The Holomorphic Inverse Function Theorem and Weierstrass Preparation

Using [A1], direct block multiplication gives $QRP=\left(\begin{array}{cc}A& 0\\ 0& \bar{A}\end{array}\right)\mathrm{.}$ Indeed the two columns of $P$ are the $(z,\bar{z})$ coordinates of a real vector, and the $\mathbb{C}$-linearity of $L$ makes the transformed action split into $A$ on the $z$ block and $\bar{A}$ on the $\bar{z}$ block.

By [L1], step 1.1, and the identity $QP=I$, one has ${\det }_{\mathbb{C}}R={\det }_{\mathbb{C}}(QRP)={\det }_{\mathbb{C}}A\cdot {\det }_{\mathbb{C}}\bar{A}={\det }_{\mathbb{C}}A\cdot \overline{{\det }_{\mathbb{C}}A}={\mid {\det }_{\mathbb{C}}A\mid }^2\mathrm{.}$ Since $R$ has real entries, the determinant polynomial gives the same real number whether computed over $\mathbb{R}$ or over $\mathbb{C}$, so ${\det }_{\mathbb{R}}L={\det }_{\mathbb{C}}R$.

The displayed formula immediately makes ${\det }_{\mathbb{R}}L$ nonzero exactly when ${\det }_{\mathbb{C}}A$ is nonzero.

By [L1], every component of $F$ is holomorphic and therefore smooth as a real-valued pair of functions on ${\mathbb{R}}^{2m}$. Hence the underlying real map $F:U\subseteq {\mathbb{R}}^{2m}\to {\mathbb{R}}^{2m}$ is $C^1$ on a neighbourhood of $a$.

The real derivative $DF(a)$ is the same linear map as the complex differential, now read over $\mathbb{R}$. Since $\det J_{\mathbb{C}}F(a)\mathrm{\ne }0$, [L2] gives ${\det }_{\mathbb{R}}DF(a)\mathrm{\ne }0$, so $DF(a)$ is invertible as a real linear map.

Apply [L3] to the real map from step 1.1 and the invertible derivative from step 1.2. This gives open neighbourhoods $U_0$ of $a$ and $V_0$ of $F(a)$ such that $F{\mathrm{\mid }}_{U_0}:U_0\to V_0$ is bijective and has a $C^1$ inverse $G:V_0\to U_0$. Moreover, $DG(w)=DF(G(w){)}^{-1}(w\in V_0)\mathrm{.}$

For each $w\in V_0$, the linear map $DF(G(w))$ is $\mathbb{C}$-linear because $F$ is holomorphic, so its inverse $DG(w)$ is also $\mathbb{C}$-linear. Step 2.1 already gives $G$ real differentiable at every point with that differential, hence the defining linear approximation for holomorphy uses a $\mathbb{C}$-linear derivative. Therefore $G$ is holomorphic on $V_0$. Together with the holomorphy of $F{\mathrm{\mid }}_{U_0}$, [F1] makes $F{\mathrm{\mid }}_{U_0}$ biholomorphic.

Define $H(z,w):=(z,f(z,w))\in {\mathbb{C}}^m\times {\mathbb{C}}^n\mathrm{.}$ Its complex differential at $(a,b)$ is $(u,v)⟼(u,{\mathrm{\partial }}_{z}f(a,b)u+{\mathrm{\partial }}_{w}f(a,b)v)\mathrm{.}$ Because ${\mathrm{\partial }}_{w}f(a,b)$ is invertible, this linear map has inverse $(\xi ,\eta )⟼(\xi ,{\mathrm{\partial }}_{w}f(a,b{)}^{-1}(\eta -{\mathrm{\partial }}_{z}f(a,b)\xi )),$ so $J_{\mathbb{C}}H(a,b)$ is invertible.

By [L1], after shrinking to neighbourhoods of $(a,b)$ and $(a,0)$ the map $H$ is biholomorphic. Write its inverse as $H^{-1}(z,\eta )=(z,\psi (z,\eta ));$ this form is forced because the first $m$ coordinates of $H$ are exactly $z$, and [L2] makes $\psi$ holomorphic.

Define $\phi (z):=\psi (z,0)$. Then $H(z,\phi (z))=H(H^{-1}(z,0))=(z,0),$ so $f(z,\phi (z))=0$. Conversely, if $(z,w)$ is in the shrunken source neighbourhood and $f(z,w)=0$, then $H(z,w)=(z,0)=H(z,\phi (z))$; injectivity of the biholomorphism from step 2.1 forces $w=\phi (z)$. This also proves the uniqueness of $\phi$.

After composing on the source and target with coordinate permutations, we may assume that the first $r\times r$ minor of $J_{\mathbb{C}}F(a)$ is invertible. Write $z=(x,y)\in {\mathbb{C}}^r\times {\mathbb{C}}^{m-r},F(z)=(F^{\mathrm{\prime }}(x,y),F^{\mathrm{\prime }\mathrm{\prime }}(x,y))\in {\mathbb{C}}^r\times {\mathbb{C}}^{n-r},$ and define $\alpha (x,y):=(F^{\mathrm{\prime }}(x,y),y)\mathrm{.}$ The complex Jacobian of $\alpha$ at $a$ is block triangular with diagonal blocks ${\mathrm{\partial }}_x{F}^{\mathrm{\prime }}(a)$ and $I_{m-r}$, so it is invertible. Hence [L1] makes $\alpha$ biholomorphic after shrinking around $a$.

Set $G:=F\circ {\alpha }^{-1}\mathrm{.}$ Because ${\alpha }^{-1}(u,v)$ has second block $v$, one has $G(u,v)=(u,h(u,v))$ for a holomorphic map $h$ into ${\mathbb{C}}^{n-r}$. Since $DG=DF\circ D{\alpha }^{-1}$ and $D{\alpha }^{-1}$ is invertible, $\mathrm{rank}DG$ is also $r$ on the shrunken neighbourhood.

In the Jacobian of $G$, the first $r$ output coordinates are exactly the coordinates $u_1,\dots ,u_r$, so the first $r$ rows already contain the $r\times r$ identity block. If some partial derivative $\mathrm{\partial }{h}_k\mathrm{/}\mathrm{\partial }{v}_j$ were nonzero at a point, adjoining the corresponding row and $v_j$ column would create an $(r+1)\times (r+1)$ minor with nonzero determinant there, contradicting $\mathrm{rank}DG=r$. Therefore every $\mathrm{\partial }{h}_k\mathrm{/}\mathrm{\partial }{v}_j$ vanishes on the neighbourhood.

Fix $u$, fix all $v$-coordinates except $v_j$, and fix a component $h_k$. By [L2], the slice $\lambda ⟼h_k(u,v_1,\dots ,v_{j-1},\lambda ,v_{j+1},\dots ,v_{m-r})$ is holomorphic on a disc, and step 3.1 says its derivative is identically $0$. Hence [L3] makes it constant. Repeating this for each $j$ shows that $h(u,v)$ is independent of $v$; writing $\varphi (u):=h(u,0)$, [L2] makes $\varphi$ holomorphic and gives $h(u,v)=\varphi (u)$.

Define the target shear $\beta (\xi ,\eta ):=(\xi ,\eta -\varphi (\xi ))\mathrm{.}$ Its inverse is $(\xi ,\eta )\mapsto (\xi ,\eta +\varphi (\xi ))$, so $\beta$ is biholomorphic near $F(a)$. Using step 4.1, $\beta \circ F\circ {\alpha }^{-1}(u,v)=\beta (u,\varphi (u))=(u,0)\mathrm{.}$ This is the claimed normal form, and when one of the dimensions $r$, $m-r$, or $n-r$ is $0$ the same formula is read with the corresponding block omitted.

Suppose $f(0)\mathrm{\ne }0$. Choose a representative, still called $f$, on a neighbourhood $U$ of $0$. By continuity from [L2], after shrinking $U$ we have $f(z)\mathrm{\ne }0$ for every $z\in U$. Then [L2] makes $1\mathrm{/}f$ holomorphic on $U$, so $[f][1\mathrm{/}f]=[1]$ in ${\mathcal{O}}_{m,0}$. Hence $[f]$ is a unit.

Suppose $f(0)=0$. For any germ $[g]\in {\mathcal{O}}_{m,0}$ one has $(fg)(0)=f(0)g(0)=0$, so $[f][g]\mathrm{\ne }[1]$. Therefore $[f]$ is not a unit.

Steps 1.1 and 1.2 show that the nonunits are exactly the germs vanishing at $0$, namely the elements of ${\mathfrak{m}}_{m,0}$ from [L1]. Any proper ideal contains no unit, so every proper ideal of ${\mathcal{O}}_{m,0}$ is contained in ${\mathfrak{m}}_{m,0}$. Since $1\mathrm{\notin }{\mathfrak{m}}_{m,0}$, this ideal is proper and therefore the unique maximal ideal. By [L3], ${\mathcal{O}}_{m,0}$ is local.

Choose a holomorphic representative of $f$ on a small polydisc, and expand it by [L1] as $f(z)={\sum }_{\alpha \in {\mathbb{N}}^m}{c}_{\alpha }{z}^{\alpha }\mathrm{.}$ Because $f$ is nonzero, some coefficient is nonzero. Let $d$ be the smallest total degree $\mathrm{\mid }\alpha \mathrm{\mid }$ for which $c_{\alpha }\mathrm{\ne }0$, and put $P_d(z):={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=d}{c}_{\alpha }{z}^{\alpha }\mathrm{.}$ Then $P_d$ is a nonzero homogeneous polynomial of degree $d$.

If $P_d(v)=0$ for every $v\in {\mathbb{C}}^m$, then the polynomial function $P_d$ vanishes identically on all of ${\mathbb{C}}^m$; applying [L2] to that finite power series would force every coefficient $c_{\alpha }$ with $\mathrm{\mid }\alpha \mathrm{\mid }=d$ to be $0$, contradicting step 1.1. Therefore choose $v\in {\mathbb{C}}^m$ with $P_d(v)\mathrm{\ne }0$.

Choose an invertible complex-linear map $T$ sending the last basis vector $e_m$ to $v$. Along the last axis one then has $f\circ T(0,\dots ,0,\zeta )=P_d(v){\zeta }^d+{\sum }_{n>d}{b}_n{\zeta }^n={\zeta }^d(P_d(v)+{\sum }_{n>d}{b}_n{\zeta }^{n-d})\mathrm{.}$ The bracketed factor is holomorphic and nonzero at $0$ because $P_d(v)\mathrm{\ne }0$. Hence [L3] makes $f\circ T$ regular in the last variable of order $d$.

By [L1], choose a representative on a neighbourhood of $0$ for which $f(0,\zeta )={\zeta }^{d}h(\zeta )$ and $h(0)\mathrm{\ne }0$. By continuity from [L2], after shrinking there is $r>0$ such that $h(\zeta )\mathrm{\ne }0$ on $\mathrm{\mid }\zeta \mathrm{\mid }\le r$. Hence the central slice has no zero on $\mathrm{\mid }\zeta \mathrm{\mid }=r$ and exactly the order-$d$ zero at $\zeta =0$ inside $\mathrm{\mid }\zeta \mathrm{\mid }<r$.

The compact set $\{0\}\times \{\mathrm{\mid }\zeta \mathrm{\mid }=r\}$ is contained in the domain of the chosen representative, and step 1.1 gives $\mathrm{\mid }f(0,\zeta )\mathrm{\mid }>0$ there. By continuity from [L2], after shrinking the $z^{\mathrm{\prime }}$-neighbourhood to some $V$ we have $\mathrm{\mid }f(z^{\mathrm{\prime }},\zeta )-f(0,\zeta )\mathrm{\mid }<\mathrm{\mid }f(0,\zeta )\mathrm{\mid }(z^{\mathrm{\prime }}\in V,\mathrm{\mid }\zeta \mathrm{\mid }=r)\mathrm{.}$ Then [L3] applied to the functions $\zeta \mapsto f(z^{\mathrm{\prime }},\zeta )$ and $\zeta \mapsto f(0,\zeta )$ shows that each nearby slice has the same zero count $d$ inside $\mathrm{\mid }\zeta \mathrm{\mid }<r$. The strict boundary inequality also makes $f(z^{\mathrm{\prime }},\zeta )\mathrm{\ne }0$ on $\mathrm{\mid }\zeta \mathrm{\mid }=r$.

By [L1], the denominator $f(z^{\mathrm{\prime }},\zeta )$ is nonzero on $V\times \{\mathrm{\mid }\zeta \mathrm{\mid }=r\}$. Hence [L2] makes ${\mathrm{\Phi }}_k(z^{\mathrm{\prime }},\zeta ):={\zeta }^k\frac{\mathrm{\partial }f\mathrm{/}\mathrm{\partial }{z}_m(z^{\mathrm{\prime }},\zeta )}{f(z^{\mathrm{\prime }},\zeta )}$ continuous on that compact cylinder. Fixing all coordinates of $z^{\mathrm{\prime }}$ except one, [L2] makes ${\mathrm{\Phi }}_k$ holomorphic in the remaining coordinate and [L3] makes the corresponding slice of $p_k$ holomorphic. The same compact continuity gives a uniform bound on ${\mathrm{\Phi }}_k$, so the ML estimate makes $p_k$ locally bounded on $V$. Therefore [L4] makes $p_k$ holomorphic on $V$.

Fix $z^{\mathrm{\prime }}\in V$ and apply [L5] to the one-variable holomorphic function $h(\zeta ):=f(z^{\mathrm{\prime }},\zeta )$ on the disc $\mathrm{\mid }\zeta \mathrm{\mid }<r$ with test function $g(\zeta )={\zeta }^k$. By [L1], the boundary circle is zero-free and the only singularities of $h^{\mathrm{\prime }}\mathrm{/}h$ inside are the zeros ${\lambda }_j(z^{\mathrm{\prime }})$, counted with their multiplicities. Thus $p_k(z^{\mathrm{\prime }})={\sum }_{j=1}^d{\lambda }_j(z^{\mathrm{\prime }}{)}^k\mathrm{.}$

By logarithmic differentiation of the polynomial in [A1], $\frac{Q^{\mathrm{\prime }}(\zeta )}{Q(\zeta )}={\sum }_{j=1}^d\frac{1}{\zeta -{\alpha }_j}\mathrm{.}$ Multiplying by $\zeta$ and expanding each summand for large $\zeta$ gives the formal Laurent identity $\zeta \frac{Q^{\mathrm{\prime }}(\zeta )}{Q(\zeta )}=d+{\sum }_{n\ge 1}{p}_n{\zeta }^{-n}\mathrm{.}$

Multiplying the identity from step 1.1 by $Q(\zeta )={\sum }_{n=0}^d{c}_n{\zeta }^{d-n}$ with $c_0=1$ and comparing the coefficient of ${\zeta }^{d-n}$ for $1\le n\le d$ yields $n{c}_n+{\sum }_{j=1}^n{c}_{n-j}{p}_j=0.$ Substituting $c_n=(-1{)}^n{e}_n$ gives $n{e}_n={\sum }_{j=1}^n(-1{)}^{j-1}{e}_{n-j}{p}_j\mathrm{.}$ Since $n\mathrm{\ne }0$ in $\mathbb{C}$, this determines $e_n$ recursively as a polynomial in $p_1,\dots ,p_n$.

The displayed recursion uses only addition, multiplication, and division by the nonzero scalar $n$. Therefore if the power sums $p_j$ vary holomorphically, then so do the $e_n$; the slice-zero case mentioned in [L1] is exactly such a holomorphic family.

Choose a representative of $f$ on a neighbourhood of the closed cylinder $V\times \{\mathrm{\mid }\zeta \mathrm{\mid }\le r\}$ given by [L2]. For each $z^{\mathrm{\prime }}\in V$, let ${\lambda }_1(z^{\mathrm{\prime }}),\dots ,{\lambda }_d(z^{\mathrm{\prime }})$ be the slice zeros in $\mathrm{\mid }\zeta \mathrm{\mid }<r$, counted with multiplicity. By [L3], the elementary symmetric functions $e_1(z^{\mathrm{\prime }}),\dots ,e_d(z^{\mathrm{\prime }})$ of those roots are holomorphic in $z^{\mathrm{\prime }}$. Define $W(z^{\mathrm{\prime }},\zeta ):={\zeta }^d-e_1(z^{\mathrm{\prime }}){\zeta }^{d-1}+\cdots +(-1{)}^d{e}_d(z^{\mathrm{\prime }})\mathrm{.}$ At $z^{\mathrm{\prime }}=0$ all slice roots equal $0$, so $e_j(0)=0$ for every $j$; therefore $W$ is a degree-$d$ Weierstrass polynomial by [L4].

For each fixed $z^{\mathrm{\prime }}\in V$, the polynomial $W(z^{\mathrm{\prime }},\zeta )$ has exactly the zeros ${\lambda }_1(z^{\mathrm{\prime }}),\dots ,{\lambda }_d(z^{\mathrm{\prime }})$ with their multiplicities. Since $f(z^{\mathrm{\prime }},\zeta )$ has the same zero multiset by construction, [L4] shows that at every slice zero $\lambda$ the quotient $f(z^{\mathrm{\prime }},\zeta )\mathrm{/}W(z^{\mathrm{\prime }},\zeta )$ extends holomorphically across $\lambda$; away from those zeros it is holomorphic by [L5]. Thus the slice quotient $q_{z^{\mathrm{\prime }}}(\zeta ):=\frac{f(z^{\mathrm{\prime }},\zeta )}{W(z^{\mathrm{\prime }},\zeta )}$ is holomorphic on $\mathrm{\mid }\zeta \mathrm{\mid }<r$.

Define $u(z^{\mathrm{\prime }},z_m):=\frac{1}{2\pi i}{\int }_{\mathrm{\mid }\zeta \mathrm{\mid }=r}\frac{f(z^{\mathrm{\prime }},\zeta )}{W(z^{\mathrm{\prime }},\zeta )(\zeta -z_m)}d\zeta \mathrm{.}$ Because $W(z^{\mathrm{\prime }},\zeta )\mathrm{\ne }0$ on $\mathrm{\mid }\zeta \mathrm{\mid }=r$, the integrand is continuous on $V\times \{\mathrm{\mid }\zeta \mathrm{\mid }=r\}\times \{\mathrm{\mid }{z}_m\mathrm{\mid }<r\}$ and holomorphic in each parameter variable. By [L6], the resulting function $u$ is separately holomorphic and locally bounded, hence holomorphic on $V\times \{\mathrm{\mid }{z}_m\mathrm{\mid }<r\}$. For fixed $z^{\mathrm{\prime }}$, the one-variable Cauchy formula [L7] applied to the holomorphic slice quotient $q_{z^{\mathrm{\prime }}}$ from step 2.1 gives $u(z^{\mathrm{\prime }},z_m)=q_{z^{\mathrm{\prime }}}(z_m),$ so $f(z^{\mathrm{\prime }},z_m)=u(z^{\mathrm{\prime }},z_m)W(z^{\mathrm{\prime }},z_m)\mathrm{.}$

On the central slice, regularity gives $f(0,\zeta )={\zeta }^{d}h(\zeta )$ with $h(0)\mathrm{\ne }0$, while step 1.1 gives $W(0,\zeta )={\zeta }^d$. Hence step 3.1 yields $u(0,0)=h(0)\mathrm{\ne }0$. By [L1], the germ of $u$ is a unit. Therefore the germs of $u$ and $W$ satisfy $f=uW$ in ${\mathcal{O}}_{m,0}$, which is the required preparation.

By [L1], after shrinking to a common neighbourhood the unit factors $u$ and $v$ are nowhere zero. Therefore for each fixed nearby parameter $z^{\mathrm{\prime }}$, the slice zeros of $f(z^{\mathrm{\prime }},\cdot )$ coincide, with multiplicity, with the slice zeros of $W(z^{\mathrm{\prime }},\cdot )$ and also with those of $P(z^{\mathrm{\prime }},\cdot )$.

Fix such a parameter $z^{\mathrm{\prime }}$. By [L2], both $W(z^{\mathrm{\prime }},\cdot )$ and $P(z^{\mathrm{\prime }},\cdot )$ are monic degree-$d$ one-variable polynomials with the same multiset of roots, counted with multiplicity. Over $\mathbb{C}$, a monic polynomial is the product of its linear factors, so these two polynomials are equal. Since this holds for every nearby $z^{\mathrm{\prime }}$, the germs satisfy $W=P$.

With $W=P$, the two preparations give $(u-v)W=0$. On the nonempty open set where $W\mathrm{\ne }0$ one therefore has $u=v$. Applying [L3] to the holomorphic function $u-v$ on the connected neighbourhood shows $u=v$ everywhere there, and hence as germs.