q)=b^3(q)+c^3(q)d^2(q)=e^2(q)+f^2(q) As well as, \vartheta_3(q)\vartheta_3(q^3)=\vartheta_4(q)\vartheta_4(q^3)+\vartheta_2(q)\vartheta_2(q^3) a(q)a(q^2)=b(q)b(q^2)+c(q)c(q^2)d(q)d(q)=e(q)e(q)+f(q)f(q) the last of which naturally leads to the third one above. Also, \vartheta_3(q^4)=\vartheta_4(q)+\vartheta_2(q^4) a(q^3)=b(q)+c(q^3)d(q^2)=e(q)+f(q^2)
\qquad\qquad d(q)= \vartheta_4^4(q)+2\,\vartheta_2^4(q) = \frac{\eta^8(\tau)}{\eta^4(2\tau)} +\frac{32\,\eta^8(4\tau)}{\eta^4(2\tau)} \quad e(q)= \vartheta_4^4(q)= \frac{\eta^8(\tau)}{\eta^4(2\tau)} \quad f(q)= \tfrac12 \vartheta_2^4(q^{1/2})= \frac{8\,\eta^8(2\tau)}{\eta^4(\tau)} These obey the beautiful relations, \vartheta_3^4(q)=\vartheta_4^4(q)+\vartheta_2^4(q) a^3(q)=b^3(q)+c^3(q)d^2(q)=e^2(q)+f^2(q) As well as, \vartheta_3(q)\vartheta_3(q^3)=\vartheta_4(q)\vartheta_4(q^3)+\vartheta_2(q)\vartheta_2(q^3) a(q)a(q^2)=b(q)b(q^2)+c(q)c(q^2)d(q)d(q)=e(q)e(q)+f(q)f(q) the last of which naturally leads to the third one above. Also, \vartheta_3(q^4)=\vartheta_4(q)+\vartheta_2(q^4) a(q^3)=b(q)+c(q^3)d(q^2)=e(q)+f(q^2)
The Jacobi thetas can be expressed in terms of each other,
\vartheta_3(q) = \vartheta_4(q)+2\,\vartheta_2(q^4) \vartheta_4(q) = 2\,\vartheta_3(q^4)-\vartheta_3(q) \vartheta_2(q) = \vartheta_3(q^{1/4})-\vartheta_3(q) similarly for the Borwein thetas, a(q) = b(q) + 3\,c(q^3) 2\,b(q) = 3a(q^3)-a(q) 2\,c(q) = a(q^{1/3})-a(q) and also similarly for the derived Jacobi thetas, d(q) = e(q) + 4\,f(q^2) 3\,e(q) = 4d(q^2)-d(q) 3\,f(q) = d(q^{1/2})-d(q)Furthermore, we have the similar, \big(\vartheta_3(q)\big)^2 = 1+4\sum_{n=0}^\infty\left(\frac{q^{4n+1}}{1-q^{4n+1}}-\frac{q^{4n+3}}{1-q^{4n+3}}\right) \quad\quad a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)
\vartheta_3(q) = \vartheta_4(q)+2\,\vartheta_2(q^4) \vartheta_4(q) = 2\,\vartheta_3(q^4)-\vartheta_3(q) \vartheta_2(q) = \vartheta_3(q^{1/4})-\vartheta_3(q) similarly for the Borwein thetas, a(q) = b(q) + 3\,c(q^3) 2\,b(q) = 3a(q^3)-a(q) 2\,c(q) = a(q^{1/3})-a(q) and also similarly for the derived Jacobi thetas, d(q) = e(q) + 4\,f(q^2) 3\,e(q) = 4d(q^2)-d(q) 3\,f(q) = d(q^{1/2})-d(q)Furthermore, we have the similar, \big(\vartheta_3(q)\big)^2 = 1+4\sum_{n=0}^\infty\left(\frac{q^{4n+1}}{1-q^{4n+1}}-\frac{q^{4n+3}}{1-q^{4n+3}}\right) \quad\quad a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)
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