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23 Weierstrass Elliptic and Modular FunctionsWeierstrass Elliptic Functions

§23.5 Special Lattices


§23.5(i) Real-Valued Functions

The Weierstrass functions take real values on the real axis iff the lattice is fixed under complex conjugation: 𝕃=𝕃¯; equivalently, when g2,g3. This happens in the cases treated in the following four subsections.

§23.5(ii) Rectangular Lattice

This occurs when both ω1 and ω3/i are real and positive. Then Δ>0 and the parallelogram with vertices at 0, 2ω1, 2ω1+2ω3, 2ω3 is a rectangle.

In this case the lattice roots e1, e2, and e3 are real and distinct. When they are identified as in (23.3.9)

23.5.1 e1 >e2>e3,
e1 >0>e3.

Also, e2 and g3 have opposite signs unless ω3=iω1, in which event both are zero.

As functions of ω3, e1 and e2 are decreasing and e3 is increasing.

§23.5(iii) Lemniscatic Lattice

This occurs when ω1 is real and positive and ω3=iω1. The parallelogram 0, 2ω1, 2ω1+2ω3, 2ω3 is a square, and

23.5.2 η1=iη3=π/(4ω1),
23.5.3 e1 =-e3=(Γ(14))4/(32πω12),
e2 =0,
23.5.4 g2 =(Γ(14))8/(256π2ω14),
g3 =0.

Note also that in this case τ=i. In consequence,

23.5.5 k2 =12,
K(k) =K(k)

§23.5(iv) Rhombic Lattice

This occurs when ω1 is real and positive, ω3>0, ω3=12ω1, and Δ<0. The parallelogram 0, 2ω1-2ω3, 2ω1, 2ω3, is a rhombus: see Figure 23.5.1.

The lattice root e1 is real, and e3=e¯2, with e2>0. e1 and g3 have the same sign unless 2ω3=(1+i)ω1 when both are zero: the pseudo-lemniscatic case. As a function of e3 the root e1 is increasing. For the case ω3=eπi/3ω1 see §23.5(v).

§23.5(v) Equianharmonic Lattice

This occurs when ω1 is real and positive and ω3=eπi/3ω1. The rhombus 0, 2ω1-2ω3, 2ω1, 2ω3 can be regarded as the union of two equilateral triangles: see Figure 23.5.2.

See accompanying text
Figure 23.5.1: Rhombic lattice. (2ω3)=ω1. Magnify
See accompanying text
Figure 23.5.2: Equianharmonic lattice. 2ω3=eπi/32ω1, 2ω1-2ω3=e-πi/32ω1. Magnify
23.5.6 η1=eπi/3η3=π23ω1,

and the lattice roots and invariants are given by

23.5.7 e1=e2πi/3e3=e-2πi/3e2=(Γ(13))6214/3π2ω12,
23.5.8 g2 =0,
g3 =(Γ(13))18(4πω1)6.

Note also that in this case τ=eiπ/3. In consequence,

23.5.9 k2 =eiπ/3,
K(k) =eiπ/6K(k)