4 Theta Functions 523Proof From the definition ofθ 00 ,
θ 00 (v;τ)θ 00 (w;τ)=∑
j,keπiτ(j(^2) +k (^2) )
e^2 πivje^2 πiwk=
∑
j+keven+
∑
j+kodd.
In the first sum on the right we can writej+k= 2 m,j−k= 2 n.Thenj=m+n,
k=m−nand
∑
j+keven=
∑
m,n∈Ze^2 πiτ(m(^2) +n (^2) )
e^2 πi(v+w)me^2 πi(v−w)n
=θ 00 (v+w; 2 τ)θ 00 (v−w; 2 τ).
In the second sum we can writej+k= 2 m+1,j−k= 2 n+1. Thenj=m+n+ 1 ,
k=m−nand
∑
j+kodd
=
∑
m,n∈Ze^2 πiτ{(m+^1 /^2 )(^2) +(n+ 1 / 2 ) (^2) }
e^2 πiv(m+n+^1 )e^2 πiw(m−n)
=θ 10 (v+w; 2 τ)θ 10 (v−w; 2 τ).
Adding, we obtain the firstrelation of the proposition.
We obtain the second relation from the first by replacingvbyv+τ/2andwby
w+τ/2. The remaining relations are obtained from the first two by increasingvand/or
wby 1/2.
By takingw=vin Proposition 4, and adding or subtracting pairs of equations
whose right sides differ only in one sign, we obtain theduplication formulas:
Proposition 5Fo r a l lv∈Candτ∈H,
θ 00 ( 2 v; 2 τ)=[θ 002 (v;τ)+θ 012 (v;τ)]/ 2 θ 00 ( 0 ; 2 τ)
=[θ 102 (v;τ)−θ 112 (v;τ)]/ 2 θ 10 ( 0 ; 2 τ),
θ 10 ( 2 v; 2 τ)=[θ 002 (v;τ)−θ 012 (v;τ)]/ 2 θ 10 ( 0 ; 2 τ)
=[θ 102 (v;τ)+θ 112 (v;τ)]/ 2 θ 00 ( 0 ; 2 τ),
θ 01 ( 2 v; 2 τ)=θ 00 (v;τ)θ 01 (v;τ)/θ 01 ( 0 ; 2 τ),
θ 11 ( 2 v; 2 τ)=θ 10 (v;τ)θ 11 (v;τ)/θ 01 ( 0 ; 2 τ).
From Proposition 4 we can also derive the followingaddition formulas:
Proposition 6Fo r a l lv,w∈Candτ∈H,
θ 012 ( 0 )θ 01 (v+w)θ 01 (v−w)
=θ 012 (v)θ 012 (w)−θ 112 (v)θ 112 (w)=θ 002 (v)θ 002 (w)−θ 102 (v)θ 102 (w),
θ 00 ( 0 )θ 01 ( 0 )θ 00 (v+w)θ 01 (v−w)
=θ 00 (v)θ 01 (v)θ 00 (w)θ 01 (w)+θ 10 (v)θ 11 (v)θ 10 (w)θ 11 (w),
θ 01 ( 0 )θ 10 ( 0 )θ 10 (v+w)θ 01 (v−w)
=θ 01 (v)θ 10 (v)θ 01 (w)θ 10 (w)+θ 00 (v)θ 11 (v)θ 00 (w)θ 11 (w),
θ 00 ( 0 )θ 10 ( 0 )θ 11 (v+w)θ 01 (v−w)
=θ 01 (v)θ 11 (v)θ 00 (w)θ 10 (w)+θ 00 (v)θ 10 (v)θ 01 (w)θ 11 (w),
where all theta functions have the same second argumentτ.