Lemma 12.28.1A3034D6EAAEF

Proof. Choose a smooth normal exponential map $\Upsilon _0$ whose first normal derivative is complex linear; the derivative of $s_ D$ identifies the normal bundle with $L|_ D$, and we use this identification to fix that first derivative. Parallel transport along the normal rays identifies $\Upsilon _0^*L$ with the pullback of $N$. In compatible parallel frames one then has

For each $p$, this is a smooth map between real two-dimensional disks whose derivative at zero is the identity. The inverse function theorem with parameters, uniformly over the compact base $D$, gives a unique small $a=a(p,w)$ with $W_ j(\Upsilon _0(p,a))=w$. It depends smoothly on $(p,w)$, and its first derivative in $w$ at $w=0$ is the identity. On an overlap of frames, $a$ and $w$ change by the same constant unitary factor, so uniqueness makes these corrections agree. After decreasing $R$, the map $(p,w)\mapsto (p,a(p,w))$ is a diffeomorphism of a tube onto its image, and its composition with $\Upsilon _0$ is the required map $\Upsilon $.

Off $D$, the function $F=\log |s_ D|_ h^2$ is pluriharmonic, and its radial derivative in the tube is nonzero. Hence every $M_ r$ is a smooth hypersurface; it is compact and connected, being a circle bundle over $D$. Locally its complex tangent space is $\ker \mathrm dW_ j$, so it is Levi-flat, with complex leaves the hypersurfaces $\{ W_ j=\text{constant}\} $. The identity $W_ j\circ \Upsilon =w$ pulls this distribution back to $\{ \mathrm dw=0\} $ in parallel frames, which is the flat horizontal foliation of the circle bundle. The map $\Upsilon $ is only a smooth tubular identification; its base projection need not be holomorphic. $\square$