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Finite Grassmann Gaussian and gauge-invariant determinant

Statement

Let $R = C ⟦ z ⟧$ , let $d \geq 1$ , and let $K \in M_{d} (R)$ be a finite matrix as in Finite matrix formal kernel and inverse series. With $I_{d}$ the top-coefficient functional of Finite Grassmann algebra and top-coefficient integral, $I_{d} ⁣ (\exp ⁣ (- \sum_{i, j = 1}^{d} {\overset{ˉ}{ψ}}_{i} K_{i j} ψ_{j})) = \det K .$ If $D = diag (u_{1}, \dots, u_{d})$ with each $u_{i} \in R^{\times}$ and $K^{'} = D K D^{- 1}$ , then the finite Grassmann integral and determinant are gauge invariant: $I_{d} (\exp (- \overset{ˉ}{ψ} K^{'} ψ)) = \det K^{'} = \det K$ . No infinite-dimensional Grassmann measure, continuum limit, or anomaly claim is made.

Facts & Assumptions

Given: A finite integer $d \geq 1$ , the commutative coefficient ring $R = C ⟦ z ⟧$ , and $K \in M_{d} (R)$ .

[F1]

Finite Grassmann algebra and top-coefficient integral defines anticommuting generators, the ordered top monomial $M_{d}$ , the coefficient functional $I_{d} (P) = (- 1)^{d (d + 1) / 2} [M_{d}] P$ , and the finite exponential.

[F2]

Finite matrix formal kernel and inverse series defines $\det K$ by the finite Leibniz sum and the diagonal vertex-phase conjugation convention.

[A1]

The integer $d \geq 1$ , commutative unital coefficient ring $R = C ⟦ z ⟧$ , matrix $K \in M_{d} (R)$ , and diagonal units $u_{i} \in R^{\times}$ are as in the Statement.

Proof

1.1A1F1algebra

Put $B_{K} = \sum_{i, j} {\overset{ˉ}{ψ}}_{i} K_{i j} ψ_{j}$ . Every factor has Grassmann degree two, so a product of more than $d$ such factors has degree greater than $2 d$ and vanishes. Hence the exponential is the finite sum $\sum_{n = 0}^{d} (- B_{K})^{n} / n!$ , and only its degree- $2 d$ term contributes to $[M_{d}]$ .

2.1A1F1F2step 1.1algebra

In $B_{K}^{d} / d!$ , any term with a repeated row or column index vanishes. The surviving terms are indexed by permutations $σ \in S_{d}$ . Reordering ${\overset{ˉ}{ψ}}_{1} ψ_{σ (1)} \dots {\overset{ˉ}{ψ}}_{d} ψ_{σ (d)}$ into $M_{d}$ contributes $(- 1)^{d (d - 1) / 2} sgn (σ)$ ; the exponential contributes $(- 1)^{d}$ . Therefore $[M_{d}] \exp (- B_{K}) = (- 1)^{d (d + 1) / 2} \sum_{σ \in S_{d}} sgn (σ) \prod_{i = 1}^{d} K_{i, σ (i)} = (- 1)^{d (d + 1) / 2} \det K .$ Applying the normalization in [F1] proves the Gaussian identity.

3.1A1F2step 2.1algebra∎

For diagonal $D$ , $K_{i j}^{'} = u_{i} K_{i j} u_{j}^{- 1}$ . In each Leibniz term, $\prod_{i} u_{i} \prod_{i} u_{σ (i)}^{- 1} = 1$ , since $σ$ permutes the indices; thus $\det K^{'} = \det K$ . The Gaussian identity from [step 2.1] then gives the same equality for the finite Grassmann integral.

Finite Grassmann Gaussian and gauge-invariant determinant

Statement

Facts & Assumptions

Proof

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