524 XII Elliptic Functions
Proof Consider the second relation. If we use the first and fourth relations of Propo-
sition 4 to evaluate the productsθ 00 (v)θ 00 (w)andθ 01 (v)θ 01 (w), we obtain
θ 00 (v)θ 01 (v)θ 00 (w)θ 01 (w)=θ 002 (v+w; 2 τ)θ^200 (v−w; 2 τ)
−θ 102 (v+w; 2 τ)θ 102 (v−w; 2 τ).Similarly, if we use the second and sixth relations of Proposition 4 to evaluate the
productsθ 10 (v)θ 10 (w)andθ 11 (v)θ 11 (w), we obtain
θ 10 (v)θ 11 (v)θ 10 (w)θ 11 (w)=θ 102 (v+w; 2 τ)θ^200 (v−w; 2 τ)
−θ 002 (v+w; 2 τ)θ 102 (v−w; 2 τ).Hence, in the second relation of the present proposition the right side is equal to
[θ 002 (v+w; 2 τ)+θ 102 (v+w; 2 τ)][θ 002 (v−w; 2 τ)−θ 102 (v−w; 2 τ)].On the other hand, if we use the first and fourth relations of Proposition 4 to evaluate
the productsθ 00 ( 0 )θ 00 (v+w)andθ 01 ( 0 )θ 01 (v−w), we see that the left side is likewise
equal to
[θ 002 (v+w; 2 τ)+θ 102 (v+w; 2 τ)][θ 002 (v−w; 2 τ)−θ 102 (v−w; 2 τ)].This proves the second relation of the proposition, and the others may be proved
similarly.
Corollary 7Fo r a l lv∈Candτ∈H,
θ 002 ( 0 )θ 012 (v)+θ 102 ( 0 )θ 112 (v)=θ 012 ( 0 )θ 002 (v), (39)
θ 102 ( 0 )θ 012 (v)+θ 002 ( 0 )θ 112 (v)=θ 012 ( 0 )θ 102 (v). (40)Moreover, for allτ∈H,
θ^400 ( 0 )=θ 014 ( 0 )+θ 104 ( 0 ). (41)Proof We get (39) and (40) from the first relation of Proposition 6 by takingw= 1 / 2
andw=( 1 +τ)/2 respectively. We obtain (41) from (39) by takingv= 1 /2.
If we regard (39) and (40) as a system of simultaneous linear equations for the
unknownsθ 012 (v),θ 112 (v), then the determinant of this system isθ 004 ( 0 )−θ 104 ( 0 )=
θ 014 ( 0 )=0. It follows that the square of any theta function may be expressed as a
linear combination of the squares of any other two theta functions.
By substituting for the theta functions their expansions as infinite products, the
formula (41) may be given the following remarkable form:
∏∞n= 1( 1 +q^2 n−^1 )^8 =∏∞
n= 1( 1 −q^2 n−^1 )^8 + 16 q∏∞
n= 1( 1 +q^2 n)^8.