The holomorphic constant-rank theorem

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.