Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22
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Dynkin duality of B_n and C_n

Example

For n2, the Bn and Cn diagrams have the same underlying chain and opposite arrows on the unique double edge; transposing the Cartan matrix exchanges them.

Facts & Assumptions

Given: An integer n2; the simple roots of Bn: α1=ε1ε2,,αn1=εn1εn,αn=εn; and of Cn: β1=ε1ε2,,βn1=εn1εn,βn=2εn.

[L1]

The Cartan matrix entry is aij=2(αj,αi)/(αi,αi), and the diagram has aijaji edges with the arrow toward the shorter root (Cartan matrix of a based root system, Dynkin diagram with edge multiplicity and arrow convention).

[L2]

Coroot duality transposes the Cartan matrix, and the dual system of Bn is Cn (Duality exchanges B and C).

Verification

technique · direct
1.1

For Bn: αi2=2 for i<n and αn2=1; (αn1,αn)=1, so an1,n=2(1)/2=1 and an,n1=2(1)/1=2; all other off-diagonal entries of adjacent pairs are 1 and the rest vanish. Thus the diagram is a chain with a double edge at the end, the arrow being governed by an,n1=2>an1,n=1 and pointing toward the shorter root αn.

L1algebra
1.2

For Cn: βi2=2 for i<n and βn2=4; (βn1,βn)=2, so an1,n=2(2)/2=2 and an,n1=2(2)/4=1; the diagram is again a chain with a double edge exactly at the end, and the arrow now points toward the shorter root βn1.

L1algebra
2.1

The matrices of steps 1.1 and 1.2 are transposes of one another, which is exactly coroot duality by [L2]; so transposing the Cartan matrix exchanges Bn and Cn, reversing the arrow while keeping the chain and the double-edge position.

L2step 1.1step 1.2algebra

Depends on

Used by

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Dependency tree · two levels

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Sources