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The canonical and quasi-canonical bases need not coincide

Statement refuted

In the four-dimensional model of The four Fock bases in a small two-factor tensor space the canonical and quasi-canonical Fock families coincide on every label.

Facts & Assumptions

Given: The rank-one Fock model with monomials M00,M01,M10,M11, the source families L,T,P of The dual-canonical, canonical and quasi-canonical Fock bases, and the computed bar matrix and vectors of The four Fock bases in a small two-factor tensor space.

[F1]

Bar and families (The tensor-product bar on the Fock module and its truncation compatibility, The dual-canonical, canonical and quasi-canonical Fock bases): ψ is triangular with ψM10=M10+(s−1−s)M01 and fixes the other monomials; the canonical family is the s−1Z[s−1]-chart and the quasi-canonical family is defined by the inverse transition matrix.

[F2]

Highest constituent (The highest weight Fock submodule carries the quantum-group global bases, Standard tableau monomials form a basis of the highest weight Fock submodule, The four Fock bases in a small two-factor tensor space): the projection π, the rectified highest label 01↑=10, and the rescaling PA=saTA↑ on the highest constituent with a=1.

Counterexample

Take I={0}, I+={0,1}, Λ=2Λ0, and retain the mixed order (M01,M10) and the notation of The dual-canonical, canonical and quasi-canonical Fock bases and The four Fock bases in a small two-factor tensor space. There P01=M01+s M10=s T10, because T10=M10+s−1M01; but P10=M10≠M10+s−1M01=T10 over Q(s). Hence the canonical and quasi-canonical families are different, and so are the dual-canonical and monomial families. The highest-component label rectification is 01↑=10 with a=1, which exhibits the genuine power and label distinction: the coincidence P01=sT10 happens on the rectified highest label, not by an identity of the two index sets. At s=1 the highest-submodule vectors P01 and T10 agree after rectification, but the full canonical and quasi-canonical families are still not interchangeable. This is no claim that every corresponding label differs at s=1.

Proof technique: direct.

1.1F1givenalgebra

The two families at label 10. In the mixed order the bar fixes M01 and sends M10 to M10+(s−1−s)M01. A vector M10+cM01 is fixed exactly when c−cˉ=s−1−s. The positive chart gives c=−s and the negative chart gives c=s−1; uniqueness follows because a bar-fixed polynomial in either sZ[s] or s−1Z[s−1] is zero. Thus L01=T01=M01, L10=M10−sM01 and T10=M10+s−1M01. Inverting the L chart gives M10=L10+sL01, so the transposed inverse chart is P01=M01+sM10 and P10=M10. Linear independence of the two monomials proves P10≠T10.

2.1F1F2step 1.1algebra

The rectified highest-label coincidence. At label 01 the same computations give P01=M01+sM10=sT10, so the two families do meet at the rectified highest label with the power a=1 of [F2]; this is an identity between specific vectors, not an identification of the two labelled families.

3.1step 1.1step 2.1∎

Conclusion. Since P10≠T10, the canonical and quasi-canonical families are different over Q(s); the dual-canonical and monomial families differ too, because L10=M10−sM01≠M10. The highest-submodule coincidence P01=sT10 is exactly the rectified normalization of [F2], and at s=1 it does not make the whole families interchangeable.

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