ScalingStacks

Proof. [04LQ]

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Proof.

Since α1​(b)=b1\alpha_{1}(b)=b_{1}, it is enough to show that, if for fixed t∈ℝt\in\mathbb{R} we let Ut={b1=t}U_{t}=\{b_{1}=t\}, then αt=α|Ut\alpha_{t}=\alpha|_{U_{t}} is a bijection onto its image. If λ2\lambda_{2} and λ3\lambda_{3} are the periods of the fibration corresponding to γ2\gamma_{2} and γ3\gamma_{3}, then αt\alpha_{t} is computed by taking primitives of λ2|Ut\lambda_{2}|_{U_{t}} and λ3|Ut\lambda_{3}|_{U_{t}}. If we let XtX_{t} denote the symplectic reduction of XX at tt and Gt:Xt→ℝ2G_{t}:X_{t}\rightarrow\mathbb{R}^{2} the reduced fibration, then it is not difficult to see that λ2|Ut\lambda_{2}|_{U_{t}} and λ3|Ut\lambda_{3}|_{U_{t}} are in fact periods of GtG_{t} (cf. [3]Lemma 5.9). Now the conclusion follows by simply observing that GtG_{t} is a proper Lagrangian submersion, i.e. an integrable system. The argument works also when t=0t=0.

An explicit computation of the periods was done in [3]Proposition 5.10 for the fibration in Example 5.8. There we found that

λ2\displaystyle\lambda_{2} =\displaystyle= β1​d​b1−e2​b2​d​b2,\displaystyle\beta_{1}\,db_{1}-e^{2b_{2}}db_{2},
λ3\displaystyle\lambda_{3} =\displaystyle= β2​d​b1−e2​b3​d​b3,\displaystyle\beta_{2}\,db_{1}-e^{2b_{3}}db_{3}, (78)

where β1\beta_{1} and β2\beta_{2} are functions depending only on b1b_{1}. The periods of the perturbed fibration obtained in Lemma 7.6 will have this same expression away from where the perturbation took place (i.e. away from the white region in Figure 15), for example in a neighborhood of the codimension 1 part of Δ\Delta. It is easy to see from this expression of the periods that α\alpha extends continuously to Δ\Delta and that it is a bijection. ∎

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