58.1. PERIODIC FUNCTIONS 1833

Lemma 58.1.16 λ (τ) = λ (τ ′) if and only if

τ′ =

aτ +bcτ +d

where 58.1.12 holds.

Proof: It only remains to verify that if ℘(w′1/2) =℘(w1/2) then it is necessary that

w′12− w1

2∈M

with a similar requirement for w2 and w′2. If w′12 −

w12 /∈ M, then there exist integers, m,n

such that−w′1

2+mw1 +nw2

is in the interior of P0, the period parallelogram whose vertices are 0,w1,w1 +w2, andw2. Therefore, it is possible to choose small a such that Pa contains the pole, 0, w1

2 , and−w′1

2 +mw1 + nw2 but no other poles of ℘ and in addition, ∂P∗a contains no zeros of z→℘(z)−℘

(w12

). Then the order of this elliptic function is 2. By assumption, and the fact

that ℘ is even,

(−w′1

2+mw1 +nw2

)=℘

(−w′1

2

)=℘

(w′12

)=℘

(w1

2

).

It follows both −w′12 +mw1 + nw2 and w1

2 are zeros of ℘(z)−℘(w1

2

)and so by Theorem

58.1.7 on Page 1820 these are the only two zeros of this function in Pa. Therefore, fromCorollary 58.1.9 on Page 1822

w1

2− w′1

2+mw1 +nw2 ∈M

which shows w12 −

w′12 ∈M. This completes the proof of the lemma.

Note the condition in the lemma is equivalent to the condition 58.1.13 because you canrelabel the coefficients. The message of either version is that the coefficient of τ in thenumerator and denominator is odd while the constant in the numerator and denominator iseven.

Next,(

1 02 1

)∼(

1 00 1

)mod2 and therefore,

λ

(2+ τ

1

)= λ (τ +2) = λ (τ) . (58.1.14)

Thus λ is periodic of period 2.Thus λ leaves invariant a certain subgroup of the unimodular group. According to the

next definition, λ is an example of something called a modular function.

58.1. PERIODIC FUNCTIONS 1833Lemma 58.1.16 A(t) = A(t’) ifand only ifve at+b~ ct+dwhere 58.1.12 holds.Proof: It only remains to verify that if ga(w’, /2) = g@(w1/2) then it is necessary thatwi WI<t_“lem2 2J ) . .with a similar requirement for w2 and w4. If 3 —“} ¢ M, then there exist integers, m,nsuch that ,Mal2is in the interior of Po, the period parallelogram whose vertices are 0,w ,w ; + w2, andw2. Therefore, it is possible to choose small a such that P, contains the pole, 0, 3, and_ yl- | + mw, +nw? but no other poles of 9 and in addition, dP} contains no zeros of z >P(z)- (4). Then the order of this elliptic function is 2. By assumption, and the factthat gis even,= _ 4 (2) = o(%) = o(™o( 5) +m +n) = 0( 5} )=0(3)=0(3).It follows both ++ + mw) +nw and “4 are zeros of 49(z) — g9(%) and so by Theorem58.1.7 on Page 1820 these are the only two zeros of this function in P,. Therefore, fromCorollary 58.1.9 on Page 1822+ mw, +nw21M wy tn, €M—=—_-- mw nw:2 2 1 2which shows +} — mt € M. This completes the proof of the lemma.Note the condition in the lemma is equivalent to the condition 58.1.13 because you canrelabel the coefficients. The message of either version is that the coefficient of t in thenumerator and denominator is odd while the constant in the numerator and denominator iseven.1 0 1 ONext, ( > 4 ) ~ ( 04 ) moa2 and therefore,(A) =2(t+2) =A(2). (58.1.14)Thus A is periodic of period 2.Thus A leaves invariant a certain subgroup of the unimodular group. According to thenext definition, A is an example of something called a modular function.