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Dynkin duality of $B_{n}$ and $C_{n}$

Example

For $n \geq 2$ , the $B_{n}$ and $C_{n}$ 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 $n \geq 2$ ; the simple roots of $B_{n}$ : $α_{1} = ε_{1} - ε_{2}, \dots, α_{n - 1} = ε_{n - 1} - ε_{n}, α_{n} = ε_{n}$ ; and of $C_{n}$ : $β_{1} = ε_{1} - ε_{2}, \dots, β_{n - 1} = ε_{n - 1} - ε_{n}, β_{n} = 2 ε_{n}$ .

[L1]

The Cartan matrix entry is $a_{i j} = 2 (α_{j}, α_{i}) / (α_{i}, α_{i})$ , and the diagram has $a_{i j} a_{j i}$ 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 $B_{n}$ is $C_{n}$ (Duality exchanges B and C).

Verification

technique · direct

1.1L1algebra

For $B_{n}$ : $∣ α_{i} ∣^{2} = 2$ for $i < n$ and $∣ α_{n} ∣^{2} = 1$ ; $(α_{n - 1}, α_{n}) = - 1$ , so $a_{n - 1, n} = 2 (- 1) / 2 = - 1$ and $a_{n, n - 1} = 2 (- 1) / 1 = - 2$ ; 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 $∣ a_{n, n - 1} ∣ = 2 > ∣ a_{n - 1, n} ∣ = 1$ and pointing toward the shorter root $α_{n}$ .

1.2L1algebra

For $C_{n}$ : $∣ β_{i} ∣^{2} = 2$ for $i < n$ and $∣ β_{n} ∣^{2} = 4$ ; $(β_{n - 1}, β_{n}) = - 2$ , so $a_{n - 1, n} = 2 (- 2) / 2 = - 2$ and $a_{n, n - 1} = 2 (- 2) / 4 = - 1$ ; the diagram is again a chain with a double edge exactly at the end, and the arrow now points toward the shorter root $β_{n - 1}$ .

2.1L2step 1.1step 1.2algebra∎

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 $B_{n}$ and $C_{n}$ , reversing the arrow while keeping the chain and the double-edge position.

Remarks

The paired diagrams below display the chain for $n \geq 4$ ; the dots abbreviate only single edges and their intervening vertices. For $n = 3$ , replace the dotted portion by the single edge joining vertices $1$ and $2 = n - 1$ . For $n = 2$ , retain only the terminal pair of vertices, numbered $1 = n - 1$ and $2 = n$ . In that rank, the two diagrams are isomorphic after interchanging the vertices, although the displayed fixed numbering still reverses the arrow.

In the row-coroot convention of [L1], the ordered terminal entries $(a_{n - 1, n}, a_{n, n - 1})$ are $(- 1, - 2)$ for $B_{n}$ and $(- 2, - 1)$ for $C_{n}$ . Thus the double-edge multiplicity is $2$ in both rows, while each arrow points to the shorter root. These are exactly the calculations in steps 1.1–2.1; the verification does not depend on the illustration.

Depends on

Used by

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Sources

Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I, Lectures 19-24 (standard reference, not scraped)

Dynkin duality of BnBn​ and CnCn​