

the second term of (2.12), the uniqueness of the Hilbert
basis decomposition implies the equivalent relationship
$$
\dbar_{z_k}\xi_\alpha=\varepsilon^{-1}
\sqrt{(\alpha_k+1)(|\alpha|+n+1)}\;\xi_{\alpha+c_k}(z,w).
\leqno(2.13)
$$
Since $\partial_{z_j}e_\alpha=0$, we have the simpler formula
$$
\partial_{z_j}\sigma(z,w)=\sum_{\alpha\in\bN^n}\partial_{z_j}\xi_\alpha(z)\;
e_\alpha(z,w).
$$
This allows us to easily compute the Chern connection
$\nabla=\nabla^{1,0}+\nabla^{0,1}$ of $B_\varepsilon$ in terms of the
orthonormal frame $(e_\alpha)$ (beware that $\nabla^{0,1}$ is not given by
$\sum_\alpha\dbar_z\xi_\alpha e_\alpha$, as
$(e_\alpha)$ is not holomorphic!).

The curvature if the hermitian metric $h$ on $B_\varepsilon$ can be computed by
evaluating
$$
\eqalign{
\partial_{z_j}\!\dbar_{\overline z_k}|\sigma|^2
&=\partial_{z_j}\!\dbar_{\overline z_k}\sum_{\alpha\in\bN^n}
\xi_\alpha\overline\xi_\alpha\cr
&=\sum_{\alpha\in\bN^n}
(\partial_{z_j}\xi_\alpha)(\overline{\partial_{z_k}\xi_\alpha})
+(\partial_{z_j}\!\dbar_{\overline z_k}\xi_\alpha)\overline\xi_\alpha
+\xi_\alpha(\partial_{z_j}\!\dbar_{\overline z_k}\overline\xi_\alpha)
+(\dbar_{z_k}\xi_\alpha)(\overline{\dbar_{z_j}\xi_\alpha})
\cr}
$$
Thanks to (2.13) we can eliminate all $\dbar$ derivatives, and we find
$$
\eqalign{
\partial_{z_j}\!\dbar_{\overline z_k}|\sigma|_h^2
&=\sum_{\alpha\in\bN^n}
(\partial_{z_j}\xi_\alpha)(\overline{\partial_{z_k}\xi_\alpha})\cr
&+\sum_{\alpha\in\bN^n}\varepsilon^{-1}
\sqrt{(\alpha_k+1)(|\alpha|+n+1)}\;(\partial_{z_j}\xi_{\alpha+c_k})
\overline\xi_\alpha\cr
&+\sum_{\alpha\in\bN^n}\varepsilon^{-1}
\sqrt{(\alpha_j+1)(|\alpha|+n+1)}\;\xi_\alpha
(\overline{\partial_{z_k}\xi_{\alpha+c_j}})\cr
&+\sum_{\alpha\in\bN^n}\varepsilon^{-2}\sqrt{(\alpha_j+1)(|\alpha|+n+1)}\;
\sqrt{(\alpha_k+1)(|\alpha|+n+1)}\;\xi_{\alpha+c_k}\overline{\xi_{\alpha+c_j}}\cr
&=\sum_{\alpha\in\bN^n}
(\partial_{z_j}\xi_\alpha)(\overline{\partial_{z_k}\xi_\alpha})\cr
&+\sum_{\alpha\in\bN^n}\varepsilon^{-1}
\sqrt{\alpha_k(|\alpha|+n)}\;(\partial_{z_j}\xi_\alpha)
\overline{\xi_{\alpha-c_k}}\cr
&+\sum_{\alpha\in\bN^n}\varepsilon^{-1}
\sqrt{\alpha_j(|\alpha|+n)}\;\xi_{\alpha-c_j}
(\overline{\partial_{z_k}\xi_\alpha})\cr
&+\sum_{\alpha\in\bN^n}\varepsilon^{-2}\sqrt{(\alpha_j-\delta_{jk})
(|\alpha|+n-1)}\;
\sqrt{(\alpha_k-\delta_{jk})(|\alpha|+n-1)}\;\xi_{\alpha-c_j}
\overline{\xi_{\alpha-c_k}}\cr}
$$
afet making changes $\alpha\mapsto\alpha-c_k$, $\alpha\mapsto\alpha-c_j$,
$\alpha\mapsto\alpha-c_j-c_k$ in the last 3 summations (and agreeing
that $\xi_\alpha=0$ if $\alpha\notin\bN^n$). This implies
$$
\eqalign{
\partial_{z_j}\!\dbar_{\overline z_k}|\sigma|_h^2
&=\sum_{\alpha\in\bN^n}
\big(\partial_{z_j}\xi_\alpha+\varepsilon^{-1}
\sqrt{\alpha_j(|\alpha|+n)}\;\xi_{\alpha-c_j}\big)
\big(\overline{\partial_{z_k}\xi_\alpha+\varepsilon^{-1}
\sqrt{\alpha_k(|\alpha|+n)}\;\xi_{\alpha-c_k}}\big)\cr
&+\sum_{\alpha\in\bN^n}\delta_{jk}\varepsilon^{-2}
\big((\alpha_j-1)(|\alpha|+n-1)-\alpha_j(|\alpha|+n)\big)
\;\xi_{\alpha-c_j}\overline{\xi_{\alpha-c_j}}\cr
&+\sum_{\alpha\in\bN^n}(1-\delta_{jk})\varepsilon^{-2}
\sqrt{\alpha_j\alpha_k}\;\big((|\alpha|+n-1)-(|\alpha|+n)\big)
\;\xi_{\alpha-c_j}\overline{\xi_{\alpha-c_k}}.\cr
&=\sum_{\alpha\in\bN^n}
\big(\partial_{z_j}\xi_\alpha+\varepsilon^{-1}
\sqrt{\alpha_j(|\alpha|+n)}\;\xi_{\alpha-c_j}\big)
\big(\overline{\partial_{z_k}\xi_\alpha+\varepsilon^{-1}
\sqrt{\alpha_k(|\alpha|+n)}\;\xi_{\alpha-c_k}}\big)\cr
&-\sum_{\alpha\in\bN^n}\varepsilon^{-2}
\sqrt{\alpha_j\alpha_k}\;\xi_{\alpha-c_j}\overline{\xi_{\alpha-c_k}}.\cr
&-\sum_{\alpha\in\bN^n}\delta_{jk}\varepsilon^{-2}
(|\alpha|+n-1)\;\xi_{\alpha-c_j}\overline{\xi_{\alpha-c_j}}.\cr}
$$
Therefore,we find a complex Hessian
$$
\eqalign{
\sum_{j,k}v_j\overline v_k\;\partial_{z_j}\!\dbar_{\overline z_k}|\sigma|_h^2
&=\sum_{\alpha\in\bN^n}
\Bigg|\sum_jv_j\Big(\partial_{z_j}\xi_\alpha+\varepsilon^{-1}
\sqrt{\alpha_j(|\alpha|+n)}\;\xi_{\alpha-c_j}\Big)\Bigg|^2\cr
&-\sum_{\alpha\in\bN^n}\varepsilon^{-2}
\Bigg|\sum_jv_j\sqrt{\alpha_j}\;\xi_{\alpha-c_j}\Bigg|^2
-\sum_{\alpha\in\bN^n}\sum_j\varepsilon^{-2}(|\alpha|+n)\;|v_j|^2|\xi_\alpha|^2.
\cr}
$$


In a related manner, the $L^2$ metric $h_0$ on the central fiber is given by
$$
\langle e_\alpha,e_\beta\rangle_{h_0}=\delta_{\alpha\beta}.
$$
We now compute a Taylor expansion of the metric $h_z$ on $B_{\varepsilon,z}$
at order $2$, when $z\to 0$.
Let $c_j=(0,\ldots,1,\ldots, 0)$ be the canonical basis of
the $\bZ$-module $\bZ^n$. We find
$$
\eqalign{
\langle w^\alpha,w^\beta\rangle_{h_z}
&=\int_{|w-\overline z|<1}w^\alpha\cdot\overline w^\beta\,
d\lambda(w)=\int_{|w|<1}
(w+\overline z)^\alpha\cdot\overline{(w+\overline z)}^\beta d\lambda(w)
\quad\hbox{on $E_z$}\cr
&=\langle w^\alpha, w^\beta\rangle_{h_0}
+\sum_j\alpha_j\overline z_j
\langle w^{\alpha-c_j}, w^\beta\rangle_{h_0}
+\sum_k\beta_k z_k
\langle w^\alpha, w^{\beta-c_k}\rangle_{h_0}\cr
&\qquad{}+\sum_{j,k}\alpha_j\beta_k\overline z_jz_k
\langle w^{\alpha-c_j}, w^{\beta-c_k}\rangle_{h_0}
+O(z^2,\overline z^2,|z|^3),\cr}
$$
therefore
$$
\eqalign{
\langle e_\alpha,e_\beta\rangle_{h_z}
=\delta_{\alpha\beta}&+\sum_j\sqrt{\alpha_j(|\alpha|+n)}
\;\overline z_j\;\delta_{\alpha-c_j,\beta}+
\sum_k\sqrt{\beta_k(|\beta|+n)}
\;z_k\;\delta_{\alpha,\beta-c_k}\cr
&+\sum_{j,k}\sqrt{\alpha_j\beta_k(|\alpha|+n)(|\beta|+n)}
\;\overline z_j z_k\;\delta_{\alpha-c_j,\beta-c_k}
+O(z^2,\overline z^2,|z|^3),\cr
}
$$
and in particular
$$
|e_\gamma|^2_{h_z}=1+\sum_j\gamma_j(|\gamma|+n)\,|z_j|^2+
O(z^2,\overline z^2,|z|^3).
$$
The modified holomorphic section
$$
\widetilde e_\alpha(z,w)=e_\alpha(w)-
\sum_k\lambda_{\alpha,k}z_k\;e_{\alpha+c_k}(w)
\leqno(2.10)
$$
then satisfies
$$
\eqalign{
\langle\widetilde e_\alpha,\widetilde e_\beta\rangle_{h_z}
=\delta_{\alpha\beta}
&-\sum_k\lambda_{\alpha,k}z_k\;\delta_{\alpha+c_k,\beta}
-\sum_j\overline\lambda_{\beta,j}
\overline z_j\;\delta_{\alpha,\beta+c_j}
+\sum_{j,k}\lambda_{\alpha,k}\overline\lambda_{\beta,j}\;
\overline z_j z_k\;\delta_{\alpha+c_k,\beta+c_j}\cr
&+\sum_j\sqrt{\alpha_j(|\alpha|+n)}
\;\overline z_j\;\delta_{\alpha-c_j,\beta}+
\sum_k\sqrt{\beta_k(|\beta|+n)}
\;z_k\;\delta_{\alpha,\beta-c_k}\cr
&+\sum_{j,k}\sqrt{\alpha_j\beta_k(|\alpha|+n)(|\beta|+n)}
\;\overline z_j z_k\;\delta_{\alpha-c_j,\beta-c_k}\cr
&-\sum_{j,k}\lambda_{\alpha,k}\sqrt{(\alpha_j+\delta_{jk})(|\alpha|+n+1)}\;
\overline z_jz_k\;\delta_{\alpha+c_k-c_j,\beta}\cr
&-\sum_{j,k}\overline\lambda_{\beta,j}
\sqrt{(\beta_k+\delta_{jk})(|\beta|+n+1)}\;
\overline z_j z_k\;\delta_{\alpha,\beta+c_j-c_k}
+O(z^2,\overline z^2,|z|^3),\cr}
$$
In the above formula, we notice that we must have $|\alpha|=|\beta|$ to get
non zero terms, and that $\alpha_j+\delta_{jk}=\beta_j+1$,
$\beta_k+\delta_{jk}=\alpha_k+1$ if
$\delta_{\alpha+c_k-c_j,\beta}=
\delta_{\alpha-c_j,\beta-c_k}$ equals~$1$.
As a consequence, the choice of positive real coefficients
$$
\lambda_{\beta-c_k,k}=\sqrt{\beta_k(|\beta|+n)},\quad\hbox{i.e.}\quad
\lambda_{\alpha,k}=\sqrt{(\alpha_k+1)(|\alpha|+n+1)}
\leqno(2.11)
$$
gives
$$
\eqalign{
\langle\widetilde e_\alpha,\widetilde e_\beta\rangle_{h_z}
=\delta_{\alpha\beta}
&+\sum_{j,k}\sqrt{\alpha_j\beta_k}\;(|\alpha|+n)
\;\overline z_j z_k\;\delta_{\alpha-c_j,\beta-c_k}\cr
&-\sum_{j,k}\sqrt{(\alpha_k+1)(\beta_j+1)}\;(|\alpha|+n+1)\;
\overline z_jz_k\;\delta_{\alpha-c_j,\beta-c_k}
+O(z^2{\!,\,}\overline z^2{\!,\,}|z|^3),\cr
=\delta_{\alpha\beta}
&-\sum_{j}\big((\alpha_j+1)(|\alpha|+n+1)-\alpha_j(|\alpha|+n)\big)
\;|z_j|^2\;\delta_{\alpha\beta}\cr
&-\sum_{j\neq k}\sqrt{\alpha_j\beta_k}\;
\big((|\alpha|+n+1)-(|\alpha|+n)\big)\;\overline z_jz_k\;
\delta_{\alpha-c_j,\beta-c_k}
+O(z^2{\!,\,}\overline z^2{\!,\,}|z|^3),\cr
=\delta_{\alpha\beta}
&-\sum_{j}(\alpha_j+|\alpha|+n+1)\;|z_j|^2\;\delta_{\alpha\beta}\cr
&-\sum_{j\neq k}\sqrt{\alpha_j\beta_k}\;
\overline z_jz_k\;\delta_{\alpha-c_j,\beta-c_k}
+O(z^2{\!,\,}\overline z^2{\!,\,}|z|^3),\cr
=\delta_{\alpha\beta}&-\sum_{j}(|\alpha|+n+1)\;|z_j|^2\;\delta_{\alpha\beta}
-\sum_{j,k}\sqrt{\alpha_j\beta_k}\;
\overline z_jz_k\;\delta_{\alpha-c_j,\beta-c_k}
+O(z^2{\!,\,}\overline z^2{\!,\,}|z|^3).\cr}
$$
If we come back to $U_\varepsilon$ for $\varepsilon>0$ not necessarily equal
to $1$ (and do not omit any more the trivial $n$-form
$dw_1\wedge\ldots\wedge dw_n$), we have
to use a rescaling $(z,w)\mapsto(\varepsilon^{-1}z,\varepsilon^{-1}w)$.
This gives for the Hilbert bundle $B_\varepsilon$ a normal coordinate frame
$$
\widetilde e_\alpha(z,w)=
\sqrt{{(|\alpha|+n)!\over\alpha_1!\,...\,\alpha_n!}}
\bigg(w^\alpha-(|\alpha|+n+1)\varepsilon^{-2}
\sum_kz_k\;w^{\alpha+c_k}\bigg)
{dw_1\wedge\ldots\wedge dw_n\over
\pi^{n/2}\varepsilon^{|\alpha|+n}}
\leqno(2.12)\phantom{\raise6pt\hbox{$\bigg|$}}
$$
such that $(\widetilde e_\alpha(0,{\scriptstyle\bullet}))_{\alpha\in\bN^n}$
is an orthonormal basis of $B_{\varepsilon,0}$ and
$$
\langle\widetilde e_\alpha,\widetilde e_\beta\rangle_{h_z}
=\delta_{\alpha\beta}-\varepsilon^{-2}\Bigg(\!
\sum_{j}(|\alpha|+n+1)\;|z_j|^2\;\delta_{\alpha\beta}
-\sum_{j,k}\sqrt{\alpha_j\beta_k}\;
\overline z_jz_k\;\delta_{\alpha-c_j,\beta-c_k}\kern-3pt\Bigg)
+O(z^2{\!,\,}\overline z^2{\!,\,}|z|^3).
\leqno(2.13)\phantom{\raise6pt\hbox{$\bigg|$}}
$$
This is in principle enough to compute the curvature of $B_\varepsilon$.
However, one has to pay attention to the fact that the
inner products $\langle\widetilde e_\alpha,\widetilde e_\beta\rangle_{h_z}$
are not 

$O(...)$ is not
uniform as $|\alpha|,|\beta|\to+\infty$, since the coefficients are not
uniformly bounded. Any germ of $L^2$ holomorphic section
$\sigma=\sum_{\alpha\in\bN^n}\xi_\alpha\widetilde e_\alpha\in
\cO(B_\varepsilon)$ at $z=0$ is defined by a family of holomorphic
coefficients $\xi_\alpha(z)$ such that
$\sum_\alpha|\xi_\alpha(0)|^2<+\infty$. If we take a germ
$\sigma\in\cO(B_{\rho\varepsilon})$ in a slightly larger neighborhood
($|\rho|>1$), we must have
$\sum_\alpha\rho^{2|\alpha|}|\xi_\alpha(0)|^2<+\infty$ an it it then
easy to see that if we have
$\sum_\alpha\rho^{\prime\,2|\alpha|}|\xi_\alpha(z)|^2<+\infty$
for some $\rho'>1$ and $z$ near~$0$, then the series
$\sum\xi_\alpha(z)\widetilde e_\alpha(z,w)$ is holomorphically
convergent.

modulo $O(z^2{\!,\,}\overline z^2{\!,\,}|z|^3)$
$$
\leqalignno{
|\sigma|^2_h&=\sum_{\alpha,\beta\in\bN^n}
\xi_\alpha\overline\xi_\beta\;\langle\widetilde e_\alpha,
\widetilde e_\beta\rangle_{h_z}\cr
&=\sum_{\alpha\in\bN^n}\big(1-(|\alpha|+n+1)|z|^2\big)|\xi_\alpha|^2
-\sum_{j,k,\gamma\in\bN^n}
\sqrt{(\gamma_j+1)(\gamma_k+1)}\;\overline z_j z_k\;
\xi_{\gamma+c_j}\overline\xi_{\gamma+c_k},&(2.12)\cr}
$$
by taking $\alpha=\gamma+c_j$ and $\beta=\gamma+c_k$.
A straightforward calculation also gives
the explicit expression
$$
\widetilde e_\alpha(z,w)=
\pi^{-n/2}
\sqrt{{(|\alpha|+n)!\over\alpha_1!\ldots\alpha_n!}}\,w^\alpha
\Big(1-\sqrt{|\alpha|+n+1}\;\sum_k z_kw_k\Big).
\leqno(2.13)
$$
