False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

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Every connected finite graph is Dynkin

Statement

False: a connected finite graph need not be the Dynkin diagram of a crystallographic root system; positive definiteness and the edge restrictions exclude most connected graphs.

Facts & Assumptions

Given: A cycle graph on $m \geq 3$ vertices and the conventions of the Dynkin diagram.

[L1]

The Dynkin diagram of a based root system has $a_{i j} a_{j i}$ edges between vertices $i, j$ and no others; a finite-type Cartan matrix is symmetrizable to a positive definite matrix. When the root system is irreducible, equivalently when its diagram is connected, that diagram is a tree (Dynkin diagram with edge multiplicity and arrow convention, Properties of finite-type Cartan matrices, Shape restrictions on Dynkin diagrams).

Proof

technique · counterexample

1.1

Let $m \geq 3$ and let $G$ be the cycle graph on $m$ vertices. A root system whose Dynkin diagram were $G$ would have Cartan matrix $A = 2 I - A d j (G)$ , since every edge corresponds to the single relation $a_{i j} = a_{j i} = - 1$ and nonedges to $0$ ; this matrix is symmetric.

L1algebra

1.2

The nonzero vector $x = (1, \dots, 1)$ satisfies $x^{T} A x = \sum_{i} 2 x_{i}^{2} + 2 \sum_{i < j} a_{i j} x_{i} x_{j} = 2 m - 2 m = 0$ , because each vertex has exactly two neighbours in a cycle; hence $A$ is not positive definite.

givenalgebra

2.1

Since a finite-type Cartan matrix must be symmetrizable to a positive definite matrix by [L1], no root system has the cycle $G$ as its Dynkin diagram, although $G$ is connected and finite. This refutes the claim that every connected finite graph is a Dynkin diagram.

L1step 1.1step 1.2∎

Depends on

Used by

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Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed., Chapter II (standard reference, not scraped)