Several Complex Variables (VI): Implicit Mapping Theorem

10 Sep 2018

• several complex variables

To conclude our discussion of the analysis aspect of the several complex variables theory, we prove the holomorphic versions of the implicit and inverse mapping theorems.

Implicit function theorem: Let $U$ be an open subset of $\bb C^n$, and let $f$ be a holomorphic function on $U$ with a zero at $\lambda$. Suppose that $\frac{\del f}{\del z_n}(\lambda)\neq0$. Then there is a polydisc

$$\Delta(\lambda,r)=\Delta(\lambda',r')\times\Delta(\lambda_n,r_n)\subseteq\bb C^{n-1}\times\bb C,$$

and a holomorphic $g:\Delta(\lambda’,r’)\to\Delta(\lambda_n,r_n)$, such that for $z=(z’,z_n)\in\Delta(\lambda,r)$,

$$f(z)=0\quad\iff\quad g(z')=z_n.$$

Proof: This is an immediate consequence of the Weierstrass preparation theorem: by the given conditions, $f$ has vanishing order 1 in $z_n$ at $\lambda$. Thus there is a polydisc $\Delta(z,r)$ and a factorisation $f=uh$, where $u$ is holomorphic and nonzero, and $h$ is a Weierstrass polynomial of degree 1, ie.

$$h(z)=z_n-g(z')$$

for some holomorphic $g$ with $g(\lambda’)=0$. Now the result follows from the fact that $f$ and $h$ have the same roots on $\Delta(\lambda,r)$. $\qed$

We will now generalise this result to holomorphic $F:U\to\bb C^m$. Any such map can be decomposed into coordinate functions $F(z)=(f_1(z),\ldots,f_m(z))$. Just as in (real) multivariable calculus, the non-degeneracy condition on the derivative generalises to a non-degeneracy condition on the $m\times n$ Jacobian matrix

$$J_F(z)\coloneqq\left(\frac{\del f_i}{\del z_j}(z)\right)_{i,j=1}^{m,n}.$$

Implicit mapping theorem: Let $U$ be an open subset of $\bb C^n$, and let $F:U\to\bb C^m$ be a holomorphic map, with coordinate functions $f_1,\ldots,f_m$ and a zero at $\lambda$. Suppose that the last $m$ columns of $J_f(z)$ form a non-singular $m\times m$ matrix. Then there is a polydisc

$$\Delta(\lambda,r)=\Delta(\lambda',r')\times\Delta(\lambda'',r'')\subseteq\bb C^{n-m}\times\bb C^m,$$

and a holomorphic $G:\Delta(\lambda’,r’)\to\Delta(\lambda^{\prime\prime},r^{\prime\prime})$, such that for $z=(z’,z^{\prime\prime})\in\Delta(\lambda,r)$,

$$F(z)=0\quad\iff\quad g(z')=z''.$$

Proof: By induction on $m$. The case $m=1$ is exactly the implicit function theorem.

We now prove the statement for $m$ from the statement for $m-1$. First, split $J_F=(J_F’,J_F^{\prime\prime})$ into its first $n-m$ and last $m$ columns, so $J_F^{\prime\prime}(\lambda)$ is non-singular. By a change of coordinates of the range $\bb C^m$, we may assume that $J_F^{\prime\prime}(\lambda)$ is the $m\times m$ identity matrix.

Now $\frac{\del f_m}{\del z_n}(\lambda)=1$, so by the implicit function theorem there exists a polydisc $\Delta(\lambda,r)$ and a holomorphic map

$$h:\Delta((\lambda_1,\ldots,\lambda_{n-1}),(r_1,\ldots,r_{n-1}))\to\Delta(\lambda_n,r_n)$$

such that for $z\in\Delta(\lambda,r)$,

$$f_m(z)=0\quad\iff\quad z_n=h(z_1,\ldots,z_{n-1}).$$

The idea is to restrict to the zero set of the last coordinate $f_m$, which is locally described by the hypersurface $(z_1,\ldots,z_{n-1},h(z_1,\ldots,z_{n-1}))$, and apply the induction hypothesis. More specifically, define the map

$$F':\Delta((\lambda_1,\ldots,\lambda_{n-1}),(r_1,\ldots,r_{n-1}))\to\bb C^{m-1}$$

by defining its coordinate functions $f’1,\ldots,f’{m-1}$ as

$$f'_i(z_1,\ldots,z_{n-1})=f_i(z_1,\ldots,z_{n-1},h(z_1,\ldots,z_{n-1})).$$

Then at the point $(\lambda_1,\ldots,\lambda_{n-1})$, we have $F’=0$, and

$$\begin{aligned} \frac{\del f_i'}{\del z_j}&=\frac{\del f_i}{\del z_j}+\frac{\del f_i}{\del z_n}\frac{\del h}{\del z_j}\\ &=\begin{cases}1,&j=i+n-m\\0,&\text{else},\end{cases} \end{aligned}$$

since $\frac{\del f_i}{\del z_n}=0$. Hence the last $m-1$ columns of $J_{F’}$ form a $(m-1)\times(m-1)$ identity matrix, so by induction hypothesis there exists (after possibly restricting to a smaller polydisc) a holomorphic map

$$G':\Delta(\lambda',r')\to\Delta((\lambda_{n-m+1},\ldots,\lambda_{n-1}),(r_{n-m+1},\ldots,r_{n-1}))$$

such that for $(z_1,\ldots,z_{n-1})\in\Delta((\lambda_1,\ldots,\lambda_{n-1}),(r_1,\ldots,r_{n-1}))$,

$$F'(z_1,\ldots,z_{n-1})=0\quad\iff\quad G'(z')=(z_{n-m+1},\ldots,z_{n-1}).$$

Hence if we define

$$G(z')\coloneqq(G'(z'),h(z',G'(z'))),$$

then for $z\in\Delta(\lambda,r)$ we have

$$\begin{aligned} F(z)=0&\iff F'(z_1,\ldots,z_{n-1})=0\\ &\qquad\text{and }z_n=h(z_1,\ldots,z_{n-1})\\ &\iff G'(z')=(z_{n-m+1},\ldots,z_{n-1})\\ &\qquad\text{and }z_n=h(z_1,\ldots,z_{n-m},G'(z'))\\ &\iff G(z')=z'', \end{aligned}$$

so $G$ satisfies the given conditions, and induction is complete. $\qed$

Now we give some important consequences of this theorem.

Inverse mapping theorem: Let $U$ be an open subset of $\bb C^n$, and let $H:U\to\bb C^n$ be a holomorphic map. Let $\lambda\in U$, and suppose that $J_H(\lambda)$ is non-singular. Then there is a neighbourhood $V\subseteq U$ of $\lambda$ such that $H$ is a biholomorphic mapping from $V$ onto some neighbourhood of $H(\lambda)$.

Proof: By the implicit function theorem applied to the function $F:\bb C^n\times U\to\bb C^n$ given by $F(z’,z^{\prime\prime})=H(z^{\prime\prime})-z’$, we get a function $G$ on some polydisc with

$$G(z')=z''\quad\iff\quad F(z',z'')=0\quad\iff\quad H(z'')=z',$$

so $G$ is an inverse for $H$ on this polydisc. $\qed$

More generally, we have the following:

Constant rank theorem: Let $U$ be an open subset of $\bb C^n$, and let $F:U\to\bb C^m$ be a holomorphic map. Suppose that $J_f$ has constant rank $k$ in $U$. Then for each $\lambda\in U$, there is a neighbourhood $U_\lambda$ of $\lambda$ in which $F$ is biholomorphically equivalent to the standard projection

$$(z_1,\ldots,z_n)\mapsto(z_1,\ldots,z_k,0,\ldots,0)$$

from a neighbourhood of $0\in\bb C^n$ to a neighbourhood of $0\in\bb C^m$.

Proof: We may assume that $\lambda=0$ and $F(\lambda)=0$, and after a change of coordinates that the upper left $k\times k$ submatrix of $J_F(z)$ is non-singular at 0, and hence in a neighbourhood $U’$ of $0\in\bb C^n$. Define $G:U’\to\bb C^n$ by

$$G(z_1,\ldots,z_n)=(f_1(z_1,\ldots,z_n),\ldots,f_k(z_1,\ldots,z_n),z_{k+1},\ldots,z_n).$$

Then $J_G$ is non-singular in $U’$, so by inverse mapping theorem $G$ is a biholomorphic map from some neighbourhood $U^{\prime\prime}\subseteq U’$ of $0\in\bb C^n$ to another.

Now $F\circ G^{-1}$ has the form $(z_1,\ldots,z_n)\mapsto(z_1,\ldots,z_k,f_{k+1}’,\ldots,f_m’)$. Since its Jacobian also has rank $k$ in $U^{\prime\prime}$, we have $\frac{\del f_j’}{\del z_i}$ are identically zero for $i>k$, ie. $f_j’$ are functions in $z_1,\ldots,z_k$ alone. Now define

$$H(z_1,\ldots,z_m)=(z_1,\ldots,z_k,z_{k+1}-f_{k+1}',\ldots,z_m-f_m').$$

Note that $H$ has inverse

$$H^{-1}(z_1,\ldots,z_m)=(z_1,\ldots,z_k,z_{k+1}+f_{k+1}',\ldots,z_m+f_m'),$$

so $H$ is biholomorphic on some neighbourhood of $0\in\bb C^m$. Now

$$H\circ F\circ G^{-1}(z_1,\ldots,z_n)=(z_1,\ldots,z_k,0,\ldots,0)$$

on a neighbourhood $U$ of $0\in\bb C^m$, as desired. $\qed$