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Basis independence of the Hattori–Stallings pseudo-trace

Statement

For a finite projective right module $P$ over a unital $C$ -algebra $A$ , an $A$ -linear endomorphism $T \in {E n d}_{A} (P)$ , and a symmetric linear functional $ω : A \to C$ , the candidate value set ${T r}_{ω, A} (T)$ defined above is a singleton. Hence the pseudo-trace is independent of the chosen finite dual basis.

Statement references

Hattori–Stallings pseudo-trace

Facts & Assumptions

Given: Two finite dual bases $(p_{i}, ϕ_{i})_{i}$ and $(q_{j}, ψ_{j})_{j}$ of $P$ , an $A$ -linear endomorphism $T$ , and a symmetric functional $ω$ .

[F1]

Hattori–Stallings pseudo-trace: each dual basis satisfies $p = \sum_{i} p_{i} ϕ_{i} (p)$ , and $ω$ vanishes on every commutator $a b - b a$ .

Proof

1.1givenF1algebra

Apply the dual-basis identity for $(q_{j}, ψ_{j})_{j}$ to each $p_{i}$ , then the identity for $(p_{i}, ϕ_{i})_{i}$ to each $T q_{j}$ . Right $A$ -linearity gives

\sum_{i} ϕ_{i} (T p_{i}) = \sum_{i, j} ϕ_{i} (T q_{j}) ψ_{j} (p_{i}), \sum_{j} ψ_{j} (T q_{j}) = \sum_{i, j} ψ_{j} (p_{i}) ϕ_{i} (T q_{j}) .

Subtracting, the difference is a finite sum of commutators $[ϕ_{i} (T q_{j}), ψ_{j} (p_{i})]$ in $A$ .

2.1givenF1step 1.1algebra∎

Apply $ω$ to the difference from step 1.1. Symmetry makes every commutator term vanish, so the two dual-basis formulas have the same scalar value. Since the bases were arbitrary, ${T r}_{ω, A} (T)$ is a singleton.

Depends on

Hattori–Stallings pseudo-trace

Dependency tree · 1 level

Sources

Bin Gui and Hao Zhang, Analytic Conformal Blocks of C2-cofinite Vertex Operator Algebras II: Convergence of Sewing and Higher Genus Pseudo-q-traces