Nahm's Conjecture: Asymptotic Computations and Counterexamples
aa r X i v : . [ m a t h . N T ] A p r NAHM’S CONJECTURE: ASYMPTOTIC COMPUTATIONSAND COUNTEREXAMPLES
MASHA VLASENKO AND SANDER ZWEGERS Introduction
Let r ≥ A a real positive definite symmetric r × r -matrix, B a vector of length r , and C a scalar. The series(1.1) F A,B,C ( q ) = X n ∈ ( Z ≥ ) r q n T An + n T B + C ( q ) n . . . ( q ) n r . converges for | q | <
1. Here we use the notation ( a ; q ) n := Q nk =1 (1 − aq k − )for n ∈ Z ≥ ∪ {∞} and the convention that the second argument is removedif it equals q (so ( q ) n = ( q ; q ) n = Q nk =1 (1 − q k )). We are concerned with thefollowing problem due to Werner Nahm [2, 3, 4]: describe all such A, B and C with rational entries for which (1.1) is a modular form. In [5, 7, 8] it wassolved by Michael Terhoeven and Don Zagier for r = 1 and the list containsseven triples ( A, B, C ) ∈ Q + × Q × Q . We develop this approach for r > r = 2.Nahm has also given a conjectural criterion for a matrix A to be such thatthere exist some B and C with modular F A,B,C (see [4]). The condition forthe matrix A is given in terms of solutions of a system of algebraic equations(1.2) 1 − Q i = r Y j =1 Q A ij j , i = 1 , . . . , r . In the last section we give several examples where the matrix A doesn’tsatisfy the condition but corresponding modular forms exist. Certainly, itdoesn’t mean that the conjecture is completely wrong, rather that its correctformulation is an interesting open question.2. Asymptotical computations
Let us explain a method to compute the asymptotics of (1.1) when q → a n ( q ). Suppose q → n i → ∞ so that q n i → Q i for some numbers Q i / ∈ { , } . Then we have a n + e i a n = q n T Ae i + e Ti Ae i + e Ti B − q n i +1 → Q A i . . . Q A ir r − Q i , where e i is a vector whose all but i th coordinates are 0 and i th coordinateis 1. We have the following statement. Date : September 13, 2018.
Lemma 2.1.
Let A be a real positive definite symmetric r × r matrix. Thenthe system of equations − Q i = r Y j =1 Q A ij j , i = 1 , . . . , r has a unique solution with Q i ∈ (0 , for all ≤ i ≤ r .Proof. We consider the function f A : [0 , ∞ ) r → R given by f A ( x ) = 12 x T Ax + r X i =1 Li (exp( − x i )) , where Li is the dilogarithm function defined by the power series Li ( z ) = P ∞ n =1 z n n for | z | <
1. It has the property zLi ′ ( z ) = − log(1 − z ).The gradient and the Hessian of f A are ∇ f A ( x ) = Ax + (log(1 − exp( − x i ))) ≤ i ≤ r ,H f A ( x ) = A + diag (cid:18) x i ) − (cid:19) ≤ i ≤ r . Using Q i = exp( − x i ), the statement of the lemma is equivalent to sayingthat f A has a unique critical point in (0 , ∞ ) r .First, f A has at least one critical point in (0 , ∞ ) r , because it takes onit’s minimum in (0 , ∞ ) r : it’s continuous, bounded from below by 0 and f A ( x ) → ∞ if || x || → ∞ , and so it takes on it’s minimum in [0 , ∞ ) r . In fact,it takes on that minimum in (0 , ∞ ) r , becauselim x i ↓ ∂f A ∂x i ( x ) = −∞ < . Second, f A has at most one critical point in (0 , ∞ ) r , because it’s differ-entiable and strictly convex on (0 , ∞ ) r : since A is positive definite, we seethat the Hessian H f A ( x ) is positive definite for all x ∈ (0 , ∞ ) r . (cid:3) Consider the unique solution Q i ∈ (0 ,
1) of (1.2) and let q = e − ε , ε > a n + e i ( q ) a n ( q ) ≈ ∀ i when n ≈ (cid:0) − log Q ε , . . . , − log Q r ε (cid:1) , and it is very likely that a n ( q ) as a function of n is maximal around this point.We will apply a version of Laplace’s method to describe the asymptotics of F A,B,C ( e − ε ) for small ε . For this we need the so called polylogarithm Li m ( z ) = ∞ X n =1 z n n m for | z | < , m ∈ Z , which satisfies the obvious relation z ddz Li m ( z ) = Li m − ( z ) . Lemma 2.2.
Let n ∈ N and q = e − ε with ε > . We fix Q ∈ (0 , andintroduce a variable ν = − log Q − nε . Then AHM’S CONJECTURE 3 (i) for all n, ε we have an inequality (2.1) log (cid:16) ( q ) ∞ ( q ) n (cid:17) < − Li ( Q ) ε + (cid:16) νε − (cid:17) log(1 − Q ) + ν Q − Q ; (ii) we have an asymptotic expansion (2.2) log (cid:16) ( q ) ∞ ( q ) n (cid:17) ∼ − X r,s ≥ Li − r − s ( Q ) B r r ! s ! ν s ε r − when ε, ν → , where B = 1 , B = − , B = , . . . are the Bernoulli numbers.Proof. log (cid:16) ( q ) ∞ ( q ) n (cid:17) = ∞ X s =1 log (cid:0) − q n + s (cid:1) = ∞ X s =1 log (cid:0) − Qe ν − sε (cid:1) = − ∞ X s =1 ∞ X p =1 Q p e p ( ν − sε ) p = − ∞ X p =1 Q p p e pν e pε − e x > x for all x = 0 and xe x − > x for x > e pν e pε − > (1 + pν ) (cid:16) pε − (cid:17) = 1 pε + (cid:16) νε − (cid:17) − p ν , and we get inequality (i) after summation in p . To prove (ii) we notice thatfor every fixed p we have an asymptotic expansion pεe pν e pε − ∼ (cid:16) ∞ X r =0 B r r ! ( pε ) r (cid:17)(cid:16) ∞ X s =0 ( pν ) s s ! (cid:17) = X r,s ≥ B r r ! s ! ( pε ) r ( pν ) s , i.e. for every fixed N and δ > δ ′ > (cid:12)(cid:12)(cid:12) pεe pν e pε − − P r + s ≤ N B r p r + s r ! s ! ε r ν s (cid:12)(cid:12)(cid:12) p N max( ε, | ν | ) N < δ whenever pε, p | ν | < δ ′ . Also we observe that when x ց x N X p> δ ′ x p a Q p → a , as well as(2.4) 1 x N X p> δ ′ x Q p p pεe pν e pε − < x N X p> δ ′ x Q p e pν < x N e δ ′ x ( ν +log Q ) − e ν +log Q → ν in small domains. Let us choose δ ′′ > a between − N − δ whenever x < δ ′′ and | ν | < δ ′′ . Now if MASHA VLASENKO AND SANDER ZWEGERS max( ε, | ν | ) < δ ′′ then (cid:12)(cid:12)(cid:12)P p ≥ Q p p (cid:16) pεe pν e pε − − P r + s ≤ N B r p r + s r ! s ! ε r ν s (cid:17)(cid:12)(cid:12)(cid:12) max( ε, | ν | ) N ≤ δ X p max( ε, | ν | ) <δ ′ p N − Q p + 1max( ε, | ν | ) N X p max( ε, | ν | ) >δ ′ Q p p pεe pν e pε − X r + s ≤ N | B r | r ! s ! ε r | ν | s ε, | ν | ) N X p max( ε, | ν | ) >δ ′ p r + s Q p ≤ (cid:16) L − N ( Q ) + 1 + X r + s ≤ N | B r | r ! s ! ( δ ′′ ) r + s (cid:17) δ and (ii) follows. (cid:3) Let B p ( X ) = P k (cid:0) pk (cid:1) B k X p − k , p ≥ D p ∈ Q [ B, X, T ], p ≥ ε / :(2.5) exp h ( B + 12 Q − Q ) T ε / − ∞ X p =3 p ! B p (cid:0) Tε / (cid:1) Li − p ( Q ) ε p − i = 1 + ∞ X p =1 D p (cid:0) B, Q − Q , T (cid:1) ε p/ . Observe that the coefficients of the series under the exponent are polynomialsin B, Q − Q and T because Li − r ( Q ) = P r − (cid:16) Q − Q (cid:17) where P r , r ≥ P ( X ) = X and P p +1 ( X ) = ( X + X ) ddX P p ( X ). Theorem 2.3.
There is an asymptotic expansion F A,B,C ( e − ε ) e − αε ∼ βe − γε (cid:16) ∞ X p =1 c p ε p (cid:17) , ε ց with the coefficients α ∈ R + , β, γ ∈ Q and c p ∈ Q , p ≥ given below.Let Q i ∈ (0 , be the solutions of (1.2) . Denote ξ i = Q i − Q i , e A = A +diag { ξ , . . . , ξ r } and let L ( x ) be the Rogers dilogarithm function. Then α = r X i =1 (cid:0) L (1) − L ( Q i ) (cid:1) > ,β = det e A − / Y i Q B i i (1 − Q i ) − / , γ = C + 124 X Q i − Q i ,c p = det e A / (2 π ) − r/ Z C p ( B, ξ, t ) e − t T e At dt , AHM’S CONJECTURE 5 where the polynomials in r variables C p ∈ Q [ B, ξ, t ] are defined as C p ( B, ξ, t ) = X p + ··· + p r = p p Y i =1 D p i ( B i , ξ i , t i ) , where D p are the polynomials in 3 variables defined by (2.5) . Recall that L ( x ) is an increasing function on R (therefore α > L ( x ) = Li ( x ) + log( x ) log(1 − x ) for x ∈ (0 ,
1) and L (1) = π . Proof.
Let α ′ = − r X i =1 L ( Q i ) , β ′ = Y i Q B i i (1 − Q i ) − / , γ ′ = C + 112 X i ξ i and t i = − log Q i ε − n i . Consider the function φ ( t, ε ) = ( q ) r ∞ a n ( q ) β ′ e α ′ ε ( q = e − ε )defined only for t ∈ t ( ε ) + Z r where t i ( ε ) is the fractional part of − log Q i ε .After a straightforward computation using (i) of Lemma 2.2 we obtain that(2.6) log φ ( t, ε ) < (cid:16) − t T At + t T (cid:0) B + 12 ξ (cid:1) − C (cid:17) ε . Then ( q ) r ∞ F A,B,C ( q ) β ′ exp( α ′ ε ) = X t ∈ t + Z r φ ( t, ε ) ∼ X t ∈ t + Z r , | t i | <ε λ φ ( t, ε )for every λ < − , where ” ∼ ” always means that the difference is o ( ε N ) forevery N . Indeed, for such λ we have P | t i | >ε λ φ ( t, ε ) = o ( ε N ) for every N due to (2.6). We can further rewrite it as X t ∈ t + Z r , | t i | <ε λ φ ( t, ε ) = X t ∈ ( t + Z r ) √ ε, | t i | <ε λ + 12 φ (cid:0) t √ ε , ε (cid:1) . Let also λ > − . Then(2.7) φ (cid:0) t √ ε , ε (cid:1) = e − t t e At − γ ′ ε (cid:16) N X p =1 C p ( t ) ε p/ (cid:17) + o ( ε N (3 λ +2) )uniformly in the domain | t i | ≤ ε λ + , and we observe that for any polynomial P (2.8) X t ∈ ( t + Z r ) √ ε, | t i | <ε λ + 12 P ( t ) e − t T e At ∼ ε − r/ Z P ( t ) e − t T e At dt (the difference is o ( ε N ) for every N ) when λ < − . Combining (2.7)and (2.8) (we will prove both facts later), we get( q ) r ∞ F A,B,C ( q ) β ′ exp( α ′ ε − γ ′ ε ) ∼ (cid:16) πε (cid:17) r/ det e A − / (cid:16) ∞ X p =1 c p ε p (cid:17) . MASHA VLASENKO AND SANDER ZWEGERS
Here we have only integer powers of ε because R C p ( t ) e − t T e At dt = 0 when p is odd. And this happens because the total t -degree of every monomialin C p has the same parity as p , which in turn follows from the definition of D p . Now, since log( q ) ∞ ∼ − π ε + 12 log (cid:16) πε (cid:17) + ε ε →
0, we obtain the statement of the theorem.To prove (2.8) we notice again that P | t i | >ε λ + 12 P ( t ) e − t t e At = o ( ε N ) forevery N , and using Poisson summation formula we have X t ∈ ( t + Z r ) √ ε P ( t ) e − t T e At = X t ∈ t + Z r P ( t √ ε ) e − ε t T e At = X s ∈ Z r g ( s ) e πis T t , where g ( s ) is the Fourier transform of P ( t √ ε ) e − ε t T e At . Then g (0) is theright-hand side of (2.8), and the sum of all remaining terms are o ( ε N ) sincefor any monomial P ′ ( t ) and g ′ ( s ) being the Fourier transform of P ′ ( t ) e − ε t t A ′ t one can check by direct computation that P s ∈ Z \{ } | g ′ ( s ) | = o ( ε N ) for any N .It remains to prove (2.7). Using (ii) of Lemma 2.2 we getlog φ ( t, ε ) ∼ − t T e At − γ ′ ε + t T (cid:0) B + 12 ξ (cid:1) − X i ∞ X p =3 p ! B p ( t i ) Li − p ( Q i ) ε p − , ε, tε → , and therefore for every N log φ (cid:0) t √ ε , ε (cid:1) = − t T e At − γ ′ ε + t T (cid:0) B + 12 ξ (cid:1) √ ε − X i N X p =3 p ! B p (cid:16) t i √ ε (cid:17) Li − p ( Q i ) ε p − + o ( ε N ( λ +1) − )uniformly in | t i | ≤ ε λ + . If we rewrite the right-hand side as P N − p =0 g p ( t ) ε p then deg g p ≤ p + 2 (because deg B p = p ). It follows that P N − p =1 g p ( t ) ε p = O ( ε λ + 2 ) uniformly in our domain since( λ + 12 )( p + 2) + p p ( λ + 1) + 2 λ + 1 ≥ λ + 2 > . Therefore we can take a sufficiently long but finite part of the standard seriesto approximate its exponent. Hence some sufficiently long but again finitepart ofexp hX i ( B i + 12 ξ ) t i √ ε − X i ∞ X p =3 p ! B p (cid:0) t i √ ε (cid:1) Li − p ( Q i ) ε p − i = 1 + ∞ X p =1 C p ( t ) ε p/ will approximate φ (cid:0) t √ ε , ε (cid:1) e t T e At + γ ′ ε . One can easily see that deg C p ( t ) ≤ p (in the variable t ). Since for p > NC p ( t ) ε p = O ( ε ( λ + )3 p + p ) = O ( ε p (3 λ +2) ) = o ( ε N (3 λ +2) ) AHM’S CONJECTURE 7 then it is sufficient to consider only the part with p ≤ N in (2.7). (cid:3) Modular functions F A,B,C
Let us search for those triples (
A, B, C ) for which F A,B,C ( q ) is a modularfunction (of any weight and any congruence subgroup). We will call such( A, B, C ) a modular triple . The idea here is that in order for F A,B,C ( q ) tobe modular, the asymptotic expansion needs to be of a special type, as wecan see from the following lemma. Lemma 3.1.
Let F ( q ) be a modular form of weight w for some subgroup offinite index Γ ⊂ SL(2 , Z ) . Then for some numbers a ∈ π Q and b ∈ C (3.1) e aε F ( e − ε ) ∼ b ε − w + o ( ε N ) ∀ N ≥ . Proof.
The group SL(2 , Z ) acts on the space M w (Γ) of modular functions ofweight w . Let S = (cid:18) − (cid:19) . Then SF ∈ M w (Γ), and in particular it hasa q -expansion (with some rational powers of q = e πiz ):1 z w F (cid:16) e − πi z (cid:17) = a q α + a q α + . . . We subsitute z = πiε and get F ( e − ε ) = (cid:16) πiε (cid:17) w h a e − π α ε + a e − π α ε + . . . i = (2 πi ) w a ε w e − π α ε h o ( ε N ) i ∀ N . (cid:3)
If we now compare the asymptotics from Theorem 2.3 with (3.1) we getthe following statement.
Corollary 3.2. If F A,B,C ( q ) is modular then (i) its weight w = 0(ii) α ∈ π Q ⇐⇒ P ri =1 L ( Q i ) ∈ π Q (iii) e − γε (cid:0) P ∞ p =1 c p ε p (cid:1) = 1 ⇐⇒ c p = γ p p ! ∀ p Condition (ii) is very interesting, we consider it in the next section. Itfollows from (iii) that modular triples satisfy an infinite number of equations(3.2) (cid:0) c p − p ! c p (cid:1)(cid:0) B, ξ, e A − (cid:1) = 0 , p = 2 , , . . . , and these equations are polynomial in the entries of B, ξ, e A − . Indeed, letus look at the expression for c p from Theorem 2.3. Since the generatingfunction for the moments of the Gaussian measure is X a ∈ ( Z ≥ ) r x a a ! . . . a r ! det e A / (2 π ) r/ Z t a e − t T e At dt = exp (cid:0) x T e A − x (cid:1) , all the moments are rational polynomials in the entries of e A − and we obtainthat c p ∈ Q [ B, ξ, e A − ]. MASHA VLASENKO AND SANDER ZWEGERS
Now let r = 1. It is easy to see that the degrees of D p ( B, X, T ) in thevariables
B, X and T are p, p and 3 p , respectively. Since c p ( B, ξ, ( A + ξ ) − )is the integral of D p ( B, ξ, t ) w.r.t. the measure ( A + ξ ) / √ π e − ( A + ξ ) t / dt and theintegral of t m is (2 m − A + ξ ) − m , the degrees of c p in the correspondingvariables are 2 p , 4 p and 3 p . It is convenient to consider the polynomials e c p ( B, ξ, A ) = ( A + ξ ) p (cid:2) c p − p ! c p (cid:3) ( B, ξ, A + ξ ) , p = 2 , , . . . Although these polynomials look rather complicated, we have found usingthe
Magma algebra system ([1]) that the ideal I = h e c , e c , e c , e c i ⊂ Q [ B, ξ, A ]contains the element ξ ( ξ + 1) A ( A − ( A + 1)( A − A − / . Consequently, if (
A, B, C ) is a modular triple then A ∈ { , , } . For each A on this list it is not hard to find the corresponding values of B , and onecan compute C from the equality γ = c . This way we obtain exactly thelist from the theorem below. Theorem 3.3 (D. Zagier [8]) . Let r = 1 . The only ( A, B, C ) ∈ Q + × Q × Q for which F A,B,C ( q ) is a modular form are given in the following table. A B C F A,B,C ( e πiz )2 0 − / θ , ( z ) /η ( z )1 11 / θ , ( z ) /η ( z )1 0 − / η ( z ) /η ( z ) η (2 z )1 / / η (2 z ) /η ( z ) − / /
24 2 η (2 z ) /η ( z )1/2 0 − / θ , ( z ) η (2 z ) /η ( z ) η (4 z )1/2 1 / θ , ( z ) η (2 z ) /η ( z ) η (4 z )Here η ( z ) = q / Q ∞ n =1 (1 − q n ) and θ ,j ( z ) = P n ≡ j − ( − [ n/ q n / .We warn the reader that if (iii) of Corollary 3.2 holds for some ( A, B, C )this does not yet imply that F A,B,C is in fact modular. To get modularityone needs to prove an identity between the corresponding q -series for eachline of the table. For example, the first two lines correspond to the wellknown Rogers–Ramanujan identities.Further computer experiments showed that e c p ∈ I for p = 6 , . . . , p . Also with the help of Magma we have got the followingdecomposition of the radical of I into prime ideals:Rad( I ) = P · ... · P where the generators of P i are given below:i generators of P i ξ ξ + 13 B − / , ξ + 2 , A B − , ξ + 2 , A B, ξ + 2 , A B + 1 / , ξ + 3 ξ + 1 , A + 17 B − / , ξ + 3 ξ + 1 , A + 18 B + 1 / , ξ − , A − B, ξ − , A − B − / , ξ − , A − B − , ξ − ξ − , A − B, ξ − ξ − , A − B − / , ξ + ξ − , A − / B, ξ + ξ − , A − / e c p ( B, ξ, A ) = 0, p =2 , , . . . is a subset of this table, and if we indeed had e c p ∈ I (or at least e c p ∈ Rad( I )) for all p then this table would be exactly the set of solutions.Let’s us now consider the case r = 2. The task of solving the system (3.2)for several small values of p becomes already very complicated. We failedto solve it with Magma in full generality for r = 2 as we did in the case r = 1. However, we can still search for modular F A,B,C , where A is of aspecial type. We will consider three families of matrices:(3.3) A = (cid:18) a − a − a a (cid:19) ⇒ ξ = ξ = √ − A = (cid:18) a − a − a a (cid:19) ⇒ ξ = ξ = √ A = (cid:18) a − a − a a (cid:19) ⇒ ξ = ξ = 12It is easy to check that (ii) of Corollary 3.2 holds for these matrices. Forthese families of matrices we can do an analysis similar to what we did for r = 1. Theorem 3.4.
Modular functions F A,B,C ( z ) with the matrix A being of theform (cid:18) a − a − a a (cid:19) exist if and only if a = 1 , a = 3 / or a = 1 / . Belowis the list of all such modular functions. A B C F A,B,C ( e πiz ) (cid:18) − − (cid:19) (cid:18) (cid:19) − ( θ , (2 z ) + θ , (2 z )) η ( z ) /η (2 z ) η ( z/ θ , (2 z ) η (2 z ) /η ( z ) (cid:18) − (cid:19) and (cid:18) − (cid:19) θ , (2 z ) η (2 z ) /η ( z ) + θ , ( z ) θ , (2 z ) η ( z ) /η ( z/ η (2 z ) η (10 z ) (cid:18) − −
14 34 (cid:19) (cid:18) − (cid:19) and (cid:18) − (cid:19) − θ , ( z ) η ( z ) /η ( z ) η (2 z ) (cid:18) (cid:19) and (cid:18) (cid:19) θ , ( z ) η ( z ) /η ( z ) η (2 z ) (cid:18) (cid:19) (cid:18) (cid:19) − ( θ , ( z ) η (2 z ) /η ( z ) η (4 z )) (cid:18) (cid:19) and (cid:18) (cid:19) θ , ( z ) θ , ( z )( η (2 z ) /η ( z ) η (4 z )) (cid:18) (cid:19) ( θ , ( z ) η (2 z ) /η ( z ) η (4 z )) Proof.
Consider the ideal I ⊂ Q [ b , b , ξ, a ] generated by ξ + ξ − ξ + 2 aξ + a − / p × (cid:2) c p − c p p ! (cid:3)(cid:16)(cid:18) b b (cid:19) , (cid:18) ξξ (cid:19) , (cid:18) a + ξ − a − a a + ξ (cid:19) − (cid:17) for p = 2 , , ,
5. We find with
Magma that the element a ( a −
14 )( a −
12 )( a −
34 )( a − a − a −
116 )belongs to I . (We ran the function GroebnerBasis( I ) which has computedthe Groebner basis for I using reversed lexicografical order on monomialswith the variables ordered as b > b > ξ > a . It took several hours, theGroebner basis contains 15 elements, and the element above is one of them.)The last term doesn’t give rational values for a , and the reason it enters hereis that we have multiplied every equation c p − c p /p ! = 0 by 3 p th power ofthe determinant ξ + 2 aξ + a − / ξ + 1 / ξ + 2 a − /
2) while preciselythe denominator of c p − c p /p ! is ( ξ + 1 / p ( ξ + 2 a − / p . Thereforeour polynomials are divisible by ( ξ + 2 a − ) p for p = 2 , , ,
5, and since ξ + ξ = 1 these factors are zero exactly when a − a = . We now havea finite list of values for a , and we plug each of them together with ξ intothe equations to find all values of b and b for which our equations vanishfor p = 2 , , ,
5. So, we get the list above. For each row we compute thecorresponding value of C from c = γ , i.e. C = c ( b , b , ξ, a ) − X i Q i − Q i = c ( b , b , ξ, a ) − ξ + 112 . AHM’S CONJECTURE 11
What remains is to prove that the F A,B,C satisfy the identities given inthe last column. For the case a = 1 /
2, this is easy, since F A,B,C splits asthe product of two rank 1 cases, for which an identity is given in Theo-rem 3.3. For the case a = 3 /
4, the identities follow directly by applyingTheorem 4.2 below, with m = 2 and A = 1 /
2, and again using identitiesfrom Theorem 3.3.Only the case a = 1 is a bit more work: using(3.4) ( − xq / ; q ) ∞ = X k ≥ q k x k ( q ) k (this is a direct consequence of (7) in Chapter 2 of [8]), with x = q − n/ , wefind X m,n ≥ q m − mn + n ( q ) m ( q ) n = X n ≥ q n ( − q − n + ) ∞ ( q ) n = X n ≥ q n ( − q − n + ) ∞ ( q ) n + X n ≥ q n +2 n + ( − q − n ) ∞ ( q ) n +1 . Now using that for n ≥ − q − n + ) ∞ = q − n ( − q ) n ( − q ) ∞ and( − q − n ) ∞ = 2 q − n − n ( − q ) n ( − q ) ∞ , this equals(3.5) ( − q ) ∞ X n ≥ q n ( − q ) n ( q ) n + 2 q ( − q ) ∞ X n ≥ q n + n ( − q ) n ( q ) n +1 =( − q ) ∞ X n ≥ q n ( q ) n ( q ; q ) n + 2 q ( − q ) ∞ X n ≥ q n + n ( q ) n ( q ; q ) n +1 . To get identities for these last two sums, we use equations (19) and (44)in [6], which (in our notation) read X n ≥ ( − n q n ( − q ; q ) n ( q ; q ) n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ , X n ≥ q n + n ( q ) n ( q ; q ) n +1 = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ) ∞ . If we use the Jacobi triple product identity ( − xq / ) ∞ ( − x − q / ) ∞ ( q ) ∞ = P n ∈ Z x n q n / on the right hand sides and replace q by − q / in the firstidentity, we get X n ≥ q n ( q ) n ( q ; q ) n = q θ , (2 z ) + θ , (2 z ) η ( z ) , X n ≥ q n + n ( q ) n ( q ; q ) n +1 = q − θ , (2 z ) η ( z ) . Further we have( − q ) ∞ = ( q ; q ) ∞ ( q ; q ) ∞ = ( q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = q η ( z ) η (2 z ) η ( z/ , ( − q ) ∞ = ( q ; q ) ∞ ( q ) ∞ = q − η (2 z ) η ( z ) , and so we get from (3.5) F A,B,C ( q ) = η ( z ) η (2 z ) η ( z/ (cid:16) θ , (2 z ) + θ , (2 z ) (cid:17) + 2 η (2 z ) η ( z ) θ , (2 z ) , where A = (cid:16) − / − / (cid:17) , B = ( ) and C = − / B = (cid:16) − / (cid:17) and C = 1 /
20 is very similar,and so we omit some of the details. We have X m,n ≥ q m − mn + n − m ( q ) m ( q ) n = 2( − q ) ∞ X n ≥ q n − n ( − q ) n ( q ) n + ( − q ) ∞ X n ≥ q n + n ( − q ) n +1 ( q ) n +1 = 2( − q ) ∞ X n ≥ q n − n ( q ) n ( q ; q ) n + ( − q ) ∞ X n ≥ q n + n ( − q ) n +1 ( q ) n +1 . Again we use two identities from Slater’s list (see [6]), namely (46) whichreads X n ≥ q n − n ( q ) n ( q ; q ) n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ) ∞ , and so we can identify it as q − / θ , (2 z ) /η ( z ), and (97), which shouldread (note that there are mistakes in some of the exponents; we have giventhe corrected version here) X n ≥ q n +2 n ( − q ; q ) n +1 ( q ; q ) n +1 = ( − q ; q ) ∞ ( q ; q ) ∞ (cid:0) ( − q ; q ) ∞ ( − q ; q ) ∞ − q ( − q ; q ) ∞ ( − q ; q ) ∞ (cid:1) ( q ; q ) ∞ = ( − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ . If we replace q by q / , we find X n ≥ q n + n ( − q ) n +1 ( q ) n +1 = q − θ , ( z ) θ , (2 z ) η ( z ) η ( z/ η (2 z ) η (10 z ) , which gives the desired result. (cid:3) In [8] one can find a list of triples (
A, B, C ) for r = 2 (Table 2 on p. 47) forwhich numerical experiments show that the condition (iii) of Corollary 3.2holds, as well as (ii). We see that the cases of Theorem 3.4 with a = 1 are AHM’S CONJECTURE 13 on this list, but the ones with a = 3 / a = 3 / Table 1.
A complete list of modular triples (
A, B, C ) withthe matrix A = (cid:18) a − a − a a (cid:19) , a > a = 2.A B C F A,B,C ( e πiz ) (cid:18)
43 2323 43 (cid:19) (cid:18) (cid:19) − ? 1 η ( z ) X n ∈ Z ( − n (cid:0) q ( n + ) + q ( n + ) − q ( n + ) (cid:1)(cid:18) − − (cid:19) and (cid:18) − − (cid:19) ? 1 η ( z ) X n ∈ Z ( − n (cid:0) q ( n + ) + q ( n + ) − q ( n + ) (cid:1)(cid:18)
32 1212 32 (cid:19) (cid:18) − (cid:19) and (cid:18) − (cid:19) − θ , ( z ) /η ( z ) (cid:18) (cid:19) and (cid:18) (cid:19) θ , ( z ) /η ( z )In Table 1, the identities for the case a = 3 / m = 2 and A = 2, and using identities from Theorem 3.3.For the case a = 4 / q .In Table 2, the identity for B = (cid:0) b − b (cid:1) is given in [8] (see (26) in Chapter2). The proof uses that for any n ∈ Z (3.6) X k,l ≥ k − l = n q kl ( q ) k ( q ) l = 1( q ) ∞ . The identity for B = (cid:18) − − (cid:19) is proven similarly, using X k,l ≥ k − l = n q kl − k − l ( q ) k ( q ) l = q n/ + q − n/ ( q ) ∞ , Table 2.
The list containing all (
B, C ) such that F A,B,C ismodular, where A = (cid:18) a − a − a a (cid:19) , a > , a = 1.B C F A,B,C ( e πiz ) (cid:18) b − b (cid:19) b a − η ( z ) P n ∈ Z + ba q an / (cid:18) − − (cid:19) a − η ( z ) P n ∈ Z + a q an / (cid:18) − a a (cid:19) and (cid:18) a − a (cid:19) a − η ( z ) P n ∈ Z + q an / for all n ∈ Z . This identity follows directly from (3.6): X k,l ≥ k − l = n q kl − k − l ( q ) k ( q ) l = X k,l ≥ k − l = n q kl − k − l (cid:0) (1 − q k ) + q k (cid:1) ( q ) k ( q ) l = X k ≥ ,l ≥ k − l = n q kl − k − l ( q ) k − ( q ) l + X k,l ≥ k − l = n q kl + k − l ( q ) k ( q ) l . If we replace k by k + 1 in the first sum on the RHS, we see that it equals q − n/ / ( q ) ∞ and the second sum equals q n/ / ( q ) ∞ .To get the identity for B = (cid:16) a − a (cid:17) we use(3.7) X k,l ≥ k − l = n q kl + l ( q ) k ( q ) l = 1( q ) ∞ ( − n q − n − n s n , with s n = P k ≥ n ( − k q k + k (this is easily obtained by checking that bothsides satisfy the recursion b n + q n +1 b n +1 = 1 / ( q ) ∞ and lim n →∞ b n = q ) ∞ )to get F A,B,C ( q ) = q a/ η ( z ) X n ∈ Z q ( a − n + n ) / s n . If we replace n by − n − s − n − = s n +1 = s n − ( − n q n + n , we easily get that P n ∈ Z q ( a − n + n ) / s n = P n ∈ Z q a ( n + n ) / ,which gives the desired result.We also checked for each matrix A in Zagier’s list for r = 2 (p. 47 in [8]) ifthe corresponding list of vectors B is complete. It appears to be complete in AHM’S CONJECTURE 15 all cases except A = (cid:18) a − a − a a (cid:19) . For such matrices only the modularforms in the first row of Table 2 were known.4. Counterexamples to Nahm’s conjecture
The Bloch group B ( K ) of a field K is an abelian group defined as thequotient of the kernel of the map(4.1) Z [ K ∗ \ → Λ K ∗ x x ∧ (1 − x )by the subgroup generated by all elements of the form(4.2) [ x ] + [1 − x ] , [ x ] + (cid:2) x (cid:3) , [ x ] + [ y ] + [1 − xy ] + (cid:2) − x − xy (cid:3) + (cid:2) − y − xy (cid:3) . If K is a number field than B ( K ) ⊗ Z Q ∼ = K ( K ) ⊗ Z Q and the regulatormap is given explicitly on B ( K ) by B ( K ) → R r x ( D ( σ ( x )) , . . . , D ( σ r ( x )))where r is the number of pairs of complex conjugate embeddings of K into C , σ , . . . , σ r is any choice of such embeddings from different pairs, and D ( x ) = ℑ (cid:0) Li ( x ) + log(1 − x ) log | x | (cid:1) is the Bloch-Wigner dilogarithm function. It vanishes on all combinationsin (4.2).Let ( Q , . . . , Q r ) be an arbitrary solution of the system of algebraic equa-tions (1.2) in some number field K . Then the element [ Q ] + · · · + [ Q r ] ∈ Z [ K ∗ \
1] belongs to the kernel of (4.1). Indeed, we have X i Q i ∧ (1 − Q i ) = X i Q i ∧ Y j Q A ij j = X i,j A ij Q i ∧ Q j = 0because of the symmetry A ij = A ji . Hence every solution of (1.2) definesan element in the Bloch group of the corresponding field.Recall that there exists the unique solution ( Q , . . . , Q r ) of (1.2) with Q i ∈ (0 , q →
1. If (1.1) is a modular function then for this solution wehave(4.3) L ( Q ) + . . . + L ( Q r ) ∈ π Q where L ( x ) is the Rogers dilogarithm function (condition (ii) of Corol-lary 3.2). Rogers dilogarithm is defined in R and takes values in π Q onall combinations of real arguments of the form (4.2). On the other hand,these are essentially all known functional equations for L ( x ). Therefore it isvery naturally to expect that [ Q ] + · · · + [ Q r ] is torsion in the correspond-ing Bloch group because of (4.3). (It is automatically torsion if the field Q ( Q , . . . , Q r ) is totally real.) Similar reasoning lead Werner Nahm to thefollowing conjecture. Conjecture 4.1.
For a positive definite symmetric r × r matrix with rationalcoefficients A the following are equivalent: (i) The element [ Q ] + · · · + [ Q r ] is torsion in the corresponding Blochgroup for every solution of (1.2) . (ii) There exist B ∈ Q r and C ∈ Q such that F A,B,C is a modularfunction.
This conjecture is true in case r = 1, and there are a lot of examplessupporting the Conjecture also for r > Q ] + · · · + [ Q r ] to be torsion, itdoesn’t actually follow from anywhere that one should consider all solutionsof (1.2) in (i). We will see soon that this requirement is indeed too strong.As an example, let us consider matrices of the form A = (cid:18) a − a − a a (cid:19) .The corresponding equations are ( − Q = Q a Q − a , − Q = Q − a Q a , hence 1 − Q Q = (cid:16) Q Q (cid:17) a = Q − Q , (1 − Q )(1 − Q ) = Q Q ,Q + Q = 1 ⇒ [ Q ] + [ Q ] = 0 in B ( C ) . This computation is the same for all values of a and we see from Table 2that indeed we have modular functions for every a .Next, let us look at the table from Theorem 3.4. One can check thatthe matrix A = (cid:18) − / − / (cid:19) satisfies condition (i) of the Conjecture.(All solutions of (1.2) are ( Q , Q ) = ( x, x ) with 1 − x = x / .) However, A = (cid:18) / − / − / / (cid:19) does not satisfy (i), and so we get a counterexampleto Nahm’s conjecture, since there do exist corresponding modular functions.Indeed, consider the corresponding equation:(4.4) ( − Q = Q / Q − / , − Q = Q − / Q / . It is algebraic equation in the variables Q / and Q / . Let t = Q / Q − / .Then we have from the above equations1 − Q Q / = t ⇒ Q / = t − (1 − Q ) , − Q Q / = t − ⇒ Q / = t (1 − Q ) , and we substitute these equalities into Q / = t Q / to get t (1 − Q ) = t t − (1 − Q ) ,t (1 − Q ) = 1 − Q = 1 − t Q ,t = 1 . AHM’S CONJECTURE 17
Consequently, all solutions of (4.4) are ( Q , Q ) = ( x, x ) where x is asolution of 1 − x = tx / for a 4th root of unity t = 1. Equivalently,(1 − x ) = x ⇔ ( x − x + 1)( x − x + 1) = 0 . We see that ( Q , Q ) = (cid:0) √− , √− (cid:1) is a solution of (4.4), and the corre-sponding element 2 (cid:2) √− (cid:3) is not torsion because D (cid:0) √− (cid:1) = 1 . ... .Here D is the Bloch-Wigner dilog (see [8, Chapter I, Section 3]) for whichit is known that D ( x ) = 0 if and only if x ∈ R .A similar thing happens in Table 1: the matrix A = (cid:18) / / / / (cid:19) satisfiesthe Conjecture while A = (cid:18) / / / / (cid:19) is a counterexample. So far wehave two counterexamples, and we notice that both matrices match into thefollowing general pattern. Theorem 4.2.
Let A be a real positive definite symmetric r × r -matrix, B a vector of length r , and C a scalar. For an arbitrary m ≥ we define A ′ = I mr + E m ⊗ ( A − I r ) , B ′ = l mr + e m ⊗ ( B − l r ) , C ′ = C/m, where E m ∈ M m × m ( Q ) such that ( E m ) ij = 1 /m , e m ∈ Q m such that ( e m ) i =1 /m and l r ∈ Q r such that ( l r ) i = i − r − r . Then F A ′ ,B ′ ,C ′ ( q ) = F A,B,C ( q /m ) . Proof.
The proof relies on the following identity q n ( q ) n = X k ∈ ( Z ≥ ) m k + ... + k m = n q m k T k + ml Tm k ( q m ; q m ) k · · · ( q m ; q m ) k m which holds for all n ≥
0. It follows directly if we use (3.4) on both sides inthe trivial identity( − xq / ; q ) ∞ = ( − xq / ; q m ) ∞ ( − xq / ; q m ) ∞ · · · ( − xq m − / ; q m ) ∞ , and compare the coefficient of x n on both sides.Using the identity we find F A,B,C ( q ) = X n ∈ ( Z ≥ ) r q n T An + n T B + C ( q ) n . . . ( q ) n r = X n ∈ ( Z ≥ ) r q n T ( A − I r ) n + n T B + C X K ∈ M r × m ( Z ≥ ) mKe m = n q m || K || + mre Tr Kl m ( q m ; q m ) K , where || K || = P ri =1 P mj =1 K ij and ( q ; q ) K = Q ri =1 Q mj =1 ( q ; q ) K ij . Nowchanging the order of summation we get that this equals X K ∈ M r × m ( Z ≥ ) q m e Tm K T ( A − I r ) Ke m + m || K || + me Tm K T B + mre Tr Kl m + C ( q m ; q m ) K . If we turn the r × m matrix K into a vector of length rm by puttingthe columns of K under each other, we can recognize this last sum as F A ′ ,B ′′ ,C ′ ( q m ), where A ′ and C ′ are as in the theorem and B ′′ = e m ⊗ B + rl m ⊗ e r . We can easily verify that rl m ⊗ e r = l mr − e m ⊗ l r , which gives B ′′ = B ′ , with B ′ as in the theorem. So we have found F A,B,C ( q ) = F A ′ ,B ′ ,C ′ ( q m ) . Now replacing q by q /m gives the desired result. (cid:3) Let us take r = 1 and m = 2. Then A = 12 A ′ = (cid:18) / − / − / / (cid:19) A = 2 A ′ = (cid:18) / / / / (cid:19) and the theorem produces modular functions for these 2 × r = 1. One can construct more counterexamples withhigher r using Theorem 4.2.Finally, we would like to give one more counterexample, this time suchthat A has integer entries. Let A = , B = 12 − , C = 115 . All solutions of (1.2) in this case are( Q , Q , Q , Q ) = (cid:16) u, u,
11 + u ,
11 + u (cid:17) with 1 − u = u and ( Q , Q , Q , Q ) = (cid:16) u, − u,
11 + u , − u (cid:17) with 1 − u = − u . It is easy to check that solutions of the first type give torsion elements inthe Bloch group, while ones of the second type give non-torsion elements.On the other hand, we have that F A,B,C ( q ) = η (2 z ) θ , ( z ) η ( z ) . We get this identity by applying the theorem below to A = (cid:16) / / / / (cid:17) , B = (cid:16) / − / (cid:17) and C = − / Theorem 4.3.
Let A be a real positive definite symmetric r × r -matrix, B a vector of length r , and C a scalar. Let A ′ , B ′ and C ′ be the symmetric r × r -matrix, the vector of length r and the scalar, resp. , given by A ′ = (cid:18) A I r I r I r (cid:19) , B ′ = B ... , C ′ = 2 C + r , AHM’S CONJECTURE 19 then F A ′ ,B ′ ,C ′ ( q ) = η (2 z ) r η ( z ) r F A,B,C ( q ) . Proof.
Using ( q ; q ) n = ( q ; q ) n ( − q ; q ) n , ( q ; q ) ∞ = ( q ; q ) ∞ ( − q ; q ) ∞ and(3.4), we see that( q ; q ) ∞ ( q ; q ) ∞ q ; q ) n = ( − q ; q ) ∞ ( q ; q ) n ( − q ; q ) n = ( − q n +1 ; q ) ∞ ( q ; q ) n = 1( q ) n X k ≥ q k + k + nk ( q ) k , and so( q ; q ) r ∞ ( q ) r ∞ F A,B,C ( q )= X n ∈ ( Z ≥ r q n T An +2 n T B +2 C ( q ) n · · · ( q ) n r X k ∈ ( Z ≥ r q k T k + n T k + ( k + k + ... + k r ) ( q ) k · · · ( q ) k r = X n,k ∈ ( Z ≥ r q n T An + k T k + n T k +2 n T B + ( k + k + ... + k r )+2 C ( q ) n · · · ( q ) n r ( q ) k · · · ( q ) k r . If we turn the two vectors n and k into one vector of length 2 r by putting k below n , we can recognize this last sum as q − r/ F A ′ ,B ′ ,C ′ ( q ), where A ′ , B ′ and C ′ are as in the theorem. So we have found( q ; q ) r ∞ ( q ) r ∞ F A,B,C ( q ) = q − r/ F A ′ ,B ′ ,C ′ ( q ) . Multiplying both sides by q r/ gives the desired result. (cid:3) References [1] W. Bosma, J. Cannon, and C. Playoust.
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Max-Planck-Institut f¨ur Mathematik, Vivatsgasse 7, 53111 Bonn, GermanyMathematisches Institut, Universit¨at zu K¨oln, Weyertal 86–90, 50931 K¨oln,Germany
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