The McKay-Thompson series of
Mathieu Moonshine modulo two
Thomas Creutzig∗, Gerald Höhn† and Tsuyoshi Miezaki‡§
Abstract
In this note, we describe the parity of the coefficients of the McKayThompson series of Mathieu moonshine. As an application, we prove
a conjecture of Cheng, Duncan and Harvey stated in connection with
Umbral moonshine for the case of Mathieu moonshine.
1
Introduction
In 2010, Eguchi, Ooguri, and Tachikawa [6] discovered a phenomenon connecting the Mathieu group M24 and the elliptic genus of a K3 surface. To
describe their observation, we introduce the following functions:
ϑ1 (z; τ ) = −
∞
X
1 2
+2πi(n+ 12 )(z+ 12 )
eπiτ (n+ 2 )
,
n=−∞
1
2πinz
ieπiz X
q 2 n(n+1)e
µ(z; τ ) =
(−1)n
,
ϑ1 (z; τ ) n∈Z
1 − q n e2πiz
∗
Technische Universität Darmstadt, Schloßgartenstrasse 7, 64289 Darmstadt, Germany.
email: [email protected]
†
Kansas State University, 138 Cardwell Hall, Manhattan, KS 66506-2602, USA. email:
[email protected]
‡
Systems Science and Information Studies, Faculty of Education, Art and Science, Yamagata University, 1-4-12 Kojirakawa, Yamagata 990-8560, Japan. email:
[email protected]
§
This work was supported by JSPS KAKENHI Grant Number 22840003, 24740031.
1
and we set Σ(τ ) as follows:
Σ(τ ) = −8(µ(1/2; τ ) + µ(τ /2; τ ) + µ((τ + 1)/2; τ )
1
= q − 8 (−2 + 90q + 462q 2 + 1540q 3 + 4554q 4 + 11592q 5 + 27830q 6 + · · · ).
Then the Mathieu moonshine phenomenon is that the first five coefficients
appearing in the Fourier expansion divided by 2,
{45, 231, 770, 2277, 5796},
are equal to dimensions of irreducible representations of M24 , and further
coefficients can be written as simple sums of dimensions of the irreducible
representations of M24 , for example 13915 = 3520 + 10395. The reason for
this mysterious phenomenon is still unknown.
In analogy to the Monstrous moonshine case [4], one can consider socalled McKay-Thompson series for any conjugacy class of M24 . We divide
the conjugacy classes into two types [7]:
type I 1A 2A 3A 5A 4B 7A 7B 8A 6A 11A 15A 15B 14A 14B 23A 23B
.
type II 12B 6B 4C 3B 2B 10A 21A 21B 4A 12A
Namely, type I classes contain a fixed point in their natural action on 24
points and the classes of type II cases do not. In [1, 9, 7], candidates for the
McKay-Thompson series for the Mathieu moonshine have been proposed. We
note that the McKay-Thompson series for the pairs, (7A, 7B), (15A, 15B),
(14A, 14B), (23A, 23B), (21A, 21B), are equal to each other. We list in the
appendix all the McKay-Thompson series.
We are using the functions
η(τ ) = q
1
24
∞
Y
(1 − q m ),
m=1
(N )
φ2 (τ )
24
d
=
q log
N − 1 dq
η(N τ )
η(τ )
∞
=1+
(N )
24 X
σ1 (k)(q k − N q N k ),
N − 1 k=0
where for N ≥ 2, φ2 (τ ) is the Eisenstein series of weight 2 for Γ0 (N ) [7].
Moreover, for the case 23A, we use the functions f (τ ) and g(τ ), which are
modular forms for Γ0 (23), explicitly given in [7].
2
The McKay-Thompson series for 7A is given as follows [7]:
(7)
1
φ (τ ) Σ(τ ) − 14 2 3
8
η(τ )
= − 2q −1/8 − q 7/8 + 4q 31/8 − 2q 47/8 + 2q 55/8 − 3q 63/8 + 6q 87/8 − 6q 103/8
Σ7A (τ ) =
− 4q 119/8 + 8q 143/8 − 6q 159/8 + 4q 167/8 − 7q 175/8 + 12q 199/8 − 10q 215/8
+ 2q 223/8 − 6q 231/8 + 18q 255/8 − 10q 271/8 + 6q 279/8 − 12q 287/8
+ 20q 311/8 − 18q 327/8 + 6q 335/8 − 13q 343/8 + 28q 367/8 + · · · .
We see that the coefficient of q n/8 in Σ7A (τ ) is odd if n = 7m2 , where m is
odd. In general we show:
Theorem 1.1. For a conjugacy class `X of M24 , the coefficient of q n/8 in
Σ`X (τ ) is odd if and only if `X ∈ {7A, 7B, 14A, 14B, 15A, 15B, 23A, 23B}
and n = `m2 , where m is odd or `X ∈ {21A, 21B} and n = `m2 , where m is
odd and m is not divisible by 3.
In this paper, we study the congruences of the Fourier coefficients of
the McKay-Thompson series modulo two. For congruences for other primes,
for Σ(τ ) we refer to the references [11, 13] and for the McKay-Thompson
series, the answer will be given in the near future by one of the authors [12].
The reason for considering the case of modulo two is that it explains the
appearance of certain irreducible representations of M24 . In [2], the authors
made the following conjecture which we state for the case of the Mathieu
moonshine only.
Conjecture 1.1 ([2], Conj. 5.11). Let n = `m2 ≡ 7 (mod 8). Then the M24 representation determined by the coefficients of q n/8 of the McKay-Thompson
series contains the following conjugate pairs of irreducible representations:
• For ` = 7, one of the pairs (χ3 , χ4 ), (χ12 , χ13 ) or (χ15 , χ16 );
• for ` = 15, the pair (χ5 , χ6 );
• for ` = 23, the pair (χ10 , χ11 ).
Here χi denotes the i-th irreducible representation as listed in the ATLAS
[3].
To see how this conjecture follows from our theorem and the Mathieu
moonshine conjecture, we formulate the latter conjecture as follows:
3
Conjecture 1.2. The McKay-Thompson series of
Mathieu moonshine deL∞
termine a virtual graded M24 -representation K = n=−1 Kn q n/8 . For n ≥ 0
the Kn are honest (and not only virtual) representations. Furthermore, each
irreducible representation λ occurs in Kn as member of a pair χ ⊕ χ̄.
This implies that if a irreducible representation λ is real, i.e. λ = λ̄, then
the multiplicity of λ in Kn is even.
Terry Gannon [8] has announced a proof of this conjecture.
Now we can can state the following corollary of our theorem.
Corollary 1.1. Assume that the Mathieu moonshine Conjecture 1.2 holds.
Let n = `m2 ≡ 7 (mod 8). Then the M24 -representation determined by
the coefficients of q n/8 of the McKay-Thompson series contains the following
conjugate pairs of irreducible representations with odd (and therefore positive)
multiplicity:
• For ` = 7, the total number of pairs (χ3 , χ4 ) and (χ12 , χ13 );
• for ` = 7 and m divisible by 3, the pair (χ15 , χ16 );
• for ` = 15, the pair (χ5 , χ6 );
• for ` = 23, the pair (χ10 , χ11 ).
Proof. Let
(1)
Kn =
26
M
mi χi
i=1
be the decomposition of the M24 -representation Kn into its irreducible constituents, i.e. mi is the multiplicity in which χi occurs in Kn . We note that if
Kn is non-zero then n is odd, i.e. we only need to consider the cases n = `m2
with m odd.
First we consider the cases ` = 15 and ` = 23 and take on both sides of
(1) the trace of an element of g of type 15A or 23A, respectively. If n is of
the form `m2 with m odd, the left-hand side is odd by Theorem 1.1. For
the right-hand side an inspection of the character table of M24 shows that if
λ 6= λ̄ one has that Tr(g|λ) = Tr(g|λ̄) is integral unless (λ, λ̄) = (χ5 , χ6 ) or
(χ10 , χ11 ), respectively, in which case Tr(g|χ + χ̄) = −1. Using Conjecture
1.2, it follows that mλ = mλ̄ has to be odd.
4
For the cases ` = 7, we take in (1) the trace of an element of type 7A.
Here, only the characters (χ3 , χ4 ) or (χ12 , χ13 ) can provide an odd contribution to the right-hand side of (1) and so in total an odd number of those
pairs has to appear. For m divisible by 3, we take in addition the trace of an
element of type 21A. Here (χ3 , χ4 ), (χ12 , χ13 ) and (χ15 , χ16 ) provide an odd
contribution. Since the total number of pairs of type (χ3 , χ4 ) and (χ12 , χ13 )
is odd and by Theorem 1.1 the left-hand side of (1) is even, it follows that
the number of pairs (χ15 , χ16 ) is odd, too.
The paper is organized as follows. It is easy to see that for the cases 12B,
6B, 4C, 3B, 2B, 10A, 4A, 12A, the coefficients of Σg (τ ) are even because
Σg (τ ) is −2 times an η-product (see Appendix A.2). Therefore, we omit the
study of these cases. The case 1A is trivial since Σ1A (τ ) = Σ(τ ) and the
definition of Σ(τ ). In Section 2, we study the cases 7A, 14A, 15A, 21A, 23A.
In Section 3, we study the remaining cases.
2
The cases 7A, 14A, 15A, 21A, 23A
Let `X ∈ {7A, 14A, 15A, 23A}, then we show that the coefficient of q n/8 in
Σ`X (τ ) is odd if and only if n = `m2 , where m is odd. For `X = 21A this
coefficient is odd if and only if n = `m2 , where m is odd and m is not divisible
by 3.
We will discuss the case of 7A in detail. Since the other cases can be
handled in complete analogy, we collect the necessary data in Table 1.
Proof. Let
E2 (τ ) = 1 + 24
∞
X
σ1 (m)q m .
m=1
By [5], we have
(2)
1
∞
X
(−1)n nq 2 n(n+1) 2 Σ(τ ) = −
E2 (τ ) + 24
.
η(τ )3
1 − qn
n=1
5
Then
(7)
φ (τ ) 1
Σ(τ ) − 14 2 3
8
η(τ )
1
∞
X
1 (−1)n nq 2 n(n+1)
(7)
=−
E
(τ
)
+
24
+
7φ
(τ
)
.
2
2
4η(τ )3
1 − qn
n=1
Σ7A (τ ) =
(3)
Define the ”characteristic function” fm (τ ) as follows:
1 mτ
mτ fm (τ ) =
ϑ3 (
) − ϑ4 (
)
4
8
8
P
P
m2
m2
where ϑ3 (τ ) = 1 + ∞
and ϑ4 (τ ) = 1 + ∞
. Then
m=1 2q
m=1 2(−q)
1 7τ
7τ f7 (τ ) =
ϑ3 ( ) − ϑ4 ( ) = q 7/8 + q 63/8 + q 175/8 + q 343/8 + · · · ,
4
8
8
η(τ )3 f7 (τ ) = q − 3q 2 + 5q 4 − 7q 7 + q 8 − 3q 9 + 14q 11 + · · · ,
(see “characteristic functions” in Table 1 for the other cases).
We have to show that all coefficients of Σ7A (τ ) + f7 (τ ) are even. This
follows, if all coefficients of (Σ7A (τ ) + f7 (τ ))η(τ )3 are even. We will show
that this is the case.
We observe that the constant term is even. Since all coefficients, except
for the constant term, of the function
1
∞
X
(−1)n nq 2 n(n+1) 1
− E2 (τ ) + 24
4
1 − qn
n=1
are even, it is enough to show that the coefficients of the following function,
again except for the constant term,
7 (7)
− φ2 (τ ) + η(τ )3 f7 (τ )
4
are even. We have the following Fourier expansion
7 (7)
7
− φ2 (τ ) + η(τ )3 f7 (τ ) = − − 6q − 24q 2 − 28q 3 − 44q 4 − 42q 5 + · · · .
4
4
Define the “sieve function” (see “sieve functions” in Table 1 for the other
cases)
7
7
ϑ3 (τ )4 = + 14q + 42q 2 + 56q 3 + 42q 4 + 84q 5 + 168q 6 + 112q 7 + · · · .
4
4
6
The coefficients of 74 ϑ3 (τ )4 without the constant term are even, hence it is
enough to show that the coefficients of the function
7 (7)
7
− φ2 (τ ) + η(τ )3 f7 (τ ) + ϑ3 (τ )4 = 8q + 18q 2 + 28q 3 − 2q 4 + 42q 5 + · · ·
4
4
are even. We are going to prove this using Sturm’s Theorem. The theta
functions ϑ3 (τ ) and ϑ4 (τ ) can be expressed as a quotient of η-functions,
namely
η(2τ )5
ϑ3 (τ ) =
η(τ )2 η(4τ )2
and
η(τ )2
.
ϑ4 (τ ) =
η(2τ )
It follows using [14, Theorem 1.64] that η(8τ )3 f7 (8τ ) is a modular form of
weight 2 for Γ0 (1344) and hence
∞
X
7 (7)
7
a7A (m)q m := − φ2 (8τ ) + η(8τ )3 f7 (8τ ) + ϑ3 (8τ )4
4
4
m=1
= 8q 8 + 18q 16 + 28q 24 + · · ·
is also a modular form of weight 2 for Γ0 (1344). Using Sturm’s theorem
[14, Theorem 2.58] and the fact that [SL2 (Z) : Γ0 (1334)] = 2160, the computer verification that a7A (m) ≡ 0 (mod 2) for m ≤ 361 shows a7A (m) ≡ 0
(mod 2) for all m.
This completes the proof of the case 7A.
The proof for the other cases is analogous. The relevant information can
be read off from Table 1.
3
The remaining cases
In this section we prove that all coefficients of the remaining cases `X ∈
{2A, 3A, 5A, 4B, 8A, 6A, 11A} are divisible by two.
3.1
The cases 2A, 3A, 5A, 4B, 8A, 6A
In this subsection, for `X ∈ {2A, 3A, 5A, 4B, 8A, 6A}, we show that the
Fourier coefficients of the McKay-Thompson series are divisible by two. We
give the detailed proof for the case of 2A. The other cases are proven in
analogy, and the necessary data is collected in Table 2.
7
Table 1: Data for the proofs in Section 2
`X
7A
14A
15A
21A
23A
characteristic functions
f7 (τ )
f7 (τ )
f15 (τ )
f7 (τ )-f63 (τ )
f23 (τ )
sieve functions
7
4
4 ϑ3 (τ )
23 (2)
12 φ2 (τ )
23 (2)
12 φ2 (τ )
2ϑ3 (τ )4
23 (2)
12 φ2 (τ )
Γ
Γ0 (1344)
Γ0 (1344)
Γ0 (4032)
Γ0 (960)
Γ0 (1472)
[SL2 (Z) : Γ]
2160
2160
9216
384
2304
Sturm bounds
361
361
1537
65
385
Proof. The McKay-Thompson series for 2A is
(2)
1
1 φ2 (τ ) (2)
3
=
Σ(τ )η(τ ) − 4φ2 (τ )
Σ2A (τ ) =
Σ(τ ) − 4
3
η(τ )3
3η(τ )3
= −2q −1/8 − 6q 7/8 + 14q 15/8 − 28q 23/8 + 42q 31/8 + · · · .
The coefficients of Σ(τ ), except for the constant term, are divisible by 24
(2)
(cf. (2)). Moreover, also the coefficients of φ2 (τ ), again except for the constant term, are divisible by 24 (see Table 2 for the other cases). The constant
(2)
term of the function Σ(τ )η(τ )3 − 4φ2 (τ ) is 6. Therefore, the coefficients of
Σ2A (τ ) are divisible by two.
Table 2: Data for the proofs in Section 3.1
(N )
m|(φ2 (τ ) − 1)
(2)
24
12
6
24
8
8
24/7
24
12
24/5
`X
φ2 (τ )
2A
3A
5A
4B
φ2 (τ )
(3)
φ2 (τ )
(5)
φ2 (τ )
(2)
φ2 (τ )
(4)
φ2 (τ )
(4)
φ2 (τ )
(8)
φ2 (τ )
(2)
φ2 (τ )
(3)
φ2 (τ )
(6)
φ2 (τ )
8A
6A
(N )
8
The proof for the other cases is analogous. The relevant information can
be read off from Table 2.
3.2
The case 11A
Finally, it remains to show that the Fourier coefficients of the McKay-Thompson
series for the element 11A are divisible by two.
Proof. The McKay-Thompson series for 11A is
264
1 (11)
3
2
Σ(τ
)η(τ
)
−
22φ
(τ
)
+
(η(τ
)η(11τ
))
.
Σ11A (τ ) =
2
12η(τ )3
5
The coefficients of Σ(τ ), except for the constant term are divisible by 24. We
need to prove that the coefficients of the following function:
∞
X
(11)
a11A (m)q m := 22φ2 (τ ) −
m=1
264
(η(τ )η(11τ ))2
5
are also (except for the constant term) divisible by 24. This is true, since
this function is a modular form for Γ0 (11) [10]. Using Sturm’s theorem [14,
Theorem 2.58] and the fact that [SL2 (Z) : Γ0 (11)] = 12, the verification that
a11A (m) ≡ 0 (mod 24) for m ≤ 13 shows a11A (m) ≡ 0 (mod 24) for all m.
Therefore, the coefficients of Σ11A (τ ) are divisible by two.
Section 2 and 3 prove Theorem 1.1.
A
McKay-Thompson series
We list all McKay-Thompson series. We refer to Section 1 for the used
notation.
A.1
McKay-Thompson series of type I
1A : Σ1A (τ ) = Σ(τ )
1 (2)
3
2A : Σ2A (τ ) =
Σ(τ
)η(τ
)
−
4φ
(τ
)
2
3η(τ )3
9
1 (3)
3
Σ(τ
)η(τ
)
−
6φ
(τ
)
2
4η(τ )3
1
(5)
3
Σ5A (τ ) =
Σ(τ )η(τ ) − 10φ2 (τ )
6η(τ )3
1
(2)
(3)
(6)
3
Σ6A (τ ) =
Σ(τ
)η(τ
)
+
2φ
(τ
)
+
6φ
(τ
)
−
30φ
(τ
)
2
2
2
12η(τ )3
1 (7)
3
Σ(τ
)η(τ
)
−
14φ
(τ
)
Σ7A (τ ) =
2
8η(τ )3
1
(4)
(8)
3
Σ8A (τ ) =
Σ(τ
)η(τ
)
+
6φ
(τ
)
−
28φ
(τ
)
2
2
12η(τ )3
1
264
(11)
2
Σ11A (τ ) =
Σ(τ ) − 22φ2 (τ ) +
(η(τ )η(11τ ))
12η(τ )3
5
1 2 (2)
182 (14)
(7)
Σ14A (τ ) =
Σ(τ )η(τ )3 + φ2 (τ ) + 14φ2 (τ ) −
φ (τ )+
3
24η(τ )
3
3 2
+112η(τ )η(2τ )η(7τ )η(14τ )
1 3 (3)
105 (15)
(5)
Σ15A (τ ) =
Σ(τ )η(τ )3 + φ2 (τ ) + 5φ2 (τ ) −
φ (τ )+
3
24η(τ )
2
2 2
+90η(τ )η(3τ )η(5τ )η(15τ )
276
1 (23)
Σ(τ )η(τ )3 − 46φ2 (τ ) +
(2f (τ ) − g(τ ))+
Σ23A (τ ) =
3
24η(τ )
11
1932
+
g(τ )
11
1
(2)
(4)
3
Σ4B (τ ) =
Σ(τ )η(τ ) + 2φ2 (τ ) − 12φ2 (τ )
6η(τ )3
3A : Σ3A (τ ) =
5A :
6A :
7A :
8A :
11A :
14A :
15A :
23A :
4B :
A.2
McKay-Thompson series of type II
η(2τ )8
η(τ )3 η(4τ )4
η(2τ )η(5τ )
10A : Σ10A (τ ) = −2
η(10τ )
η(4τ )2 η(6τ )3
12A : Σ12A (τ ) = −2
η(2τ )η(3τ )η(12τ )2
4A : Σ4A (τ ) = −2
10
21A :
2B :
3B :
6B :
12B :
4C :
1
7η(7τ )3
η(τ )3
Σ21A (τ ) = −
−
3 η(3τ )η(21τ ) η(3τ )2
η(τ )5
Σ2B (τ ) = −2
η(2τ )4
η(τ )3
Σ3B (τ ) = −2
η(3τ )2
η(2τ )2 η(3τ )2
Σ6B (τ ) = −2
η(τ )η(6τ )2
η(τ )η(4τ )η(6τ )
Σ12B (τ ) = −2
η(2τ )η(12τ )
η(τ )η(2τ )2
Σ4C (τ ) = −2
η(4τ )2
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12