Notes — Book IV: Combinatorics of Dürer’s Nets, $\sqrt[3]{2}$ & Projectivity

1. Exact Algebraic & Topological Audit of Dürer’s Polyhedral Nets

Book IV (pp. 142–157) marks a watershed between Luca Pacioli's Divina Proportione (pacioli:1509)—where Leonardo da Vinci rendered polyhedra as perspective wooden frames—and Johannes Kepler's Harmonices Mundi Book II (kepler:1619), which completed the enumeration of all thirteen Archimedean solids.

In geometor.durer (construct_durer_polyhedral_nets()), every unfolded planar net is modeled as a connected spanning tree of exact regular polygons with unit edge quadrance $Q = 1$ across three nested number fields:

  • Cube ($F = 6, E = 12, V = 8$): pure rational lattice $\mathbb{Q}$, $5$ interior fold hinges + $14$ open cut boundary edges ($5 + \frac{14}{2} = 12$ edges in 3D).

  • Tetrahedron ($4, 6, 4$), Octahedron ($8, 12, 6$), Icosahedron ($20, 30, 12$), and Archimedean Cuboctahedron ($14, 24, 12$): quadratic field $\mathbb{Q}(\sqrt{3})$, verified to have pairwise disjoint face interiors (non_overlapping_verified = True, zero self-overlap—an early constructive witness to Dürer’s Conjecture on edge-unfolding convex polyhedra!).

  • Regular Dodecahedron ($12, 30, 20$, Fig. 33): cyclotomic golden-ratio field $\mathbb{Q}(\sqrt{5}, \sin 36^\circ) = \mathbb{Q}\!\left(\sqrt{5},\, \frac{\sqrt{10 - 2\sqrt{5}}}{4}\right)$, unfolded into two six-pentagon rosettes joined across a shared hinge ($11$ interior fold hinges + $38$ open cut edges $\implies 11 + \frac{38}{2} = 30$ edges, Euler characteristic $\chi = V - E + F = 20 - 30 + 12 = 2$).

Philological Note on Figure 35 (Truncated Tetrahedron, SLUB N1525 p. 149)

On page 149, introducing the first semi-regular solid (Truncated Tetrahedron, [3.6.6]), Dürer’s 1525 prose inadvertently reads “vier sechs ecketter vnd **drey* dryangletter ebner felder”* (“four hexagonal and three triangular plane faces”), even though he immediately states the true 3D counts (zwelf eck vnnd achtzehen scharpfer seyten: $V = 12, E = 18 \implies F = 2 - 12 + 18 = 8$) and his own woodcut Figure 35 accurately draws four hexagons and four triangles! Joachim Camerarius <camerarius> silently restored quatuor triangulis in the 1532 Latin edition (durer:1532:camerarius), and Dürer’s posthumous 1538 edition (durer:1538) added two further Archimedean truncations (bringing Dürer’s total to nine of Archimedes’s thirteen solids).

2. Two Mean Proportionals & Wantzel’s Cubic Impossibility

In Plato’s mechanical carpenter’s square (Fig. 50, p. 163), setting right-angled legs $gb = g$ and $be = e$ along orthogonal axes with right angles at $c = (0, c)$ and $d = (-d, 0)$ forces altitude-on-hypotenuse similarity (Euclid VI.8) in $\triangle gcd$ and $\triangle cde$:

$$\frac{g}{c} = \frac{c}{d} = \frac{d}{e} \implies c^2 = gd, \quad d^2 = ce \implies c^3 = g^2 e.$$

Taking $g = 1$ and $e = 2$ yields $c = \sqrt[3]{2}$ and $d = \sqrt[3]{4}$, so $operatorname{Vol}(c) / operatorname{Vol}(g) = c^3 / g^3 = 2$ identically! Because $x^3 - 2$ is irreducible over $\mathbb{Q}$ of degree $3 \neq 2^k$, Pierre Laurent Wantzel (wantzel:1837) and Felix Klein (klein:1895) proved why all three of Dürer’s Book IV constructions (Sporus’s graduated ruler, Plato’s sliding square, and Hero’s pivoting straightedge) necessarily employ neusis or mechanical linkage beyond unmarked straightedge and compass.

3. Homogeneous Central Projection & Cross-Ratio Invariance

In construct_durer_perspective_apparatus(), let the artist’s fixed eye point be $E = (-12, 3, 5)$ looking orthogonally toward the vertical glass picture frame $x = 0$ (principal vanishing point $V_x = (3, 5)$ on the glass, focal distance $f = 12$). Every vertex $P = (X, Y, Z) \in \mathbb{Q}^3$ projects along rectilinear sight-ray $EP$ to exact rational picture-plane coordinates:

$$u(P) = 3 + \frac{12\,(Y - 3)}{X + 12}, \qquad v(P) = 5 + \frac{12\,(Z - 5)}{X + 12}.$$

For the cube $[4, 8] \times [-2, 2] \times [0, 4]$:

  1. Exact Convergence at the Principal Vanishing Point: All four receding depth edges $A'E', B'F', C'G', D'H'$ intersect on the picture plane at the single point $V_x = (3, 5)$—corresponding to the ideal point at infinity $(1 : 0 : 0 : 0) \in \mathbb{P}^3(\mathbb{Q})$.

  2. Foreshortening & Harmonic Cross-Ratio ($\operatorname{CR} = -1$): Along the lower-left depth edge from $A = (4, -2, 0)$ to $E_{\text{back}} = (8, -2, 0)$, the true 3D midpoint $M = (6, -2, 0)$ projects to $M' = \left(-\frac{1}{3},\, \frac{5}{3}\right)$, which is strictly closer to the back corner $E' = (0, 2)$ than to the front corner $A’ = left(-frac{3}{4},, frac{5}{4}right)$—yet its projective cross-ratio with vanishing point $V_x = (3, 5)$ (the harmonic fourth of the midpoint and ideal point at infinity, cf. geometor/pappus pappus:synagoge) is preserved with 100% symbolic exactness:

    $$\operatorname{CR}(A', E';\, M', V_x) = \frac{(u_{A'} - u_{M'})\,(u_{E'} - u_{V_x})}{(u_{A'} - u_{V_x})\,(u_{E'} - u_{M'})} = \frac{\left(-\frac{3}{4} + \frac{1}{3}\right)(0 - 3)}{\left(-\frac{3}{4} - 3\right)\left(0 + \frac{1}{3}\right)} = \frac{\left(-\frac{5}{12}\right)(-3)}{\left(-\frac{15}{4}\right)\left(\frac{1}{3}\right)} = -1.$$