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Act I gave us a finite root system Δ in Euclidean space 𝔼 with a Weyl group acting by reflections. The Cartan integer n_{β, α} = 2(β, α)/(α, α) must be an integer for every ordered pair (α, β), by sl₂ representation theory.

The geometric consequence: integrality plus Cauchy–Schwarz restricts the angle between two roots to exactly four values, 90°, 120°, 135°, 150°. This is the crystallographic restriction. Every subsequent classification argument is a combinatorial walk through those four options.

From integrality to a Diophantine identity

Take any two linearly independent roots α, β ∈ Δ. Both of the Cartan integers

are integers, by explainer 3. Their product is

where θ is the angle between α and β. The right side is symmetric in the pair, and 4 cos²θ is an integer because both factors are.

Since cos²θ ∈ [0, 1], 4 cos²θ ∈ [0, 4]. Integrality forces

five integer values. 4 cos²θ = 4 means parallel roots, which we excluded. For linearly independent roots the product is in {0, 1, 2, 3}: four possibilities, four allowed angles.

The four legal angles

Converting each of the four integer values back to an angle:

• Product 0 ⇒ cos²θ = 0 ⇒ θ = 90°. The roots are orthogonal, both Cartan integers are zero, and bracketing them gives zero by the bracket-additivity rule from explainer 3 (since α + β would have to live in an unrelated direction, but by orthogonality it cannot).
• Product 1 ⇒ cos²θ = 1/4 ⇒ θ = 60° or 120°. One Cartan integer is ±1, the other is ±1. The roots have the same squared length: (α, α) = (β, β). This is the simply-laced case.
• Product 2 ⇒ cos²θ = 1/2 ⇒ θ = 45° or 135°. One Cartan integer is ±1, the other is ±2. The squared lengths are in ratio 2:1, so |α|/|β| = √2 (or its reciprocal).
• Product 3 ⇒ cos²θ = 3/4 ⇒ θ = 30° or 150°. One Cartan integer is ±1, the other is ±3. The squared lengths are in ratio 3:1, so |α|/|β| = √3 (or its reciprocal).

The Cartan integers are asymmetric precisely when the roots have different lengths. The "more negative" integer comes from the longer root, and the length ratio is √k where k is the product.

Simple roots are obtuse. For simple roots (positive, part of a minimal generating set, see next explainer), (α, β) ≤ 0. This cuts each pair to its obtuse member: 120°, 135°, 150°. So between two distinct simple roots the angle is exactly 90°, 120°, 135°, or 150°.

Angle picker

All four types of Dynkin edge

Each allowed configuration corresponds to a different kind of edge in a Dynkin diagram, recording how strongly two nodes interact:

The triple edge is so restrictive it only appears in G₂ (rank 2). No simple Lie algebra of rank 3 or more contains one. The double edge appears in B_n, C_n, F₄, and only at the end of a chain (B_n, C_n) or between simply-laced arms (F₄). The simple edge (120°) builds the entire ADE family.

Why the classification is now a combinatorics problem

The crystallographic restriction in one sentence: any two roots are either orthogonal or meet at 120°, 135°, or 150°, with non-orthogonal pairs in length ratio 1, √2, or √3. The classification reduces from vector configurations to graphs.

Once we know the angles and lengths between every pair of simple roots, we can reconstruct the full root system by Weyl reflections. The data lives in a Cartan matrix, an n × n integer matrix whose off-diagonals encode the edge type. Each Cartan matrix determines a Dynkin diagram, and the classification becomes: which connected graphs can be Dynkin diagrams of a positive-definite Cartan matrix?

The proof is short. Every admissible Dynkin diagram must be a tree with valency at most 3. Counting trees with valency ≤ 3 reduces to the Diophantine inequality 1/p + 1/q + 1/r > 1 on the arm lengths. It has two infinite families of solutions — linear chains (giving A_n) and one-fork trees (giving D_n) — plus exactly three exceptional solutions with all arms of length ≥ 2: (3, 3, 2), (4, 3, 2), (5, 3, 2), corresponding to E₆, E₇, E₈.

What discreteness gives us. The classification problem becomes "which integer matrices of a specific form have strictly positive eigenvalues?" — a finite combinatorial question. Integrality collapses the continuum of angles to four, and that is what makes the classification finite.

Takeaways

• The Diophantine identity. n_{α, β} · n_{β, α} = 4 cos²θ, where both factors are integers and cos²θ ∈ [0, 1]. The product is therefore an integer in {0, 1, 2, 3, 4}, and ruling out linearly-dependent roots (product 4) leaves {0, 1, 2, 3}.
• Four legal angles. 90°, 120°, 135°, 150°. No other angle is compatible with integrality of the Cartan numbers for a pair of simple roots.
• Length ratios. When the product is 2 the roots are in ratio 1 : √2; when 3, in ratio 1 : √3. Same length corresponds to product 0 (orthogonal) or product 1 (120° simple edge).
• Simply-laced vs non-simply-laced. ADE types are simply-laced (all roots the same length, all edges single). BCFG types have two length classes, and their Dynkin diagrams contain at least one double or triple edge.
• Handoff to Dynkin diagrams. The four legal configurations correspond exactly to the four kinds of Dynkin edge: no edge, single edge, double edge with arrow, triple edge with arrow. The next explainer uses these as the building blocks of a graph-theoretic encoding.

The next explainer turns the combinatorics into a single diagram per algebra.