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Local Type Iia Potential Model Detecting the RR Bianchi Sign

Physical scope

A smooth local potential configuration on a contractible coordinate patch that distinguishes the correct negative massless RR Bianchi sign from the inconsistent positive sign. It is a kinematic sign check only; it asserts no solution of the supergravity equations, global string background, or empirical realization.

Example

Let $ℓ > 0$ have length dimension and $U = (- ℓ, ℓ)^{10}$ have coordinates $x^{0}, \dots, x^{9}$ of length dimension, orientation $d x^{0} \land \dots \land d x^{9}$ , flat Lorentzian metric $η = diag (- 1, 1, \dots, 1)$ , and constant dilaton $Φ = 0$ . Set $B_{2} = \frac{x^{1}}{ℓ} d x^{2} \land d x^{3}, C_{1} = \frac{x^{4}}{ℓ} d x^{5}, C_{3} = 0.$ These smooth potentials give $H_{3} = ℓ^{- 1} d x^{1} \land d x^{2} \land d x^{3}, G_{2} = ℓ^{- 1} d x^{4} \land d x^{5}, G_{4} = 0, {\tilde{G}}_{4} = \frac{x^{4}}{ℓ^{2}} d x^{1} \land d x^{2} \land d x^{3} \land d x^{5} .$ The potential components are dimensionless, while the field-strength components have dimension length $^{- 1}$ . Direct differentiation yields $d {\tilde{G}}_{4} = - ℓ^{- 2} d x^{1} \land d x^{2} \land d x^{3} \land d x^{4} \land d x^{5} = - H_{3} \land G_{2}$ , a nonzero five-form. Thus the plus-sign formula $d {\tilde{G}}_{4} = + H_{3} \land G_{2}$ fails for this configuration.

Example references

M. J. D. Hamilton, §6.1, Theorem 6.2(e)

Joseph Polchinski, §12.1.5 and §12.1.8

Type Iia RR Bianchi From Potentials

Facts & Assumptions

Given: $U$ , its coordinates and orientation, the flat metric and constant dilaton above, and the three smooth potentials above.

[A1]

Type Iia Two Derivative Supergravity Action defines $H_{3} = d B_{2}$ , $G_{2} = d C_{1}$ , $G_{4} = d C_{3}$ , and ${\tilde{G}}_{4} = G_{4} - C_{1} \land H_{3}$ in the massless, source-free convention. This adopted postulate supplies the declared field definitions; no deductive proof of the action is used.

[F1]

Type Iia RR Bianchi From Potentials: for these local potentials with $d H_{3} = 0$ , the correct identity is $d {\tilde{G}}_{4} = - H_{3} \land G_{2}$ . Its local proof is complete and the example is conditional on this adopted massless convention.

[F2]

The exterior derivative is a graded derivation and The exterior derivative squares to zero: $d$ obeys the signed product rule and $d^{2} = 0$ .

Verification

1.1givenA1F2algebra

Differentiating the potentials gives $H_{3} = ℓ^{- 1} d x^{1} \land d x^{2} \land d x^{3}$ , $G_{2} = ℓ^{- 1} d x^{4} \land d x^{5}$ , $G_{4} = 0$ , $d H_{3} = 0$ , and $d G_{2} = 0$ . Since $C_{1} \land H_{3} = - x^{4} ℓ^{- 2} d x^{1} \land d x^{2} \land d x^{3} \land d x^{5}$ (moving $d x^{5}$ past three one-forms changes sign), the definition in [A1] gives the displayed positive ${\tilde{G}}_{4}$ .

2.1givenF1step 1.1algebra

Differentiating ${\tilde{G}}_{4}$ gives $d {\tilde{G}}_{4} = ℓ^{- 2} d x^{4} \land d x^{1} \land d x^{2} \land d x^{3} \land d x^{5} = - ℓ^{- 2} d x^{1} \land d x^{2} \land d x^{3} \land d x^{4} \land d x^{5} .$ The minus sign occurs when $d x^{4}$ is moved across $d x^{1}, d x^{2}, d x^{3}$ . On the other hand $H_{3} \land G_{2} = + ℓ^{- 2} d x^{1} \land d x^{2} \land d x^{3} \land d x^{4} \land d x^{5}$ , so $d {\tilde{G}}_{4} = - H_{3} \land G_{2} \neq + H_{3} \land G_{2}$ . The coordinate five-form is nonzero because these coordinate one-forms are linearly independent throughout $U$ .

3.1givenA1F1step 1.1step 2.1∎

The potentials are smooth on the contractible patch and $d H_{3} = 0$ , so they meet the hypotheses of [F1] and realize its negative-sign identity explicitly. This example distinguishes the erroneous plus sign; it does not assert that these fields solve the Euler–Lagrange equations.

Local Type Iia Potential Model Detecting the RR Bianchi Sign

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources