58.1. PERIODIC FUNCTIONS 1847
Now as discussed earlier
λ(τ′) = λ (φ (τ))≡
℘
(w′1+w′2
2
)−℘
(w′22
)℘
(w′12
)−℘
(w′22
)=
℘
(1+τ ′
2
)−℘
(τ ′2
)℘( 1
2
)−℘
(τ ′2
)These numbers in the above fraction must be the same as ℘
( 1+τ
2
),℘(
τ
2
), and ℘
( 12
)but
they might occur differently. This is because ℘ does not change and these numbers arethe zeros of a polynomial having coefficients involving only numbers and ℘(z) . It couldhappen for example that ℘
(1+τ ′
2
)=℘
(τ
2
)in which case this would change the value of
λ . In effect, you can keep track of all possibilities by simply permuting the ei in the formulafor λ (τ) given by e3−e2
e1−e2. Thus consider the following permutation table.
1 2 32 3 13 1 22 1 31 3 23 2 1
.
Corresponding to this list of 6 permutations, all possible formulas for λ (φ (τ)) can beobtained as follows. Letting τ ′ = φ (τ) where φ is a unimodular matrix corresponding to achange of basis,
λ(τ′)= e3− e2
e1− e2= λ (τ) (58.1.34)
λ(τ′)= e1− e3
e2− e3=
e3− e2 + e2− e1
e3− e2= 1− 1
λ (τ)=
λ (τ)−1λ (τ)
(58.1.35)
λ(τ′) =
e2− e1
e3− e1=−
[e3− e2− (e1− e2)
e1− e2
]−1
= − [λ (τ)−1]−1 =1
1−λ (τ)(58.1.36)
λ(τ′) =
e3− e1
e2− e1=−
[e3− e2− (e1− e2)
e1− e2
]= − [λ (τ)−1] = 1−λ (τ) (58.1.37)
λ(τ′)= e2− e3
e1− e3=
e3− e2
e3− e2− (e1− e2)=
11− 1
λ (τ)
=λ (τ)
λ (τ)−1(58.1.38)
λ(τ′)= e1− e3
e3− e2=
1λ (τ)
(58.1.39)