ScalingStacks

Example 1.6 . [03YX]

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Example 1.6.

(Constant solution) The simplest solution is where Vi​jV^{ij} and Wp​q¯W^{p\bar{q}} are independent of the base variables and satisfy (1.9). We shall see that many interesting solutions can be thought heuristically as perturbation of the constant solution after introducing some topology. Some important special cases for us are:

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    N=2,𝔫=1N=2,\mathfrak{n}=1, V=W=A>0V=W=A>0. The subcase where η\eta takes value in ℂ\mathbb{C} is relevant for the Taub-NUT metric (cf. Example 1.8), and the subcase where η\eta is a periodic variable is relevant for the Ooguri-Vafa metric (cf. Section 1.3). In the periodic case the choice of the connection ϑ\vartheta is parametrised by H1​(S1×ℝ2,S1)≃S1H^{1}(S^{1}\times\mathbb{R}^{2},S^{1})\simeq S^{1}.

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    N=3,𝔫=2N=3,\mathfrak{n}=2, Vi​j=ai​jV^{ij}=a_{ij} is symmetric positive definite, and W=A=det(ai​j)W=A=\det(a_{ij}). We write ga=ai​j​d​μi​d​μj+A​|d​η|2g_{a}=a_{ij}d\mu_{i}d\mu_{j}+A|d\eta|^{2}. The subcase where η\eta takes value in ℂ\mathbb{C} will be relevant for constructing new Taub-NUT type Calabi-Yau metrics on ℂ3\mathbb{C}^{3}, and the subcase where η\eta is a periodic variable will be relevant for the positive vertex. In the periodic case the choice of the connection ϑ\vartheta is parametrised by H1​(S1×ℝ3,T2)≃T2H^{1}(S^{1}\times\mathbb{R}^{3},T^{2})\simeq T^{2}.

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    N=3,𝔫=1N=3,\mathfrak{n}=1, Wp​q¯=ap​q¯W^{p\bar{q}}=a_{p\bar{q}} is Hermitian, and V=A=det(ap​q¯)V=A=\det(a_{p\bar{q}}). We write ga=Re​(ap​q¯​d​ηp​d​η¯q)+A​d​μ⊗d​μg_{a}=\text{Re}(a_{p\bar{q}}d\eta_{p}d\bar{\eta}_{q})+Ad\mu\otimes d\mu. Here η1,η2\eta_{1},\eta_{2} are periodic coordinates with period 1. The choice of the connection ϑ\vartheta is parametrised by H1​(T2×ℝ3,S1)≃T2H^{1}(T^{2}\times\mathbb{R}^{3},S^{1})\simeq T^{2}. This case will be relevant for the negative vertex. Notice that if we demand that the fibration on MM induced by μ,Im​(η1),Im​(η2)\mu,\text{Im}(\eta_{1}),\text{Im}(\eta_{2}) is a special Lagrangian fibration with phase zero, then we would need a1​2¯=a2​1¯a_{1\bar{2}}=a_{2\bar{1}}, namely ap​q=ap​q¯a_{pq}=a_{p\bar{q}} is symmetric.

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