2 A Theorem in Number Theory

Let 0< ϕ < ψ < π/2 denote two angles. Then 0<cot(ψ)<cot(ϕ)<∞. Is it possible that

cot(ϕ) = 2 cot(ψ), (1)

and yet ϕ and ψ are both rational multiples of π? We prove the following theorem. It says that, with exactly one obvious exception, it is not possible that both the ratio of the cotangent values of ϕ and ψ is rational, and the two angles are rational multiples of π. In particular (1) is not possible when both ϕ and ψ are rational multiples of π. This incommensurability will be used to prove a key complexity reduction to reach our complexity trichotomy classification.

I Theorem 1. Suppose 0 < ϕ < ψ < π/2, and cot(ϕ) = rcot(ψ), for some r ∈ Q and r6= 3. Thenϕandψ are not both rational multiples ofπ.

Proof. We first note that the exceptionr= 3 is witnessed by cot(π/6) = 3 cot(π/3).

We write r= ^a_b for relatively prime integersa andb. We are given cot(ϕ) = ^a_bcot(ψ).

For a contradiction, supposeϕandψare both rational multiples ofπ, and we writeϕ= ^kπ_n andψ=^k_n^0π0 , where 1≤k < ⁿ₂, 1≤k⁰<ⁿ₂⁰, and gcd(k, n) = gcd(k⁰, n⁰) = 1.

Letζn= exp(2πi/n) be a primitive root of unity. Then it is easy to verify thaticot(ϕ) =

1+ζ^k_n

1−ζ^k_n. If we writet =icot(ϕ), then t ∈ Φ_n = Q(ζ_n), the n-th cyclotomic field (the field extension by adjoiningζn to Q). Also ζ_n^k = ^t−_t+1¹. As gcd(k, n) = 1, we have Φn =Q(ζ_n^k)⊆ Q(t)⊆Φn, and soQ(t) = Φn.

By cot(ϕ) = ^a_bcot(ψ), we have Φn⁰ = Φn. It is well known that this implies that either n=n⁰, or nis odd andn⁰= 2n, orn⁰ is odd andn= 2n⁰.

I T C S 2 0 1 8

2:4 A Complexity Trichotomy fork-Regular Asymmetric Spin Systems

We first consider the casen=n⁰. This case actually follows from Siegel’s theorem [12].

For a uniform treatment we give a direct proof here.

For n = n⁰, we have 1 ≤ k < k⁰ < ⁿ₂, and so n ≥5. A Dirichlet character χ to the modulusn is a function fromZ to Cthat is multiplicative, has period n, andχ(j)6= 0 iff gcd(j, n) = 1. Soχis extended from a group character onZ^×n (i.e., a multiplicative function taking nonzero values in C on Z^×n ={(jmodn) ∈ Zn | gcd(j, n) = 1}), and all nonzero values ofχare roots of unity. A Dirichlet characterχis said to be odd ifχ(−1) =−1.

We need to take an odd Dirichlet character χ to the modulus n. An odd Dirichlet character χ (for n > 2) exists: The group of Dirichlet characters mod n is isomorphic to Z^×n. Since n > 2, {1,−1} is a subgroup of order two in Z^×n. The subgroup of even characters is isomorphic to Z^×n/{1,−1}. Hence, not every Dirichlet character mod n is even. A more constructive proof is as follows: The character group of Z^×n, by Chinese remaindering, is a direct product of the character groups of Z^×_pei

i

according to the prime factorization n = Q

ip^e_iⁱ. For each odd prime p_i, the group Z^×_pei i

is cyclic of even order m=φ(p^e_iⁱ) = (p_i−1)p^e_iⁱ⁻¹. Let ρbe a generator, thenρ^m/2=−1. Thus we can define a characterχ onZ^×_pei

i

byχ(ρ^j) = ζ_m^j . Thenχ(−1) =−1. If 4 |n, thenZ^×n has a factorZ^×₂^e fore≥2, which is isomorphic to Z2⊕Z₂^e−2 as an additive group, with generators {−1,5}.

Thus everyj∈Z^×₂^eis uniquely expressed as (−1)^u5^vmod 2^e, foru= 0,1, and 0≤v <2^e−2. Then an odd characterχ onZ^×₂^e can be defined by χ((−1)^u5^v) = (−1)^u. If n≡2 mod 4, thenZ^×n has a trivial factorZ^×₂ = 1, and the character group ofZ^×n is isomorphic to that of Z^×_n/2, wheren/2>1 is odd. Hence by Chinese remaindering, we can define an odd Dirichlet characterχonZ^×n.

An important notion we will use in this proof is that of Leopoldt’s character coordin- ates [21, 15]. In our case, for any odd Dirichlet character χ to the modulus n, the fol- lowing can be taken as the definition of Leopoldt’s character coordinatesy(χ| t)∈C, for t=icot(^kπ_n )∈Φ_n,

y(χ|t)τ(χd|1) =

n

X

j=1

χ(j)σj(t), (2)

wheredis theconductor ofχ, ² χ_d is the induced primitive character ofχ modd, overline denotes complex conjugation, the valueτ(χd | 1) =Pd

j=1χd(j)e^−2πij/d is the Gauss sum, andσj is the automorphism in the Galois groupG= Gal(Φn/Q) that maps ζn to ζ_n^j. The sumPn

j=1χ(j)σj(t) is actually over relatively prime integers j ∈ Z^×n, since otherwise the Dirichlet character χ(j) = 0. In the expressiont=icot(^kπ_n )∈Φn, we have gcd(k, n) = 1, and soσ_k ∈G, andt=σ_k(t₁) wheret₁=icot(^π_n). For any fixedk∈Z^×n,σ_j◦σ_k =σ_jkruns through allGwhenj runs throughZ^×n. Then

n

X

j=1

χ(j)σj(t) =χ(k) X

j∈Z^×n

χ(k)⁻¹χ(j)σj(σk(t1)) =χ(k) X

j∈Z^×n

χ(kj)σjk(t1) =χ(k) X

j∈Z^×n

χ(j)σj(t1).

Hence y(χ | t) = χ(k) y(χ | t₁), as the Gauss sum τ(χd | 1) 6= 0. Similarly for t⁰ = icot(^k_n^0π), (recall we have n⁰ = n), y(χ | t⁰) = χ(k⁰) y(χ | t₁). Hence the norm of the two Leopoldt’s character coordinates are equal, |y(χ | t)| = |y(χ | t⁰)|. However if

2 See [1], p. 167–p. 171 for the definitions of induced characters and the conductor of a character. In number theory it is traditional to denote the conductor of a character byf as is written in [15]; we use dhere in order not to confuse it with the constraint functionf(u, v) in Section 1.

J. Y. Cai, Z. G. Fu, K. Girstmair, and M. Kowalczyk 2:5

t⁰ = rt, where 0 < r ∈ Q, then y(χ | t⁰) = ry(χ | t) by (2), and so r = 1 (we will see that y(χ | t) = χ(k) y(χ | t₁) 6= 0). This contradicts r > 1, which is a consequence of cot(ϕ)>cot(ψ) by 0< ϕ < ψ < π/2.

Now we assumenis odd andn⁰ = 2n. We want to take an odd Dirichlet characterχto the modulus 2n. Sincenis odd, the character groups ofZ^×n andZ^×₂nare isomorphic, namely for everyj∈Z^×n exactly one ofj orj+nis odd and so belongs toZ^×₂n. Sincen >1 is odd, in its prime factorizationn=Q

ip^e_iⁱ, everypi is odd. Then Z^×n ∼=Y

i

Z^×_pei i

, and everyZ^×_pei i

is cyclic of even orderφ(p^e_iⁱ) = (pi−1)p^e_iⁱ⁻¹. So we can define an odd Dirichlet characterχ onZ^×n by Chinese remaindering, by defining it to be odd on eachZ^×_pei

i

, namelyχ(−1) =−1.

In particular there is an odd characterχto the modulus 2n. Sincenis an induced modulus, and odd, the conductordofχis also odd.

Take any odd Dirichlet character χ mod 2n. It is proved in [15] (Theorem 2, p. 380) that

y(χ|icot(π

2n)) = 4n d

Y

p|2n

1−χ_d(p) p

!

B_χd, (3)

and

y(χ|icot(π

n)) = 2n d

Y

p|n

1−χ_d(p) p

!

B_χd. (4)

Here B_χd is the generalized Bernoulli number. (Eqn. (4) is proved in Theorem 2 of [15]

for any non-principalχmodnwithout requiring nbeing odd, and so the proof below that y(χ|icot(^π_n))6= 0 is also valid for the previous casen=n⁰.)

By definition the Bernoulli polynomial B(Z) is the firstB⁽¹⁾(Z) defined by te^Zt

e^t−1 =

∞

X

m=0

B^(m)(Z)t^m/m!. (5)

And the generalized Bernoulli numberBχ_d is defined by

d

X

j=1

χd(j) te^jt e^dt−1 =

∞

X

m=0

B_χ^(m_d⁾t^m/m!, (6)

withB_χd=B⁽¹⁾χ_d. It follows immediately from (5) (and is also well known) thatB(Z) =Z−¹₂. SubstitutingtbydtandZ byj/din (5), we get the following equality from (5) and (6)

Bχ_d=

d

X

j=1

χd(j)B(j/d). (7)

It follows easily from the definition thatPd

j=1χd(j) = 0. (This uses the fact thatχd is not principal, namely not identically 1 onZ^×d ={jmodd|gcd(j, d) = 1}; indeedχ_d(−1) =−1, and χd(j) = 0 if gcd(j, d) >1, and so P0

jχd(j) = P0

jχd(−j) = −P0

jχd(j), where each sumP0

j is overZ^×d.) It follows thatBχ_d=Pd

j=1χd(j)j/d.

It is a nontrivial fact that Pd

j=1χd(j)j 6= 0 for any odd characterχd (see [28] Theorem 4.9, p. 37). HenceBχ_d 6= 0, and therefore alsoy(χ|icot(^π_n))6= 0 and y(χ|icot(₂^π_n))6= 0.

I T C S 2 0 1 8

2:6 A Complexity Trichotomy fork-Regular Asymmetric Spin Systems

Fort⁰ =icot(^k₂⁰_n^π) andt=icot(^kπ_n), it follows from (3) and (4) that y(χ|t⁰) = χ(k⁰)y(χ|icot(π

2n))

= χ(k⁰)2 1−χd(2) 2

! 2n d

Y

p|n

1−χd(p) p

! Bχ_d

= χ(k⁰)

2−χd(2)

y(χ|icot(π n))

= χ(k⁰)χ(k)

2−χd(2) y(χ|t)

On the other hand, since by assumptiont=^a_bt⁰ for integersaandb, we have y(χ|t) =a

b y(χ|t⁰).

Hence, by being nonzero, and taking the norm squared, we get

b²=a²· |2−χ_d(2)|². (8)

Sinceχdis primitive modd, anddis odd, we haveρ=χd(2)6= 0. Denote this root of unity byρ=χ_d(2). We have

b²=a²[5−2(ρ+ρ)].

If we started with n⁰ odd and n= 2n⁰, we would have the same equation withaandb exchanged.

a²=b²[5−2(ρ+ρ)].

Ifρ= 1 thena=b, this is a contradiction toϕ6=ψ. Ifρ=−1 thenb²= 9a² or a²= 9b². This gives us the unique exceptional caseϕ=π/6 andψ=π/3.

Back to n⁰ = 2n with n odd; the other case being symmetric. Suppose ρ 6=±1, then it is a nonreal algebraic integer, and satisfies the equation 2a²(ρ²+ 1) = ρ(5a²−b²). Its minimal polynomial is monic with integer coefficients. Hence 2a² |(5a²−b²). Hencea|b.

Since gcd(a, b) = 1, we get a= 1. Back to (8) we get b <3, sinceρ6=±1. And sob = 2.

But in this case the solution (1±√

15i)/4 to 2(ρ²+ 1) =ρis not a root of unity. J We will use Theorem 1 to prove a key complexity reduction, stated in Lemma 18, after we formally define Holant problems and reductions in Section 3.

No documento 9th Innovations in Theoretical Computer Science (páginas 34-37)