Part 7 of 15. Two short vector-norm arguments rule out every Dynkin diagram with a cycle and every node of valency > 3, collapsing the classification to a handful of shapes. A single Diophantine inequality, 1/p + 1/q + 1/r > 1, then pins the branched ones to Dn, E6, E7, E8. The list closes here.
The classification problem is now: which connected edge-labelled graphs are Dynkin diagrams of positive-definite Cartan matrices? Two constraints do nearly all the work. An admissible diagram is a tree. And every node has total edge weight at most 3 (single=1, double=2, triple=3). Together these rule out 4-arm stars, cycles, triple edges adjacent to anything, and every dense configuration.
What survives is a shortlist: linear chains (An), chains with a double edge at the end (Bn/Cn), chains with a middle double edge (F4), the isolated triple edge (G2), and valency-3 trees whose arms need a length bound (D and E candidates). A single Diophantine inequality on the three arm lengths (p, q, r) gives four solution classes: Dn, E6, E7, E8. The (6, 3, 2) "E9" hits boundary equality and describes an affine root system. E8 is the ceiling.
Both pruning proofs are short vector-norm arguments. At each node, pick a simple root vi normalised to unit length. Rescaling preserves angles and edge structure. Pairwise inner products satisfy
because (vi, vj)² = (nαβ·nβα)/4, so 2(vi, vj) = −√(nαβ · nβα). Single edge: −1; double: −√2; triple: −√3.
Claim: a connected admissible Dynkin diagram must be a tree, with no cycles, and therefore the number of edges equals the number of nodes minus one.
Proof sketch. Let v = v1 + ⋯ + vn. Positive definiteness gives v ≠ 0 and
Expanding the squared length:
Diagonals sum to n. Off-diagonals 2(vi, vj) are zero without an edge, and ≤ −1 with one. So
so the edge count is strictly less than n. For a connected graph this gives exactly n − 1 edges, a tree. One extra edge would make (v, v) ≤ 0.
What trees buy us. Unique paths between nodes, which lets the Diophantine argument below reason about arms radiating from one branch node.
Claim: in an admissible Dynkin diagram, every node has total edge weight at most 3, where a single edge counts as weight 1, a double as 2, and a triple as 3.
Proof sketch. Take a node v with k neighbours u1, …, uk. In a tree the ui are pairwise orthogonal: any path between them must pass through v, so they are non-adjacent.
Project v onto the span of its neighbours. In the orthonormal basis u1, …, uk, the projection has squared length
The projection has squared length ≤ (v, v) = 1, with equality iff v is in the subspace. v is not in the span of its neighbours (it is linearly independent), so the inequality is strict:
Each (v, ui)² = (edge-multiplicity)/4. Summing gives (total edge-multiplicity at v)/4 < 1, so total edge weight is at most 3.
The bound is sharp. A node can have three single edges (1+1+1 = 3) but not four; one single plus one double (1+2 = 3) but not two doubles; one triple alone but nothing alongside. With tree-ness, the shortlist collapses to:
Figure 1. The diagram pruning sandbox. Build any graph by placing nodes and drawing edges; the two indicators on the right report which constraints the current graph satisfies. A shape can only be a valid Dynkin diagram if it is a tree (no cycles, |E| = |V| − 1 per connected component) and every node has single-edge valency at most 3. The presets cover both legal examples (A5, D5, E6, E8) and deliberately illegal ones (triangle, 4-star, 5-cycle, two-branch tree) so you can see the indicators flip.
Pruning gives a shortlist of candidate shapes. The valency-3 branched trees still admit arbitrary arm lengths. A spherical Diophantine inequality restricts them sharply.
A valency-3 branched tree is a "Y" with arm lengths (p, q, r), each counting nodes including the branch node. Sort p ≥ q ≥ r ≥ 2 (an arm of length 1 means no actual arm).
The total node count is p + q + r − 2.
Use unit simple roots as before. Label the p-arm vectors w1, …, wp−1 from tip to branch; same for q- and r-arms. Let c be the branch node.
Define three weighted arm sums with weights increasing toward the branch:
and form P = Pp + Pq + Pr − c (subtracting the branch node, shared by all arms). Compute (P, P).
The arithmetic is elementary: each arm contributes a term depending only on its length, the branch contributes 1, and cross-terms between arms vanish by tree orthogonality. The result:
and positive-definiteness of the Cartan matrix forces (P, P) > 0. Rearranging gives the spherical Diophantine inequality:
The classification inequality. Every branched-tree Dynkin diagram must satisfy it, and every integer solution (p ≥ q ≥ r ≥ 2) corresponds to a simple Lie algebra.
Why spherical? 1/p + 1/q + 1/r > 1 is the condition for a spherical triangle with angles (π/p, π/q, π/r) to have angle sum > π — positive curvature. The same inequality classifies finite Coxeter groups, Platonic solids, and spherical tilings.
Enumerate by smallest arm length r ≥ 2. For each r, the inequality constrains (p, q):
| r | Remaining constraint | q | p range | Solutions (p, q, r) | Algebra |
|---|---|---|---|---|---|
| 2 | 1/p + 1/q > 1/2 | 2 | p ≥ 2, any | (p, 2, 2), p ≥ 2 | Dp+2, infinite family |
| 2 | 1/p + 1/3 > 1/2 ⇒ 1/p > 1/6 ⇒ p ≤ 5 | 3 | p ∈ {3, 4, 5} | (3, 3, 2) | E6 |
| 2 | 3 | (4, 3, 2) | E7 | ||
| 2 | 3 | (5, 3, 2) | E8 | ||
| 2 | 1/p + 1/4 > 1/2 ⇒ 1/p > 1/4 ⇒ p < 4, and p ≥ q = 4 fails | ≥ 4 | no solutions | (4, 4, 2) is boundary, fails (1/4 + 1/4 + 1/2 = 1 exactly) | forbidden |
| 3 | 1/p + 1/q > 2/3, p ≥ q ≥ 3 ⇒ 2/p ≥ 1/p + 1/q ≤ 2/3 conflicts | ≥ 3 | no solutions | (3, 3, 3) is boundary, fails (1/3 + 1/3 + 1/3 = 1 exactly) | forbidden |
| ≥ 4 | 1/p + 1/q + 1/r ≤ 3/4 < 1 | ≥ 4 | no solutions | — | forbidden |
Figure 2. Enumeration of integer solutions to 1/p + 1/q + 1/r > 1 with p ≥ q ≥ r ≥ 2. The only solutions are the infinite Dn family (q = r = 2, any p ≥ 2) and three singleton exceptions (5, 3, 2), (4, 3, 2), (3, 3, 2) corresponding to E8, E7, E6. The (4, 4, 2) and (3, 3, 3) configurations are boundary cases where the inequality becomes equality; those correspond to affine Kac-Moody algebras rather than finite-dimensional simple Lie algebras.
So the complete solution set is:
That is all the branched-tree Dynkin diagrams.
Figure 3. Live Diophantine solver. The three sliders set the arm lengths (p, q, r), and the bar chart shows 1/p + 1/q + 1/r against the red threshold at 1. Green regions (sum strictly > 1) correspond to finite Lie algebras, and the legal answers are Dn, E6, E7, or E8. The (6, 3, 2) point (drag p to 6, q to 3, r to 2) sits exactly on the boundary, sum = 1, and corresponds to the affine extension of E8, not a finite simple Lie algebra. Every setting with sum strictly less than 1 is a hyperbolic or Lorentzian Kac-Moody algebra, also not on the finite list.
Putting together everything from Act II:
The nine families A, B, C, D, E6, E7, E8, F4, G2 are the complete list. Every other connected Dynkin-diagram shape fails one of the constraints.
(6, 3, 2) is close to working: 1/6 + 1/3 + 1/2 = exactly 1. The Diophantine inequality demands strict inequality, so the Cartan matrix is positive semi-definite with one zero eigenvalue. The associated object is not a finite simple Lie algebra.
It is the affine Kac-Moody algebra Ẽ8, an infinite-dimensional Lie algebra obtained by adjoining an affine node to E8's diagram. Affine Kac-Moody algebras are central to 2D conformal field theory, integrable systems, and moonshine; the Moonshine act (Parts 16–20) of the Modular Forms series picks up this thread. They sit on the boundary between the finite classification and infinity.
For Cartan–Killing we stay on the strict-greater-than side, and E8 is the ceiling.
Act II is complete. The full proof, from Lie algebra to root system to crystallographic restriction to Dynkin diagram to Diophantine inequality, is on the page. The finite-dimensional simple Lie algebras over ℂ are exactly An, Bn, Cn, Dn, G2, F4, E6, E7, E8.
Append a ninth node α0 = −αmax to E8's diagram and you get the affine E8 diagram Ẽ8. Each node carries a mark, the coefficient of its simple root in the highest root: marks (2, 3, 4, 6, 5, 4, 3, 2) on (α1, …, α8), and 1 on α0.
The affine diagram also gives every maximal proper subalgebra of E8: delete one node from Ẽ8 and read the connected pieces. This is the Borel–de Siebenthal rule.
Figure 5. The extended Dynkin diagram Ẽ8 with marks. Toggle between the finite E8 (8 nodes) and the affine Ẽ8 (9 nodes, with α0 attached to α8). Click any node to delete it; the status bar reports the Lie-type of each connected component that remains. Deleting α0 recovers finite E8; deleting α1 (the branch tip) gives D8; deleting α2 gives A8; deleting α4 (the valency-3 branch node) fragments into three classical pieces.
The Borel–de Siebenthal rule. Deleting one node from Ẽ8 gives a maximal equal-rank semisimple subalgebra. The eight non-trivial deletions give: D8 (α1), A8 (α2), A1 ⊕ A7 (α3), A1 ⊕ A2 ⊕ A5 (α4), A4 ⊕ A4 (α5), D5 ⊕ A3 (α6), E6 ⊕ A2 (α7), E7 ⊕ A1 (α8).