Part 4 of 9. Which graphs can be Dynkin diagrams? Two vector arguments rule out cycles and crowded nodes, and one inequality on arm lengths settles the branched trees, so the list of simple Lie algebras ends at E8.
Part 3 reduced every simple Lie algebra to a connected Dynkin diagram. Which connected graphs with single, double and triple bonds come from positive-definite Cartan matrices? Two short arguments show that the diagram is a tree with bond weight at most 3 at each node, and one Diophantine inequality settles the branched trees.
At each node take the simple root rescaled to a unit vector vi. Then
because (vi, vj)² = nαβnβα/4: −1 for a single bond, −√2 for a double, −√3 for a triple.
No cycles. Let v = v1 + ⋯ + vn. Positive definiteness gives
and expanding,
Each edge contributes at most −1, so
A connected graph on n nodes with fewer than n edges is a tree.
Weight at most 3 per node. Count a single bond as weight 1, a double as 2 and a triple as 3. Take a node v with neighbours u1, …, uk; in a tree they are pairwise orthogonal. Projecting v onto their span gives
which is less than (v, v) = 1, because the simple roots are independent and v is not in that span:
Each (v, ui)² is the bond weight over 4, so the total weight at v is below 4. A node can carry three single bonds but not four, a single and a double but not two doubles, and a triple only on its own, so G2 is the only diagram with a triple bond. The test in the next section then leaves chains (An), chains with one double bond (Bn and Cn with it at the end, F4 with it in the middle of four nodes), and trees of single bonds with one node of valency 3.
Both arguments are cases of one test. Scale every node's vector to squared length 2; the Gram matrix B then has 2 on the diagonal and −√m for a bond of multiplicity m, and the vectors exist exactly when B is positive definite. A graph that passes is of finite type. One whose B is positive semidefinite with a one-dimensional kernel is affine, the diagram of an affine Kac–Moody algebra. Anything else is indefinite.
With single bonds B = 2I − A, where A is the adjacency matrix, so a simply-laced graph passes exactly when the largest eigenvalue of A is below 2. That makes it easy to count how rarely a graph passes.
Figure 2. Largest adjacency eigenvalue of every tree on up to 12 vertices, one dot each. Below the line at 2 sit only paths, D shapes and E6, E7, E8; on it, the affine D̃ and Ẽ. Verdicts come from B's minors in exact integers. Click a dot to draw its tree.
A branched tree is a Y with arm lengths (p, q, r), each counting the branch node c, so it has p + q + r − 2 nodes; sort them so that p ≥ q ≥ r ≥ 2. On the p-arm label the unit vectors u1, …, up−1 from the tip toward c and weight them:
The three arm vectors are mutually orthogonal and c is not in their span, so, as in the valency argument, the squared cosines between c and them sum to less than 1:
Rearranging gives the inequality every branched diagram must satisfy:
Conversely every solution gives a positive definite matrix. The same inequality says that a triangle with angles π/p, π/q, π/r lives on the sphere, which is how it also picks out the finite triangle reflection groups and the Platonic solids.
| r | Remaining constraint | q | p range | Solutions (p, q, r) | Algebra |
|---|---|---|---|---|---|
| 2 | 1/p + 1/q > 1/2 | 2 | any p ≥ 2 | (p, 2, 2) | Dp+2 |
| 2 | 1/p > 1/6, so p ≤ 5 | 3 | 3, 4, 5 | (3, 3, 2), (4, 3, 2), (5, 3, 2) | E6, E7, E8 |
| 2 | 1/p + 1/q ≤ 1/4 + 1/4 = 1/2 | ≥ 4 | none | (4, 4, 2) gives equality | none |
| 3 | 1/p + 1/q ≤ 1/3 + 1/3 = 2/3 | ≥ 3 | none | (3, 3, 3) gives equality | none |
| ≥ 4 | 1/p + 1/q + 1/r ≤ 3/4 | ≥ 4 | none | none |
Figure 3. Solutions of 1/p + 1/q + 1/r > 1 with p ≥ q ≥ r ≥ 2, enumerated by r. (4, 4, 2), (3, 3, 3) and (6, 3, 2), just past E8, reach equality: they are affine diagrams.
So the branched diagrams are Dn = (n − 2, 2, 2) for n ≥ 4, E6 = (3, 3, 2), E7 = (4, 3, 2) and E8 = (5, 3, 2). With the chains that is the complete list: An, Bn, Cn, Dn and the five exceptions G2, F4, E6, E7, E8.
Figure 4. Arm lengths (p, q, r) and the sum 1/p + 1/q + 1/r against the threshold 1. Above it the tree is Dn, E6, E7 or E8; exactly on it, as at (6, 3, 2), it is affine; below it, indefinite.
(6, 3, 2) fails by the smallest margin: 1/6 + 1/3 + 1/2 = 1, B is positive semidefinite with a one-dimensional kernel, and the diagram is Ẽ8, the affine extension of E8. The other equality cases are Ẽ7 = (4, 4, 2) and Ẽ6 = (3, 3, 3).
The determinant makes the threshold literal. Expanding along the branch node, whose arms are chains with determinants p, q and r,
so det B has the sign of the inequality. With the branch node ordered last, the other leading minors belong to the three chains and are positive, so a negative determinant means exactly one negative eigenvalue. These Lorentzian diagrams, such as E10 = (7, 3, 2), define Kac–Moody algebras too, and none is finite-dimensional.
Figure 5. det B for every T-shaped tree with arms up to 12, computed exactly from its matrix. Teal cells are positive (finite), purple zero (affine), grey negative; the curve is 1/p + 1/q + 1/r = 1. Along q = 3, r = 2 the determinant falls by one per node.
Append α0 = −αmax to E8's diagram to get Ẽ8. Each node carries a mark, the coefficient of its simple root in the highest root: (2, 3, 4, 6, 5, 4, 3, 2) on α1, …, α8, and 1 on α0. Deleting one node of Ẽ8 leaves a subalgebra of E8 of full rank, and Borel and de Siebenthal showed that every maximal one arises this way, from a node whose mark is prime.
Figure 6. The extended diagram Ẽ8 with its marks; α0 attaches to α8. Click a node to delete it, and the status bar names the pieces. Deleting α0 recovers E8; deleting the branch node α4 leaves three chains.
Five of the eight other deletions give maximal subalgebras: D8, A8, A4 ⊕ A4, E6 ⊕ A2 and E7 ⊕ A1. The marks of α3, α4 and α6 are 4, 6 and 4, and their pieces sit inside others: A1 ⊕ A7 in E7 ⊕ A1, A1 ⊕ A2 ⊕ A5 in E6 ⊕ A2, and D5 ⊕ A3 in D8. Part 5 finds Ẽ6, Ẽ7 and Ẽ8 again, through the double covers of the tetrahedral, octahedral and icosahedral rotation groups.