LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6-luna)

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Local complex Hessian of the multi-modulus O3/O7 coordinates

Statement

Let $r \geq 1$ and $s \geq 0$ , let $N = (N^{1}, \dots, N^{s})$ be local complex coordinates on a patch in $C^{s}$ , and let $K_{0} (N, \overset{ˉ}{N})$ be a real $C^{2}$ function there. Let $v^{α}, ρ_{α} \in R$ for $1 \leq α \leq r$ , let $κ_{α β γ}$ be a real symmetric tensor with indices in ${1, \dots, r}$ , and set $P (v) = κ_{α β γ} v^{α} v^{β} v^{γ}, P_{α} = κ_{α β γ} v^{β} v^{γ}, H_{α β} = κ_{α β γ} v^{γ} .$ Assume $P > 0$ and $H_{α β}$ is invertible, with inverse $H^{α β}$ . For smooth complex functions $ζ_{α} (N, \overset{ˉ}{N})$ define $ζ_{α}^{R} = ζ_{α} + {\overset{ˉ}{ζ}}_{α}$ and $T_{α} = \frac{3 i}{2} ρ_{α} + \frac{3}{4} P_{α} (v) - \frac{3}{2} ζ_{α} (N, \overset{ˉ}{N}) .$ On any patch where $(T_{α}, N^{A})$ are valid complex coordinates, define $K = K_{0} - 2 \log P (v (T, \overset{ˉ}{T}, N, \overset{ˉ}{N}))$ and $G_{α β} = - \frac{3}{2} (\frac{H_{α β}}{P} - \frac{3}{2} \frac{P_{α} P_{β}}{P^{2}}), G^{α β} = - \frac{2}{3} P H^{α β} + 2 v^{α} v^{β} .$ Then $G^{α β}$ is the inverse of $G_{α β}$ , and the nonzero complex Hessian blocks of $K$ are $K_{T_{α} {\overset{ˉ}{T}}_{β}} = \frac{G^{α β}}{P^{2}}, K_{T_{α} {\overset{ˉ}{N}}^{\overset{ˉ}{B}}} = \frac{3}{2} \frac{G^{α β}}{P^{2}} \partial_{{\overset{ˉ}{N}}^{\overset{ˉ}{B}}} ζ_{β}^{R},$ $K_{N^{A} {\overset{ˉ}{N}}^{\overset{ˉ}{B}}} = K_{0, A \overset{ˉ}{B}} - \frac{3 v^{α}}{P} \partial_{A} \partial_{\overset{ˉ}{B}} ζ_{α}^{R} + \frac{9}{4} \frac{G^{α β}}{P^{2}} (\partial_{A} ζ_{α}^{R}) (\partial_{\overset{ˉ}{B}} ζ_{β}^{R}) .$ Equivalently, after the coordinate change to $(v^{α}, ρ_{α}, N^{A})$ , the associated real-coordinate Hermitian quadratic form is $G_{α β} d v^{α} d v^{β} + \frac{1}{4} (d \log P)^{2} + (K_{0, A \overset{ˉ}{B}} - \frac{3 v^{α}}{P} \partial_{A} \partial_{\overset{ˉ}{B}} ζ_{α}^{R}) d N^{A} d {\overset{ˉ}{N}}^{\overset{ˉ}{B}} + \frac{9}{4} \frac{G^{α β}}{P^{2}} Ξ_{α} Ξ_{β},$ where $Ξ_{α} = d ρ_{α} - i ((\partial_{A} {\overset{ˉ}{ζ}}_{α}) d N^{A} - (\partial_{\overset{ˉ}{A}} ζ_{α}) d {\overset{ˉ}{N}}^{\overset{ˉ}{A}})$ and products denote the corresponding symmetric quadratic form.

Statement references

Grimm and Louis, §3.2 and Appendix C.1

Facts & Assumptions

Given: The real cubic $P$ , its derivatives, the invertible matrix $H_{α β}$ , the real variables $v^{α}, ρ_{α}$ , the complex coordinates $N^{A}$ , the smooth functions $ζ_{α}$ , and the local-coordinate hypothesis in the Statement.

[A1]

The map $(v, ρ, N) \mapsto (T, N)$ is a valid local coordinate map on the chosen patch; no global invertibility is assumed.

Proof

technique · implicit differentiation and matrix multiplication

Given: The data and local coordinate patch in the Statement.

1.1givenA1algebra

Write $ζ_{α}^{R} = ζ_{α} + {\overset{ˉ}{ζ}}_{α}$ . The real part of the coordinate definition is $T_{α} + {\overset{ˉ}{T}}_{α} = \frac{3}{2} (P_{α} - ζ_{α}^{R})$ . Differentiating it with respect to $T_{β}$ at fixed $(\overset{ˉ}{T}, N, \overset{ˉ}{N})$ gives $δ_{α}^{β} = 3 H_{α γ} \partial_{T_{β}} v^{γ}$ , hence $\partial_{T_{β}} v^{α} = \frac{1}{3} H^{α β}$ . Differentiating at fixed $(T, \overset{ˉ}{T}, \overset{ˉ}{N})$ with respect to $N^{A}$ gives $0 = 3 H_{α γ} \partial_{A} v^{γ} - \frac{3}{2} \partial_{A} ζ_{α}^{R}$ , hence $\partial_{A} v^{α} = \frac{1}{2} H^{α β} \partial_{A} ζ_{β}^{R}$ . Conjugation gives the barred derivatives.

2.1givenstep 1.1algebra

The cubic identities $P_{α} v^{α} = P$ and $H_{α β} v^{β} = P_{α}$ imply $P_{β} H^{β α} = v^{α}$ . Hence $\partial_{T_{α}} P = 3 P_{β} (\partial v^{β} / \partial T_{α}) = v^{α}$ and $\partial_{N^{A}} P = 3 P_{β} (\partial v^{β} / \partial N^{A}) = \frac{3}{2} v^{α} \partial_{A} ζ_{α}^{R}$ . Differentiating $K = K_{0} - 2 \log P$ gives $K_{T_{α}} = - 2 v^{α} / P$ and $K_{N^{A}} = K_{0, A} - 3 v^{α} \partial_{A} ζ_{α}^{R} / P$ .

3.1givenstep 2.1algebra

Multiplying the proposed inverse by $G_{α β}$ , use $H_{α β} v^{β} = P_{α}$ , $P_{α} v^{α} = P$ , and $H_{α β} H^{β γ} = δ_{α}^{γ}$ . The coefficient of $P_{α} v^{γ} / P$ in the product is $- 3 - \frac{3}{2} + \frac{9}{2} = 0$ , leaving $δ_{α}^{γ}$ . Thus $G^{α β}$ is the inverse of $G_{α β}$ .

4.1givenstep 1.1step 2.1step 3.1algebra

Differentiating $K_{T_{α}} = - 2 v^{α} / P$ in a barred coordinate gives $K_{T_{α} {\overset{ˉ}{T}}_{β}} = - \frac{2 H^{α β}}{3 P} + \frac{2 v^{α} v^{β}}{P^{2}} = \frac{G^{α β}}{P^{2}}, K_{T_{α} {\overset{ˉ}{N}}^{\overset{ˉ}{B}}} = - \frac{H^{α γ} ζ_{γ, \overset{ˉ}{B}}^{R}}{P} + \frac{3 v^{α} v^{γ} ζ_{γ, \overset{ˉ}{B}}^{R}}{P^{2}} = \frac{3}{2} \frac{G^{α γ}}{P^{2}} ζ_{γ, \overset{ˉ}{B}}^{R} .$ Differentiating $K_{N^{A}} = K_{0, A} - 3 v^{α} ζ_{α, A}^{R} / P$ gives $K_{N^{A} {\overset{ˉ}{N}}^{\overset{ˉ}{B}}} = K_{0, A \overset{ˉ}{B}} - \frac{3 v^{α}}{P} ζ_{α, A \overset{ˉ}{B}}^{R} - \frac{3 H^{α γ}}{2 P} ζ_{α, A}^{R} ζ_{γ, \overset{ˉ}{B}}^{R} + \frac{9 v^{α} v^{γ}}{2 P^{2}} ζ_{α, A}^{R} ζ_{γ, \overset{ˉ}{B}}^{R} .$ The last two terms equal $\frac{9}{4} G^{α γ} ζ_{α, A}^{R} ζ_{γ, \overset{ˉ}{B}}^{R} / P^{2}$ by the displayed formula for $G^{α γ}$ .

5.1givenstep 1.1step 3.1step 4.1algebra∎

In the pullback of the Hessian blocks from step 4.1, the terms involving $d T$ and the derivatives of $ζ^{R}$ factor as $\frac{G^{α β}}{P^{2}} (d T_{α} + \frac{3}{2} ζ_{α, A}^{R} d N^{A}) (d {\overset{ˉ}{T}}_{β} + \frac{3}{2} ζ_{β, \overset{ˉ}{B}}^{R} d {\overset{ˉ}{N}}^{\overset{ˉ}{B}}) .$ Using the coordinate definition and its conjugate, the two factors are $\frac{3}{2} (H_{α γ} d v^{γ} + i Ξ_{α})$ and $\frac{3}{2} (H_{β δ} d v^{δ} - i Ξ_{β})$ , respectively. Their product has no mixed $d v Ξ$ term because these are complex conjugates and $H, G^{α β}, v, Ξ$ are real. The $Ξ$ coefficient is $9 G^{α β} / (4 P^{2})$ . For the $d v$ coefficient, the inverse formula in step 3.1 gives $\frac{9}{4 P^{2}} G^{α β} H_{α γ} H_{β δ} = - \frac{3 H_{γ δ}}{2 P} + \frac{9 P_{γ} P_{δ}}{2 P^{2}} = G_{γ δ} + \frac{1}{4} \partial_{γ} \log P \partial_{δ} \log P,$ since $\partial_{γ} P = 3 P_{γ}$ . The remaining Hessian block is exactly $(K_{0, A \overset{ˉ}{B}} - 3 v^{α} ζ_{α, A \overset{ˉ}{B}}^{R} / P) d N^{A} d {\overset{ˉ}{N}}^{\overset{ˉ}{B}}$ . This proves the stated real-coordinate form.

Remarks

The coordinate map is only local. The lemma does not assert that the cubic-volume Hessian is invertible on every Kähler cone, or that one global prepotential or Kähler potential exists. The hypotheses also do not imply positivity: with $r = 2$ , $κ_{111} = κ_{122} = 1$ (all other independent coefficients zero), and $v = (1, 0)$ , one has $P = 1$ and $H = I$ , but $G_{22} = - 3 / 2$ . Thus the result is a Hessian identity; additional cone hypotheses are needed to call this quadratic form a positive Kähler metric.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Thomas W. Grimm and Jan Louis, The effective action of N=1 Calabi-Yau orientifolds, arXiv:hep-th/0403067v3