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A symmetrizable indefinite rank two gcm
Example
The symmetric GCM is indefinite. The nonzero vector has imaginary root with squared length for .
Facts & Assumptions
Given: The displayed rank-two matrix and D=I.
The positive half is free modulo its positive Serre ideal. (Serre presentation of a kac moody algebra).
The root metric has entries d_i a_ij. (Invariant bilinear form for a symmetrizable kac moody algebra).
Imaginary means a root outside W Pi. (Real and imaginary kac moody roots).
A positive vector with negative image characterizes indefinite type. (Finite affine indefinite trichotomy for indecomposable gcms).
Verification
The matrix meets every GCM condition, is connected and symmetric, has determinant , and . Hence it is indefinite by F4.
The two positive Serre generators have degrees and , each of total height five. Every element of their generated positive ideal is a linear combination of these and positive adjoints, so has no component below height five. The free bracket is nonzero: its tensor image is the difference of the distinct words . F1 therefore ensures its degree-(1,1) class survives. It is a root vector of weight .
F2 gives . For each reflection, and make expansion of equal to . Thus every Weyl translate of a simple root has squared length 2. The root from step 1.2 has length −2, so cannot be such a translate and is imaginary by F3.
Sources
Source comparison: Kleshchev, §4.1 and Theorem 9.3.5, pp.50–57 and 125–126; local degree-(1,1) calculation.
Depends on
Used by
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Dependency tree · two levels
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Sources
- Kleshchev, Lectures on Infinite Dimensional Lie Algebras — §4.1 and Theorem 9.3.5, pp.50–57 and 125–126; local degree-(1,1) calculation (standard reference, not scraped)