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Proof.
To see Ξ 1 \Theta_{1} π \bm{e} N 0 N_{0} π ~ = e β 1 β k β Ο β π \tilde{\bm{e}}=e^{\sqrt{-1}k\phi}\bm{e}
(4.55) y ~ = e β ( k β β k + ) β β 1 β Ο β y ; u ~ 1 = e k + β β 1 β Ο β u 1 , u ~ 2 = e β k β β β 1 β Ο β u 2 . \tilde{y}=e^{-(k_{-}-k_{+})\sqrt{-1}\phi}y;\ \ \tilde{u}_{1}=e^{k_{+}\sqrt{-1}\phi}u_{1},\tilde{u}_{2}=e^{-k_{-}\sqrt{-1}\phi}u_{2}.
for some local real-valued function Ο \phi H H
(4.56) u Β― 1 β d β u 1 β u 1 β d β u Β― 1 \displaystyle\bar{u}_{1}du_{1}-u_{1}d\bar{u}_{1} = u ~ Β― 1 β d β u ~ 1 β u ~ 1 β d β u ~ Β― 1 β 2 β k + β β 1 β | u 1 | 2 β d β Ο , \displaystyle=\bar{\tilde{u}}_{1}d\tilde{u}_{1}-\tilde{u}_{1}d\bar{\tilde{u}}_{1}-2k_{+}\sqrt{-1}|u_{1}|^{2}d\phi,
(4.57) u Β― 2 β d β u 2 β u 2 β d β u Β― 2 \displaystyle\bar{u}_{2}du_{2}-u_{2}d\bar{u}_{2} = u ~ Β― 2 β d β u ~ 2 β u ~ 2 β d β u ~ Β― 2 + 2 β k β β β 1 β | u 2 | 2 β d β Ο , \displaystyle=\bar{\tilde{u}}_{2}d\tilde{u}_{2}-\tilde{u}_{2}d\bar{\tilde{u}}_{2}+2k_{-}\sqrt{-1}|u_{2}|^{2}d\phi,
(4.58) Ξ \displaystyle\Gamma = Ξ ~ β k β β k + 2 β d β Ο . \displaystyle=\widetilde{\Gamma}-\frac{k_{-}-k_{+}}{2}d\phi.
Then it is a straightforward to compute that Ξ ~ 1 = Ξ 1 \widetilde{\Theta}_{1}=\Theta_{1} Ξ 1 \Theta_{1}
Now we consider the local expansion of Ξ₯ \Upsilon Ο \psi 3.24
(4.59) β z Ο ~ = β β 1 β z 2 β r 3 β d β y β§ d β y Β― β z 2 β r 3 β ( y β d β y Β― + y Β― β d β y ) β§ Ξ + z r β d β Ξ + O β² β ( 1 ) β d β y + O β² β ( 1 ) β d β y Β― + O β² β ( r ) . \partial_{z}\tilde{\omega}=-\sqrt{-1}\frac{z}{2r^{3}}dy\wedge d\bar{y}-\frac{z}{2r^{3}}(yd\bar{y}+\bar{y}dy)\wedge\Gamma+\frac{z}{r}d\Gamma+O^{\prime}(1)dy+O^{\prime}(1)d\bar{y}+O^{\prime}(r).
Next, applying Proposition 3.28 3.26
(4.60) d D c β h = d D c β ( 1 2 β r + O β² β ( r ) ) = β 1 4 β r 3 β d D c β | y | 2 + O β² β ( 1 ) = β β 1 β ( y β d β y Β― β y Β― β d β y ) + 4 β | y | 2 β Ξ 4 β r 3 + O β² β ( 1 ) . d_{D}^{c}h=d_{D}^{c}(\frac{1}{2r}+O^{\prime}(r))=-\frac{1}{4r^{3}}d_{D}^{c}|y|^{2}+O^{\prime}(1)=-\frac{\sqrt{-1}(yd\bar{y}-\bar{y}dy)+4|y|^{2}\Gamma}{4r^{3}}+O^{\prime}(1).
Putting together these, and noting that d β Ξ 0 d\Theta_{0} 2.50
(4.61) d β Ξ 1 β Ξ₯ = O β² β ( 1 ) β d β y + O β² β ( 1 ) β d β y Β― + O β² β ( r ) + O β² β ( 1 ) β d β z . d\Theta_{1}-\Upsilon=O^{\prime}(1)dy+O^{\prime}(1)d\bar{y}+O^{\prime}(r)+O^{\prime}(1)dz.
Now translating into the coordinates u 1 , u 2 u_{1},u_{2} π \mathbb{L}