Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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Properties of finite-type Cartan matrices

Statement

Let A=(aij) be the Cartan matrix of a reduced crystallographic root system Φ relative to a base (Cartan matrix of a based root system). Then:

  1. aii=2 for all i, and aij is a nonpositive integer for ij;
  2. aij=0 if and only if aji=0;
  3. aijaji{0,1,2,3} for ij;
  4. there is a diagonal matrix D with positive diagonal entries such that DAD1 is symmetric and positive definite.

Facts & Assumptions

Given: A reduced crystallographic root system Φ with base Δ={α1,,αr}, base entries aij=2(αj,αi)/(αi,αi), and the inner product (,) on E.

[L1]

aii=2 and aij=2(αj,αi)/(αi,αi) is an integer (Cartan matrix of a based root system, Reduced crystallographic Euclidean root system).

[L2]

For distinct simple roots, (αi,αj)0 (Distinct simple roots have nonpositive inner product).

[L3]

For nonproportional roots α,β the product of Cartan integers is 4cos2θ{0,1,2,3} and the angle is 90, 60/120, 45/135 or 30/150 (Rank-two root-system classification).

[L4]

The simple roots form a basis of E, so their Gram matrix G=((αi,αj)) is symmetric and positive definite, and any symmetric matrix representing the inner product in a basis is positive definite (Simple roots form a signed integral basis).

Proof

technique · direct
1.1

aii=2(αi,αi)/(αi,αi)=2 and aijZ; for ij one has aij0 because (αi,αj)0 by [L2] and (αi,αi)>0.

L1L2algebra
1.2

aij=0 if and only if aji=0: both entries are nonzero exactly when (αi,αj)0, since the denominators (αi,αi),(αj,αj) are positive.

L1algebra
1.3

For ij, the simple roots αi,αj are nonproportional, so aijaji=4cos2θ{0,1,2,3} by [L3], giving the third assertion.

L3algebra
1.4

Let D=diag(α1,,αr); then DAD1 has entries αiaijαj1=2(αj,αi)/(αiαj), which is symmetric in i,j because the inner product is symmetric.

L1algebra
2.1

The matrix in step 1.4 is positive definite: it is twice the Gram matrix ((αi/αi,αj/αj)) of the normalized simple roots, and those vectors are a basis of E, so their Gram matrix is positive definite by [L4]. Discarding the factor 2 preserves positive definiteness.

L4step 1.4algebra

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