Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Kähler-metric F-term scalar potential in the rigid $N = 1$ model

Physical scope

Conditional classical elimination of the chiral auxiliary fields in the bosonic zero-derivative sector of the adopted rigid N=1 Kähler sigma-model action on flat Minkowski spacetime. The result excludes gauge D-terms, supergravity, quantum corrections, and empirical claims.

Statement

Use the rigid $N = 1$ Kähler sigma-model action with scalar coordinates $z = (z^{1}, \dots, z^{n}) \in U \subseteq C^{n}$ , smooth real Kähler potential $K$ , positive-definite Hermitian metric $g_{i \overset{ˉ}{j}} = \partial_{i} \partial_{\overset{ˉ}{j}} K$ , holomorphic superpotential $W$ , and complex algebraic auxiliary fields $F^{i}$ . Use the ordinary complex bosonic sector of that postulate: all positive-degree coefficient-algebra components and fermions are set to zero, and the full formal even auxiliary coefficient is restricted to the complex field $F^{i}$ . This restriction is for the potential calculation and does not assert supersymmetry closure on the restricted slice. With spacetime derivatives set to zero, write $W_{i} = \partial_{i} W (z)$ and ${\overset{ˉ}{W}}_{\overset{ˉ}{j}} = \partial_{\overset{ˉ}{j}} \overline{W (z)}$ . The auxiliary Lagrangian is

$L_{F} = g_{i \overset{ˉ}{j}} F^{i} {\overset{ˉ}{F}}^{\overset{ˉ}{j}} + F^{i} W_{i} + {\overset{ˉ}{F}}^{\overset{ˉ}{j}} {\overset{ˉ}{W}}_{\overset{ˉ}{j}} .$

Let $g^{i \overset{ˉ}{j}}$ be the inverse Hermitian matrix. The algebraic field equation is $F^{i} = - g^{i \overset{ˉ}{j}} {\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ , and eliminating $F$ gives the nonnegative scalar-potential contribution

$V_{F} = g^{i \overset{ˉ}{j}} W_{i} {\overset{ˉ}{W}}_{\overset{ˉ}{j}} \geq 0.$

For the canonical Kähler potential $K = \sum_{i} {\overset{ˉ}{z}}^{\overset{ˉ}{i}} z^{i}$ , this becomes $V_{F} = \sum_{i} ∣ W_{i} ∣^{2}$ . The simple sum is specific to that metric; for a noncanonical Kähler metric the inverse matrix in the displayed formula is required. This statement concerns the classical F-term sector only and does not include gauge D-terms or quantum corrections.

Statement references

Stephen P. Martin, A Supersymmetry Primer, §§3.1–3.2 and 4.7.

Adel Bilal, Introduction to Supersymmetry, §7.1.

Rigid four-dimensional $N = 1$ Wess–Zumino sigma-model action

Facts & Assumptions

Given: A point $z \in U$ in the Kähler coordinate patch of the adopted model and its bosonic zero-derivative auxiliary fields.

[A1]

Rigid four-dimensional $N = 1$ Wess–Zumino sigma-model action: in its ordinary complex bosonic sector, $g_{i \overset{ˉ}{j}} (z, \overset{ˉ}{z})$ is positive-definite Hermitian and the zero-derivative auxiliary Lagrangian, obtained with $F_{r a w} = 2 F$ , $[U]_{D} = - I_{D} (U) / 4$ and $[Y]_{F} = I_{F} (Y) / 2$ , is $g_{i \overset{ˉ}{j}} F^{i} {\overset{ˉ}{F}}^{\overset{ˉ}{j}} + F^{i} W_{i} + {\overset{ˉ}{F}}^{\overset{ˉ}{j}} {\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ , with no derivatives of $F$ .

Proof

Given: [A1], with $W_{i}$ and ${\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ evaluated at the same scalar value $z$ .

1.1A1algebra

Since $F^{i}$ occurs algebraically and without derivatives, its Euler–Lagrange equation is the ordinary partial derivative of $L_{F}$ with respect to the independent conjugate coordinate ${\overset{ˉ}{F}}^{\overset{ˉ}{j}}$ . Differentiating [A1] gives $g_{i \overset{ˉ}{j}} F^{i} + {\overset{ˉ}{W}}_{\overset{ˉ}{j}} = 0$ .

1.2A1algebra

Expand the Hermitian square $g_{i \overset{ˉ}{j}} (F^{i} + g^{i \overset{ˉ}{k}} {\overset{ˉ}{W}}_{\overset{ˉ}{k}}) ({\overset{ˉ}{F}}^{\overset{ˉ}{j}} + g^{ℓ \overset{ˉ}{j}} W_{ℓ})$ . The two cross terms are $F^{i} W_{i}$ and ${\overset{ˉ}{F}}^{\overset{ˉ}{j}} {\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ because $g_{i \overset{ˉ}{j}} g^{ℓ \overset{ˉ}{j}} = δ_{i}^{ℓ}$ and $g_{i \overset{ˉ}{j}} g^{i \overset{ˉ}{k}} = δ_{\overset{ˉ}{j}}^{\overset{ˉ}{k}}$ ; the final term is $g^{ℓ \overset{ˉ}{k}} W_{ℓ} {\overset{ˉ}{W}}_{\overset{ˉ}{k}}$ . Therefore $L_{F}$ equals this square minus $g^{ℓ \overset{ˉ}{k}} W_{ℓ} {\overset{ˉ}{W}}_{\overset{ˉ}{k}}$ .

2.1A1step 1.1algebra

Positive definiteness makes $g$ invertible. Multiply the equation in step 1.1 by $g^{k \overset{ˉ}{j}}$ and use $g^{k \overset{ˉ}{j}} g_{i \overset{ˉ}{j}} = δ_{i}^{k}$ to obtain $F^{k} = - g^{k \overset{ˉ}{j}} {\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ .

3.1A1step 2.1step 1.2algebra∎

Substitute the solution from step 2.1: the square in step 1.2 vanishes, so the on-shell auxiliary Lagrangian is $- g^{i \overset{ˉ}{j}} W_{i} {\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ . To verify the sign directly, set $v^{i} = g^{i \overset{ˉ}{j}} {\overset{ˉ}{W}}_{\overset{ˉ}{j}}$ . Hermitian symmetry and the inverse identities give $g^{i \overset{ˉ}{j}} W_{i} {\overset{ˉ}{W}}_{\overset{ˉ}{j}} = g_{i \overset{ˉ}{j}} v^{i} {\overset{ˉ}{v}}^{\overset{ˉ}{j}}$ , which is positive when $v \neq 0$ and zero when $v = 0$ . Thus the contraction is nonnegative; for $g_{i \overset{ˉ}{j}} = δ_{i \overset{ˉ}{j}}$ it is $\sum_{i} ∣ W_{i} ∣^{2}$ . This is the classical scalar potential in the stated complex F-term sector; no order or positivity statement is made for general nilpotent coefficient-algebra values.

Depends on

Rigid four-dimensional $N = 1$ Wess–Zumino sigma-model action

Kähler-metric F-term scalar potential in the rigid $N = 1$ model

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Kähler-metric F-term scalar potential in the rigid N=1N=1 model

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Kähler-metric F-term scalar potential in the rigid $N = 1$ model