Pruning and the Diophantine Threshold

Part 3 reduced every simple Lie algebra to a connected Dynkin diagram and listed nine kinds. This part asks whether that list is complete: which connected graphs with single, double and triple bonds come from positive-definite Cartan matrices? Two constraints do nearly all the work: the diagram is a tree, and every node has total bond weight at most 3. What they leave is a short list of chains and one family of branched trees, whose arm lengths a single Diophantine inequality then settles.

Unit simple roots

Both pruning proofs are short vector-norm arguments. At each node take the simple root rescaled to a unit vector v_i, which keeps the angles. Pairwise inner products satisfy

because (v_i, v_j)² = (n_αβ·n_βα)/4, so 2(v_i, v_j) = −√(n_αβ · n_βα). Single edge: −1; double: −√2; triple: −√3.

No cycles

A connected Dynkin diagram is a tree. To see it, let v = v₁ + ⋯ + v_n. Positive definiteness gives v ≠ 0 and

Expanding,

Diagonals sum to n. Off-diagonals 2(v_i, v_j) are zero without an edge, and ≤ −1 with one. So

so there are fewer than n edges. A connected graph on n nodes with at most n − 1 edges is a tree; one extra edge would make (v, v) ≤ 0. In a tree, paths between nodes are unique, which lets the argument below treat a branched diagram as arms radiating from one node.

Valency at most 3

Count a single bond as weight 1, a double as 2 and a triple as 3. Every node has total weight at most 3. Take a node v with neighbours u₁, …, u_k. In a tree the u_i are pairwise non-adjacent, so pairwise orthogonal.

Project v onto the span of its neighbours. In the orthonormal basis u₁, …, u_k, the projection has squared length

This is at most (v, v) = 1, with equality only if v lies in the span of its neighbours. It does not, since the simple roots are linearly independent, so

Each (v, u_i)² is the bond multiplicity over 4, so the total weight at v is less than 4.

A node can therefore carry three single bonds but not four, a single and a double but not two doubles, and a triple bond only on its own, which makes G₂ the only diagram with a triple bond. Together with the tree condition this leaves plain chains (A_n), chains with a double bond at the end (B_n and C_n) or in the middle (at most four nodes, F₄), and trees of single bonds with one node of valency 3, whose arm lengths are still open.

A positivity test for any graph

Both arguments are cases of one test. Each node stands for a vector, and each bond fixes the angle between two of them at 120°, 135° or 150°; the question is whether such vectors can form a basis of Euclidean space. Scale every node's vector to squared length 2. The Gram matrix then has 2 on the diagonal and 2 cos θ = −√m for a bond of multiplicity m. Such vectors exist exactly when that matrix is positive definite.

Almost every graph fails the test. A graph that passes is of finite type, the Dynkin diagram of a simple Lie algebra. A graph whose Gram matrix is positive semidefinite with a one-dimensional kernel sits exactly on the boundary: these are the affine diagrams, the Dynkin diagrams of affine Kac–Moody algebras. Anything else is indefinite.

Several presets sit on the boundary. The four-armed star is D̃₄, one arm more than D₄ can carry. The triangle is Ã₂: a cycle of single bonds lands exactly on the boundary, and any extra edge pushes it past. Lengthening the long leg of E₈ by one node (the E₈ + 1 preset) gives Ẽ₈, and the determinant lands on zero. A failing diagram's verdict names one of the two reasons proved above, a cycle or too much bond weight at a node.

The T-shape

A branched tree is a "Y" with arm lengths (p, q, r), each counting the branch node, so it has p + q + r − 2 nodes. Sort them so that p ≥ q ≥ r ≥ 2 (an arm of length 1 is no arm).

The arm-length inequality

Use unit vectors as before. On the p-arm, label the vectors u₁, …, u_p−1 from the tip toward the branch node c, and add them with weights that grow toward c:

Adjacent vectors have inner product −½ and all others are orthogonal, which gives the two values on the right. Only u_p−1 touches c. No edge joins two different arms, so the three arm vectors ε_p, ε_q, ε_r are mutually orthogonal, and c is not in their span. As in the valency argument, the squared cosines of the angles between c and the three arm vectors sum to less than 1:

Rearranging gives the inequality every branched diagram must satisfy:

Conversely, every solution with p ≥ q ≥ r ≥ 2 gives a positive definite matrix; build any T-shape in Figure 1 to check.

The inequality 1/p + 1/q + 1/r > 1 also says that a triangle with angles π/p, π/q, π/r has angle sum above π, so it lives on the sphere. The same condition picks out the finite triangle reflection groups, and with them the Platonic solids.

Solving the inequality

Enumerate by smallest arm length r ≥ 2. For each r, the inequality constrains (p, q):

So the branched diagrams are D_n for n ≥ 4, from (n − 2, 2, 2), so that D₄ is (2, 2, 2) and D₅ is (3, 2, 2); E₆ from (3, 3, 2); E₇ from (4, 3, 2); and E₈ from (5, 3, 2). E₈, of rank 8 and dimension 248, is the largest of the five exceptional algebras, while A_n and D_n grow without bound.

The Diophantine solver

The complete list

Chains of single bonds give A_n for every n ≥ 1. A chain with a double bond at one end gives B_n (n ≥ 2), and reversing its arrow gives C_n (n ≥ 3). A double bond in the middle gives only F₄, since longer such chains fail the positivity test. The lone triple bond gives G₂. The branched trees give D_n, E₆, E₇ and E₈. Every other connected graph fails one of the constraints, so these nine, four families and five exceptions, are the complete list.

What lies past E₈

(6, 3, 2) fails by the smallest possible margin: 1/6 + 1/3 + 1/2 = 1 exactly. The matrix is then positive semidefinite with a one-dimensional kernel, so no finite root system has this diagram. It is the diagram Ẽ₈ of an affine Kac–Moody algebra, an infinite-dimensional Lie algebra built from E₈, and the other two equality cases, (4, 4, 2) and (3, 3, 3), are Ẽ₇ and Ẽ₆.

Past the boundary, every T-shaped diagram with 1/p + 1/q + 1/r < 1 has exactly one negative eigenvalue. Deleting the branch node leaves three chains, a positive definite matrix one size smaller, and eigenvalue interlacing then allows at most one eigenvalue that is not positive. These Lorentzian diagrams define Kac–Moody algebras too, and a few of them, such as E₁₀ = (7, 3, 2), are hyperbolic. None is finite-dimensional.

Affine extension and node-deletion surgery

Append a ninth node α₀ = −α_max 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.

Deleting one node from Ẽ₈ and reading off the connected pieces gives subalgebras of E₈ of full rank. Borel and de Siebenthal showed that every maximal semisimple subalgebra of full rank arises this way, from deleting a node whose mark is prime.

The eight deletions other than α₀ give: D₈ (α₁), A₈ (α₂), A₁ ⊕ A₇ (α₃), A₁ ⊕ A₂ ⊕ A₅ (α₄), A₄ ⊕ A₄ (α₅), D₅ ⊕ A₃ (α₆), E₆ ⊕ A₂ (α₇), E₇ ⊕ A₁ (α₈). The marks of α₃, α₄ and α₆ are 4, 6 and 4, and their pieces are not maximal: A₁ ⊕ A₇ sits inside E₇ ⊕ A₁, A₁ ⊕ A₂ ⊕ A₅ inside E₆ ⊕ A₂, and D₅ ⊕ A₃ inside D₈. The other five are maximal.

The same inequality picks out the Platonic solids. Part 5 finds the affine diagrams Ẽ₆, Ẽ₇ and Ẽ₈ again, through the double covers of their rotation groups.