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A nonzero Q-exact deformation on the four-torus

Example

Let $T^{4} = (R / 2 π Z)^{4}$ have global angular coordinates $x^{1}, \dots, x^{4}$ and orientation $ω = d x^{1} \land d x^{2} \land d x^{3} \land d x^{4}$ . Use the graded algebra $Ω^{∙} (T^{4}; C)$ and integrate an inhomogeneous form by integrating its degree-four component over $T^{4}$ . Let $Q = d$ be the exterior derivative, determined by $d f = \sum_{i} (\partial_{i} f) d x^{i}$ , $d (d x^{i}) = 0$ , and the graded Leibniz rule. Set $O = d x^{1} \land d x^{2}$ , $S = d x^{3} \land d x^{4}$ , and $V = \sin (x^{3}) d x^{4}$ . For $t \in R$ , define $Z (t) = \int_{T^{4}} (O e^{- S - t Q V})_{[4]}$ . The degree-four density is $- (1 + t \cos (x^{3})) ω$ , which depends on $t$ at points where $\cos (x^{3}) \neq 0$ , but $Z (t) = - (2 π)^{4}$ for every $t \in R$ .

Facts & Assumptions

Given: $T^{4} = (R / 2 π Z)^{4}$ , its oriented volume form $ω$ , the exterior derivative $Q = d$ , and the forms $O, S, V$ specified in the Example.

[F1]

Finite-Dimensional Q-Exact Localization Data: the graded integral extracts the top-degree form and integrates it over the oriented cycle; an odd derivation is $Q$ -invariant when the integral of every admissible $Q$ -derivative vanishes.

[F2]

Finite-Dimensional Q-Exact Deformation Invariance: if $Q S = Q O = 0$ , $(Q^{2} V) e^{- S - t Q V} = 0$ , the $Q$ -boundary contribution vanishes, and the integrand and parameter derivative have an integrable majorant on compact parameter intervals, then $I_{Γ} (O e^{- S - t Q V})$ is constant in $t$ .

Verification

1.1givenF1

For a smooth periodic three-form $η = \sum_{i = 1}^{4} (- 1)^{i - 1} a_{i} (x) d x^{1} \land \dots \hat{d x^{i}} \dots \land d x^{4}$ , the degree-four coefficient of $d η$ is $\sum_{i} \partial_{i} a_{i}$ . For each $i$ , integrating $\partial_{i} a_{i}$ over $x^{i} \in [0, 2 π]$ gives $a_{i} ∣_{x^{i} = 2 π} - a_{i} ∣_{x^{i} = 0} = 0$ by periodicity; iterated integration over the other three coordinates therefore gives $\int_{T^{4}} (d η)_{[4]} = 0$ . Thus $I_{Γ} (Q F) = 0$ for every $F$ .

2.1givenF2step 1.1algebra

Directly, $Q O = Q S = 0$ , $Q V = d V = \cos (x^{3}) d x^{3} \land d x^{4}$ , and $Q^{2} V = d (d V) = - \sin (x^{3}) d x^{3} \land d x^{3} \land d x^{4} = 0$ . For $t$ in any compact interval $K$ , the coefficients of $O e^{- S - t Q V}$ and its $t$ -derivative are bounded on the compact torus by $1 + \sup_{t \in K} ∣ t ∣$ ; this constant is integrable on $T^{4}$ . Therefore [F2] applies.

3.1step 2.1algebra

Put $H_{t} = S + t Q V = (1 + t \cos (x^{3})) d x^{3} \land d x^{4}$ . Since $H_{t} \land H_{t} = 0$ , $e^{- H_{t}} = 1 - H_{t}$ . Multiplication by $O$ gives $(O e^{- H_{t}})_{[4]} = - (1 + t \cos (x^{3})) ω$ .

4.1step 3.1algebra∎

The integral of $\cos (x^{3})$ over $[0, 2 π]$ is zero, while each of the four coordinate intervals has length $2 π$ . Hence $Z (t) = - \int_{[0, 2 π]^{4}} (1 + t \cos (x^{3})) d^{4} x = - (2 π)^{4}$ , independent of $t$ ; the pointwise $t$ -dependent term integrates to zero by the same periodic boundary cancellation as in step 1.1.

A nonzero Q-exact deformation on the four-torus

Example

Facts & Assumptions

Verification

Depends on

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