File:Symmetric Hesse pencil.svg

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Original file (SVG file, nominally 400 × 400 pixels, file size: 40 KB)

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Summary

Description
English: Curves in the Hesse pencil of cubic curves , as seen in a symmetric view of the projective plane in which the lines form an equilateral triangle with the point 1:1:1 at its center, and the line forms the line at infinity.

Different colors represent different curves in the pencil, corresponding to different choices of the parameter :

  • The three black lines forming an equilateral triangle represent the degenerate case corresponding to
  • The black point in the center is an isolated point from the degenerate case , which factors as . It has as its roots the point 1:1:1 (shown, from the quadratic factor) and the line at infinity (not shown, from the linear factor).
  • The blue and green curves represent choices of with , forming two ovals in the real projective plane: a central oval inside the equilateral triangle, and an external oval with three inflection points at infinity, separating it into three real branches. The choices of for the blue curves are (from lighter to darker) 36, 12, 6, 4, and 3. For the green curves, only the central oval is visible; their parameters are (from lighter to darker) 2, 1.5, 1.25, and 1.1
  • The red curves represent choices of with , having only a single oval with the same three inflection points and three real branches. Their choices of are (from lighter to darker) –8, –2, –1, –0.5, 0.
Date
Source Own work
Author David Eppstein
SVG development
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 This vector image was created with Python.

Source code

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Deutsch  English  +/−

# Draw curves in the Hesse pencil x^3 + y^3 + z^3 = 3kxyz
#
# To show off the curve symmetry we use symmetric projective coordinates:
# lines x=0, y=0, and z=0 form an equilateral triangle
# with 0:0:1, 0:1:0, and 1:0:0 as vertices and 1:1:1 at its center.
# For this view the line at infinity is x+y+z = 0.

from PADS.SVG import *
from math import cos,sin,pi
import sys

o = 200+200j
bbox = 400+400j
r = 45
a = o + 0.5j*r + 3**0.5*r/2
b = o + 0.5j*r - 3**0.5*r/2
c = o + -1j*r

def p2c(p): # projective to complex, unused
    px,py,pz = p
    return (px*a+py*b+pz*c)/(px+py+pz)

def c2p(q): # complex to projective
    q = (q-o)/r     # normalize
    z = (-q.imag * 2 + 1)/3
    x = q.real/3**0.5 + (1-z)/2
    y = -q.real/3**0.5 + (1-z)/2
    return (x,y,z)

def between(p,q,s): return (1-s)*p+s*q

def root(p,q,f): # find root of function by binary search
    fp = f(p)
    fq = f(q)
    if fp > fq:
        p,q,fp,fq = q,p,fq,fp
    if fp > 0: return p
    if fq < 0: return q
    for i in range(40):
        mid = (p+q)/2
        if f(mid) < 0:
            p = mid
        else: q = mid
    return between(p,q,mid)

svg = SVG(bbox,sys.stdout)

def hesse(k):
    def thefun(p):
        x,y,z = c2p(p)
        return x**3+y**3+z**3-3*k*x*y*z
    return thefun

def onside(p,q,n):
    return [between(p,q,j/n) for j in range(n)]

def beyond(p,q,n):
    return [between(p,q,1-5**(j/n)) for j in range(1,n)]

# degenerate case, k=infinity, three black lines
svg.group(fill=colors.none,stroke=colors.black)
for p,q in [(a,b),(b,c),(c,a)]:
    svg.segment(6*p-5*q,6*q-5*p)
svg.ungroup()

# other degenerate case, k=1, isolated point (+ line @ infty not shown)
svg.circle(o,3,fill=colors.black,stroke=colors.none)

# case k > 1 has two lobes, the outer one with three real components
def innerlobe(k,pc):
    h = hesse(k)
    curve = []
    for i in 0,1,2:
        p,q = [a,b,c][i-1], [a,b,c][i]
        for tripoint in onside(p,q,9):
            curve.append(root(o,tripoint,h))
    curve = curve + curve[:3]
    svg.polycurve(curve,opacity=pc)

def outerlobe(k,pc):
    h = hesse(k)
    for i in 0,1,2:
        p,q,r = [a,b,c][i],[a,b,c][i-1],[a,b,c][i-2]
        onlobe = root(p,8*p-7*o,h)
        focus = 2*onlobe-p
        wedge = list(reversed(beyond(p,q,20)))+[p]+beyond(p,r,20)
        svg.polycurve([root(focus,w,h) for w in wedge],opacity=pc)

# k >> 1, draw both lobes
svg.group(fill=colors.none,stroke=colors.blue)
# outerlobe(1,"100%")
inners = [3,4,6,12,36]
opacities = ["100%", "85%","70%","55%", "40%"]
for i in range(5):
    innerlobe(inners[i],opacities[i])
    outerlobe(inners[i],opacities[i])
svg.ungroup()

# k > 1 but close to 1, only inner lobe visibl
svg.group(fill=colors.none,stroke=colors.green)
for i in range(4):
    innerlobe([2,1.5,1.25,1.1][i],["40%","60%","80%","100%"][i])
svg.ungroup()

# k < 1, one lobe
def singlelobe(k,pc):
    h = hesse(k)
    for i in 0,1,2:
        p,q,r = [a,b,c][i],[a,b,c][i-1],[a,b,c][i-2]
        midpoint = (p+q)/2
        focus = 15*midpoint-14*o
        border = list(reversed(beyond(p,r,12)))+onside(p,q,12)+[q]+beyond(q,r,12)
        svg.polycurve([root(focus,w,h) for w in border],opacity=pc)

svg.group(fill=colors.none,stroke=colors.red)
outers = [0,-0.5,-1,-2,-8]
opacities = ["100%", "85%","70%","55%", "40%"]
for i in range(5):
    singlelobe(outers[i],opacities[i])
svg.ungroup()
svg.close()

Licensing

I, the copyright holder of this work, hereby publish it under the following license:
Creative Commons CC-Zero This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication.
The person who associated a work with this deed has dedicated the work to the public domain by waiving all of their rights to the work worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law. You can copy, modify, distribute and perform the work, even for commercial purposes, all without asking permission.

Captions

Symmetric view of the curves in the Hesse pencil of cubic curves

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27 October 2025

40,981 byte

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1b96de628250ebf4d0127ef8887b1d18f0b6511d

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Date/TimeThumbnailDimensionsUserComment
current22:28, 27 October 2025Thumbnail for version as of 22:28, 27 October 2025400 × 400 (40 KB)wikimediacommons>David EppsteinUploaded own work with UploadWizard

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