Vanishing Sums of Small Length

In document Complex hyperbolic triangle groups (Page 96-103)

2.4 Vanishing Sums of Roots of Unity

2.4.2 Vanishing Sums of Small Length

To prove Theorem 2.4.3.1 we first prove the following theorem, which is an exten- sion of Theorem 6 of [2] and Theorem A.3.1 of [21].

Notation. From this point on we shall denote the map Lα ∈ V by α.

Theorem 2.4.2.1. Let α, β, γ, δ and ξ be primitive roots of unity of orders 3, 5, 7, 11 and 13 respectively. Let S be a non-empty vanishing sum of length at most 15. Then one of the following is true:

a) For some root of unity θ, the sum S involves one of the five sets of roots of unity:

{θ, αθ, α2θ}, {θ, βθ, β2θ, β3θ, β4θ}, {αθ, α2θ, βθ, β2θ, β3θ, β4θ},

2.4. VANISHING SUMS OF ROOTS OF UNITY 97

b) The sum S is similar to one of the following sums:

1 +γ2+γ3+γ4+γ5−(α+α2)(γ+γ6) −(β+β2 +β3+β4)+γ+γ2+γ3+γ4+γ5+γ6 1 +δ+δ2+δ3+δ4+δ5+δ6+δ7+δ8+δ9+δ10 1 +γ+γ6−(α+α2)(γ2+γ3 +γ4 +γ5) −(α+α2)+δ+δ2+δ3+δ4+δ5+δ6+δ7+δ8+δ9+δ10 −(β+β2 +β3+β4)+γ2+γ3 +γ4+γ5−(α+α2)(γ +γ6) (α+α2)(β+β4)−(β2+β3)+γ +γ2+γ3 +γ4 +γ5+γ6 1−(α+α2)(γ+γ2+γ3+γ4+γ5+γ6) 1 +ξ+ξ2+ξ3+ξ4+ξ5+ξ6+ξ7+ξ8+ξ9+ξ10+ξ11+ξ12 1 +γ2+γ3 +γ4+γ5−(β+β2 +β3+β4)(γ+γ6) 1 +δ2+δ3+δ4+δ5+δ6+δ7+δ8+δ9−(α+α2)(δ+δ10) −(β+β2 +β3 +β4) +γ+γ6 (α+α2)(γ2 +γ3+γ4 +γ5) −(α+α2) +γ2 +γ3 +γ4+γ5 (β+β2 +β3+β4)(γ+γ6) (α+α2)(β+β2 +β3+β4) +γ+γ2 +γ3+γ4 +γ5 +γ6 −(α+α2)(1 +δ+δ10) +δ2+δ3+δ4+δ5+δ6+δ7+δ8+δ9 −(β+β2 +β3 +β4) +δ +δ2 +δ3 +δ4 +δ5 +δ6 +δ7+δ8+δ9+δ10 −(α+α2) +ξ+ξ2+ξ3 +ξ4 +ξ5+ξ6 +ξ7+ξ8 +ξ9 +ξ10+ξ11+ξ12 1 +β2 +β3(β+β4)(γ +γ2+γ3 +γ4 +γ5+γ6) 1−(α+α2)(γ+γ6)(β+β2 +β3 +β4)(γ2 +γ5) +γ3 +γ4 1−(α+α2)(γ+γ6)(β+β2 +β3 +β4)(γ3 +γ4) +γ2 +γ5 1 +δ3 +δ4+δ5+δ6+δ7+δ8(α+α2)(δ+δ2+δ9+δ10) 1 +δ2 +δ4+δ5+δ6+δ7+δ9(α+α2)(δ+δ3+δ8+δ10) 1 +ξ2 +ξ3 +ξ4+ξ5 +ξ6+ξ7 +ξ8 +ξ9+ξ10+ξ11(α+α2)(ξ +ξ12).

98CHAPTER 2. TRIANGLE GROUPS WITH TWO ELLIPTIC ELEMENTS

Proof. First suppose thatS is minimal. ThenS is similar to a minimal vanishing sum with e(S) = r(S). In this case we can assume that its reduced exponent is square-free and odd, by Proposition 2.4.1.1 and the ensuing observation. Letpbe the largest prime divisor ofr(S) so that by Proposition 2.4.1.4, p≤13. Let ω be a primitivepth root of unity and expressS as,

S0+ωS1 +...+ωp−1Sp−1,

whereSi have exponents prime to pand remembering that ω=Lω.

If p= 3, then S can involve only 1, α, α2 and so it must be that S is similar toS= 1 +α+α2.

If p≥5, there are two cases, either all of the Si have more than one term or

there is at least one Si with length one.

Let us first assume that all of the Si have more than one term. If p = 5

we haveS= S0+βS1 +...+β4S4, therefore either all Si have length three or

there is an Si with length two, otherwise l(S) > 15. Since Si all have the same

value, by Proposition 2.4.1.3, no Si has value zero otherwise S is not minimal.

Therefore we have five non-vanishing formal sums Si, all of length two or three,

only involving 1, α, α2.

If one of 1,α,α2is involved in all five sums thenSinvolves{θ, βθ, β2θ, β3θ, β4θ} for some rootθ. If none of 1, α, α2 is involved in all five sums, then at least one is involved in four of them. By multiplying by the appropriate root of unity, we may assume that 1 is involved inS1,S2,S3 and S4. Thenα andα2 are involved

inS0 and hence S involves {αθ, α2θ, βθ, β2θ, β3θ, β4θ} for some rootθ.

If p = 7 and all Si have more than one term, then either all Si have length

two or there is one Si with length three and the rest have length two, otherwise

l(S) > 15. Since the Si all have the same value, by Proposition 2.4.1.3, no Si

has value zero otherwise S is not minimal. If some Si 6=Sj, then Si−Sj is a

2.4. VANISHING SUMS OF ROOTS OF UNITY 99

then S involves {θ, γθ, γ2θ, γ3θ, γ4θ, γ5θ, γ6θ}, for some root θ. If l(S

i −Sj) = 5

then Si −Sj is

kε(1 +β+β2 +β3+β4),

for some k ∈ Q and a root of unity ε. Suppose kε 6= 1, then S would involve εωi,βεωi, β2εωi,β3εωi,β4εωi i.e. S involves{θ, βθ, β2θ, β3θ, β4θ}for θ =εωi. If kε= 1 then S is similar to

1 +β2+β3−(β+β4)(γ+γ2+γ3+γ4+γ5+γ6).

If p ≥ 11 and all of the Si have more than one term, then l(S) ≥ 2p ≥ 22.

Therefore it must be that some Si has at most one term. We may now therefore

assume that there is at least one Si with length one. Since Si all have the same

value, by Proposition 2.4.1.3, noSihas value zero otherwiseSis not minimal and,

without loss of generality, we may assume that S0 = 1 and hence all theSi have

value 1. If some Si 6= 1, then 1−Si is a vanishing sum with reduced exponent

prime to p. Note that all Si are of length at least one, hence for each Si we have

l(Si)≤l(S)−(p−1)≤16−p.

Therefore

l(1−Si)≤16−p+ 1, if Si is not monic and hence1−Si is monic

and l(1−Si)≤16−p−1, if Si is monic and hence 1−Si is not monic.

If p= 5 then it is clear that r(1−Si) = 3. Therefore1−Si is

100CHAPTER 2. TRIANGLE GROUPS WITH TWO ELLIPTIC ELEMENTS

for some k ∈ Q and a root of unity ε. If p = 7, then l(1−Si) ≤ 10 if 1−Si

is monic and l(1−Si) ≤ 8 if 1−Si is not monic with r(1−Si) = 3, 5 or 15.

Therefore1−Si is one of the following sums:

kε(1 +α+α2)

kε(1 +β+β2+β3 +β4)

or kε(1 +β2 +β3(α+α2)(β+β4))

for somek ∈Q and a root of unity ε. Otherwise S involves {θ, αθ, α2θ},

{θ, βθ, β2θ, β3θ, β4θ} or {αθ, α2θ, βθ, β2θ, β3θ, β4θ} for some root θ.

If p= 11, thenl(1−Si)≤6 if 1−Si is monic and l(1−Si)≤4 if 1−Si

is not monic. Vanishing sums of length less than or equal to six are covered by Conway and Jones and we see that 1−Si is one of the following sums:

kε(1 +α+α2)

or kε(1 +β+β2+β3 +β4)

for some k ∈ Q and a root of unity ε. If p= 13, then l(1−Si) ≤4 if 1−Si is

monic and l(1−Si) ≤ 2 if 1−Si is not monic. Vanishing sums of length less

than or equal to four are covered by Conway and Jones and we see that 1−Si

must be

kε(1 +α+α2) for somek ∈Q and a root of unity ε.

2.4. VANISHING SUMS OF ROOTS OF UNITY 101

Therefore 1−Si can only be one of the following sums:

kε(1 +α+α2)

kε(1 +β+β2 +β3+β4)

or kε(1 +β2 +β3 (α+α2)(β+β4))

for some k ∈ Q and a root of unity ε. Suppose 1−Si =kε(1 +α+α2) and

kε 6= 1, then S would involve εωi, αεωi, α2εωi i.e. S involves {θ, αθ, α2θ} for θ = εωi. A similar argument can be used if 1−Si is equal to one of the other

two equations, so that ifkε6= 1 then S is covered in part a) of the theorem. If kε = 1, we substitute our possible Si into the sum S for p = 5, 7, 11 and

p = 13 and obtain sums similar to those given in part b) of the theorem. For example, when p= 13 the sum S is of the form

S= 1 +ξS1+ξ2S2 +...+ξ12S10,

with l(S)≤ 15 and r(1−Si) = 3, hence Si = −α−α2 for at most two Si. If

we first consider the case when Si = 1 for every i, then we just have the sum

1 +ξ+ξ2+ξ3+ξ4+ξ5+ξ6+ξ7+ξ8+ξ9+ξ10+ξ11+ξ12. If Si =−α−α2 for one suchi then S is similar to

−(α+α2)+ξ+ξ2+ξ3+ξ4+ξ5+ξ6+ξ7+ξ8+ξ9+ξ10+ξ11+ξ12.

Finally if Si =−α−α2 for two i then it is similar to

1+ξ2+ξ3+ξ4+ξ5+ξ6+ξ7+ξ8+ξ9+ξ10+ξ11−(α+α2)(ξ+ξ12).

102CHAPTER 2. TRIANGLE GROUPS WITH TWO ELLIPTIC ELEMENTS

theorem. At least one such subsum has at most seven terms, and thereforeSmust involve one of the following sets of roots:

{θ, αθ, α2θ}, {θ, βθ, β2θ, β3θ, β4θ}, {αθ, α2θ, βθ, β2θ, β3θ, β4θ},

{θ, γθ, γ2θ, γ3θ, γ4θ, γ5θ, γ6θ}, {θ, θβ2, θβ3, θαβ, θαβ4, θα2β, θα2β4}

2.4. VANISHING SUMS OF ROOTS OF UNITY 103

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