# Holomorphic Factorization of Mappings Into $\SpC$

Josua Schott  Josua Schott

(June 23, 2022)

###### Abstract

It is shown that any symplectic $2n\times 2n$-matrix, whose entries are
complex holomorphic functions on a reduced Stein space, can be decomposed into
a finite product of elementary symplectic matrices if and only if it is null-
homotopic. Moreover, if this is the case, the number of factors can be bounded
by a constant depending only on $n$ and the dimension of the space.

##  1 Introduction

It is well known that over any field, in particular over the field of complex
numbers, a matrix in the Special Linear Group $\mathrm{SL}_{n}(\mathbb{C})$ is
a product of elementary matrices. The proof is usually an application of a
Gauss or a Gauss-Jordan process. The same question for $\mathrm{SL}_{n}(R)$
over a commutative ring $R$ is much more difficult and has been studied a lot.
For the ring $R=\mathbb{C}[z]$ of polynomials in one variable, it is true
since $R$ is an Euclidean ring. For $n=2$ and the ring $R=\mathbb{C}[z,w]$,
Cohn [Co66] found the following counterexample: the matrix

$\begin{pmatrix}1-zw&z^{2}\\\ -w^{2}&1+zw\end{pmatrix}\in\mathrm{SL}_{2}(R)$

does not decompose as a finite product of unipotent matrices. In the 1970s,
Quillen [Qu76] and Suslin [Sus77] independently gave a positive answer in the
case $\mathrm{SL}_{n}(\mathbb{C}[z_{1},...,z_{m}])$, for $n\geq 3$ and $m\geq
1$. For the ring $R$ of complex-valued continuous functions on a normal
topological space, Vaserstein [Vas88] showed, that a matrix in
$\mathrm{SL}_{n}(R)$ decomposes into a product of elementary matrices if and
only it is null-homotopic. In Gromov’s seminal paper on the Oka principle, the
starting point of modern Oka theory, he asks this question for the ring of
complex-valued holomorphic functions (he calls it the Vaserstein problem, see
[Gro89, p. 886]). Ivarsson and Kutzschebauch [IK12] were able to give a
positive answer to this problem in full generality.

The same question for the symplectic group $\mathrm{Sp}_{2n}(R)$ hasn’t been
studied to the same degree. Again, there is a positive answer for the ring
$\mathbb{C}[z]$, since this is an Euclidean ring. For $n\geq 2$, Kopeiko
[Kop78] proved it for the polynomial ring $R=k[z_{1},...,z_{m}]$; and
Grunewald, Mennicke and Vaserstein [GMV91] for the ring
$R=\mathbb{Z}[z_{1},...,z_{m}]$. In [IKL20a], Ivarsson, Kutzschebauch and Løw
prove it for every commutative ring $R$ with identity and bass stable rank
$sr(R)=1$. Furthermore, they show that, for the ring $R=C(X)$ of complex-
valued continuous functions on a normal topological space, a matrix in
$\mathrm{Sp}_{2n}(C(X))$ can be decomposed into a product of elementary
matrices if and only it is null-homotopic; we call this the Continuous
Vaserstein problem for symplectic matrices. The same authors [IKL20b] show a
similar result in the case of $\mathrm{Sp}_{4}(\mathcal{O}(X))$, where
$\mathcal{O}(X)$ denotes the ring of holomorphic functions on a reduced Stein
space $X$. In this paper, we give a full solution for
$\mathrm{Sp}_{2n}(\mathcal{O}(X))$ for $n\geq 1$, the Holomorphic Vaserstein
problem for symplectic matrices.

We call a $(2n\times 2n)$-matrix $M$ symplectic if it satisfies $M^{T}JM=J$
with respect to the skew-symmetric matrix

$J=\begin{pmatrix}0&I_{n}\\\ -I_{n}\end{pmatrix},$

where $I_{n}$ denotes the $n\times n$-identity matrix and $0$ the $n\times
n$-zero matrix. For symmetric matrices $A\in\mathbb{C}^{n\times n}$, i.e.
$A^{T}=A$, matrices of the form

$\begin{pmatrix}I_{n}&A\\\
0&I_{n}\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}I_{n}&0\\\
A&I_{n}\end{pmatrix}$

are symplectic and we call them elementary symplectic matrices. For simplicity
reasons, let’s identify the space of symmetric matrices
$\mathrm{Sym}_{n}(\mathbb{C})=\\{A\in\mathbb{C}^{n\times n}:A^{T}=A\\}$ with
$\mathbb{C}^{\frac{n(n+1)}{2}}$.

######  Theorem 1.1 (Main theorem).

There exists a natural number $K=K(n,d)$ such that given any finite
dimensional reduced Stein space $X$ of dimension $d$ and any null-homotopic
holomorphic mapping $f:X\rightarrow\textnormal{Sp}_{2n}(\mathbb{C})$ there
exist a holomorphic mapping

$G=(G_{1},...,G_{K}):X\rightarrow(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$

such that

$f(x)=\begin{pmatrix}I_{n}&0\\\
G_{1}(x)&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&G_{2}(x)\\\
0&I_{n}\end{pmatrix}\cdots\begin{pmatrix}I_{n}&0\\\
G_{K-1}(x)&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&G_{K}(x)\\\
0&I_{n}\end{pmatrix},\quad x\in X.$

It is immediate that any product of elementary symplectic matrices is
connected by a path to the constant identity matrix $I_{2n}$, by multiplying
the off-diagonal entries by $t$, i.e.

$\begin{pmatrix}I_{n}&tA\\\
0&I_{n}\end{pmatrix}\quad\text{or}\quad\begin{pmatrix}I_{n}&0\\\
tA&I_{n}\end{pmatrix}.$

Therefore the requirement of null-homotopy of the map $f$ is neccessary.

###  1.1 Strategy of proof

There are basically two main ingredients for the proof. In a first step, one
proves a continuous version of the theorem (we also call this the Continuous
Vaserstein problem for symplectic matrices \- see Theorem 2.1). Once this is
shown, we want to apply an Oka principle, which allows us, very roughly
speaking, to deform the continuous solution into a holomorphic one. But one by
one. Let’s introduce the elementary symplectic matrix mapping
${M_{k}:\mathbb{C}^{\frac{n(n+1)}{2}}\to\SpC}$ by

$M_{k}(Z)=\begin{cases}\begin{pmatrix}I_{n}&Z\\\ 0&I_{n}\end{pmatrix}&\text{if
}k=2l\\\ \begin{pmatrix}I_{n}&0\\\ Z&I_{n}\end{pmatrix}&\text{if
}k=2l+1\end{cases}$

and then define the mapping
$\Psi_{K}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to\SpC$ by

$\Psi_{K}(Z_{1},...,Z_{K})=M_{1}(Z_{1})M_{2}(Z_{2})\cdots M_{K}(Z_{K}).$

As already said, Ivarsson, Kutzschebauch and Løw [IKL20a] solved the
(Continuous Vaserstein problem for symplectic matrices).. In fact, there is a
natural number $K=K(n,d)$ such that given any finite dimensional normal
topological space $Y$ of dimension $d$ and any null-homotopic continuous
mapping $f:Y\to\SpC$ there exists a continuous mapping
$F:Y\to(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ such that the diagram

${(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}}$${Y}$${\SpC}$$\scriptstyle{\Psi_{K}}$$\scriptstyle{F}$$\scriptstyle{f}$

commutes. Unfortunately, the mapping $\Psi_{K}$ is not a submersion and, in
addition, the fibers are difficult to analyse. We therefore consider
$\Phi_{K}:=\pi_{2n}\circ\Psi_{K}$, where $\pi_{2n}$ denotes the projection of
a $(2n\times 2n)$-matrix to its last row. We obtain the commutative diagram

${(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}}$${Y}$${\mathbb{C}^{2n}\setminus\\{0\\}.}$$\scriptstyle{\Phi_{K}=\pi_{2n}\circ\Psi_{K}}$$\scriptstyle{F}$$\scriptstyle{\pi_{2n}\circ
f}$

The mapping $\Phi_{K}$ is surjective for $K\geq 3$ (see Theorem 3.5) and it is
submersive outside some set of singularities
$S_{K}\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ (see Theorem 3.4). We will
find an open submanifold
$E_{K}\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\setminus S_{K}$ such that
$\Phi_{K}|_{E_{K}}:E_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$ is a stratified
elliptic submersion (see section 3). Further, the  (Continuous Vaserstein
problem for symplectic matrices). allows us, for a given any null-homotopic
continuous mapping $f:Y\to\SpC$, to construct a continuous section $F:Y\to
E_{K}$ such that

${E_{K}}$${Y}$${\mathbb{C}^{2n}\setminus\\{0\\}}$$\scriptstyle{\Phi_{K}}$$\scriptstyle{F}$$\scriptstyle{\pi_{2n}\circ
f}$

commutes (see Theorem 2.11). Then, an application of the Oka principle allows
us, roughly speaking, to homotopically deform $F$ into a holomorphic mapping,
such that the above diagram commutes.

The organisation of this paper is as follows. In section 2, we’re going to
state the Continuous Vaserstein Problem. Then we introduce the notion of a
dominating spray and the appropriate version of the Oka-principle (Theorem
2.8). Finally, we’ll prove the main theorem by induction on $n$.

In section 3, it remains to be shown that
$\Phi_{K}:E_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$ is stratified elliptic so
that the assumptions of the Oka principle are met. However, it turns out to be
the most difficult part of the article because at first glance it seems like
we don’t have enough complete holomorphic vector fields to define a spray.
Luckily, we are able to develop some rudimentary machinery to overcome this
hurdle.

###  1.2 Acknowledgements

First of all, I would like to thank my supervisor Prof. Dr. Frank
Kutzschebauch for the opportunity to work on such an interesting problem and
for the many exciting and supportive discussions. I am also very thankful to
Dr. George Ionita for proofreading and his helpful feedback. Last but not
least, I would like to thank my family, especially my wife Anna, who always
supported me and believed in me.

##  2 Holomorphic Vaserstein problem for symplectic matrices

Let’s start with the so-called Continuous Vaserstein problem for symplectic
matrices. This has been proved by Ivarsson, Kutzschebauch and Løw ([IKL20a,
Theorem 1.3]) and is one of the key ingredients.

######  Theorem 2.1 (Continuous Vaserstein problem for symplectic matrices).

There exists a natural number $K(n,d)$ such that given any finite dimensional
normal topological space $X$ of (covering) dimension $d$ and any null-
homotopic continuous mapping $M\colon X\to\operatorname{Sp}_{2n}(\mathbb{C})$
there exist $K$ continuous mappings

$G_{1},\dots,G_{K}\colon X\to\mathbb{C}^{n(n+1)/2}$

such that

$M(x)=M_{1}(G_{1}(x))\dots M_{K}(G_{K}(x)).$

###  2.1 Oka principle

The Oka principle is a powerful tool. Roughly speaking, it allows us to
homotopically deform a continuous mapping into a holomorphic one in certain
situations. We want to make this more precise in this section and therefore
introduce some definitions and terminologies.

Let’s start with the notion of a stratified elliptic submersion $h:Z\to X$
from a complex space $Z$ onto a complex space $X$, following [Gro89] and
[For10].

######  Definition 2.2.

Let $X$ and $Z$ be complex spaces. A holomorphic map $h\colon Z\to X$ is said
to be a submersion if for each point $z_{0}\in Z$ it is locally equivalent via
a fiber-preserving biholomorphic map to a projection $p\colon U\times V\to U$,
where $U\subset X$ is an open set containing $h(z_{0})$ and $V$ is an open set
in some $\mathbb{C}^{d}$.

Let $h\colon Z\to X$ be a holomorphic submersion of a complex manifold $Z$
onto a complex manifold $X$. For any $x\in X$ the fiber over $x$ of this
submersion will be denoted by $Z_{x}$. At each point $z\in Z$ the tangent
space $T_{z}Z$ contains the vertical tangent space $VT_{z}Z=\ker Dh$. For
holomorphic vector bundles $p\colon E\to Z$ we denote the zero element in the
fiber $E_{z}$ by $0_{z}$.

######  Definition 2.3.

Let $h\colon Z\to X$ be a holomorphic submersion of a complex manifold $Z$
onto a complex manifold $X$. A spray on $Z$ associated with $h$ is a triple
$(E,p,s)$, where $p\colon E\to Z$ is a holomorphic vector bundle and $s\colon
E\to Z$ is a holomorphic map such that for each $z\in Z$ we have

  * (i)

$s(E_{z})\subset Z_{h(z)}$,

  * (ii)

$s(0_{z})=z$, and

  * (iii)

the derivative $Ds(0_{z})\colon T_{0_{z}}E\to T_{z}Z$ maps the subspace
$E_{z}\subset T_{0_{z}}E$ surjectively onto the vertical tangent space
$VT_{z}Z$.

######  Remark 2.4.

We will also say that the submersion admits a spray. A spray associated with a
holomorphic submersion is sometimes called a (fiber) dominating spray.

One way of constructing dominating sprays, as pointed out by Gromov, is to
find finitely many $\mathbb{C}$-complete vector fields that are tangent to the
fibers and span the tangent space of the fibers at all points in $Z$. One can
then use the flows $\varphi_{j}^{t}$ of these vector fields $V_{j}$ to define
$s\colon Z\times\mathbb{C}^{N}\to Z$ via
$s(z,t_{1},\dots,t_{N})=\varphi_{1}^{t_{1}}\circ\dots\circ\varphi_{N}^{t_{N}}(z)$
which gives a spray.

######  Definition 2.5.

We say that a submersion $h\colon Z\to X$ is stratified elliptic if there is a
descending chain of closed complex subspaces $X=X_{m}\supset\cdots\supset
X_{0}$ such that each stratum $Y_{k}=X_{k}\setminus X_{k-1}$ is regular and
the restricted submersion $h\colon Z|_{Y_{k}}\to Y_{k}$ admits a spray over a
small neighborhood of any point $x\in Y_{k}$.

######  Remark 2.6.

We say that the submersion admits stratified sprays and that the
stratification $X=X_{m}\supset\cdots\supset X_{0}$ is associated with the
stratified spray.

Let’s consider the following diagram

${P_{0}\times X}$${E}$${P\times
X}$${B}$$\scriptstyle{\mathrm{incl}}$$\scriptstyle{F}$$\scriptstyle{\pi}$$\scriptstyle{F}$$\scriptstyle{f}$

Here $\pi:E\to B$ is a holomorphic submersion of a complex space $E$ onto a
complex space $B$, $X$ is a Stein space, $P_{0}\subset P$ are compact
Hausdorff spaces (the parameter spaces), and $f:P\times X\to B$ is an
$X$-holomorphic map, meaning that $f(p,\cdot):X\to B$ is holomorphic on $X$
for every fixed $p\in P$. A map $F:P\times X\to E$ such that $\pi\circ F=f$ is
said to be a lifting of $f$; such $F$ is $X$-holomorphic on $P_{0}$ if
$F(p,\cdot)$ is holomorphic for every $p\in P_{0}$.

######  Definition 2.7.

A holomorphic map $\pi:E\to B$ between reduced complex spaces enjoys the
Parametric Oka Property (POP) if for any collection
$(X,X^{\prime},K,P,P_{0},f,F_{0})$ where $X$ is a reduced Stein space,
$X^{\prime}$ is a closed complex subvariety of $X$, $P_{0}\subset P$ are
compact Hausdorff spaces, $f:P\times X\to B$ is an $X$-holomorphic map, and
$F_{0}:P\times X\to E$ is a continuous map such that $\pi\circ F=f$, the map
$F_{0}(p,\cdot)$ is holomorphic on $X$ for all $p\in P_{0}$ and is holomorphic
on $K\cup X^{\prime}$ for all $p\in P$, there exists a homotopy $F_{t}:P\times
X\to E$ such that the following hold for all $t\in[0,1]$:

  1. (i)

$\pi\circ F_{t}=f$,

  2. (ii)

$F_{t}=F_{0}$ on $(P_{0}\times X)\cup(P\times X^{\prime})$,

  3. (iii)

$F_{t}$ is $X$-holomorphic on $K$ and uniformly close to $F_{0}$ on $P\times
K$, and

  4. (iv)

the map $F_{1}:P\times X\to E$ is $X$-holomorphic.

An important cornerstone in the proof of holomorphic factorization is the
following very general version of the Oka principle (see [For17, Corollary
6.14.4 (i)])

######  Theorem 2.8 (Oka principle).

Every stratified elliptic submersion enjoys POP.

###  2.2 Holomorphic Vaserstein problem for symplectic matrices

Recall the mapping $\Psi_{K}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to\SpC$ given
by $\Psi_{K}(Z_{1},...,Z_{K})=M_{1}(Z_{1})\cdots M_{K}(Z_{K})$, where
$M_{k}:\mathbb{C}^{\frac{n(n+1)}{2}}\to\SpC$ is defined by

$M_{k}(Z)=\begin{cases}\begin{pmatrix}I_{n}&Z\\\ 0&I_{n}\end{pmatrix}&\text{if
}k=2l\\\ \begin{pmatrix}I_{n}&0\\\ Z&I_{n}\end{pmatrix}&\text{if
}k=2l+1.\end{cases}$

Further, recall the map
$\Phi_{K}=\pi_{2n}\circ\Psi_{K}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$,
where $\pi_{2n}:\mathbb{C}^{2n\times 2n}\to\mathbb{C}^{2n}$ denotes the
projection of a $2n\times 2n$-matrix to its last row.

Define the set

$\mathcal{W}_{K}:=\left\\{(Z_{1},...,Z_{K})\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}:Z_{2i-1}e_{n}\neq
0\text{ for some }1\leq i\leq\lceil\tfrac{K-1}{2}\rceil\right\\}$

and let $S_{K}\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ denote the set of
points, where the mapping $\Phi_{K}$ is not submersive.

The following is another cornerstone in the proof of the main theorem and we
will prove it in the next section.

######  Theorem 2.9.

For $K\geq 3$, there exists an open submanifold
$E_{K}\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ satisfying

  1. (i)

$\mathcal{W}_{K}\subset
E_{K}\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\setminus S_{K}$

  2. (ii)

the restriction $\Phi_{K}|_{E_{K}}:E_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$ is
a stratified elliptic submersion.

The  (Oka principle). tells us, that $\Phi_{K}|_{E_{K}}$ has the Parametric
Oka Property, i.e. we get the following

######  Corollary 2.10 (Application of the Oka principle).

Let $P$ be an Euclidean compact, $X$ a finite dimensional reduced Stein space
and $f:P\times X\to\SpC$ a null-homotopic, continuous $X$-holomorphic mapping.
Assume there is a natural number $K$ and a continuous map $F:P\times X\to
E_{K}$ such that

${E_{K}}$${P\times
X}$${\mathbb{C}^{2n}\setminus\\{0\\}}$$\scriptstyle{\Phi_{K}}$$\scriptstyle{F}$$\scriptstyle{\pi_{2n}\circ
f}$

is commutative. Then there exists a continuous homotopy $F_{t}:P\times X\to
E_{K}$ with $F_{0}=F$, $\pi_{2n}\circ f=\Phi_{K}\circ F_{t}$ and such that
$F_{1}:P\times X\to E_{K}$ is $X$-holomorphic.

In a next step, we prove the existence of a natural number $K$ and a
continuous lifting $F:P\times X\to E_{K}$ such that the diagram in the
corollary commutes.

######  Theorem 2.11.

There exists a natural number $L(n,d)$ such that given any Euclidean compact
$P$, any finite dimensional reduced Stein space $X$, such that $P\times X$ has
covering dimension $d$, and any null-homotopic, continuous $X$-holomorphic
mapping $f:P\times X\to\SpC$, there exists a continuous lifting $F:P\times
X\to E_{L}$ of $\pi_{2n}\circ f$. In particular, there exists a continuous
homotopy $F_{t}:P\times X\to E_{L}$ of liftings of $\pi_{2n}\circ f$, such
that $F_{0}=F$ and $F_{1}$ is $X$-holomorphic.

###### Proof.

The  (Continuous Vaserstein problem for symplectic matrices). provides us with
a natural number $K=K(n,d)$ such that given any normal topological space $Y$
of dimension $d$ and any null-homotopic continuous mapping $f:Y\to\SpC$ there
exists a continuous lifting
$G=(G_{1},...,G_{K}):Y\to(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ with
$f=\Psi_{K}\circ G$. For $L=K+2$, we define the mapping
$F:Y\to(\mathbb{C}^{\frac{n(n+1)}{2}})^{L}$ by

$F=(I_{n},0,G_{1}-I_{n},G_{2},...,G_{K})$

where $I_{n}$ denotes the $n\times n$-identity matrix and $0$ the $n\times
n$-zero matrix. Observe that

$\begin{pmatrix}I_{n}&0\\\ G_{1}&I_{n}\end{pmatrix}=\begin{pmatrix}I_{n}&0\\\
I_{n}&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\ G_{1}-I_{n}&I_{n}\end{pmatrix}$

and therefore $F$ is a lifting of $f$, more precisely, $f=\Psi_{L}\circ F$.
The image of $F$ is contained in $\mathcal{W}_{L}$, since $I_{n}e_{n}\neq 0$.
By Theorem 2.9, we have $\mathcal{W}_{L}\subset E_{L}$ which means that $F$ is
a mapping $F:Y\to E_{L}$ and we obtain a commuting diagram

${E_{L}}$${Y}$${\SpC}$${\mathbb{C}^{2n}\setminus\\{0\\}.}$$\scriptstyle{\Psi_{L}}$$\scriptstyle{\Phi_{L}}$$\scriptstyle{F}$$\scriptstyle{f}$$\scriptstyle{\pi_{2n}}$

Choose $Y=P\times X$, where $P$ is an Euclidean compact and $X$ a reduced
Stein space such that $Y$ has dimension $d$. An application of Corollary 2.10
completes the proof. ∎

Before we come to the proof of the main theorem, we need one more result also
known as Whitehead’s lemma. For some commutative ring $R$ with identity, the
symplectic group $\mathrm{Sp}_{2n}(R)$ is a subgroup of $\mathrm{SL}_{2n}(R)$.
We shall write matrices with block notation

$\begin{pmatrix}A&B\\\ C&D\end{pmatrix}$

where $A,B,C$ and $D$ are $n\times n$ matrices with entries in $R$ satisfying
the symplectic conditions

$\displaystyle A^{T}C=C^{T}A$ (1) $\displaystyle B^{T}D=D^{T}B$ (2)
$\displaystyle A^{T}D-C^{T}B=I_{n}$ (3)

where $I_{n}$ is the $n\times n$ identity matrix.

An elementary symplectic matrix is either of the form

$\begin{pmatrix}I_{n}&B\\\ 0&I_{n}\end{pmatrix}$

where $B$ is symmetric ($B^{T}=B$) or of the form

$\begin{pmatrix}I_{n}&0\\\ C&I_{n}\end{pmatrix}$

where $C$ is symmetric. Products of matrices of the first type are additive in
$B$ and of the second type in $C$. Special cases are the matrices $E_{ij}(a)$
when $B$ is the matrix with $a$ in position $ij$ and $ji$ and otherwise zero.
For $F_{ij}(a)$ the roles of $B$ and $C$ are changed. Clearly any elementary
matrix of the first type is a product of matrices $E_{ij}(b_{ij})$ for $i\leq
j$ and similarly for the second type.

We also introduce the symplectic matrices $K_{ij}(a)$ defined by $B=C=0$ and
$A=I_{n}$ except in position $ij$ where there is an $a$. Finally,
$D=(A^{T})^{-1}$. This equals $I$ except in position $ji$ where there is $-a$
if $i\neq j$ and $a^{-1}$ if $i=j$ (this requires $a\in R^{*}$).

######  Lemma 2.12 (Whitehead’s lemma).

There holds

$K_{ii}(a)=E_{ii}(a-1)F_{ii}(1)E_{ii}(a^{-1}-1)F_{ii}(-a),\quad a\in R^{*},$

and if $i\neq j$, then

$K_{ij}(a)=F_{jj}(-a)E_{ij}(1)F_{jj}(a)E_{ii}(a)E_{ij}(-1),\quad a\in R.$

###  2.3 Proof of the  (Main theorem).

The proof is by induction on $n$. Note that the theorem is true for $n=1$,
since $\mathrm{Sp}_{2}(\mathbb{C})=\mathrm{SL}_{2}(\mathbb{C})$ (see [IK12]),
that is, the base case is fine.

For the induction step, we first observe that, according to Theorem 2.11,
there exists a natural number $L(n,d)$ such that given any Euclidean compact
$P$, any finite dimensional Stein space $X$, such that $P\times X$ has
covering dimension $d$, and any null-homotopic $X$-holomorphic mapping
$f:P\times X\to\SpC$, there exists a continuous homotopy $F_{t}:P\times X\to
E_{L}$ of liftings of $\pi_{2n}\circ f$ (i.e. $\pi_{2n}\circ f=\Phi_{L}\circ
F_{t}$ for all $0\leq t\leq 1$), such that $F_{1}$ is $X$-holomorphic. In
particular, this implies that $\Phi_{L}\circ F_{t}$ doesn’t depend on $t$,
hence we get

$\Psi_{L}(F_{t}(p,x))f(p,x)^{-1}=\begin{pmatrix}A_{1,t}(p,x)&a_{2,t}(p,x)&B_{1,t}(p,x)&b_{2,t}(p,x)\\\
a_{3,t}(p,x)&a_{4,t}(p,x)&b_{3,t}(p,x)&b_{4,t}(p,x)\\\
C_{1,t}(p,x)&c_{2,t}(p,x)&D_{1,t}(p,x)&d_{1,t}(p,x)\\\ 0&0&0&1\end{pmatrix},$

where $A_{1,t}(p,x),B_{1,t}(p,x),C_{1,t}(p,x)$ and $D_{1,t}(p,x)$ are
$(n-1)\times(n-1)$ matrices, and the remaining mappings are vectors of
appropriate dimension. Since $\Psi_{L}(F_{t}(p,x))f(p,x)^{-1}$ are symplectic
matrices for all $0\leq t\leq 1$, there follows $a_{2,t}(p,x)\equiv
0,a_{4,t}(p,x)\equiv 1$ and $c_{2,t}(p,x)\equiv 0$, so that

$\Psi_{L}(F_{t}(p,x))f(p,x)^{-1}=\begin{pmatrix}A_{1,t}(p,x)&0&B_{1,t}(p,x)&b_{2,t}(p,x)\\\
a_{3,t}(p,x)&1&b_{3,t}(p,x)&b_{4,t}(p,x)\\\
C_{1,t}(p,x)&0&D_{1,t}(p,x)&d_{1,t}(p,x)\\\ 0&0&0&1\end{pmatrix}$

and in addition

$\tilde{f}_{t}(p,x)=\begin{pmatrix}A_{1,t}(p,x)&B_{1,t}(p,x)\\\
C_{1,t}(p,x)&D_{1,t}(p,x)\end{pmatrix}\in\mathrm{Sp}_{2n-2}(\mathbb{C}).$

Since $\Psi_{L}(F_{0}(p,x))=f(p,x)$, there holds $\tilde{f}_{0}=I_{2n-2}$ and
thus the holomorphic map
${\tilde{f}:=\tilde{f}_{1}:P\times X\to\mathrm{Sp}_{2n-2}(\mathbb{C})}$ is
null-homotopic. Let $\psi$ be the standard inclusion of $\mathrm{Sp}_{2n-2}$
into $\mathrm{Sp}_{2n}$ given by

$\psi\begin{pmatrix}A&B\\\ C&D\end{pmatrix}=\begin{pmatrix}A&0&B&0\\\
0&1&0&0\\\ C&0&D&0\\\ 0&0&0&1\end{pmatrix}.$

By the induction hypothesis,

$\psi(\tilde{f}(p,x)^{-1})=\begin{pmatrix}\tilde{A}(p,x)&0&\tilde{B}(p,x)&0\\\
0&1&0&0\\\ \tilde{C}(p,x)&0&\tilde{D}(p,x)&0\\\ 0&0&0&1\end{pmatrix}$

is a finite product of holomorphic elementary symplectic matrices. Then the
matrix
${M(p,x):=\Psi_{L}(F_{1}(p,x))f(p,x)^{-1}\psi(\tilde{f}(p,x)^{-1})}$ is given
by

$\displaystyle M(p,x)=\begin{pmatrix}I_{n-1}&0&0&b_{2,1}(p,x)\\\
a_{3,1}(p,x)\tilde{A}(p,x)+b_{3,1}(p,x)\tilde{C}(p,x)&1&a_{3,1}(p,x)\tilde{B}(p,x)+b_{3,1}(p,x)\tilde{D}(p,x)&b_{4,1}(p,x)\\\
0&0&I_{n-1}&d_{2,1}(p,x)\\\ 0&0&0&1\end{pmatrix}.$

Since $M(p,x)$ is a symplectic matrix, there holds

$a_{3,1}(p,x)\tilde{A}(p,x)+b_{3,1}(p,x)\tilde{C}(p,x)=-d_{2,1}(p,x)^{T}$

and

$a_{3,1}(p,x)\tilde{B}(p,x)+b_{3,1}(p,x)\tilde{D}(p,x)=b_{2,1}(p,x)^{T},$

so that

$\Psi_{L}(F_{1}(p,x))f(p,x)^{-1}\psi(\tilde{f}(p,x)^{-1})=\begin{pmatrix}I_{n-1}&0&0&b_{2,1}(p,x)\\\
-d_{2,1}(p,x)^{T}&1&b_{2,1}(p,x)^{T}&b_{4,1}(p,x)\\\
0&0&I_{n-1}&d_{2,1}(p,x)\\\ 0&0&0&1\end{pmatrix}.$

Then

$\Psi_{L}(F_{1}(p,x))f(p,x)^{-1}\psi(\tilde{f}(p,x)^{-1})\begin{pmatrix}I_{n}&-B(p,x)\\\
0&I_{n}\end{pmatrix}=\begin{pmatrix}I_{n-1}&0&0&0\\\
-d_{2,1}(p,x)^{T}&1&0&0\\\ 0&0&I_{n-1}&d_{2,1}(p,x)\\\ 0&0&0&1\end{pmatrix},$

where

$B(p,x)=\begin{pmatrix}0&-b_{2,1}(p,x)\\\
-b_{2,1}(p,x)^{T}&-b_{4,1}(p,x)-d_{2,1}(p,x)^{T}b_{2,1}(p,x)\end{pmatrix}$

is holomorphic. By making the substitution $y:=-d_{2,1}$, the matrix on the
right hand side equals the product

$K_{n1}(y_{1}(p,x))K_{n2}(y_{2}(p,x))\cdots K_{n,n-1}(y_{n-1}(p,x))$

which is a finite product of elementary symplectic matrices due to
(Whitehead’s lemma).. In summary, this proves that $f(p,x)$ is indeed a finite
product of elementary symplectic matrices.

So far, the number of factors depends on the mapping $f:P\times X\to\SpC$.
Assume there is no uniform bound $K(n,d)$, that is, for each natural number
$i$, there is an Euclidean compact $P_{i}$ and a reduced Stein space $X_{i}$,
such that $Y_{i}=P_{i}\times X_{i}$ has dimension $d$, and a null-homotopic
$X$-holomorphic mapping $f_{i}:Y_{i}\to\SpC$ which does not factor into less
than $i$ elementary matrix factors. Set $Y=\bigcup_{i}Y_{i}$ and let
$F:Y\to\SpC$ be the null-homotopic mapping, which equals $f_{i}$ on $Y_{i}$.
We just proved the existence of a constant $K$ which bounds the number of
elementary factors in which $F$ decomposes. But then $K$ is an upper bound for
each $f_{i}$ which contradicts the assumption. Hence there is a uniform bound
$K(n,d)$ of factors and this complectes the proof.

Actually, we’ve shown a generalized version of the  (Main theorem)..

######  Theorem 2.13 (Generalized version of main theorem).

There is a natural number $K=K(n,d)$ such that given any Euclidean compact
$P$, any finite dimensional reduced Stein space $X$, such that $P\times X$ has
covering dimension $d$, and any null-homotopic $X$-holomorphic mapping
$f:P\times X\to\SpC$ there exist $K$ $X$-holomorphic mappings

$G_{1},...,G_{K}:P\times X\to\mathbb{C}^{\frac{n(n+1)}{2}}$

with

$f(p,x)=M_{1}(G_{1}(p,x))M_{2}(G_{2}(p,x))\cdots M_{K}(G_{K}(p,x)).$

##  3 Stratified ellipticity of the mapping $\Phi_{K}$

Recall that we identified
$\mathrm{Sym}_{n}(\mathbb{C})=\\{Z\in\mathbb{C}^{n\times n}:Z^{T}=Z\\}$ with
$\mathbb{C}^{\frac{n(n+1)}{2}}$ for simplicity. Furthermore, the elementary
symplectic matrix mapping ${M_{K}:\mathbb{C}^{\frac{n(n+1)}{2}}\to\SpC}$ we
defined by

$M_{K}(Z)=\begin{cases}\begin{pmatrix}I_{n}&Z\\\ 0&I_{n}\end{pmatrix}&\text{if
}K=2k+1\\\ \begin{pmatrix}I_{n}&0\\\ Z&I_{n}\end{pmatrix}&\text{if
}K=2k.\end{cases}$

Let’s write
$\vec{Z}_{K}:=(Z_{1},...,Z_{K})\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$. Then
the mapping
$\Phi_{K}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$
is defined by

$\Phi_{K}(\vec{Z}_{K}):=M_{K}(Z_{K})\cdots M_{1}(Z_{1})e_{2n}.$

######  Remark 3.1.

The attentive reader may have noticed that we now define $\Phi_{K}$ in a
transposed way compared to the previous sections. This is purely for aesthetic
reasons.

The following recursive formula can be derived immediately from the
definition.

######  Corollary 3.2 (Recursive formula of $\Phi_{K}$).

For $K\geq 1$, the mapping
$\Phi_{K}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$
satisfies

$\displaystyle\Phi_{K}(\vec{Z}_{K})=M_{K}(Z_{K})\Phi_{K-1}(\vec{Z}_{K-1}),$
(4)

with the convention $\Phi_{0}:=e_{2n}$.

The main goal of this section is to prove

######  Theorem 3.3.

For $K\geq 3$, there is an open submanifold $E_{K}$ of
$(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ such that

$\Phi_{K}|_{E_{K}}:E_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$

is a stratified elliptic submersion.

In subsection 3.1 we will classify the points for which $\Phi_{K}$ is not
submersive. We will also show that $\Phi_{K}$ is surjective.

The subsequent subsections are then devoted to the task of finding stratified
sprays. In subsection 3.2, we stratify $\mathbb{C}^{2n}\setminus\\{0\\}$
suitably. In 3.3, we find formulas of holomorphic vector fields which are
fiber-preserving for $\Phi_{K}$. Unfortunately, some of those fields aren’t
$\mathbb{C}$-complete. We therefore classify some complete vector fields in
subsection 3.4. In subsection 3.5, we analyze the fibers of $\Phi_{K}$ from a
topological point of view. In subsection 3.6 we lay the mathematical basis for
the construction of the sprays. And finally we carry out all the necessary
calculations in 3.7.

###  3.1 Notations and basic properties

Let’s start with some notations. Let $E_{ij}$ be the $n\times n$ matrix having
a $1$ at entry $(i,j)$ and is zero elsewhere. Then
$\tilde{E}_{ij}=\frac{1}{1+\delta_{ij}}(E_{ij}+E_{ji})$ is an elementary
symmetric matrix.

In the following we will make the identification of
$\mathrm{Sym}_{n}(\mathbb{C})$ and $\mathbb{C}^{\frac{n(n+1)}{2}}$ more
precise. The set $\\{\tilde{E}_{ij}:1\leq i\leq j\leq n\\}$ forms a basis of
$\mathrm{Sym}_{n}(\mathbb{C})$. The sets $I:=\\{i\in\mathbb{N}:1\leq
i\leq\frac{n(n+1)}{2}\\}$ and $J:=\\{(i,j)\in\mathbb{N}^{2}:1\leq i\leq j\leq
n\\}$ have the same order. Hence there is a bijection $\alpha:I\to J$, which
induces an isomorphism
$S:\mathbb{C}^{\frac{n(n+1)}{2}}\to\mathrm{Sym}_{n}(\mathbb{C})$ by defining
$S(e_{i})=\tilde{E}_{\alpha(i)}$ for all $1\leq i\leq\tfrac{n(n+1)}{2}$. By an
abuse of notation, $Z\in\mathbb{C}^{\frac{n(n+1)}{2}}$ denotes both, the
vector and the corresponding symmetric matrix $S(Z)$, depending on the
corresponding context, of course.

In this very first subsection, we will compute the set of points where
$\Phi_{K}$ is not submersive. Also we’ll give a proof for the surjectivity of
$\Phi_{K}$. But let’s spell that out in more detail first.

We denote the set of points in $(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ where
$\Phi_{K}$ is not submersive by $S_{K}$. We also define the open set

$\mathcal{W}_{K}:=\left\\{\vec{Z}_{K}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}:Z_{2i-1}e_{n}\neq
0\text{ for some }1\leq i\leq\lceil\tfrac{K-1}{2}\rceil\right\\}$

and $\mathcal{W}_{K}^{c}$ denotes its complement in
$(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$.

######  Theorem 3.4 (Singularity set of $\Phi_{K}$).

For $K\geq 2$, the set $S_{K}$ is given by

$S_{K}=\left\\{\vec{Z}_{K}\in\mathcal{W}_{K}^{c}:\mathrm{rank}(W_{K}(\vec{Z}_{K}))<n\right\\}$

where $W_{K}(\vec{Z}_{K})$ is the augmented matrix
$(Z_{2}|Z_{4}|\cdots|Z_{2k})$ for $k=\lfloor\tfrac{K-1}{2}\rfloor$.

######  Theorem 3.5.

For $K\geq 3$, the mapping
$\Phi_{K}|_{\mathcal{W}_{K}}:\mathcal{W}_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$
is surjective.

A direct consequence of these two statements is

######  Corollary 3.6.

For $K\geq 3$ and for any open submanifold $E$ in
$(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ with

$\mathcal{W}_{K}\subset E\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\setminus
S_{K},$

the mapping $\Phi_{K}|_{E}:E\to\mathbb{C}^{2n}\setminus\\{0\\}$ is a
surjective submersion.

####  3.1.1 Singularity set of $\Phi_{K}$

In order to compute the singularity set $S_{K}$, we need to know the Jacobian
of $\Phi_{K}$, denoted by $J\Phi_{K}$. For the computations, we need some
auxiliary tools. For a fixed $1\leq i\leq n$, let’s define the mapping
$F_{i}:\mathbb{C}^{n}\to\mathbb{C}^{n\times n}$ by

$F_{i}(v):=\left[\tilde{E}_{i1}v\cdots\tilde{E}_{in}v\right]=\begin{pmatrix}v_{i}&&&&\\\
&\ddots&&&\\\ v_{1}&\cdots&v_{i}&\cdots&v_{n}\\\ &&&\ddots&\\\
&&&&v_{i}\end{pmatrix}.$

Furthermore, we define
$F:\mathbb{C}^{n}\to\mathbb{C}^{n\times\frac{n(n+1)}{2}}$ by

$F(v):=\left[\tilde{E}_{\alpha(1)}v\cdots\tilde{E}_{\alpha(\frac{n(n+1)}{2})}v\right].$

Observe that the matrix $F_{i}(v)$ is a submatrix of $F(v)$, for every $1\leq
i\leq n$. This implies that $F(v)$ is surjective if and only if $v\neq 0$. And
if $F(v)$ is not surjective, then we even have $F(v)=0$.

######  Corollary 3.7.

For $u,v\in\mathbb{C}^{n}$ such that $\begin{pmatrix}u\\\
v\end{pmatrix}\in\mathbb{C}^{2n}\setminus\\{0\\}$, the Jacobian of the mapping
${\mathbb{C}^{\frac{n(n+1)}{2}}\to\mathbb{C}^{2n}\setminus\\{0\\}},{Z\mapsto
M_{K}(Z)\begin{pmatrix}u\\\ v\end{pmatrix}}$ is given by

$A_{K}(u,v):=\begin{cases}\begin{pmatrix}F(v)\\\ 0\end{pmatrix}&\text{if
}K=2k+1\\\ \begin{pmatrix}0\\\ F(u)\end{pmatrix}&\text{if }K=2k.\end{cases}$

In particular, there holds $M_{K}(Z)A_{K}(u,v)=A_{K}(u,v)$ for all
$Z\in\mathbb{C}^{\frac{n(n+1)}{2}}$.

###### Proof.

Note that each symmetric matrix $Z\in\mathbb{C}^{\frac{n(n+1)}{2}}$ can be
written as a sum $Z=\sum_{1\leq i\leq j\leq n}z_{ij}\tilde{E}_{ij}$. Hence
$\frac{\partial}{\partial{{z}_{ij}}}Z=\tilde{E}_{ij}$. Further, note

$\frac{\partial}{\partial{{z}_{ij}}}M_{2k+1}(Z)\begin{pmatrix}u\\\
v\end{pmatrix}=\begin{pmatrix}0&\tilde{E}_{ij}\\\
0&0\end{pmatrix}\begin{pmatrix}u\\\
v\end{pmatrix}=\begin{pmatrix}\tilde{E}_{ij}v\\\ 0\end{pmatrix}$

and

$\frac{\partial}{\partial{{z}_{ij}}}M_{2k}(Z)\begin{pmatrix}u\\\
v\end{pmatrix}=\begin{pmatrix}0&0\\\
\tilde{E}_{ij}&0\end{pmatrix}\begin{pmatrix}u\\\
v\end{pmatrix}=\begin{pmatrix}0\\\ \tilde{E}_{ij}u\end{pmatrix},$

respectively. From here, the claim follows by definition of the mapping $F$. ∎

We are now ready to compute the Jacobian of $\Phi_{K}$. By the recursive
formula (4), the product rule and the previous corollary, we obtain

######  Corollary 3.8 (Jacobian of $\Phi_{K}$).

The Jacobian $J\Phi_{1}$ is given by $A_{1}(e_{2n})$. For $K\geq 2$, the
Jacobian $J\Phi_{K}$ of $\Phi_{K}$ is given by

$J\Phi_{K}(\vec{Z}_{K})=\begin{pmatrix}M_{K}(Z_{K})J\Phi_{K-1}(\vec{Z}_{K-1})&|&A_{K}(\Phi_{K-1}(\vec{Z}_{K-1}))\end{pmatrix}.$

If the Jacobian $J\Phi_{K-1}$ is surjective, then so is $J\Phi_{K}$, since
$M_{K}(Z_{K})$ is a regular matrix. Or, equivalently, if
$\vec{Z}_{K}=(\vec{Z}_{K-1},Z_{K})\in S_{K}$ is a singular point for
$\Phi_{K}$, then $\vec{Z}_{K-1}\in S_{K-1}$ is a singular point for
$\Phi_{K-1}$. This observation suggests that we will compute $S_{K}$
recursively.

######  Lemma 3.9.

Let $\vec{Z}_{K}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ and assume that there
is $1\leq k\leq\big{\lceil}\frac{K-1}{2}\big{\rceil}$ such that
${Z_{2i-1}e_{n}=0}$ for all $1\leq i\leq k$. Then
$\Phi_{j}(\vec{Z}_{j})=e_{2n}$ for all $1\leq j\leq 2k$.

###### Proof.

We prove this by induction on $j$. For the base step, observe that

$\Phi_{1}(\vec{Z}_{1})=M_{1}(Z_{1})e_{2n}=\begin{pmatrix}Z_{1}e_{n}\\\
e_{n}\end{pmatrix}$

by definition. Since we assume $Z_{1}e_{n}=0$, $\Phi_{1}(\vec{Z}_{1})=e_{2n}$
follows immediately.

For the induction step, let $1<j\leq 2k$. There holds
$\Phi_{j-1}(\vec{Z}_{j-1})=e_{2n},$ by the induction hypothesis. Then we
obtain

$\Phi_{j}(\vec{Z}_{j})=M_{j}(Z_{j})\Phi_{j-1}(\vec{Z}_{j-1})=M_{j}(Z_{j})e_{2n}$

by the recursive formula (4). If $j$ is even, then we’re done, by definition
of $M_{j}$. Let’s assume that $j=2l-1$ for some integer $l$. Observe that
$l\leq k$ is satisfied, hence $Z_{j}e_{n}=0$ by assumption. This implies

$\Phi_{j}(\vec{Z}_{j})=\begin{pmatrix}I_{n}&Z_{j}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}0\\\
e_{n}\end{pmatrix}=\begin{pmatrix}Z_{j}e_{n}\\\ e_{n}\end{pmatrix}=e_{2n}.$

This completes the proof. ∎

######  Lemma 3.10.

For $K\geq 2$, there holds
$\mathcal{W}_{K}\subset(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\setminus S_{K}$,
that is, the restriction $\Phi_{K}|_{\mathcal{W}_{K}}$ is a submersion.

###### Proof.

Consider $\vec{Z}_{K}\in\mathcal{W}_{K}$. There is a smallest index $1\leq
k\leq\big{\lceil}\frac{K-1}{2}\big{\rceil}$ such that $Z_{2k-1}e_{n}\neq 0$
and $Z_{2i-1}e_{n}=0$ for all $1\leq i<k$. Setting $L:=2k-1$, the Jacobian
$J\Phi_{L}$ is of the form

$\begin{pmatrix}*&F(P_{s}^{L-1})\\\ *&0\end{pmatrix}$

and

$J\Phi_{L+1}=\begin{pmatrix}M_{L+1}(Z_{L+1})J\Phi_{L}&A_{L+1}(\Phi_{L})\end{pmatrix}=\begin{pmatrix}*&F(P_{s}^{L-1})&0\\\
*&0&F(P_{f}^{L})\end{pmatrix}.$

By definition of $F$, the Jacobian $J\Phi_{L+1}$ has full rank, if
$P_{s}^{L-1}\neq 0$ and $P_{f}^{L}\neq 0$. By Lemma 3.9, there holds
$\Phi_{L-1}=e_{2n}$. Hence

$P_{s}^{L-1}=\begin{pmatrix}0&I_{n}\end{pmatrix}\Phi_{L-1}=e_{n}\neq 0.$

Furthermore,

$P_{f}^{L}=\begin{pmatrix}I_{n}&0\end{pmatrix}\Phi_{L}=\begin{pmatrix}I_{n}&0\end{pmatrix}M_{L}(Z_{L})\Phi_{L-1}=\begin{pmatrix}I_{n}&Z_{L}\end{pmatrix}e_{2n}=Z_{L}e_{n}\neq
0.$

This showes that the Jacobian $J\Phi_{L+1}$ has full rank. By construction,
there holds $L+1\leq K$. By the recursive formula of the Jacobian and
regularity of $M_{i}(Z_{i})$, $1\leq i\leq K$, we conclude that $J\Phi_{K}$
has full rank, too. ∎

Let’s write $C_{K}:=\begin{pmatrix}I_{n}&0\end{pmatrix}J\Phi_{K}$ and
$D_{K}:=\begin{pmatrix}0&I_{n}\end{pmatrix}J\Phi_{K}$.

######  Lemma 3.11.

For a point
$\vec{Z}_{2k+2}=(\vec{Z}_{2k+1},Z_{2k+2})\in\mathcal{W}_{2k+2}^{c}$ the
following statements are equivalent.

  1. (i)

The Jacobian $J\Phi_{2k+2}(\vec{Z}_{2k+2})$ is surjective in $\vec{Z}_{2k+2}$.

  2. (ii)

The Jacobian $J\Phi_{2k+1}(\vec{Z}_{2k+1})$ is surjective in $\vec{Z}_{2k+1}$.

  3. (iii)

$\mathrm{rank}(D_{2k+1})=\mathrm{rank}(D_{2k})=n.$

###### Proof.

Note that
$\mathcal{W}_{K}^{c}\subset\mathcal{W}_{K-1}^{c}\times\mathbb{C}^{\frac{n(n+1)}{2}}$
by definition. Moreover, there holds

$A_{2k+1}(e_{2n})=\begin{pmatrix}F(e_{n})\\\
0\end{pmatrix}\quad\text{and}\quad A_{2k+2}(e_{2n})=0,$

for a point $\vec{Z}_{2k+2}\in\mathcal{W}_{2k+2}^{c}$, by Lemma 3.9 and
definition of $F$. We conclude

$J\Phi_{2k+2}(\vec{Z}_{2k+2})=M_{2k+2}(Z_{2k+2})\begin{pmatrix}J\Phi_{2k+1}(\vec{Z}_{2k+1})&0\end{pmatrix}$

and this shows equivalence of $(i)$ and $(ii)$, since $M_{2k+2}(Z_{2k+2})$ is
a regular matrix.

To show equivalence of $(ii)$ and $(iii)$ observe that

$\displaystyle
J\Phi_{2k+1}(\vec{Z}_{2k+1})=\begin{pmatrix}C_{2k}+Z_{2k+1}D_{2k}&F(e_{n})\\\
D_{2k}&0\end{pmatrix}.$

Since $F(e_{n})$ is surjective by definition of $F$, the claim follows
immediately. ∎

In order to prove Theorem 3.4, it remains to show the next lemma.

######  Lemma 3.12.

For a point $\vec{Z}_{2k+2}\in\mathcal{W}_{2k+2}^{c}$ there holds

$\mathrm{im}(D_{2k+2})=\mathrm{im}(D_{2k}|Z_{2k+2}).$ ($\star$)

In particular, the singularity set of $\Phi_{K}$, $K\geq 2$, is given by

$S_{K}:=\\{\vec{Z}_{K}\in\mathcal{W}_{K}^{c}:\mathrm{rank}(W_{K}(\vec{Z}_{K}))<n\\},$

where $W_{K}$ is an augmented matrix
$W_{K}(\vec{Z}_{K}):=(Z_{2}|Z_{4}|\cdots|Z_{2k}),$ for
$k=\lfloor\frac{K-1}{2}\rfloor$.

###### Proof.

By the previous lemma, it is enough to show ($\star$ ‣ 3.12). There holds

$\displaystyle J\Phi_{2k+2}(\vec{Z}_{2k+2})$
$\displaystyle=\begin{pmatrix}C_{2k+1}&0\\\
D_{2k+1}+Z_{2k+2}C_{2k+1}&0\end{pmatrix}=\begin{pmatrix}C_{2k}+Z_{2k+1}D_{2k}&F(e_{n})&0\\\
D_{2k}+Z_{2k+2}(C_{2k}+Z_{2k+1}D_{2k})&Z_{2k+2}F(e_{n})&0\end{pmatrix}.$

Since $F(e_{n})$ is surjective, we get
$\mathrm{im}(Z_{2k+2})=\mathrm{im}(Z_{2k+2}F(e_{n}))$ and thus
$\mathrm{im}(D_{2k+2})=\mathrm{im}(D_{2k}|Z_{2k+2})$. ∎

####  3.1.2 Surjectivity of $\Phi_{K}$

The proof of surjectivity is strongly based on this lemma.

######  Lemma 3.13.

For $a\in\mathbb{C}^{n}\setminus\\{0\\}$ fixed, the mapping
$\varphi_{a}:\mathrm{Sym}_{n}(\mathbb{C})\to\mathbb{C}^{n},Z\mapsto Za$ is
surjective.

###### Proof.

Let $a\neq 0$ being fixed. Then the linear mapping
$F(a):\mathbb{C}^{\frac{n(n+1)}{2}}\to\mathbb{C}^{n}$ is surjective, by
definition of $F$. Therefore it is enough to show that the following diagram

${\mathbb{C}^{\frac{n(n+1)}{2}}}$${\mathrm{Sym}_{n}(\mathbb{C})}$${\mathbb{C}^{n}.}$$\scriptstyle{F(a)}$$\scriptstyle{S^{-1}}$$\scriptstyle{\varphi_{a}}$

is commutative. This is the case if and only if $F(a)v=S(v)a$ for all
$v\in\mathbb{C}^{\frac{n(n+1)}{2}}$. Write
$v=\sum_{i=1}^{\frac{n(n+1)}{2}}v_{i}e_{i}$. By definition of $F$ and $S$, we
get

$F(a)v=\sum_{i=1}^{\frac{n(n+1)}{2}}v_{i}F(a)e_{i}=\sum_{i=1}^{\frac{n(n+1)}{2}}v_{i}\tilde{E}_{\alpha(i)}a=\left(\sum_{i=1}^{\frac{n(n+1)}{2}}v_{i}\tilde{E}_{\alpha(i)}\right)a=S(v)a.$

This completes the proof. ∎

We are now ready for the

###### Proof of Theorem 3.5.

In a first step, let $K=3$ and consider $\begin{pmatrix}a\\\
b\end{pmatrix}\in\mathbb{C}^{2n}\setminus\\{0\\}$. Observe that $M_{K}(-Z)$ is
the inverse of $M_{K}(Z)$ for every $K$ and every $Z$. Then
$\Phi_{3}(\vec{Z}_{3})=\begin{pmatrix}a\\\ b\end{pmatrix}$ if and only if

$M_{2}(Z_{2})M_{1}(Z_{1})e_{2n}=M_{3}(-Z_{3})\begin{pmatrix}a\\\
b\end{pmatrix}.$

The left-hand-side is given by

$\begin{pmatrix}I_{n}&0\\\
Z_{2}&I_{n}\end{pmatrix}\begin{pmatrix}Z_{1}e_{n}\\\
e_{n}\end{pmatrix}=\begin{pmatrix}Z_{1}e_{n}\\\
(I_{n}+Z_{2}Z_{1})e_{n}\end{pmatrix}$

and the right-hand-side by

$\begin{pmatrix}I_{n}&-Z_{3}\\\ 0&I_{n}\end{pmatrix}\begin{pmatrix}a\\\
b\end{pmatrix}=\begin{pmatrix}a-Z_{3}b\\\ b\end{pmatrix}.$

The symmetric matrix $Z_{3}$ can be chosen such that $a-Z_{3}b\neq 0$. To see
this, observe that it is automatically satisfied if $b=0$, since $a$ and $b$
aren’t simultaneously zero. On the other hand, if $b\neq 0$, then an
application of Lemma 3.13 provides the existence of such $Z_{3}$.

The vector $e_{n}$ is obviously non-zero, hence another application of Lemma
3.13 yields the existence of a symmetric matrix $Z_{1}$ such that

$Z_{1}e_{n}=a-Z_{3}b.$

We have $Z_{1}e_{n}\neq 0$ by construction, which enables a third application
of Lemma 3.13 and proves the existence of a symmetric matrix $Z_{2}$ such that
$(I_{n}+Z_{2}Z_{1})e_{n}=b$. Thus we’ve found
$\vec{Z}_{3}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{3}$ such that

$\Phi_{3}(\vec{Z}_{3})=\begin{pmatrix}a\\\ b\end{pmatrix}.$

Moreover, since $Z_{1}e_{n}\neq 0$, we even have
$\vec{Z}_{3}\in\mathcal{W}_{3}$, which completes the proof for $K=3$.

For $K>3$ and $x\in\mathbb{C}^{2n}\setminus\\{0\\}$ we find
$\vec{Z}_{3}\in\mathcal{W}_{3}$ such that $\Phi_{3}(\vec{Z}_{3})=x$. Now, we
set $Z_{K}:=(\vec{Z}_{3},0,...,0)\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$. Then
we have $\Phi_{K}(\vec{Z}_{K})=\Phi_{3}(\vec{Z}_{3})=x$ and, moreover,
$\vec{Z}_{K}\in\mathcal{W}_{K}$ by definition. This completes the proof. ∎

###  3.2 Stratification of $\mathbb{C}^{2n}\setminus\\{0\\}$

Consider a fixed point $y:=(a,b)\in\mathbb{C}^{2n}\setminus\\{0\\}$ and let
$\mathcal{F}^{K}_{y}:=\mathcal{F}^{K}_{a,b}:=\Phi_{K}^{-1}(y)$ denote the
fiber of $\Phi_{K}$ over $y$. By the recursive formula (4) of $\Phi_{K}$ we
can write the $K$-fiber $\mathcal{F}^{K}_{y}$ as a union of $(K-1)$-fibers

$\mathcal{F}^{K}_{y}=\bigcup_{Z\in\mathbb{C}^{\frac{n(n+1)}{2}}}\mathcal{F}^{K-1}_{M_{K}(Z)y}$

For an “appropriate” stratification of $\mathbb{C}^{2n}\setminus\\{0\\}$, we
will use the projection
$\pi_{K}:\mathbb{C}^{n}\times\mathbb{C}^{n}\to\mathbb{C}^{n}$ given by

$\pi_{K}(u,v):=\begin{cases}u&\text{if }K=2k\\\ v&\text{if
}K=2k+1.\end{cases}$

The following stratification turns out to be a natural one. Let

${Y^{K}_{ng}=\\{y\in\mathbb{C}^{2n}\setminus\\{0\\}:\pi_{K}(y)=0\\}}$

denote the non-generic stratum and its complement
$Y^{K}_{g}=(\mathbb{C}^{2n}\setminus\\{0\\})\setminus Y^{K}_{ng}$ the generic
stratum.

######  Lemma 3.14.

For a point $y\in Y_{ng}^{K}$ in the non-generic stratum, the corresponding
non-generic fiber $\mathcal{F}^{K}_{y}$ satisfies

$\mathcal{F}^{K}_{y}=\mathcal{F}^{K-1}_{y}\times\mathbb{C}^{\frac{n(n+1)}{2}},$

where $\mathcal{F}^{K-1}_{y}$ is a generic $(K-1)$-fiber.

###### Proof.

We carry out the proof only for $K=2k+1$, since it applies equally to $K=2k$
for reasons of symmetry. Let $y=(a,b)\in Y^{K}_{ng}$, that is, $\pi_{K}(y)=0$.
Observe that $y=(\pi_{K-1}(y),\pi_{K}(y))$. On the one hand, this means
$b=\pi_{K}(y)=0$ and on the other $a=\pi_{K-1}(y)\neq 0$, since $y\neq 0$ by
definition. This implies that $\mathcal{F}^{K-1}_{y}$ is a generic
$(K-1)$-fiber.

The non-generic $K$-fiber $\mathcal{F}^{K}_{y}$ is given by the defining
equations $\Phi_{K}(\vec{Z}_{K})=y$. By the recursive formula (4) of
$\Phi_{K}$ this system of equations is equivalent to
$\Phi_{K-1}(\vec{Z}_{K-1})=M_{K}(-Z_{K})y$. But there holds

$M_{K}(-Z_{K})y=\begin{pmatrix}I_{n}&-Z_{K}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}a\\\ 0\end{pmatrix}=\begin{pmatrix}a\\\
0\end{pmatrix}=y,$

which means that the defining equations are independent of the matrix
$Z_{K}\in\mathbb{C}^{\frac{n(n+1)}{2}}$. In fact, we obtain

$\mathcal{F}^{K}_{y}=\mathcal{F}^{K-1}_{y}\times\mathbb{C}^{\frac{n(n+1)}{2}}.$

∎

Informally, the next statement tells us that for fibers in the generic stratum
we can reduce the number of defining equations from $2n$ to $n$.

First we introduce the following convention. Let
$\pi:\mathbb{C}^{k}\to\mathbb{C}^{l}$ be the standard projection
$(z_{1},...,z_{l},...,z_{k})\mapsto(z_{1},...,z_{l})$. For a continuous
mapping $f:\mathbb{C}^{l}\to\mathbb{C}^{m}$ its pullback $\pi^{*}f$ is a
mapping $\pi^{*}f:\mathbb{C}^{k}\to\mathbb{C}^{m}$ and by an abuse of
notation, we just write $f$ instead of $\pi^{*}f$.

Also define $\tilde{P}^{K}:=\pi_{K+1}\circ\Phi_{K}$. Then, $\tilde{P}^{K}_{j}$
and $\pi_{K}(y)_{j}$ denote the $j$-th component of $\tilde{P}^{K}$ and
$\pi_{K}(y)$, respectively.

######  Lemma 3.15.

Set
$X:=(\mathbb{C}^{\frac{n(n+1)}{2}})^{K-1}\times\mathbb{C}^{\frac{n(n-1)}{2}}$
and define the variety

$\mathcal{G}_{\pi_{K}(y)}:=\\{\vec{Z}\in
X:\tilde{P}^{K-1}(\vec{Z})=\pi_{K}(y)\\}.$

Then there are $n$ meromorphic mappings $\psi_{j}^{K}:X\to
X\times\mathbb{C}^{n}$, $1\leq j\leq n$, such that each generic fiber
$\mathcal{F}^{K}_{y}$ is biholomorphic to $\mathcal{G}_{\pi_{K}(y)}$ via
$\psi_{j}^{K}$ for some $1\leq j\leq n$.

###### Proof.

Let $y\in Y^{K}_{g}$ be a point in the generic stratum, i.e.
$\pi_{K}(y)_{j}\neq 0$ for some $1\leq j\leq n$. From the definition of
$\pi_{K}$ and $\tilde{P}^{K}$ on the one hand and the recursive formula (4) on
the other hand, it follows that the fiber $\mathcal{F}^{K}_{y}$ is given by

$\mathcal{F}^{K}_{y}=\\{\vec{Z}_{K}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}:\tilde{P}^{K}(\vec{Z}_{K})=\pi_{K+1}(y),\tilde{P}^{K-1}(\vec{Z}_{K})=\pi_{K}(y)\\}$

and further there holds

$\displaystyle\tilde{P}^{K}=\tilde{P}^{K-2}+Z_{K}\tilde{P}^{K-1}.$ (5)

Hence $\tilde{P}^{K}_{j}\neq 0$ is satisfied on the fiber
$\mathcal{F}^{K}_{y}$ and the latter equation can be rearranged to

$Z_{K}e_{j}=\frac{1}{\tilde{P}^{K-1}_{j}}\left(\tilde{P}^{K}-\tilde{P}^{K-2}-\sum_{k=1,k\neq
j}^{n}\tilde{P}^{K-1}_{k}Z_{K}e_{k}\right).$

We obtain

$\displaystyle f_{ij}:=z_{K,ij}$
$\displaystyle=\frac{1}{\tilde{P}^{K-1}_{j}}\left(\tilde{P}^{K}_{i}-\tilde{P}^{K-2}_{i}-\sum_{k=1,k\neq
j}^{n}\tilde{P}^{K-1}_{k}z_{K,ik}\right)\quad 1\leq i\leq n,i\neq j,$ (6)
$\displaystyle f_{jj}:=z_{K,jj}$
$\displaystyle=\frac{1}{\tilde{P}^{K-1}_{j}}\left(\tilde{P}^{K}_{j}-\tilde{P}^{K-2}_{j}-\frac{1}{\tilde{P}^{K-1}_{j}}\sum_{k=1,k\neq
j}^{n}b_{k}\left(\tilde{P}^{K}_{k}-\tilde{P}^{K-2}_{k}-\sum_{l=1,l\neq
j}^{n}\tilde{P}^{K-1}_{l}z_{K,kl}\right)\right).$ (7)

Set $f_{j}:=(f_{1j},...,f_{nj}):X\to\mathbb{C}^{n}$ and $\psi_{j}:X\to
X\times\mathbb{C}^{n}$, $\Psi_{j}(x)=(x,f_{j}(x))$. By construction, the
variety $\mathcal{G}_{\pi_{K}(y)}$ is mapped biholomorphically onto
$\mathcal{F}^{K}_{y}$ by $\psi_{j}$. ∎

####  3.2.1 On the singularities of the fibers

In this short section we will classify the fibers; we distinguish between
smooth and singular fibers. In fact, most of the fibers $\mathcal{F}^{K}_{y}$
are completely contained in $\mathcal{W}_{K}$ and therefore smooth, by Lemma
3.10.

######  Lemma 3.16.

A fiber $\mathcal{F}^{K}_{y}$ contains singularities if and only if
$\pi_{1}(y)=e_{n}$.

###### Proof.

We start with the case $K=2k$. Suppose there is a singularity
$\vec{Z}_{K}\in\mathcal{F}^{K}_{y}\cap S_{K}$. Then there holds
$Z_{2i-1}e_{n}=0$ for all $1\leq i\leq k$, by Lemma 3.12. Lemma 3.9 implies
$\Phi_{K}(\vec{Z}_{K})=e_{2n}$ and thus
$\pi_{1}(y)=\pi_{1}(\Phi_{K}(\vec{Z}_{K}))=\pi_{1}(e_{2n})=e_{n}$.

Now let $K=2k+1$ and suppose again there is a singularity
$\vec{Z}_{K}\in\mathcal{F}^{K}_{y}\cap S_{K}$. Again there holds
$Z_{2i-1}e_{n}=0$ for all $1\leq i\leq k$, by Lemma 3.12. From the even case,
we know that $\Phi_{K-1}(\vec{Z}_{K-1})=e_{2n}$. The recursive formula (4) of
$\Phi_{K}$ implies

$y=\Phi_{K}(\vec{Z}_{K})=M_{K}(Z_{K})\Phi_{K-1}(\vec{Z}_{K-1})=\begin{pmatrix}I_{n}&Z_{K}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}0\\\
e_{n}\end{pmatrix}=\begin{pmatrix}Z_{K}e_{n}\\\ e_{n}\end{pmatrix},$

and therefore $\pi_{1}(y)=e_{n}$. This completes the proof of necessary
condition.

For the proof of the sufficient condition, we consider a fiber
$\mathcal{F}^{K}_{y}$ with $\pi_{1}(y)=e_{n}$ and set
$k:=\big{\lceil}\frac{K-1}{2}\big{\rceil}$. Let
$\vec{Z}_{K}\in\mathcal{F}^{K}_{y}$ and recall that
$\vec{Z}_{K}=(Z_{1},Z_{2},...,Z_{K})\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$.
The matrices $Z_{2},Z_{4},...,Z_{2k}\in\mathbb{C}^{\frac{n(n+1)}{2}}$ can take
any value, since $Z_{1}e_{n}=Z_{3}e_{n}=...=Z_{2k-1}e_{n}=0$ by assumption and
therefore $\Phi_{i}(\vec{Z}_{i})=e_{2n}$ for all $1\leq i\leq 2k$ by Lemma
3.9. Hence we set $Z_{2}=Z_{4}=...=Z_{2k}=0$ and then
$\vec{Z}_{K}\in\mathcal{F}^{K}_{y}\cap S_{K}$ is a singularity by Lemma 3.12.
∎

###  3.3 Holomorphic vector fields tangential to the fibers

We will construct stratified sprays using $\mathbb{C}$-complete holomorphic
vector fields. The main goal of this subsection is therefore to find enough
vector fields that are holomorphic on $(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$,
complete and, in particular, tangential to the fibers $\mathcal{F}^{K}_{y}$.

####  3.3.1 Fiber-preserving vector fields

Let $X$ be some Stein manifold and $f=(f_{1},...,f_{n}):X\to\mathbb{C}^{n}$ a
holomorphic mapping. A holomorphic vector field ${V:X\to
TX},{x\mapsto(x,V_{x})}$ is fiber-preserving for $f$, if it is tangential to
the fibers of $f$. This is the case if and only if $V$ is in the kernel of the
tangent map $df$, that is, $df_{x}(V_{x})=0$. This is equivalent to say that
the Lie derivative
$\mathcal{L}_{V_{x}}(f_{i})=V_{x}(f_{i})=(df_{i})_{x}(V_{x})=0$ vanishes for
all $1\leq i\leq n$.

######  Lemma 3.17.

For $N>n$, let $P\in\mathbb{C}[z_{1},...,z_{N}]^{n}$ be a polynomial mapping
$P:\mathbb{C}^{N}\to\mathbb{C}^{n}$ and let
$x=(z_{\alpha_{0}},...,z_{\alpha_{n}})$, with
$1\leq\alpha_{0}<...<\alpha_{n}\leq N$. Let’s write
$\frac{\partial}{\partial{z}}P:=(\frac{\partial}{\partial{z}}P_{1},...,\frac{\partial}{\partial{z}}P_{n})^{T}$.
Then

$D_{x}(P):=\det\begin{pmatrix}\frac{\partial}{\partial{{z}_{\alpha_{0}}}}&\cdots&\frac{\partial}{\partial{{z}_{\alpha_{n}}}}\\\
\frac{\partial}{\partial{{z}_{\alpha_{0}}}}P&\cdots&\frac{\partial}{\partial{{z}_{\alpha_{n}}}}P\end{pmatrix}$

is a holomorphic vector field on $\mathbb{C}^{N}$ which is fiber-preserving
for $P$.

###### Proof.

The Lie derivative $\mathcal{L}_{D_{x}(P)}(P_{i})=D_{x}(P)(P_{i})=0$ vanishes
for each $1\leq i\leq n$, since the first and the $(i+1)$-th row of
$D_{x}(P)(P_{i})$ are the same. ∎

We now introduce a few more notations. For a fixed natural number $K$, the
mapping $\tilde{P}^{K}$ (defined before Lemma 3.15) is a polynomial mapping in
$\mathbb{C}[z_{1},...,z_{N_{K}}]^{n}$ with $N_{K}:=K\tfrac{n(n+1)}{2}$. Since
we’ll only be interested in $K>1$, the constraint $N_{K}>n$ is given.

There are $\binom{N_{K}}{n+1}$ possibilities to choose $(n+1)$ of the $N_{K}$
variables. Let $\mathcal{T}_{K}$ denote the set of all such possible choices.
Recall that we see $\mathbb{C}^{N_{K}}$ as a product of $K$ copies of
$\mathbb{C}^{\frac{n(n+1)}{2}}$ and we write
$\vec{Z}_{K}=(Z_{1},...,Z_{K})\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$. With
this convention, the set $\mathcal{T}_{K}$ can be given by

$\mathcal{T}_{K}:=\\{x=(z_{i_{0},j_{0}k_{0}},...,z_{i_{n},j_{n}k_{n}}):1\leq
i_{0}\leq...\leq i_{n}\leq K,1\leq j_{r}\leq k_{r}\leq n,0\leq r\leq n\\}.$

The vector fields
$\partial_{x}^{K}:=D_{x}(\tilde{P}^{K}),x\in\mathcal{T}_{K}$, are fiber-
preserving for $\tilde{P}^{K}$, by Lemma 3.17; or, equivalently, they’re
tangential to the variety $\mathcal{G}_{\pi_{K+1}(y)}$ by construction (c.f.
Lemma 3.15).

The following lemma collects some interesting examples of fiber-preserving
vector fields for $\Phi_{K}$. In fact, they’ll play a cruical role in the
construction of a dominating spray (see Theorem 3.36)

######  Lemma 3.18.

Let $1\leq j^{*}\leq n$, $x\in\mathcal{T}_{K-1}$ and
$u:=(\partial_{x}^{K-1}(\tilde{P}^{K-2}_{1}),...,\partial_{x}^{K-1}(\tilde{P}^{K-2}_{n}))^{T}$.
Then the vector field

$\displaystyle\varphi^{K}_{x,j^{*}}$
$\displaystyle=(\tilde{P}^{K-1}_{j^{*}})^{2}\partial_{x}^{K-1}-\tilde{P}^{K-1}_{j^{*}}\sum_{\begin{subarray}{c}i=1\\\
i\neq
j^{*}\end{subarray}}^{n}u_{i}\frac{\partial}{\partial{{z}_{K,j^{*}i}}}+\left(\sum_{\begin{subarray}{c}i=1\\\
i\neq
j^{*}\end{subarray}}^{n}\tilde{P}^{K-1}_{i}u_{i}-\tilde{P}^{K-1}_{j^{*}}u_{j^{*}}\right)\frac{\partial}{\partial{{z}_{K,j^{*}j^{*}}}}$

is holomorphic on $(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ and fiber-preserving
for $\Phi_{K}$. Moreover, $\varphi^{K}_{x,j^{*}}$ is complete if and only if
$\partial_{x}^{K-1}$ is complete.

For $1\leq i\leq j\leq n,i\neq j^{*},j\neq j^{*}$, the vector field

$\displaystyle\gamma_{ij,j^{*}}^{K}$
$\displaystyle=(\tilde{P}^{K-1}_{j^{*}})^{2}\frac{\partial}{\partial{{z}_{K,ij}}}+\frac{1}{1+\delta_{ij}}\left(2\tilde{P}^{K-1}_{i}\tilde{P}^{K-1}_{j}\frac{\partial}{\partial{{z}_{K,j*j*}}}-\tilde{P}^{K-1}_{i}\tilde{P}^{K-1}_{j^{*}}\frac{\partial}{\partial{{z}_{K,j^{*}j}}}-\tilde{P}^{K-1}_{j}\tilde{P}^{K-1}_{j^{*}}\frac{\partial}{\partial{{z}_{K,j^{*}i}}}\right)$

is complete, holomorphic on $(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ and fiber-
preserving for $\Phi_{K}$.

###### Proof.

Observe that $\tilde{P}^{K-1}\equiv 0$ on non-generic fibers
$\mathcal{F}^{K}_{y}$. Hence the above fields are trivial and there is nothing
to show. We therefore consider $y\in Y_{g}^{K}$ in the generic stratum, i.e.
$\pi_{K}(y)\neq 0$. Without loss of generality assume $\pi_{K}(y)_{1}\neq 0$.
According to Lemma 3.15, the mapping $\psi_{1}^{K}$ is defined by
$\Psi_{1}(x)=(x,f_{1}(x))$ for some meromorphic map
$f_{1}=(f_{11},...,f_{n1})$ and it maps $\mathcal{G}_{\pi_{K}(y)}$
biholomorphically onto the generic fiber $\mathcal{F}^{K}_{y}$.

Consider a vector field $V$ tangential to $\mathcal{G}_{\pi_{K}(y)}$. Then the
push-forward $W:=(\Psi_{1})_{*}(V)$ is given by

$W=V+\sum_{i=1}^{n}V(f_{i1})\frac{\partial}{\partial{{z}_{K,i1}}}.$

On the one hand $W$ is tangential to the fiber $\mathcal{F}^{K}_{y}$ and on
the other hand it is complete if and only if $V$ is complete.

Let’s write
$W(\tilde{P}^{K}):=(W(\tilde{P}^{K}_{1}),...,W(\tilde{P}^{K}_{n}))^{T}$ and
$W(Z_{K}):=\sum_{1\leq i\leq j\leq n}W(z_{K,ij})\tilde{E}_{ij}$. By the
recursive formula (5) and since $W(\tilde{P}^{K-1})=0$, we get

$\displaystyle 0$
$\displaystyle=W(\tilde{P}^{K})=W(\tilde{P}^{K-2})+W(Z_{K})\tilde{P}^{K-1}=u+W(Z_{K})\tilde{P}^{K-1}.$

In the special case, where $V=\partial_{x}^{K-1}$ for some
$x\in\mathcal{T}_{K-1}$, there holds

$\displaystyle W(Z_{K})\tilde{P}^{K-1}$
$\displaystyle=\left(\sum_{i=1}^{n}V(f_{i1})\tilde{E}_{i1}\right)\tilde{P}^{K-1}=\begin{pmatrix}\sum_{i=1}^{n}V(f_{i1})\tilde{P}^{K-1}_{i}\\\
\tilde{P}^{K-1}_{1}V(f_{21})\\\ \vdots\\\
\tilde{P}^{K-1}_{1}V(f_{n1})\end{pmatrix}$
$\displaystyle=\underbrace{\begin{pmatrix}\tilde{P}^{K-1}_{1}&\cdots&\tilde{P}^{K-1}_{n}\\\
&\ddots&\\\
&&\tilde{P}^{K-1}_{1}\end{pmatrix}}_{=:A}\underbrace{\begin{pmatrix}V(f_{11})\\\
\vdots\\\ V(f_{n1})\end{pmatrix}}_{=:b},$

where $A$ is a regular matrix, since $\tilde{P}^{K-1}_{1}\neq 0$. Therefore we
obtain $b=-A^{-1}u$ with

$A^{-1}=\frac{1}{(\tilde{P}^{K-1}_{1})^{2}}\begin{pmatrix}\tilde{P}^{K-1}_{1}&-\tilde{P}^{K-1}_{2}&-\tilde{P}^{K-1}_{3}&\cdots&-\tilde{P}^{K-1}_{n}\\\
&\tilde{P}^{K-1}_{1}&&&\\\ &&\ddots&&\\\ &&&\ddots&\\\
&&&&\tilde{P}^{K-1}_{1}\end{pmatrix}.$

The vector field $\varphi_{x,1}^{K}:=(\tilde{P}^{K-1}_{1})^{2}W$ is
holomorphic on $(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ and fiber-preserving for
$\Phi_{K}$. Moreover, it is complete if and only if $W$ is complete, since
$\tilde{P}^{K-1}$ is in the kernel of $W$. This proves the first part of the
lemma.

In the special case, where $V=\frac{\partial}{\partial{{z}_{K,ij}}},1<i\leq
j\leq n$, there holds $u=W(\tilde{P}^{K-2})=0$ and

$W(Z_{K})\tilde{P}^{K-1}=\tilde{E}_{ij}\tilde{P}^{K-1}+\left(\sum_{i=1}^{n}V(f_{i1})\tilde{E}_{i1}\right)\tilde{P}^{K-1}=\tilde{E}_{ij}\tilde{P}^{K-1}+Ab.$

Therefore

$\displaystyle b$
$\displaystyle=-A^{-1}\tilde{E}_{ij}\tilde{P}^{K-1}=-\frac{1}{1+\delta_{ij}}A^{-1}(\tilde{P}^{K-1}_{i}e_{j}+\tilde{P}^{K-1}_{j}e_{i})$
$\displaystyle=\frac{1}{1+\delta_{ij}}\frac{1}{(\tilde{P}^{K-1}_{1})^{2}}\left(2\tilde{P}^{K-1}_{i}\tilde{P}^{K-1}_{j}e_{1}-\tilde{P}^{K-1}_{1}\tilde{P}^{K-1}_{i}e_{j}-\tilde{P}^{K-1}_{1}\tilde{P}^{K-1}_{j}e_{i}\right).$

As before, we multiply $W$ by $(\tilde{P}^{K-1}_{1})^{2}$ to obtain a fiber-
preserving vector field $\gamma^{K}_{ij,1}:=(\tilde{P}^{K-1}_{1})^{2}W$ which
is holomorphic on $(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$. Note that
$\frac{\partial}{\partial{{z}_{K,ij}}}$ is complete on
$\mathcal{G}_{\pi_{K}(y)}$ by definition, hence $\gamma_{ij,1}^{K}$ is
complete. This proves the second part of the lemma. ∎

###  3.4 Complete holomorphic vector fields tangent to the fibers

In the previous subsection, we’ve constructed vector fields tangent to the
fibers. Unfortunately, some of those fields aren’t complete (see Example 3.23
below). In addition, it is quite laborious to decide whether a given field is
complete. The goal of this subsection is to build machinery that will make
this decision easier. Furthermore, we will list the most important examples of
complete fields.

For $K>L$, let
$f_{K,L}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to(\mathbb{C}^{\frac{n(n+1)}{2}})^{L},(Z_{1},...,Z_{K})\mapsto(Z_{1},...,Z_{L})$
be the standard projection. Given a vector field $V$ on
$(\mathbb{C}^{\frac{n(n+1)}{2}})^{L}$ tangential to the fibers
$\mathcal{F}^{L}_{y}$, its pullback $f_{K,L}^{*}V$ is a vector field on
$(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ tangential to the fibers
$\mathcal{F}^{K}_{y^{\prime}}$, by the recursive formula (4). By an abuse of
notation, we just write $V$ instead of $f_{K,L}^{*}V$. Furthermore, set

$\mathcal{T}_{K}^{C}:=\\{x\in\mathcal{T}_{K}:\partial_{x}^{K}\text{ is a
complete vector field}\\}.$

######  Definition 3.19.

For $K\geq 3$, define the collection

$\mathcal{V}_{K}:=\bigcup_{L=3}^{K}\left(\bigcup_{j=1}^{n}\left\\{\varphi_{x,j}^{L}:x\in\mathcal{T}_{L-1}^{C}\right\\}\cup\left\\{\gamma^{L}_{rs,j}:1\leq
r,s\leq n\right\\}\right)$

of principal vector fields for $\Phi_{K}$.

######  Remark 3.20.

We can consider $\mathcal{V}_{K-1}$ as a subset of $\mathcal{V}_{K}$ using the
convention introduces just before the definition.

We are now working on a machinery that should make it easier for us to decide
whether a tuple $x$ corresponds to a $\mathbb{C}$-complete vector field
$\partial_{x}^{K}$. Let $x=(x_{1},...,x_{m})\in\mathbb{C}^{m}$ and
$P:\mathbb{C}^{m}\to\mathbb{C}^{m-1}$ be a polynomial mapping. We define an
equivalence relation on $\mathbb{C}^{l}$ in the following way

$u,v\in\mathbb{C}^{l},u\sim_{x}v\quad:\Leftrightarrow\quad
u_{i}-v_{i}\in\bigcap_{k=1}^{m}\ker\left(\frac{\partial}{\partial
x_{k}}\right),\text{ for all }1\leq i\leq l.$

A vector $v\in\mathbb{C}^{l}$ is called constant in $x$ if $v\sim_{x}0$.

######  Lemma 3.21.

Let $x=(x_{1},...,x_{m})\in\mathbb{C}^{m}$ and
$P:\mathbb{C}^{m}\to\mathbb{C}^{m-1},x\mapsto P(x)$ be a polynomial mapping.
Assume there exists $v\in\mathbb{C}^{m-1}$, $v\sim_{x}0$, and
$\lambda_{ij}\in\mathbb{C}$, $\lambda_{ij}\sim_{x}0$, for all $1\leq i,j\leq
m$, such that

$\frac{\partial}{\partial{{x}_{i}}}\frac{\partial}{\partial{{x}_{j}}}P(x)=\lambda_{ij}v.$
(8)

Then $V=D_{x}(P)$ is $\mathbb{C}$-complete.

###### Proof.

We consider the vector field
$V=D_{x}(P)=\sum_{j=1}^{m}V_{j}\frac{\partial}{\partial{{x}_{j}}}$, where
$V_{j}$ is given by

$V_{j}=\det\left(\frac{\partial P}{\partial x_{1}},...,\frac{\partial
P}{\partial x_{j-1}},\frac{\partial P}{\partial x_{j+1}},...,\frac{\partial
P}{\partial x_{m}}\right).$

For $1\leq j\leq m$, define $f_{j}(x)=\sum_{k=1}^{m}\lambda_{kj}x_{k}$. Then
$\frac{\partial}{\partial{{x}_{j}}}P(x)\sim_{x}f_{j}(x)v$ by construction,
this means

$\frac{\partial}{\partial{{x}_{j}}}P(x)=c_{j}+f_{j}(x)v,$

for some $c_{j}\in\mathbb{C},c_{j}\sim_{x}0$. Obviously, $f_{k}(x)v$ and
$f_{l}(x)v$ are linearly dependent, therefore

$\displaystyle V_{j}$
$\displaystyle=\det\left(c_{1}+f_{1}v,...,c_{j-1}+f_{j-1}v,c_{j+1}+f_{j+1}v,...,c_{m}+f_{m}v\right)$
$\displaystyle=\underbrace{\det(c_{1},...,c_{j-1},c_{j+1},...,c_{m})}_{=:\alpha_{j0}}+f_{1}\underbrace{\det(v,c_{2},...,c_{j-1},c_{j+1},...,c_{m})}_{=:\alpha_{j1}}+$
$\displaystyle+\cdots+f_{n}\underbrace{\det(c_{1},...,c_{j-1},c_{j+1},...,c_{m-1},v)}_{=:\alpha_{jm}}.$

With the convention $\alpha_{jj}:=0$, we obtain

$\displaystyle V_{j}$
$\displaystyle=\alpha_{j0}+\sum_{k=1}^{m}\alpha_{jk}f_{k}=\alpha_{j0}+\sum_{k=1}^{m}\alpha_{jk}\sum_{l=1}^{m}\lambda_{lk}x_{l}=\alpha_{j0}+\sum_{l=1}^{m}x_{l}\underbrace{\sum_{k=1}^{m}\lambda_{lk}\alpha_{jk}}_{=:a_{jl}}.$

Set $b:=(\alpha_{10},...,\alpha_{m0})^{T}$. Then we just proved that
$V(x)=Ax+b$, where $A=(a_{jl})_{1\leq j,l\leq m}$ is a $m\times m$-matrix with
$a_{jl}\sim_{x}0$.

Let $\gamma$ be a flow curve, i.e. a holomorphic map
$\gamma:\mathbb{C}\to\mathbb{C}^{m}$ with
$\frac{d}{dt}\gamma(t)=V(\gamma(t))$. This leads to the system

$\frac{d}{dt}\gamma(t)=A\gamma(t)+b,$

which implies that $\gamma$ exitsts for all time $t\in\mathbb{C}$. ∎

The mapping
$\tilde{P}^{K}:(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}\to\mathbb{C}^{n}$ does not
a priori fit into the setting of the previous lemma, but this problem can be
solved with a simple trick. By fixing all but $(n+1)$ of the
$\frac{n(n+1)}{2}K$ variables, we may interpret $\tilde{P}^{K}$ as a
polynomial mapping $\mathbb{C}^{n+1}\to\mathbb{C}^{n}$. More precisely, each
$(n+1)$-tupel $x\in\mathcal{T}_{K}$ corresponds to a natural inclusion map
$i_{x}:\mathbb{C}^{n+1}\to(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$. Then
$\tilde{P}^{K}\circ i_{x}$ is a polynomial mapping
$\mathbb{C}^{n+1}\to\mathbb{C}^{n}$.

######  Proposition 3.22.

(List of complete vector fields)

  1. (Type 1)

For $1\leq m\leq n$,
$x=(z_{k-1,mm},z_{k,11},...,z_{k,nn})\in\mathcal{T}_{K}^{C},\quad K\geq k.$

  2. (Type 2)

For $l\neq m$,
$x=(z_{k-1,mm},z_{k,l1},...,z_{k,ln})\in\mathcal{T}_{K}^{C},\quad K\geq k.$

  3. (Type 3)

For $(n+1)$ distinct pairs of indices $(i_{0},j_{0}),...,(i_{n},j_{n})$,
${x=(z_{k,i_{0}j_{0}},...,z_{k,i_{n}j_{n}})\in\mathcal{T}_{K}^{C}},\ {K\geq
k}.$

  4. (Type 4)

For $1\leq i^{*}\leq n$,
$x=(z_{1,ni^{*}},z_{2,11},...,z_{2,nn})\in\mathcal{T}_{K}^{C},K\geq 2.$

  5. (Type 5)

Let $k<l$, $1\leq j^{*}\leq n$ and let $(i_{1},j_{1}),...,(i_{n},j_{n})$ be
$n$ distinct pairs of indices. Then

$x=(z_{k,i_{1}j_{1}},...,z_{k,i_{n}j_{n}},z_{l,j^{*}j^{*}})\in\mathcal{T}_{K}^{C},K\geq
l.$

  6. (Type 6)

For $1\leq r\leq n$ consider the partition
$\\{1,...,n\\}=\\{i_{1},...,i_{r}\\}\dot{\cup}\\{j_{1},...,j_{n-r}\\}$. Let
$i^{*}\in\\{i_{1},...,i_{r}\\}$ and
$j^{*},j^{\prime}\in\\{j_{1},...,j_{n-r}\\}$. Then

$x=(z_{k-1,j^{*}j_{1}},...,z_{k-1,j^{*}j_{n-r}},z_{k,i^{*}i_{1}},...,z_{k,i^{*}i_{r}},z_{k,i^{*}j^{\prime}})\in\mathcal{T}_{K}^{C},K\geq
k.$

  7. (Type 7)

For $0\leq r\leq n$ consider the partition
$\\{1,...,n\\}=\\{i_{1},...,i_{r}\\}\dot{\cup}\\{j_{1},...,j_{n-r}\\}$. Let
$i^{*}\in\\{i_{1},...,i_{r}\\}$ and
$j^{*},j^{\prime}\in\\{j_{1},...,j_{n-r}\\}$. Then

$x=(z_{k,j^{*}j_{1}},...,z_{k,j^{*}j_{n-r}},z_{k+1,i^{*}i_{1}},...,z_{k+1,i^{*}i_{r}},z_{k+2,j^{\prime}j^{\prime}})\in\mathcal{T}_{K}^{C},K\geq
k+2.$

  8. (Type 8)

For $1\leq i\leq n,i\neq j$,
$x=(z_{1,in},z_{2,j1},...,z_{2,jn})\in\mathcal{T}_{K}^{C},K\geq 2.$

###### Proof.

For the proof of (Type 1), we consider the mapping
$P:\mathbb{C}^{n+1}\to\mathbb{C}^{n}$ given by

$x:=(z_{k-1,mm},z_{k,11},...,z_{k,nn})\mapsto\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{k-1}&I_{n}\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix},$

where $A$ and $B$ are arbitrary $n\times n$-matrices, both constant in $x$;
whereas $c$ and $d$ are arbitrary vectors in $\mathbb{C}^{n}$ both constant in
$x$. Observe that

$\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{k-1}&I_{n}\end{pmatrix}\begin{pmatrix}c\\\
d\end{pmatrix}=\begin{pmatrix}B&A\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{k}&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k-1}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}d\\\ c\end{pmatrix}.$

Thanks to this symmetry condition, we don’t need to make a case distinction
between even and odd $K$. In fact, it is enough to prove that $D_{x}(P)$ is a
$\mathbb{C}$-complete vector field. At first, note that

$\frac{\partial^{2}}{\partial z_{k-1,mm}^{2}}P(x)\equiv
0,\quad\frac{\partial}{\partial{{z}_{k,ii}}}\frac{\partial}{\partial{{z}_{k,jj}}}P(x)\equiv
0,\quad 1\leq i,j\leq n.$

Hence most of the $\lambda$’s in Lemma 3.21 can be chosen to be zero. It
remains to consider
$\frac{\partial}{\partial{{z}_{k,ii}}}\frac{\partial}{\partial{{z}_{k-1,mm}}}P(x)$,
$1\leq i\leq n$. We get

$\displaystyle\frac{\partial}{\partial{{z}_{k,ii}}}\frac{\partial}{\partial{{z}_{k-1,mm}}}P(x)$
$\displaystyle=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{ii}\\\
0&0\end{pmatrix}\begin{pmatrix}0&0\\\
\tilde{E}_{mm}&0\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix}$
$\displaystyle=\begin{pmatrix}0&A\tilde{E}_{ii}\end{pmatrix}\begin{pmatrix}0\\\
\tilde{E}_{mm}c\end{pmatrix}$
$\displaystyle=A\tilde{E}_{ii}\tilde{E}_{mm}c=c_{m}\delta_{mi}Ae_{m},$

for all $1\leq i\leq n$. Since $v:=Ae_{m}$ is independent of $i$, the
conditions of Lemma 3.21 are satisfied. Therefore $D_{x}(P)$ is a complete
field and we conclude that vector fields $\partial_{x}^{K}$, $K\geq k$, of
(Type 1) are complete.

For the proof of (Type 2) we choose $P$ as before. Again we have
$\frac{\partial^{2}}{\partial z_{k-1,mm}^{2}}P\equiv 0$ and
$\frac{\partial}{\partial{{z}_{k,li}}}\frac{\partial}{\partial{{z}_{k,lj}}}P\equiv
0$, $1\leq i,j\leq n$. Further, we compute

$\displaystyle\frac{\partial}{\partial{{z}_{k,li}}}\frac{\partial}{\partial{{z}_{k-1,mm}}}P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{li}\\\
0&0\end{pmatrix}\begin{pmatrix}0&0\\\
\tilde{E}_{mm}&0\end{pmatrix}\begin{pmatrix}c\\\
d\end{pmatrix}=A\tilde{E}_{li}\tilde{E}_{mm}c=c_{m}A\tilde{E}_{li}e_{m}.$

Observe that $\tilde{E}_{li}e_{m}=\delta_{im}e_{l}$, since we assume $l\neq
m$. Therefore we obtain

$\frac{\partial}{\partial{{z}_{k,li}}}\frac{\partial}{\partial{{z}_{k-1,mm}}}P(x)=\delta_{im}c_{m}Ae_{l}$

for all $1\leq i\leq n$. This proves that the conditions of Lemma 3.21 are
satisfied and hence the field $D_{x}(P)$ is complete. We conclude that the
fields $\partial_{x}^{K}$, $K\geq k$, of (Type 2) are complete.

For the proof of (Type 3), we set
$P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix}$. Observe that
${\frac{\partial}{\partial{{z}_{k,i_{r}j_{r}}}}P(x)\sim_{x}0}$ for all
$r=0,...,n$. Hence the conditions of Lemma 3.21 are trivially satisfied and we
conclude that the vector fields $\partial_{x}^{K}$ of (Type 3) are complete.

For the proof of (Type 4) we set
$P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{2}&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&Z_{1}\\\
0&I_{n}\end{pmatrix}e_{2n}$. Note that $\frac{\partial^{2}}{\partial
z_{1,ni^{*}}^{2}}P\equiv 0$ and
$\frac{\partial}{\partial{{z}_{2,ii}}}\frac{\partial}{\partial{{z}_{2,jj}}}P\equiv
0$, $1\leq i,j\leq n$. Furthermore, we compute

$\frac{\partial}{\partial{{z}_{2,jj}}}\frac{\partial}{\partial{{z}_{1,ni^{*}}}}P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&0\\\
\tilde{E}_{jj}&0\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{ni^{*}}\\\
0&0\end{pmatrix}\begin{pmatrix}0\\\
e_{n}\end{pmatrix}=B\tilde{E}_{jj}\underbrace{\tilde{E}_{ni^{*}}e_{n}}_{=e_{i^{*}}}=\delta_{ji^{*}}Be_{i^{*}}.$

Hence we can apply Lemma 3.21 and conclude that vector fields
$\partial_{x}^{K}$, $K\geq 2$, of (Type 4) are complete.

For the proof of (Type 5) we first consider

$P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{l}&I_{n}\end{pmatrix}\begin{pmatrix}U&V\\\
W&X\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix},$

where $U,V,W$ and $X$ are arbitrary $n\times n$-matrices constant in $x$. As
in the previous cases, we have $\frac{\partial^{2}}{\partial
z_{l,j^{*}j^{*}}^{2}}P\equiv 0$ and
$\frac{\partial}{\partial{{z}_{k,i_{r}j_{r}}}}\frac{\partial}{\partial{{z}_{k,i_{s}j_{s}}}}P\equiv
0$, $1\leq r,s\leq n$. Further, let’s compute

$\displaystyle\frac{\partial}{\partial{{z}_{l,j^{*}j^{*}}}}\frac{\partial}{\partial{{z}_{k,ij}}}P(x)$
$\displaystyle=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&0\\\
\tilde{E}_{j^{*}j^{*}}&0\end{pmatrix}\begin{pmatrix}U&V\\\
W&X\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{ij}\\\
0&0\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix}$
$\displaystyle=B\tilde{E}_{j^{*}j^{*}}U\tilde{E}_{ij}d=\underbrace{(e_{j^{*}}^{T}U\tilde{E}_{ij}d)}_{\sim_{x}0}Be_{j^{*}}.$

If we replace $\begin{pmatrix}I_{n}&0\\\ Z_{l}&I_{n}\end{pmatrix}$ by
$\begin{pmatrix}I_{n}&Z_{l}\\\ 0&I_{n}\end{pmatrix}$ in $P$, we obtain

$\frac{\partial}{\partial{{z}_{l,j^{*}j^{*}}}}\frac{\partial}{\partial{{z}_{k,ij}}}P(x)=A\tilde{E}_{j^{*}j^{*}}W\tilde{E}_{ij}d=(e_{j^{*}}^{T}W\tilde{E}_{ij}d)Ae_{j^{*}}.$

In both cases, Lemma 3.21 implies that $D_{x}(P)$ is complete and in
conclusion, the vector fields $\partial_{x}^{K}$, $K\geq l$, of (Type 5) are
complete.

For the proof of (Type 6), we set $P$ as for (Type 1). Observe that
$\tilde{E}_{i^{*}i}\tilde{E}_{j^{*}j}=0$ for $i^{*},i\in\\{i_{1},...,i_{r}\\}$
and $j^{*},j\in\\{j_{1},...,j_{n-r}\\}$. Then we get

$\displaystyle\frac{\partial}{\partial{{z}_{k-1,j^{*}j}}}\frac{\partial}{\partial{{z}_{k,i^{*}i}}}P(x)$
$\displaystyle=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{i^{*}i}\\\
0&0\end{pmatrix}\begin{pmatrix}0&0\\\
\tilde{E}_{j^{*}j}&0\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix}$
$\displaystyle=A\tilde{E}_{i^{*}i}\tilde{E}_{j^{*}j}c=0,$

for all $i\in\\{i_{1},...,i_{r}\\}$ and $j\in\\{j_{1},...,j_{n-r}\\}$.
Furthermore, there holds

$\displaystyle\frac{\partial}{\partial{{z}_{k-1,j^{*}j}}}\frac{\partial}{\partial{{z}_{k,i^{*}j^{\prime}}}}P(x)$
$\displaystyle=A\tilde{E}_{i^{*}j^{\prime}}\tilde{E}_{j^{*}j}c\sim_{x}0,$

and hence the vector fields $\partial_{x}^{K}$, $K\geq k$, of (Type 6) are
complete, by Lemma 3.21.

For the proof of (Type 7) we set
$P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k+2}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{k+1}&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&Z_{k}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix}$. By the same argument
as in (Type 6) we obtain
$\frac{\partial}{\partial{{z}_{k+2,j^{\prime}j^{\prime}}}}\frac{\partial}{\partial{{z}_{k+1,i^{*}i}}}P(x)=0$
for all $i\in\\{i_{1},...,i_{r}\\}$ and
$\frac{\partial}{\partial{{z}_{k+1,i^{*}i}}}\frac{\partial}{\partial{{z}_{k,j^{*}j}}}P(x)=0$
for all $i\in\\{i_{1},...,i_{r}\\}$, $j\in\\{j_{1},...,j_{n-r}\\}$. It remains
to compute

$\displaystyle\frac{\partial}{\partial{{z}_{k+2,j^{\prime}j^{\prime}}}}\frac{\partial}{\partial{{z}_{k,j^{*}j}}}P(x)$
$\displaystyle=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{j^{\prime}j^{\prime}}\\\
0&0\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{k+1}&I_{n}\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{j^{*}j}\\\
0&0\end{pmatrix}\begin{pmatrix}c\\\ d\end{pmatrix}$
$\displaystyle=A\tilde{E}_{j^{\prime}j^{\prime}}\begin{pmatrix}Z_{k+1}&I_{n}\end{pmatrix}\begin{pmatrix}\tilde{E}_{j^{*}j}d\\\
0\end{pmatrix}$
$\displaystyle=A\tilde{E}_{j^{\prime}j^{\prime}}(Z_{k+1}\tilde{E}_{j^{*}j}d)=(e_{j^{\prime}}^{T}Z_{k+1}\tilde{E}_{j^{*}j}d)Ae_{j^{\prime}}.$

Observe that $e_{j^{\prime}}^{T}Z_{k+1}$ is constant in $x$, since we assume
$j^{\prime}\in\\{j_{1},...,j_{n-r}\\}$. Hence the conditions of Lemma 3.21 are
satisfied and we conclude that the fields $\partial_{x}^{K}$, $K\geq k+2$, of
(Type 7) are complete.

For the proof of (Type 8), let
$P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{2}&I_{n}\end{pmatrix}\begin{pmatrix}Z_{1}e_{n}\\\ e_{n}\end{pmatrix}$. For
$1\leq r\leq n$, we compute

$\frac{\partial}{\partial{{z}_{1,in}}}\frac{\partial}{\partial{{z}_{2,jr}}}P(x)=\begin{pmatrix}A&B\end{pmatrix}\begin{pmatrix}0&0\\\
\tilde{E}_{jr}&0\end{pmatrix}\begin{pmatrix}e_{i}\\\
0\end{pmatrix}=\delta_{ir}Be_{j}.$

Hence the conditions of Lemma 3.21 are met and the fields $\partial_{x}^{K}$,
$K\geq 2$, of (Type 8) are complete. ∎

In the following, we find an example of an incomplete vector field
$\partial_{x}^{2}$. As a direct consequence of this, we don’t know how large
the collection $\mathcal{V}_{K}$ of principal vector fields is. We shall see
later, however, that it is powerful enough to construct stratified sprays.

######  Example 3.23.

Consider the tupel $x=(z_{1,n1},...,z_{1,nn},z_{2,n1})$ and the mapping
$P(x)=e_{n}+Z_{2}Z_{1}e_{n}.$ The Jacobian $JP$ is given by
$\begin{pmatrix}Z_{2}&z_{1,nn}e_{1}+z_{1,n1}e_{n}\end{pmatrix}$ and we get the
vector field

$\displaystyle\partial_{x}^{2}=\det\begin{pmatrix}\partial/\partial
z_{1,n1}&\cdots&\partial/\partial z_{1,nn}&\partial/\partial z_{2,n1}\\\
z_{2,11}&\cdots&z_{2,1n}&z_{1,nn}\\\ z_{2,21}&\cdots&z_{2,2n}&0\\\
\vdots&&\vdots&\vdots\\\ z_{2,n1}&\cdots&z_{2,nn}&z_{1,n1}\end{pmatrix}.$

Therefore
$\partial_{x}^{2}(z_{2,n1})=\pm\det(Z_{2})=\pm(\alpha_{1}z_{2,n1}^{2}+\alpha_{2}z_{2,n1}+\alpha_{3})$
for some
$\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{C}\cap\ker(\partial_{x}^{2})$. In
fact, $\alpha_{1}$ is the principal minor of order $n-2$ obtained by removing
the first and last rows and columns from $Z_{2}$. Hence $\alpha_{1}\not\equiv
0$ on $(\mathbb{C}^{\frac{n(n+1)}{2}})^{2}$, which means that the variable
$z_{2,n1}$ occurs quadratically. We conclude that $\partial_{x}^{2}$ is
incomplete.

###  3.5 Topological analysis of the fibers

So far we haven’t learned anything about the fibers $\mathcal{F}^{K}_{y}$ from
a topological perspective. In this subsection we will show that all fibers are
connected for $K\geq 3$. In fact, all fibers are irreducible, except the
singular fibers $\mathcal{F}^{3}_{a,e_{n}^{T}}$ and
$\mathcal{F}^{4}_{0,e_{n}^{T}}$ which consist of two irreducible components,
with the smooth part breaking down in two connected components.

######  Lemma 3.24.

The fibers $\mathcal{F}^{3}_{a,b}$ are connected. The singular fibers, i.e.
$\mathcal{F}^{3}_{a,e_{n}^{T}}$ consists of two irreducible components.

###### Proof.

We start with the non-generic fibers, i.e. we assume $b=0$. In this case, we
have
$\mathcal{F}^{3}_{a,0}=\mathcal{F}^{2}_{a,0}\times\mathbb{C}^{\frac{n(n+1)}{2}}$
with $a\neq 0$. The defining equations of $\mathcal{F}^{2}_{a,0}$ are given by

$\begin{pmatrix}a\\\ 0\end{pmatrix}=\begin{pmatrix}I_{n}&Z_{2}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}Z_{1}e_{n}\\\
e_{n}\end{pmatrix}=\begin{pmatrix}Z_{1}e_{n}\\\
(I_{n}+Z_{2}Z_{1})e_{n}\end{pmatrix},$

and therefore
$\mathcal{F}^{2}_{a,0}\cong\mathbb{C}^{\frac{n(n-1)}{2}}\times\\{Z\in\mathbb{C}^{\frac{n(n+1)}{2}}:e_{n}+Za=0\\}\cong\mathbb{C}^{n(n-1)}$.
In conclusion, the non-generic fiber $\mathcal{F}^{3}_{a,0}$ is biholomorphic
to some $\mathbb{C}^{N}$ and hence connected.

We continue with the generic fibers $\mathcal{F}^{3}_{a,b}$, i.e. $b\neq 0$.
In the first step, we consider the smooth fibers, that is, we assume $b\neq
e_{n}$ in addition. Due to Lemma 3.15, the fiber $\mathcal{F}^{3}_{a,b}$ is
biholomorphic to
$\mathcal{G}_{\pi_{3}(a,b)}\times\mathbb{C}^{\frac{n(n-1)}{2}}$, where

$\mathcal{G}_{\pi_{3}(a,b)}=\\{\vec{Z}_{2}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{2}:b=P_{s}(\vec{Z}_{2})=e_{n}+Z_{2}Z_{1}e_{n}\\}.$

Define $C_{j}:=\\{\vec{Z}_{2}\in\mathcal{G}_{\pi_{3}(a,b)}:z_{1,nj}\neq 0\\},\
1\leq j\leq n.$ Similar as in Lemma 3.15, we can express the variables
$z_{2,j1},...,z_{2,jn}$, which proves that $C_{j}$ is biholomorphic to
$\mathbb{C}^{*}\times\mathbb{C}^{N}$ for some natural number $N$ and therefore
connected. Since we assume $b\neq e_{n}$, $Z_{1}e_{n}=0$ is not possible in
$\mathcal{G}_{\pi_{3}(a,b)}$, which means that $\mathcal{G}_{\pi_{3}(a,b)}$ is
covered by $\cup_{j=1}^{n}C_{j}$. It remains to show, that the intersection
$\cap_{j=1}^{n}C_{j}\neq\emptyset$ is non-empty. Choose a symmetric matrix
$Z_{1}^{*}$ with $Z_{1}^{*}e_{n}=(1,...,1)^{T}$. By Lemma 3.13, there is a
symmetric matrix $Z_{2}$ with $b-e_{n}=Z_{2}(Z_{1}^{*}e_{n})$. This shows that
the intersection is indeed non-empty. In conclusion,
$\mathcal{G}_{\pi_{3}(a,b)}$ and $\mathcal{F}^{3}_{a,b}$ are connected.

Finally, let’s consider the singular fibers $\mathcal{F}^{3}_{a,e_{n}^{T}}$.
By Lemma 3.15, such fibers are biholomorphic to
$\mathcal{G}_{\pi_{3}(a,b)}\times\mathbb{C}^{\frac{n(n-1)}{2}}$ where

$\mathcal{G}_{\pi_{3}(a,b)}=\\{\vec{Z}_{2}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{2}:Z_{2}Z_{1}e_{n}=0\\}.$

Observe that this variety has two irreducible components
$A_{1}=\\{\vec{Z}_{2}\in\mathcal{G}_{\pi_{3}(a,e_{n}^{T})}:Z_{1}e_{n}=0\\}$
and
$A_{2}=\\{\vec{Z}_{2}\in\mathcal{G}_{\pi_{3}(a,e_{n}^{T})}:\det(Z_{2})=0\\}.$
This proves that singular fibers $\mathcal{F}^{3}_{a,e_{n}^{T}}$ have two
irreducible components. Since the intersection of these components equals the
singularity set $S_{3}$, $\mathcal{F}^{3}_{a,e_{n}^{T}}$ is connected. ∎

######  Theorem 3.25.

The fibers $\mathcal{F}^{K}_{y}$ are connected for $K\geq 3$. Moreover, the
smooth part of the singular fibers is connected for $K\geq 5$.

###### Proof.

We prove this theorem by induction on $K$. Note that we’ve shown the base case
$K=3$ in the previous lemma. Let $K\geq 4$ and assume that the $(K-1)$ fibers
$\mathcal{F}^{K-1}_{y}$ are connected. Recall that

$\mathcal{F}^{K}_{y}=\bigcup_{Z\in\mathbb{C}^{\frac{n(n+1)}{2}}}\mathcal{F}^{K-1}_{M_{K}(Z)y}.$
(9)

We will now introduce the following auxiliary function. Let
$\rho:\mathcal{F}^{K}_{y}\to\mathbb{C}^{\frac{n(n+1)}{2}}$ denote the
restriction of the projection $(\vec{Z}_{K-1},Z_{K})\mapsto Z_{K}$ to the
fiber $\mathcal{F}^{K}_{y}$. We’ll show some useful facts.

  1. (i)

$\rho$ is surjective: Observe that $\vec{Z}_{K}\in\mathcal{F}^{K}_{y}$ if and
only if $\vec{Z}_{K-1}\in\mathcal{F}^{K-1}_{M_{K}(-Z_{K})y}$, by (9). Hence
the $\rho$-fibers are given by
$\rho^{-1}(Z_{K}^{*})=\mathcal{F}^{K-1}_{M_{K}(-Z_{K}^{*})y}$. The
$(K-1)$-fibers $\mathcal{F}^{K-1}_{M_{K}(Z)y}$ are non-empty for all
$Z\in\mathbb{C}^{\frac{n(n+1)}{2}}$, by Theorem 3.5.

  2. (ii)

$\rho$ is submersive in $\vec{Z}_{K}$ if $\vec{Z}_{K-1}\not\in S_{K-1}$:
Observe that $T_{\vec{Z}_{K}}(\mathcal{F}^{K}_{y}\cap S_{K}^{c})=\ker
J\Phi_{K}(\vec{Z}_{K})$. By assumption, the Jacobian
$J\Phi_{K-1}(\vec{Z}_{K-1})$ is surjective. Given
$W_{K}\in\mathbb{C}^{\frac{n(n+1)}{2}}$, we find
$\vec{W}_{K-1}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K-1}$ such that
$J\Phi_{K-1}(\vec{Z}_{K-1})\vec{W}_{K-1}=-M_{K}(-Z_{K})A_{K}(\Phi_{K-1}(\vec{Z}_{K-1}))W_{K}$.
By the recursive formula of the Jacobian (see Corollary 3.8), we get

$J\Phi_{K}(\vec{Z}_{K})\begin{pmatrix}\vec{W}_{K-1}\\\
W_{K}\end{pmatrix}=M_{K}(Z_{K})J\Phi_{K-1}(\vec{Z}_{K-1})\vec{W}_{K-1}+A_{K}(\Phi_{K-1}(\vec{Z}_{K-1}))W_{K}=0$

and $d\rho_{\vec{Z}_{K}}(\vec{W}_{K-1},W_{K})=W_{K}$.

  3. (iii)

Each connected component $A\subset\mathcal{F}^{K}_{y}$ is $\rho$-saturated,
that is, $A=\rho^{-1}(\rho(A))$: It is enough to show “$\supset$”, by
definition of the preimage. Each $\rho$-fiber is connected and therefore we
have either $\rho^{-1}(b)\subset A$ or $\rho^{-1}(b)\cap A=\emptyset$.

  4. (iv)

$\rho(A)$ is open for each connected component $A\subset\mathcal{F}^{K}_{y}$:
Given a point $b\in\rho(A)$ we find a regular point $a\in\rho^{-1}(b)$. To see
this, notice that in every $(K-1)$ fiber we find points with $Z_{1}e_{n}\neq
0$, by the previous lemma and by (9). Submersivity is a local property, hence
there exists an open neighborhood $U\subset A$ of $a$ in which $\rho$ is
submersive. Furthermore, $U$ is mapped openly, that is, $\rho(U)$ is an open
neighborhood of $b$ in $\rho(A)$.

We can write $\mathcal{F}^{K}_{y}$ as a disjoint union of connected components
$\dot{\bigcup}_{i\in I}A_{i}$. Then

$\displaystyle\mathbb{C}^{\frac{n(n+1)}{2}}\underset{(i)}{=}\rho(\mathcal{F}^{K}_{y})\underset{(iii)}{=}\dot{\bigcup}_{i\in
I}\rho(A_{i})$

can be written as the disjoint union of open sets. Since
$\mathbb{C}^{\frac{n(n+1)}{2}}$ is connected, we conclude that
$\mathcal{F}^{K}_{y}$ has to be connected too.

It remains to show, that the smooth part of the fibers $\mathcal{F}^{K}_{a,b}$
is connected for $K\geq 5$. Let’s start with the case $K=2k+1$. Then the
singular fibers are $\mathcal{F}^{K}_{a,e_{n}^{T}}$, by Lemma 3.16. If we now
also note (9), then the singularities are all in the subfiber
$\mathcal{F}^{2k}_{0,e_{n}^{T}}$. Since
$\mathcal{F}^{2k}_{0,e_{n}^{T}}\subset\mathcal{F}^{2k+1}_{a,e_{n}^{T}}$ has
codimension $n$, the complement
$U:=\mathcal{F}^{2k+1}_{a,e_{n}^{T}}\setminus\mathcal{F}^{2k}_{0,e_{n}^{T}}$
is connected. If we consider a smooth point
$p\in\mathcal{F}^{2k}_{0,e_{n}^{T}}$, then each open neighborhood of $p$
intersects with $U$. Hence $p$ has to be in the same connected component as
$U$. This proves that the smooth part
$\mathcal{F}^{K}_{a,e_{n}^{T}}\setminus\mathrm{Sing}(\mathcal{F}^{K}_{a,e_{n}^{T}})$
is connected.

In the case $K=2k$, observe that $\mathcal{F}^{2k}_{0,e_{n}^{T}}$ is the only
singular fiber. Furthermore, there holds

$\mathcal{F}^{2k}_{0,e_{n}^{T}}\setminus\mathrm{Sing}(\mathcal{F}^{2k}_{0,e_{n}^{T}})=\left(\mathcal{F}^{2k-1}_{0,e_{n}^{T}}\setminus\mathrm{Sing}(\mathcal{F}^{2k-1}_{0,e_{n}^{T}})\right)\times\mathbb{C}^{\frac{n(n+1)}{2}},$

and this is connected by the induction hypothesis. This completes the proof of
the theorem. ∎

######  Theorem 3.26.

The fibers of the submersion
$\Phi_{K}:\mathcal{W}_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$ are connected for
$K\geq 3$.

###### Proof.

We want to show that the intersection $\mathcal{F}^{K}_{y}\cap\mathcal{W}_{K}$
is connected. Smooth fibers $\mathcal{F}^{K}_{y}$ are contained in
$\mathcal{W}_{K}$, hence we only need to consider the case when
$\mathcal{F}^{K}_{y}$ is a singular fiber. For $K=3$ we can apply the strategy
from Lemma 3.24 for smooth fibers and cover
$\mathcal{F}^{3}_{a,e_{n}^{T}}\cap\mathcal{W}_{3}$ by $n$ intersection
connected charts.

Next, assume the claim to be true for $K-1=2k-1$. For

$\mathcal{R}:=\\{\vec{Z}_{2k-1}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{2k-1}:Z_{1}e_{n}=...=Z_{2k-3}e_{n}=0,Z_{2k-1}e_{n}\neq
0\\},$

we can rewrite

$\mathcal{W}_{2k}=\mathcal{W}_{2k-1}\times\mathbb{C}^{\frac{n(n+1)}{2}}\dot{\cup}\mathcal{R}\times\mathbb{C}^{\frac{n(n+1)}{2}}.$

Observe that $\mathcal{F}^{2k-1}_{0,e_{n}^{T}}\cap\mathcal{R}=\emptyset$, by
definition. Therefore, the non-generic singular fiber
$\mathcal{F}^{2k}_{0,e_{n}^{T}}$ satisfies

$\mathcal{F}^{2k}\cap\mathcal{W}_{2k}=\left(\mathcal{F}^{2k-1}_{0,e_{n}^{T}}\cap\mathcal{W}_{2k-1}\right)\times\mathbb{C}^{\frac{n(n+1)}{2}},$

which is connected by the induction hypothesis.

Finally, we assume the claim to be true for $K-1=2k$. Let
$\rho:\mathcal{F}^{K}_{a,e_{n}^{T}}\cap\mathcal{W}_{K}\to\mathbb{C}^{\frac{n(n+1)}{2}}$
be the restriction of the projection $(\vec{Z}_{K-1},Z_{K})\mapsto Z_{K}$ to
$\mathcal{F}^{K}_{a,e_{n}^{T}}\cap\mathcal{W}_{K}$. The fiber
$\rho^{-1}(Z_{K})$ is given by
$\mathcal{F}^{K-1}_{M_{K}(-Z_{K})y}\cap\mathcal{W}_{K-1}$. From here we can
argue as in the proof of Theorem 3.25. ∎

######  Lemma 3.27.

Each generic fiber $\mathcal{F}^{3}_{y}$ containes points $\vec{Z}_{3}$ with
$Q_{f}^{1}(\vec{Z}_{3})=z_{1,n1}\cdots z_{1,nn}\neq 0$. Moreover, each generic
fiber $\mathcal{F}^{K}_{y}$ containes points $\vec{Z}_{K}$ with

${Q_{f}^{K-2}(\vec{Z}_{K})Q_{s}^{K-2}(\vec{Z}_{K})=P^{K-2}_{1}(\vec{Z}_{K-2})\cdots
P^{K-2}_{n}(\vec{Z}_{K-2})P^{K-2}_{n+1}(\vec{Z}_{K-2})\cdots
P^{K-2}_{2n}(\vec{Z}_{K-2})\neq 0}.$

###### Proof.

Let’s start with the case $K=3$. We consider a generic fiber
$\mathcal{F}^{3}_{a,b}$, $b\neq 0$. Recall that
$\vec{Z}_{3}\in\mathcal{F}^{3}_{a,b}$ if and only if
$\vec{Z}_{2}\in\mathcal{F}^{2}_{a-bZ_{3},b}$ by the recursive formula (4).
Since we assume $b\neq 0$, there is a symmetric matrix
$Z_{3}\in\mathbb{C}^{\frac{n(n+1)}{2}}$ with $a-Z_{3}b=(1,...,1)^{T}$, by
Lemma 3.13. The fiber $\mathcal{F}^{2}_{(1,...,1),b}\neq\emptyset$ is non-
empty by Theorem 3.5. Furthermore, there holds $P_{f}^{2}\equiv P_{f}^{1}$, by
the recursive formula (4). This implies $Q_{f}^{1}\equiv 1$ on the fiber
$\mathcal{F}^{2}_{(1,...,1),b}\subset\mathcal{F}^{3}_{a,b}$ and we’re done
with the case $K=3$.

We now consider the case $K\geq 4$. The fiber
$\mathcal{F}^{K-2}_{(1,...,1),(1,...,1)}\neq\emptyset$ is non-empty, by
Theorem 3.5, and there clearly holds $Q_{f}^{K-2}Q_{s}^{K-2}\equiv 1$ on the
whole fiber. Therefore it is enough to show, that this fiber sits inside each
generic $K$-fiber, that is,
$\mathcal{F}^{K-2}_{(1,...,1),(1,...,1)}\subset\mathcal{F}^{K}_{y}$. We prove
this claim only for $K=2k+1$, since the proof is symmetric for $K=2k$. In this
case, a point $y=(a,b)\in Y_{g}^{K}$ in the generic stratum satisfies $b\neq
0$. Let’s write $v:=(1,...,1)^{T}$. Then the fiber $\mathcal{F}^{K-2}_{v,v}$
sits inside $\mathcal{F}^{K}_{a,b}$ if and only if we find symmetric matrices
$Z_{K-1},Z_{K}\in\mathbb{C}^{\frac{n(n+1)}{2}}$ with

$\begin{pmatrix}a\\\ b\end{pmatrix}=\begin{pmatrix}I_{n}&Z_{K}\\\
0&I_{n}\end{pmatrix}\begin{pmatrix}I_{n}&0\\\
Z_{K-1}&I_{n}\end{pmatrix}\begin{pmatrix}v\\\ v\end{pmatrix},$

or equivalently,

$\begin{pmatrix}a-Z_{K}b\\\ b\end{pmatrix}=\begin{pmatrix}v\\\
Z_{K-1}v+v\end{pmatrix}.$

We can split up this system of equations into two independent systems
$a-Z_{K}b=v$ and $b=Z_{K-1}v+v$. An application of Lemma 3.13 to both systems
yields the existence of such matrices $Z_{K-1}$ and $Z_{K}$, since $b\neq 0$
and $v\neq 0$. This completes the proof. ∎

###  3.6 Construction of stratified sprays

The most convenient way to construct a dominating spray associated to a
submersion is to define some finite composition of flow maps of complete
fiber-preserving vector fields. The collection of complete vector fields in
Proposition 3.22 does not span the tangent space of the fibers
$\mathcal{F}^{K}_{y}$ in every point. This is the reason why these are a
priori not sufficient to define a dominating spray. However, we can enlarge
this collection until it spans the tangent space in a sufficiently large set
of points. In the following section we discuss the meaning of ’sufficiently
large’ and the mathematical details for the enlargement.

####  3.6.1 Strategy for the construction

Let $M$ be a Stein manifold and let $\mathcal{VC}_{hol}(M)$ denote the set of
$\mathbb{C}$-complete holomorphic vector fields on $M$. For a vector field
$V\in\mathcal{VC}_{hol}(M)$ its corresponding flow $\alpha^{V}_{t}$,
$t\in\mathbb{C}$, is a one-parameter subgroup of $\mathrm{Aut}_{hol}(M)$, the
holomorphic automorphism group on $M$. For a set
$A\subset\mathcal{VC}_{hol}(M)$ of complete holomorphic vector fields on $M$
we define

$S_{A}:=\bigcup_{V\in
A}\\{\alpha^{V}_{t}:t\in\mathbb{C}\\}\subset\mathrm{Aut}_{hol}(M).$

Let $G_{A}:=\langle S_{A}\rangle$ denote the subgroup generated by $S_{A}$.
Furthermore, we denote the pull-back of an automorphism $\alpha$ by
$\alpha^{*}$.

######  Definition 3.28.

For a set $A\subset\mathcal{VC}_{hol}(M)$ of complete holomorphic vector
fields on $M$, define

$\Gamma(A):=\\{\alpha^{*}X:\alpha\in G_{A},X\in A\\}$

the collection of complete holomorphic vector fields generated by $A$, and for
an open set $Y\subset M$ define

$C_{A}(Y):=\\{\alpha(y):\alpha\in G_{A},y\in Y\\}$

the $G_{A}$-closure of $Y$.

Some basic properties follow directly from the definition. Let $Y\subset M$ be
an open set. Then $C_{A}(Y)$ is the smallest set containing $Y$, which is
invariant under $G_{A}$. Moreover, $C_{A}(Y)$ is open in $M$, hence, for a
fixed collection $A\subset\mathcal{VC}_{hol}(M)$, $C_{A}$ can be interpreted
as a map $C_{A}:\mathcal{T}_{M}\to\mathcal{T}_{M}$, where $\mathcal{T}_{M}$
denotes the natural topology on $M$. In particular, $C_{A}$ satisfies the
conditions of a topological closure operator.

The following lemma describes another basic property.

######  Lemma 3.29.

Let $A,B\subset\mathcal{VC}_{hol}(M)$ be finite collections of complete
holomorphic vector fields on $M$ with $A\subset B\subset\Gamma(A)$. Then
$C_{A}(X)=C_{B}(X)$ for all open subsets $X\subset M$.

###### Proof.

We’re going to prove that $G_{A}=G_{B}$. To do this, it suffices to show that
$G_{B}\subset G_{A}$, since the reverse inclusion trivially holds by
assumption $A\subset B$.

At first, we consider $V\in B$. There is $\beta\in G_{A}$ and $W\in A$ with
$V=\beta^{*}W$, since $B\subset\Gamma(A)$. The flow of $V$ satisfies

$\alpha_{t}^{V}=\alpha_{t}^{\beta^{*}W}=\beta\circ\alpha_{t}^{W}\circ\beta^{-1}\in
G_{A},$

since $\beta,\alpha_{t}^{W}\in G_{A}$.

In a next step, let $\beta\in G_{B}$ be any automorphism. By definition of
$G_{B}$, there are vector fields $V_{i_{1}},...,V_{i_{m}}\in B$ and times
$t_{1},...,t_{m}\in\mathbb{C}$ with
$\beta=\alpha_{t_{1}}^{V_{i_{1}}}\circ\cdots\circ\alpha_{t_{m}}^{V_{i_{m}}}$.
From the previous step we know that each $\alpha_{t_{j}}^{V_{i_{j}}}\in G_{A}$
is an automorphism in $G_{A}$ and hence so is $\beta$, i.e. $\beta\in G_{A}$
and this proves the claim. ∎

Here comes the main step for the enlargement of the collection of complete
fields.

######  Lemma 3.30.

Let $M$ be a Stein manifold, $X\subset M$ an open subset and
$A\subset\mathcal{VC}_{hol}(M)$ a finite set of $\mathbb{C}$-complete
holomorphic vector fields on $M$ which spans the tangent bundle $TX$. Then
there is a finite subset $A\subset B\subset\Gamma(A)$ which spans the tangent
bundle $TC_{A}(X)$.

###### Proof.

For each field $V\in A$ let $\alpha^{V}_{t},t\in\mathbb{C},$ be the
corresponding vector flow. Let $N_{0}$ be the set of points $x\in C_{A}(X)$
where the fields of $A$ don’t span the tangent space $T_{x}M$. This is an
analytic subset $N_{0}\subset C_{A}(X)\setminus X$. Further, we define

$N_{k}:=\\{x\in N_{k-1}:\alpha^{V}_{t}(x)\in N_{k-1},\forall V\in A,\forall
t\in\mathbb{C}\\},\quad k\geq 1.$

Let $k\geq 0$ be arbitrary but fixed. Then $N_{k}$ has at most countably many
connected components. Denote by $A_{i}^{k},i\in I_{k}$, those connected
components of $N_{k}$ which aren’t entirely contained in $N_{k+1}$ and let
$a_{k}:=\max_{i\in I_{k}}\dim A_{i}^{k}$ be the maximal dimension of them.
Choose a point $x_{i}^{k}\in A_{i}^{k}\cap N_{k}\setminus N_{k+1},i\in I_{k},$
of each such component. By definition of the sets $N_{k}$ and $N_{k+1}$, there
is a field $V\in A$ for each point $x$ in the sequence $\\{x_{i}^{k}\\}_{i\in
I}$, such that $\alpha^{V}_{t}(x)\not\in N_{k}$ for some $t\in\mathbb{C}$. For
$V\in A$ define

$u_{V}^{k}:=\\{x\in\\{x_{i}^{k}\\}_{i\in I_{k}}:\alpha^{V}_{t}(x)\not\in
N_{k},\text{ for some }t\in\mathbb{C}\\}.$

Then this yields

$\\{x_{i}^{k}\\}_{i\in I_{k}}=\bigcup_{V\in A}u_{V}^{k}.$

For each point $x\in u_{V}^{k}$ the set
$\\{t\in\mathbb{C}:\alpha^{V}_{t}(x)\in N_{k}\\}$ is discrete and hence

$\bigcup_{x\in u_{V}^{k}}\\{t\in\mathbb{C}:\alpha^{V}_{t}(x)\in N_{k}\\}$

is a meagre set in $\mathbb{C}$. This implies the existence of a time
$t_{V}^{k}\in\mathbb{C}$ such that $\alpha^{V}_{t_{V}^{k}}(x)\not\in N_{k}$
for all $x\in u_{V}^{k}$. Define

$\tilde{N}_{k+1}:=\\{x\in N_{k}:\alpha^{V}_{t_{V}^{k}}(x)\in N_{k},\forall
V\in A\\}.$

Clearly, there holds

$N_{k+1}\subset\tilde{N}_{k+1}\subset N_{k}.$

The set $\tilde{N}_{k+1}$ has at most countably many connected components. Let
$\tilde{a}_{k+1}$ denote the maximal dimension of them. There holds
$a_{k+1}\leq\tilde{a}_{k+1}<a_{k}$ by construction. Since $M$ is finite
dimensional, this implies that there is $L\in\mathbb{N}$ such that
$N_{k}=\emptyset$ for all $k>L$.

Let $B_{k}$ be the set of pullbacks $(\alpha_{t_{V}^{k}}^{V})^{*}(W)$ for
$V,W\in A$ and set

$B:=\bigcup_{k\geq 0}^{L}A\cup B_{k}.$

Again by construction, the collection $B\subset\Gamma(A)$ is a finite set of
$\mathbb{C}$-complete holomorphic vector fields on $M$ that spans the tangent
bundle $TC_{A}(X)$. Moreover, $C_{A}(X)=C_{B}(X)$ by the previous lemma, which
implies that $C_{A}(X)$ is invariant under the flows of $B$. This finishes the
proof. ∎

In a next step, we want to adapt this argument to our setting so that we can
apply it to every fiber simultaneously, so to speak. Moreover, we are ready to
define a dominating spray.

######  Lemma 3.31.

Let $M$ be a Stein manifold, $\pi:M\to Y$ a holomorphic mapping and $X\subset
M$ a (connected) open subset such that the restriction $\pi|_{X}:X\to Y$ is a
surjective submersion with connected fibers.

Suppose that there is a finite set $A\subset\mathcal{VC}_{hol}(M)$ of
$\mathbb{C}$-complete fiber-preserving holomorphic vector fields on $M$ which
spans the tangent bundle $T(M_{y}\cap X)$ of each fiber $M_{y}:=\pi^{-1}(y)$.
Then there is a finite set $B\subset\Gamma(A)$ of $\mathbb{C}$-complete fiber-
preserving holomorphic vector fields which spans the tangent bundle
$T(C_{A}(M_{y}\cap X))$ of each fiber $M_{y}$. In particular, the surjective
submersion $\pi|_{C_{A}(X)}$ admits a spray.

###### Proof.

We can proceed similarly as in the previous lemma to obtain a finite
collection $B\subset\Gamma(A)$ which spans the tangent bundle
$T(C_{A}(M_{y}\cap X))$ of each fiber $M_{y}$. This follows from the
assumption that all fields in $A$ are fiber-preserving.

Moreover, the map $\pi|_{X}:X\to Y$ is a surjective submersion, hence
$\pi|_{C_{A}(X)}:C_{A}(X)\to Y$ is also a surjective submersion. Write
$B=\\{W_{1},...,W_{L}\\}$. Then the map $s:C_{A}(X)\times\mathbb{C}^{L}\to
C_{A}(X)$ given by

$s(z,t_{1},...,t_{L})=\alpha_{t_{1}}^{W_{1}}\circ\cdots\circ\alpha_{t_{L}}^{W_{L}}(z)$

is a dominating spray associated to $\pi|_{C_{A}(X)}$, since $C_{A}(X)$ is
invariant with respect to the flows of $W_{1},...,W_{L}$ by Lemma 3.29. ∎

This corollary shows that we can relax the assumptions of the previous lemma
and we will apply it in this form later.

######  Corollary 3.32.

Let $M$ be a Stein manifold, $\pi:M\to Y$ a holomorphic map and $X\subset M$ a
connected open set such that the restriction $\pi|_{X}$ is a surjective
submersion with connected fibers. Furthermore, we are given an open subset
$W\subset X$ and a finite collection $A\subset\mathcal{VC}_{hol}(M)$ of
complete fiber-preserving holomorphic vector fields on $M$ which spans the
tangent bundle $T(M_{y}\cap W)$ of each fiber $M_{y}$.

Suppose that there is a finite collection $B\subset\mathcal{VC}_{hol}(M)$ of
complete fiber-preserving holomorphic vector fields on $M$ such that $X\subset
C_{B}(W)$ and $A\subset B$. Then $\pi|_{C_{B}(W)}$ admits a spray.

###### Proof.

Since $B$ contains $A$ by assumption, $B$ spans the tangent bundle
$T(M_{y}\cap W)$ of each fiber, hence there is a finite collection
$\tilde{B}\subset\Gamma(B)$ which spans the tangent bundle $T(M_{y}\cap
C_{B}(W))$ of each fiber. Since $X\subset C_{B}(W)$ by assumption, $\tilde{B}$
spans the tangent bundle $T(M_{y}\cap X)$ for each fiber. Now, we apply Lemma
3.31 to finish the proof. ∎

######  Remark 3.33.

Given a finite collection $A\subset\mathcal{VC}_{hol}(M)$ and two open sets
$X,W\subset M$ with $W\subset X$, there holds $X\subset C_{A}(W)$ if and only
if $X\setminus W\subset C_{A}(W)$. This follows from the fact that $C_{A}$ is
extensive.

This lemma will help us decide if we have $X\subset C_{A}(X)$ for a suitable
finite collection $A$.

######  Lemma 3.34.

Let $M$ be a Stein manifold and $N\subset M$ an analytic subvariety given by

$N:=\\{x\in M:f(x)=0\\}$

for some holomorphic mapping $f:M\to\mathbb{C}$. Assume that there are
complete holomorphic vector fields $V_{1},...,V_{k}$ on $M$ (and we denote the
respective flows by $\alpha^{1}_{t},...,\alpha^{k}_{t}$) such that

$V_{i_{n}}\circ\cdots\circ V_{i_{1}}(f(x))\neq 0,\forall x\in N,$

for some finite sequence $\\{i_{1},...,i_{n}\\}\subset\\{1,...,k\\}$. Then
there is a composition of the flows $\alpha^{1}_{t},...,\alpha^{k}_{t}$ which
leaves the subvariety $N$. More precisely, there holds

$\\{\alpha^{i_{n}}_{t_{n}}\circ\cdots\circ\alpha^{i_{1}}_{t_{1}}(x):t_{1},...,t_{n}\in\mathbb{C}\\}\not\subset
N,\quad\forall x\in N.$

###### Proof.

Define the subvariety $N_{1}:=\\{x\in N:V_{i_{1}}(f(x))=0\\}$. The orbit of
$\alpha_{t}^{i_{1}}$ through points of $N\setminus N_{1}$ is leaving $N$.
Next, define the subvariety $N_{2}:=\\{x\in
N_{1}:V_{i_{2}}(V_{i_{1}}(f(x)))=0\\}$. Then the orbit of $\alpha_{t}^{i_{2}}$
through points of $N_{1}\setminus N_{2}$ is leaving $N_{1}$. We proceed
inductively and set $N_{l}:=\\{x\in N_{l-1}:V_{i_{l}}\circ\cdots\circ
V_{i_{1}}(f(x))=0\\}$. Then the orbit of $\alpha_{t}^{i_{l}}$ through points
of $N_{l-1}\setminus N_{l}$ is leaving $N_{l-1}$. This is true for all $1\leq
l\leq n-1$, which implies that an invariant set with respect to the fields
$V_{1},...,V_{k}$ has to be in the set $N_{n}$. There holds
$V_{i_{n}}\circ\cdots\circ V_{i_{1}}(f(x))\neq 0$, by assumption. Hence the
set $N_{n}$ is empty and there is no invariant set in $N$ with respect to the
fields $V_{1},...,V_{k}$. This proves the claim. ∎

####  3.6.2 Construction of a spray over the generic stratum

We begin this subsection with a definition. Recall the collection
$\mathcal{V}_{K}$ of $\mathbb{C}$-complete fiber-preserving holomorphic vector
fields from Definition 3.19. These collections are defined for $K\geq 3$.
According to Lemma 3.15, there are $n$ meromorphic mappings
$\psi_{1},...,\psi_{n}$ such that each generic fiber $\mathcal{F}^{2}_{y}$ is
biholomorphic to $\mathcal{G}_{\pi_{2}(y)}\cong\mathbb{C}^{n(n-1)}$ via
$\psi_{j}$ for some $1\leq j\leq n$. Let $\frac{\partial}{\partial{x_{i}}}$,
$1\leq i\leq n(n-1)$ denote the standard vector fields on
$\mathbb{C}^{n(n-1)}$. Then

$\mathcal{V}_{2}=\bigcup_{j=1}^{n}\left\\{(\psi_{j})_{*}\left(\frac{\partial}{\partial{x_{i}}}\right):1\leq
i\leq n(n-1)\right\\}$

is the collection obtained by pushing forward the standard vector fields via
$\psi_{1},...,\psi_{n}$.

######  Definition 3.35.

For $K\geq 2$, we define $\mathcal{Q}_{K}:=\Gamma(\mathcal{V}_{K})$ the
collection of $\mathbb{C}$-complete $\mathcal{F}^{K}_{y}$-fiber-preserving
holomorphic vector fields generated by $\mathcal{V}_{K}$.

Recall the set $\mathcal{W}_{K}$ which is given by

$\mathcal{W}_{K}:=\left\\{\vec{Z}_{K}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}:Z_{2i-1}e_{n}\neq
0,\text{ for some }1\leq
i\leq\big{\lceil}\tfrac{K-1}{2}\big{\rceil}\right\\},$

where $\lceil x\rceil$ is the ceiling function, which maps $x$ to the least
integer greater than or equal to $x$. The set $\mathcal{W}_{K}$ is open and
connected.

The following result, from now on we will call it Spanning theorem, is cruical
for the construction of a spray over the generic stratum.

######  Theorem 3.36 (Spanning theorem).

Let $K\geq 2$. Then there is a finite set $A_{K}\subset\mathcal{Q}_{K}$ of
$\mathbb{C}$-complete fiber-preserving holomorphic vector fields on
$(\mathbb{C}^{\frac{n(n+1)}{2}})^{K}$ spanning the tangent bundle
$T(\mathcal{F}^{K}_{y}\cap\mathcal{W}_{K})$ of every generic fiber
$\mathcal{F}^{K}_{y}$. In particular, there holds $A_{K}\subset A_{K+1}$, when
considering $A_{K}$ as a subset of $\mathcal{Q}_{K+1}$.

Let $K\geq 3$. Then the mapping
$\Phi_{K}|_{\mathcal{W}_{K}}:\mathcal{W}_{K}\to\mathbb{C}^{2n}\setminus\\{0\\}$
is a surjective submersion with connected fibers $\mathcal{F}^{K}_{y}$ (see
Theorem 3.5 and Theorem 3.26). By the  (Spanning theorem)., the conditions of
Lemma 3.31 are satisfied. Hence the submersion

$\Phi_{K}:C_{A_{K}}(\mathcal{W}_{K})|_{Y_{g}^{K}}\to Y_{g}^{K}$

over the generic stratum $Y_{g}^{K}\subset\mathbb{C}^{2n}\setminus\\{0\\}$
admits a spray.

######  Remark 3.37 (Application of the  (Spanning theorem).).

Recall that, by Lemma 3.16, a fiber $\mathcal{F}^{K}_{y}$ contains
singularities if and only if $y=(y_{1},...,y_{n},0,...,0,1)$ for
$y_{1},...,y_{n}\in\mathbb{C}$. In particular, the intersection of singular
fibers and $\mathcal{W}_{K}^{c}$, the complement of $\mathcal{W}_{K}$, is non-
empty. We do not know, whether the collection $A_{K}$ from the  (Spanning
theorem). can be supplemented by finitely many fields in $\mathcal{Q}_{K}$ so
that $A_{K}$ spans the tangent space $T_{\vec{Z}_{K}}\mathcal{F}^{K}_{y}$ for
a smooth point $\vec{Z}_{K}$ in $\mathcal{W}_{K}^{c}$. However, since smooth
fibers $\mathcal{F}^{K}_{y}$ are completely contained in $\mathcal{W}_{K}$,
the collection $A_{K}$ spans the tangent bundle $T\mathcal{F}^{K}_{y}$ of each
smooth generic fiber $\mathcal{F}^{K}_{y}$.

####  3.6.3 Construction of a spray over the non-generic stratum

In this subsection, we show that the submersion

$\Phi_{K}:C_{A_{K}}(\mathcal{W}_{K})|_{Y_{ng}^{K}}\to Y_{ng}^{K}$

over the non-generic stratum
$Y_{ng}^{K}\subset\mathbb{C}^{2n}\setminus\\{0\\}$ admits a spray.

We need the following result.

######  Lemma 3.38.

Let $K\geq 3$ and let $A\subset\mathcal{Q}_{K}$ be a finite collection of
complete holomorphic fiber preserving vector fields. Then there holds

$C_{A}(\mathcal{W}_{K}\cap\mathcal{F}^{K}_{y})=C_{A}(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y})\times\mathbb{C}^{\frac{n(n+1)}{2}}$

for each non-generic fiber $\mathcal{F}^{K}_{y}$.

###### Proof.

In a first step, we prove that

$\displaystyle\mathcal{W}_{K}\cap\mathcal{F}^{K}_{y}=(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y})\times\mathbb{C}^{\frac{n(n+1)}{2}}$
(10)

for each non-generic fiber $\mathcal{F}^{K}_{y}$ and $K\geq 3$. From the
definition of the set $\mathcal{W}_{K}$ we directly conclude
$\mathcal{W}_{2k+1}=\mathcal{W}_{2k}\times\mathbb{C}^{\frac{n(n+1)}{2}}$.
Since each non-generic fiber satisfies
$\mathcal{F}^{K}_{y}=\mathcal{F}^{K-1}_{y}\times\mathbb{C}^{\frac{n(n+1)}{2}}$
by Lemma 3.14, equation (10) follows for $K=2k+1$.

Consider now $K=2k$ even and a non-generic fiber $\mathcal{F}^{K}_{y}$, that
is, $y=(0,b)^{T}$ for some non-zero $b\in\mathbb{C}^{n}$. Further, observe
that
$\mathcal{W}_{2k}=\mathcal{W}_{2k-1}\times\mathbb{C}^{\frac{n(n+1)}{2}}\cup\mathcal{R},$
where

$\mathcal{R}:=\\{\vec{Z}_{2k}\in(\mathbb{C}^{\frac{n(n+1)}{2}})^{2k}:\pi_{n}(Z_{2i-1})=0,1\leq
i\leq k-1,\pi_{n}(Z_{2k-1})\neq 0\\}.$

It suffices to show that $\mathcal{F}^{K}_{y}\cap\mathcal{R}=\emptyset$, in
order to prove (10). Assume for contradiction there is
$\vec{Z}_{2k}\in\mathcal{F}^{K}_{y}\cap\mathcal{R}$. Lemma 3.9 implies
$\Phi_{2k-2}(Z_{2k-2})=e_{2n}$ and
$\Phi_{2k-1}(\vec{Z}_{2k-1})=\begin{pmatrix}0&b\end{pmatrix}^{T}$ follows by
Lemma 3.14. According to the recursive formula (4), we obtain

$\begin{pmatrix}0\\\ b\end{pmatrix}=\begin{pmatrix}I_{n}&Z_{2k-1}\\\
0&I_{n}\end{pmatrix}e_{2n}=\begin{pmatrix}\pi_{n}(Z_{2k-1})\\\
e_{n}\end{pmatrix}$

contradicting assumption $\pi_{n}(Z_{2k-1})\neq 0$. This proves equation (10).

In a second step, we show that, over the non-generic stratum, none of the
vector fields $V\in\mathcal{Q}_{K}$ flows in a new direction. It suffices to
prove the claim for the generating set $\mathcal{V}_{K}$. Note that there is
nothing to show for vector fields in $\mathcal{V}_{K-1}$. And fields in
$\mathcal{V}_{K}\setminus\mathcal{V}_{K-1}$ vanish over the non-generic
stratum, by Lemma 3.18. This proves the claim. In particular, the vector flow
$\alpha_{t}^{V}$ of $V\in\mathcal{Q}_{K}$ fixes the new directions, i.e.
$\alpha_{t}^{V}(\vec{Z}_{K-1},Z_{K})=(f(\vec{Z}_{K-1},Z_{K}),Z_{K})$ for some
suitable holomorphic function $f$.

In the last step, we apply equation (10) and step two. We get

$\displaystyle
C_{A}(\mathcal{W}_{K}\cap\mathcal{F}^{K}_{y})=C_{A}(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y}\times\mathbb{C}^{\frac{n(n+1)}{2}})=C_{A}(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y})\times\mathbb{C}^{\frac{n(n+1)}{2}}.$

This finishes the proof. ∎

######  Lemma 3.39.

Let $K\geq 3$ and $A_{K}\subset\mathcal{Q}_{K}$ be the finite collection
provided by the  (Spanning theorem).such that the tangent bundle
$T(\mathcal{F}^{K}_{y}\cap\mathcal{W}_{K})$ of every generic fiber
$\mathcal{F}^{K}_{y}$ is spanned by $A_{K}$. Then the restricted submersion

$\Phi_{K}:C_{A_{K}}(\mathcal{W}_{K})|_{Y_{ng}^{K}}\to Y_{ng}^{K}$

over the non-generic stratum
$Y_{ng}^{K}\subset\mathbb{C}^{2n}\setminus\\{0\\}$ admits a spray.

###### Proof.

Let $A_{K-1}$ and $A_{K}$ be the collections from the  (Spanning theorem). and
let $\mathcal{F}^{K}_{y}$ be a non-generic fiber. There holds
$\mathcal{F}^{K}_{y}=\mathcal{F}^{K-1}_{y}\times\mathbb{C}^{\frac{n(n+1)}{2}}$
by Lemma 3.14, where $\mathcal{F}^{K-1}_{y}$ is a generic $(K-1)$-fiber.
According to Lemma 3.14, the vector fields from $A_{K}\setminus A_{K-1}$
vanish over $\mathcal{F}^{K}_{y}$ and we get

$C_{A_{K}}(\mathcal{W}_{K}\cap\mathcal{F}^{K}_{y})=C_{A_{K-1}}(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y})\times\mathbb{C}^{\frac{n(n+1)}{2}}.$

Collection $A_{K-1}$ spans the tangent bundle
$T(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y})$ for every generic fiber
$\mathcal{F}^{K-1}_{y}$, by the  (Spanning theorem).. By Theorem 3.31, there
is a finite collection $B\subset\Gamma(A_{K-1})$ spanning the tangent bundle
$T(C_{A_{K-1}}(\mathcal{W}_{K-1}\cap\mathcal{F}^{K-1}_{y}))$ for every generic
fiber $\mathcal{F}^{K-1}_{y}$. We add the vector fields

$\left(\frac{\partial}{\partial z_{K,ij}}\right)_{1\leq i\leq j\leq n}$

which span $\mathbb{C}^{\frac{n(n+1)}{2}}$ to the collection $B$. This new
collection, let’s call it $\tilde{B}$, spans the tangent bundle
${T(C_{A_{K}}(\mathcal{W}_{K}\cap\mathcal{F}^{K}_{y}))}$ for every non-generic
fiber $\mathcal{F}^{K}_{y}$. Similar as in Theorem 3.31, we can use the vector
flows $\alpha_{t}^{V}$, $V\in\tilde{B}$, to construct a dominating spray
associated to the submersion
$\Phi_{K}:C_{A_{K}}(\mathcal{W}_{K})|_{Y_{ng}^{K}}\to Y_{ng}^{K}$. ∎

###  3.7 The Spanning theorem

In this subsection we prove the  (Spanning theorem)., which we will do by
induction on the number of factors $K$. As it turns out, for various reasons,
it requires several base steps before we get to the actual induction step. It
makes sense to explain the proof strategy continuously. So let’s first
introduce or recall some notations and then we start with the first base step,
$K=2$.

We write $\Phi_{K}=(P_{1}^{K},...,P_{2n}^{K})^{T}$ as well as
${P_{f}^{K}=(P_{1}^{K},...,P_{n}^{K})^{T}}$ and
${P_{s}^{K}=(P_{n+1}^{K},...,P_{2n}^{K})^{T}}$. Similarly, we write
${Q_{f}^{K}=P_{1}^{K}\cdots P_{n}^{K}}$ and ${Q_{s}^{K}=P_{n+1}^{K}\cdots
P_{2n}^{K}}$. For $y\in\mathbb{C}^{2n}\setminus\\{0\\}$ we sometimes write
$y=(a,b)$ with $a=(a_{1},...,a_{n})$ and $b=(b_{1},...,b_{n})$.

######  Lemma 3.40 (Spanning theorem for $K=2$).

Let $\mathcal{F}^{2}_{y}$ be a generic fiber, i.e.
$\pi_{2}(y)_{j^{*}}=P^{1}_{j^{*}}\neq 0$ for some $1\leq j^{*}\leq n$. Then
the collection

$A_{2}=\left\\{(P^{1}_{j^{*}})^{2}\frac{\partial}{\partial{{z}_{1,ij}}}:1\leq
i\leq j<n\right\\}\cup\left\\{\gamma^{2}_{ij,j^{*}}:{1\leq i\leq j\leq
n},{i\neq j^{*}},{j\neq j^{*}}\right\\}$

consist of complete holomorphic vector fields which are fiber-preserving for
$\Phi_{2}$. Moreover, $A_{2}$ spans the tangent bundle $T\mathcal{F}^{2}_{y}$.

###### Proof.

According to Lemma 3.15, we are able to express the variables
$z_{2,1j^{*}},...,z_{2,nj^{*}}$. This gives us a meromorphic mapping
$\psi_{j^{*}}$ which maps $\mathcal{G}_{\pi_{2}(y)}\cong\mathbb{C}^{n(n-1)}$
biholomorphic to $\mathcal{F}^{2}_{y}$. In particular, the vector fields
$\frac{\partial}{\partial{{z}_{1,ij}}},1\leq i\leq j<n$ and
$\frac{\partial}{\partial{{z}_{2,ij}}},{1\leq i\leq j\leq n},{i\neq
j^{*}},{j\neq j^{*}}$ are complete holomorphic and tangential to
$\mathcal{G}_{\pi_{2}(y)}$. Moreover, they span the tangent bundle
$T\mathcal{G}_{\pi_{2}(y)}$. The collection $A_{2}$ is obtained by computing
the push-forwards with respect to the mapping $\psi_{j^{*}}$. ∎

####  3.7.1 Preparation and explanation of the induction step

Recall that each $K$-fiber $\mathcal{F}^{K}_{y}$ can be written as a union of
$(K-1)$-fibers, that is,

$\mathcal{F}^{K}_{y}=\bigcup_{Z\in\mathbb{C}^{\frac{n(n+1)}{2}}}\mathcal{F}^{K-1}_{M_{K}(Z)y}.$

Equivalently, ${\vec{Z}_{K}=(\vec{Z}_{K-1},Z_{K})\in\mathcal{F}^{K}_{y}}$ if
and only if $\vec{Z}_{K-1}\in\mathcal{F}^{K-1}_{M_{K}(-Z_{K})y}$. The
directions
$\left(\frac{\partial}{\partial{{z}_{K,ij}}}\right)_{1\leq i\leq j\leq n}$ are
somehow transversal to the ’fibration’ and we use to speak of the new
directions. The following statement is very useful for the induction step.

######  Lemma 3.41.

For $K=3$, there are finitely many vector fields from $\mathcal{V}_{3}$
spanning the new directions in a generic fiber $\mathcal{F}^{3}_{y}$ in points
with

$Q_{f}^{1}\neq 0.$

For $K\geq 4$, there are finitely many vector fields from $\mathcal{V}_{K}$
spanning the new directions in a generic fiber $\mathcal{F}^{K}_{y}$ in points
with

$Q_{f}^{K-2}\neq 0\quad\text{and}\quad Q_{s}^{K-2}\neq 0.$

Before we prove this lemma, let’s apply it. We therefore define the sets

$\mathcal{U}_{3}:=\\{\vec{Z}_{3}\in\mathcal{W}_{3}:Q_{f}^{1}(\vec{Z}_{3})\neq
0\\}$

and for $K\geq 4$,

$\mathcal{U}_{K}:=\\{\vec{Z}_{K}\in\mathcal{W}_{K}:Q_{f}^{K-2}(\vec{Z}_{K})Q_{s}^{K-2}(\vec{Z}_{K})\neq
0\\}.$

What now follows is one argument of the induction step.

######  Lemma 3.42.

Let $K\geq 3$ and suppose that the  (Spanning theorem). is true for $K-1$.
Then there is a finite collection $A\subset\mathcal{Q}_{K}$ spanning the
tangent bundle $T(\mathcal{F}^{K}_{y}\cap\mathcal{U}_{K})$ for every generic
fiber $\mathcal{F}^{K}_{y}$.

###### Proof.

In a first step, assume $K=3$ and consider a point
$\vec{Z}_{3}\in\mathcal{U}_{3}$ such that $y:=\Phi_{3}(\vec{Z}_{3})\in
Y^{3}_{g}$ is in the generic stratum, i.e. $b=P_{s}^{3}(\vec{Z}_{3})\neq 0$.
By definition of the set $\mathcal{U}_{3}$, there holds
$Q_{f}^{1}(\vec{Z}_{3})\neq 0$. According to Lemma 3.41, there is a finite
collection $A\subset\mathcal{Q}_{3}$ which spans the new directions in points
with $Q_{f}^{1}\neq 0$. Hence $\vec{Z}_{3}$ is a point, where the new
directions are spanned.

By the recursive formula (4), there holds
$\Phi_{2}(\vec{Z}_{2})=M_{3}(-Z_{3})y$. Moreover,

$Z_{1}e_{n}=\pi_{2}\circ\Phi_{2}(\vec{Z}_{2})=\pi_{2}\circ
M_{3}(-Z_{3})y=a-Z_{3}b$

implies that $\vec{Z}_{2}\in\mathcal{F}^{2}_{M_{3}(-Z_{3})y}$ is contained a
generic $2$-fiber, since $\vec{Z}_{3}=(\vec{Z}_{2},Z_{3})\in\mathcal{U}_{3}$.
Therefore the tangent space
${T_{\vec{Z}_{3}}\mathcal{F}^{2}_{M_{3}(-Z_{3})y}\subset
T_{\vec{Z}_{3}}\mathcal{F}^{3}_{y}}$ is spanned by
$A_{2}\subset\mathcal{Q}_{2}$ according to Lemma 3.40. Since the new
directions are complementary, the collection $A\cup A_{2}$ spans the tangent
space $T_{\vec{Z}_{3}}\mathcal{F}^{3}_{y}$. This holds for every generic fiber
$\mathcal{F}^{3}_{y}$, hence there is a finite collection in $\mathcal{Q}_{3}$
which spans the tangent bundle $T(\mathcal{F}^{3}_{y}\cap\mathcal{U}_{3})$ for
every generic fiber.

In the case $K\geq 4$ we’ll argue similarly. Let
$\vec{Z}_{K}\in\mathcal{U}_{K}$ be a point such that
$y:=\Phi_{K}(\vec{Z}_{K})\in Y_{g}^{K}$ is in the generic stratum, i.e.
$\pi_{K}(y)\neq 0$. Write $\vec{Z}_{K}=(\vec{Z}_{K-1},Z_{K})$ and
$\vec{Z}_{K-1}=(\vec{Z}_{K-2},Z_{K-1})$. Note that

$\Phi_{K-1}(\vec{Z}_{K-1})=M_{K}(-Z_{K})y\quad\text{ and
}\quad\Phi_{K-1}(\vec{Z}_{K-1})=M_{K-1}(Z_{K-1})\Phi_{K-2}(\vec{Z}_{K-2})$

by the recursive formula (4). By defintion of $\pi_{K}$ and $M_{K}$, there
holds $\pi_{K}\circ M_{K}=\pi_{K}$, hence
$\pi_{K-1}\circ\Phi_{K-1}=\pi_{K-1}\circ\Phi_{K-2}$. Moreover, by definition
of $\mathcal{U}_{K}$, we have $Q_{f}^{K-2}Q_{s}^{K-2}(\vec{Z}_{K})\neq 0$,
which implies $\pi_{K-1}\circ\Phi_{K-1}(\vec{Z}_{K-1})\neq 0$, that is,
$\vec{Z}_{K-1}$ is contained in the generic fiber
$\mathcal{F}^{K-1}_{M_{K}(-Z_{K})y}$. In addition, we have
$\vec{Z}_{K-1}\in\mathcal{W}_{K-1}$, since otherwise this would contradict
$Q_{f}^{K-2}Q_{s}^{K-2}\neq 0$ by Lemma 3.9. Therefore the tangent space
$T_{\vec{Z}_{K-1}}\mathcal{F}^{K-1}_{M_{K}(-Z_{K})y}$ is spanned by the finite
collection in $\mathcal{Q}_{K-1}$ provided from the  (Spanning theorem). for
$K-1$. Similarly as in the previous case, $\vec{Z}_{K}$ is a point where the
new directions are spanned, according to Lemma 3.41. Again, the new directions
are complementary and we’ve shown that there is a finite collection in
$\mathcal{Q}_{K}$ spanning the tangent space
$T_{\vec{Z}_{K}}\mathcal{F}^{K}_{y}$ for each generic fiber. And this proves
the lemma. ∎

######  Remark 3.43.

An application of Corollary 3.32 and the former lemma leads us to the
following observation. To complete the induction step, it suffices to find a
finite collection $A\subset\mathcal{Q}_{K}$ satisfying

$\displaystyle\mathcal{F}^{K}_{y}\cap\mathcal{W}_{K}\subset
C_{A}(\mathcal{F}^{K}_{y}\cap\mathcal{U}_{K}),\quad\text{for each generic
fiber }\mathcal{F}^{K}_{y}.$ (11)

The idea of the proof is to stratify $\mathcal{W}_{K}$ suitably, i.e. find a
finite descending chain of subspaces

$\mathcal{W}_{K}=:X_{N}\supset...\supset X_{0}=\emptyset,$

where the spaces $X_{0},...,X_{N-1}$ are closed. In a first step, we’ll find a
finite collection $A_{N}\subset\mathcal{Q}_{K}$ such that the stratum
$S_{N}:=X_{N}\setminus X_{N-1}$ satisfies

$\mathcal{F}^{K}_{y}\cap S_{N}\subset
C_{A_{N}}(\mathcal{F}^{K}_{y}\cap\mathcal{U}_{K})$

for each generic fiber $\mathcal{F}^{K}_{y}$. Then, we find a finite
collection $A_{k}\subset\mathcal{Q}_{K}$, for each stratum
$S_{k}:=X_{k}\setminus X_{k-1}$, $1\leq k\leq N-1$, such that

$\mathcal{F}^{K}_{y}\cap S_{k}\subset C_{A_{k}}(\mathcal{F}^{K}_{y}\cap
S_{k+1})$

for each generic fiber $\mathcal{F}^{K}_{y}$. Define the finite collection
$A:=\bigcup_{k=1}^{N}A_{k}\subset\mathcal{Q}_{K}.$ Then we get

$\mathcal{F}^{K}_{y}\cap S_{1}\subset C_{A}(\mathcal{F}^{K}_{y}\cap
S_{2})\subset\cdots\subset C_{A}(\mathcal{F}^{K}_{y}\cap S_{N})\subset
C_{A}(\mathcal{F}^{K}_{y}\cap\mathcal{U}_{K})$

for each generic fiber $\mathcal{F}^{K}_{y}$. As it turns out, the
stratification of $\mathcal{W}_{K}$ in the cases $K=3$, $K=4$ and $K\geq 5$
differ in some elementary ways.

###### Proof of Lemma 3.41.

Let $\mathcal{F}^{K}_{y}$ be a generic fiber, that is, there holds
$\tilde{P}^{K}_{j^{*}}\neq 0$ for some $1\leq j^{*}\leq n$. Recall that this
fiber is biholomorphic to
$\mathcal{G}_{\pi_{K}(y)}\times\mathbb{C}^{\frac{n(n-1)}{2}}$ with the
biholomorphism obtained by expressing the $n$ variables
$z_{K,j^{*}1},...,z_{K,j^{*}n}$ in terms of the remaining variables
$\left(z_{K,ij}\right)_{\begin{subarray}{c}1\leq i\leq j\leq n;i,j\neq
j^{*}\end{subarray}}$. Clearly, the fields
$\left(\frac{\partial}{\partial{{z}_{K,ij}}}\right)_{\begin{subarray}{c}1\leq
i\leq j\leq n;i,j\neq j^{*}\end{subarray}}$ are tangential to
$\mathcal{G}_{\pi_{K}(y)}\times\mathbb{C}^{\frac{n(n-1)}{2}}$ and moreover,
they span $\\{0\\}\times\mathbb{C}^{\frac{n(n-1)}{2}}$. Hence the
corresponding lifts span the new directions
$\left(\frac{\partial}{\partial{{z}_{K,ij}}}\right)_{\begin{subarray}{c}1\leq
i\leq j\leq n;i,j\neq j^{*}\end{subarray}}$ in $\mathcal{F}^{K}_{y}$. It
remains to find $n$ vector fields
$\partial_{x_{1}}^{K-1},...,\partial_{x_{n}}^{K-1}$ tangent to
$\mathcal{G}_{\pi_{K}(y)}\times\\{0\\}$ such that the projection of the
corresponding lifts $\varphi_{x_{1},j^{*}}^{K},...,\varphi_{x_{n},j^{*}}^{K}$
to the new directions
$\left(\frac{\partial}{\partial{{z}_{K,j^{*}j}}}\right)_{1\leq j\leq n}$ are
linearly independent. Let $\alpha_{j^{*},1},...,\alpha_{j^{*},n}$ denote the
component vectors of such projections in the frame
$\left(\frac{\partial}{\partial{{z}_{K,j^{*}j}}}\right)_{1\leq j\leq n}$. We
need to show that the matrix $A:=(\alpha_{j^{*},1}|\cdots|\alpha_{j^{*},n})$
is regular. Set
$u_{m}:=(\partial_{x_{m}}^{K-1}(\tilde{P}^{K-2}_{1}),...,\partial_{x_{m}}^{K-1}(\tilde{P}^{K-2}_{n}))^{T}$
and recall from the proof of Lemma 3.18, that there is a regular matrix
$B:=F_{j^{*}}(\tilde{P}^{K})^{-1}$ with $\alpha_{j^{*},m}=Bu_{m}$. Therefore
$A$ is regular if and only if $U:=(u_{1}|\cdots|u_{n})$ is regular.

In a first step, we assume $K\geq 4$ and consider the tupels of (Type 1)

$x_{m}:=(z_{K-2,mm},z_{K-1,11},...,z_{K-1,nn}),1\leq m\leq n.$

Without loss of generality, let $K=2k+1$. The entries of $U$ are given by
$u_{ij}:=\partial_{x_{i}}^{2k}(P_{j}^{2k})$. Let’s compute the following
derivatives

$\frac{\partial}{\partial{{z}_{K-2,ii}}}P_{s}^{2k}=\begin{pmatrix}Z_{K-1}&I_{n}\end{pmatrix}\begin{pmatrix}0&\tilde{E}_{ii}\\\
0&0\end{pmatrix}\begin{pmatrix}P_{f}^{2k-2}\\\
P_{s}^{2k-2}\end{pmatrix}=P_{n+i}^{2k-2}Z_{K-1}e_{i},$

and

$\frac{\partial}{\partial{{z}_{K-1,mm}}}P_{s}^{2k}=\begin{pmatrix}\tilde{E}_{mm}&0\end{pmatrix}\begin{pmatrix}P_{f}^{2k-1}\\\
P_{s}^{2k-1}\end{pmatrix}=P_{m}^{2k-1}e_{m}.$

Furthermore, we have $\frac{\partial}{\partial{{z}_{K-1,mm}}}P_{f}^{2k}\equiv
0$ and

$\frac{\partial}{\partial{{z}_{K-2,ii}}}P_{f}^{2k}=\tilde{E}_{ii}P_{s}^{2k-1}=P_{n+i}^{2k-1}e_{i}.$

We obtain

$\displaystyle u_{ij}=\partial_{x_{i}}^{2k}(P_{j}^{2k})$
$\displaystyle=\det\begin{pmatrix}\delta_{ij}P_{n+i}^{2k-1}&0&\cdots&0\\\
P_{n+i}^{2k-1}Z_{K-1}e_{i}&P_{1}^{2k-1}e_{1}&\cdots&P_{n}^{2k-1}e_{n}\end{pmatrix}=\delta_{ij}P_{n+i}^{2k-1}Q_{f}^{2k-1},$

and hence the matrix

$U=\begin{pmatrix}P_{n+1}^{2k-1}Q_{f}^{2k-1}&&\\\ &\ddots&\\\
&&P_{2n}^{2k-1}Q_{f}^{2k-1}\end{pmatrix}$

is regular, provided $Q_{f}^{2k-1}Q_{s}^{2k-1}\neq 0$.

In a second step, let $K=3$ and consider the tupels of (Type 4)

$x_{m}:=(z_{1,n1},z_{2,11},...,z_{2,nn}),1\leq m\leq n.$

We obtain derivatives
$\frac{\partial}{\partial{{z}_{1,ni}}}P_{s}^{2}=Z_{2}e_{i},\frac{\partial}{\partial{{z}_{2,ii}}}P_{s}^{2}=z_{1,ni}e_{i}$
and
$\frac{\partial}{\partial{{z}_{1,ni}}}P_{f}^{2}=e_{i},\frac{\partial}{\partial{{z}_{2,ii}}}P_{f}^{2}\equiv
0$. Hence we get

$u_{ij}=\det\begin{pmatrix}\delta_{ij}&0&\cdots&0\\\
Z_{2}e_{i}&z_{1,n1}e_{1}&\cdots&z_{1,nn}e_{n}\end{pmatrix}=\delta_{ij}Q_{f}^{1}$

and $U=Q_{f}^{1}I_{n}$. This completes the proof. ∎

####  3.7.2 Spanning theorem for $K=3$

We want to find a finite collection $A\subset\mathcal{Q}_{3}$ such that (11)
is fulfilled for $K=3$.

Let $1\leq i,j\leq n$, $i\neq j$, and consider the tupels of (Type 8)

${x_{i,j}:=(z_{1,ni},z_{2,j1},...,z_{2,jn}).}$

The corresponding vector fields $\partial_{x_{i,j}}^{2}$ are of the form

$\partial^{2}_{x_{i,j}}=\det\begin{bmatrix}\partial/\partial
z_{1,ni}&\partial/\partial z_{2,j1}&\cdots&\partial/\partial
z_{2,jj}&\cdots&\partial/\partial z_{2,jn}\\\ z_{2,1i}&z_{1,nj}&&0&&0\\\
\vdots&&\ddots&&\\\ z_{2,ji}&z_{1,n1}&\cdots&z_{1,nj}&\cdots&z_{1,nn}\\\
\vdots&&&&\ddots&&\\\ z_{2,ni}&0&&0&&z_{1,nj}\end{bmatrix}$

and they satisfy

$\partial_{x_{i,j}}^{2}(z_{1,kn})=\delta_{ik}(z_{1,nj})^{n},$

for $1\leq i,j,k\leq n$.

By definition, points in the variety
$\mathcal{N}:=\mathcal{W}_{3}\setminus\mathcal{U}_{3}$ satisfy
$Q_{f}^{1}=z_{1,1n}\cdots z_{1,nn}=0$ and $P_{f}^{1}=Z_{1}e_{n}\neq 0$. Hence
there are indices $1\leq i_{1}<...<i_{k}\leq n$ and $1\leq
j_{1}<...<j_{n-k}\leq n$ satisfying

  1. (i)

$\\{1,...,n\\}=\\{i_{1},...,i_{k}\\}\dot{\cup}\\{j_{1},...,j_{n-k}\\}$,

  2. (ii)

$z_{1,in}=0$, for all $i\in\\{i_{1},...,i_{k}\\}$,

  3. (iii)

$z_{1,jn}\neq 0$, for all $j\in\\{j_{1},...,j_{n-k}\\}$.

Fix an index $j\in\\{j_{1},...,j_{n-k}\\}$. Then observe

$\partial_{x_{i_{1},j}}^{2}\circ\cdots\circ\partial_{x_{i_{k},j}}^{2}(z_{1,i_{1}n}\cdots
z_{1,i_{k}n})=(z_{1,jn})^{n}\partial_{x_{i_{1},j}}^{2}\circ\cdots\circ\partial_{x_{i_{k-1},j}}^{2}(z_{1,i_{1}n}\cdots
z_{1,i_{k-1}n}),$

which inductively implies

$\partial_{x_{i_{1},j}}^{2}\circ\cdots\circ\partial_{x_{i_{k},j}}^{2}(z_{1,i_{1}n}\cdots
z_{1,i_{k}n})=(z_{1,jn})^{kn}\neq 0.$

Let $\vec{Z}_{3}\in\mathcal{N}$ be a point over the generic stratum, that is,
$\Phi_{3}(\vec{Z}_{3})\in Y_{g}^{3}$. According to Lemma 3.34, we find
finitely many of the above fields $\partial_{x_{i,j}}^{2}$ such that a
suitable finite composition of the respective flows moves $\vec{Z}_{3}$ away
from $\mathcal{N}$. More precisely, we have a finite collection
$A\subset\mathcal{Q}_{3}$ and an automorphism $\alpha\in G_{A}$ such that
$\alpha(\vec{Z}_{3})\in C_{A}(\mathcal{U}_{3})$. That’s exactly what we need
to get (11).

####  3.7.3 The Spanning theorem for $K=4$

As in the previous step, we want to find a suitable finite collection $A$ such
that $(\ref{equation:inductionStep})$ is satisfied. In order to do this, we’ll
do a divide and conquer with the variety
$\mathcal{N}_{4}:=\mathcal{W}_{4}\setminus\mathcal{U}_{4}.$

######  Lemma 3.44.

Define the set

$\mathcal{N}_{1}:=\\{\vec{Z}_{4}\in\mathcal{W}_{4}:Q_{f}^{2}(\vec{Z}_{4})Q_{s}^{2}(\vec{Z}_{4})=0,P_{s}^{2}(\vec{Z}_{4})\neq
0\\}.$

Then there is a finite collection $A\subset\mathcal{Q}_{4}$ such that

$\mathcal{F}^{4}_{y}\cap\mathcal{N}_{1}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\mathcal{U}_{4})$

for each generic fiber $\mathcal{F}^{4}_{y}$.

###### Proof.

In a first step, recall that $P_{f}^{2}=P_{f}^{1}$ by the recursive formula
(4). Consider a point $\vec{Z}_{4}$ in the set

$X_{1}=\\{\vec{Z}_{4}\in\mathcal{N}_{1}:Q_{f}^{1}\neq 0\\}.$

By definition of this set, we find indices $1\leq i^{*},i\leq n$ with
$P_{n+i^{*}}^{2}(\vec{Z}_{4})\neq 0$ and $P_{n+i}^{2}(\vec{Z}_{4})=0$. Pick
the vector field $\partial_{x_{i}}^{3}$ corresponding to the tupel
$x_{i}=(z_{2,ii},z_{3,i^{*}1},...,z_{3,i^{*}n})$ of (Type 2). It is of the
form

$\pm(P_{n+i^{*}}^{2})^{n}\frac{\partial}{\partial{{z}_{2,ii}}}+\sum_{l=1}^{n}\alpha_{l}\frac{\partial}{\partial{{z}_{3,i^{*}l}}}$

for some suitable holomorphic functions $\alpha_{1},...,\alpha_{n}$. Then

$\partial_{x_{i}}^{3}(P_{n+i}^{2})=\partial_{x_{i}}^{3}(\delta_{in}+e_{i}^{T}Z_{2}Z_{1}e_{n})=z_{1,in}(P_{n+i^{*}}^{2})^{n}\neq
0,$

and this applies for every $1\leq i\leq n$ with $P_{n+i}^{2}(\vec{Z}_{4})=0$.
Hence there is a finite collection $A\subset\mathcal{Q}_{4}$ such that

$\mathcal{F}^{4}_{y}\cap X_{1}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\mathcal{U}_{4})$

for each generic fiber $\mathcal{F}^{4}_{y}$.

In a second step, let’s consider a point $\vec{Z}_{4}=(\vec{Z}_{3},Z_{4})$ in
the set

$X_{2}=\\{\vec{Z}_{4}:Q_{f}^{1}=0,P_{f}^{1}\neq 0,P_{s}^{2}\neq
0\\}\subset\mathcal{N}_{1}.$

Observe that the projection of $\vec{Z}_{4}$ to the first component
$\vec{Z}_{3}$ is contained in a generic $3$-fiber
$\mathcal{F}^{3}_{\tilde{y}}$ and in $\mathcal{W}_{3}$. From the case $K=3$,
we know the existence of a finite collection $B\subset\mathcal{Q}_{3}$ such
that $\vec{Z}_{3}\in C_{B}(\mathcal{F}^{3}_{\tilde{y}}\cap\mathcal{U}_{3})$.
Put the corresponding pullbacks into the collection $A\subset\mathcal{Q}_{4}$.
Then we’ve found a finite collection $A$ such that

$\mathcal{F}^{4}_{y}\cap X_{2}\subset C_{A}(\mathcal{F}_{y}^{4}\cap
X_{1})\subset C_{A}(\mathcal{F}^{4}_{y}\cap\mathcal{U}_{4})$

for each generic fiber $\mathcal{F}^{4}_{y}$.

In a third step, let’s consider a point $\vec{Z}_{4}$ in the set

$X_{3}=\\{\vec{Z}_{4}:P_{f}^{1}=0\\}\subset\mathcal{N}_{1},$

that is, $Z_{1}e_{n}=0$. The vector field $\partial_{x}^{3}$ corresponding to
the tupel $x=(z_{1,1n},...,z_{1,nn},z_{3,nn})$ of (Type 7), satisfies

$\partial_{x}^{3}(z_{1,nn})=\det\left(e_{1}+Z_{3}Z_{2}e_{1},...,e_{n-1}+Z_{3}Z_{2}e_{n-1},e_{n}\right)$

in $\vec{Z}_{4}$. Hence there is a finite collection $A\subset\mathcal{Q}_{4}$
such that

$\mathcal{F}^{4}_{y}\cap\underbrace{\\{\vec{Z}_{4}:P_{f}^{1}(\vec{Z}_{4})=0,\partial_{x}^{3}(z_{1,nn})\neq
0\\}}_{=:\tilde{N}}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\\{\vec{Z}_{4}:P_{f}^{1}\neq 0,P_{s}^{2}\neq
0\\})$

for each generic fiber $\mathcal{F}^{4}_{y}$.

Next, we consider the vector fields $\partial_{x_{i}}^{3}$ and
$\partial_{y_{i}}^{3}$, $i=1,...,n-1$, corresponding to the tupels
${x_{i}=(z_{2,ii},z_{3,1n},...,z_{3,nn})}$ and
${y_{i}=(z_{3,ii},z_{3,1n},...,z_{nn})}$ of (Type 2) and (Type 3),
respectively. They are of the form
$\partial_{x_{i}}^{3}=\frac{\partial}{\partial{{z}_{2,ii}}}$ and
$\partial_{y_{i}}^{3}=\frac{\partial}{\partial{{z}_{3,ii}}}$ on
$\mathcal{N}^{4}(P_{f}^{2})$. The formula

$\displaystyle\frac{d}{dt}\det(A_{1}(t),...,A_{n}(t))=\sum_{l=1}^{n}\det(A_{1}(t),...,A_{l}^{\prime}(t),...,A_{n}(t))$
(12)

is sort of a product rule, where $A_{1}(t),...,A_{n}(t)$ denote columns of a
$n\times n$-matrix $A(t)$ depending on $t$. In our case, we consider a matrix
$A$, where the first $(n-1)$ columns are given by

$A_{i}=e_{i}+\sum_{l=1}^{n}z_{2,li}Z_{3}e_{l},\quad 1\leq i\leq n-1,$

and the $n$-th column $A_{n}=e_{n}$. Observe that

$\frac{\partial}{\partial{{z}_{2,ii}}}A_{j}=\frac{\partial}{\partial{{z}_{2,ii}}}(e_{j}+\sum_{l=1}^{n}z_{2,lj}Z_{3}e_{l})=\delta_{ij}Z_{3}e_{i}.$

Hence

$\frac{\partial}{\partial{{z}_{2,11}}}\cdots\frac{\partial}{\partial{{z}_{2,(n-1)(n-1)}}}\det(A)=\det(Z_{3}e_{1},...,Z_{3}e_{n-1},e_{n}).$

Furthermore, we obtain

$\frac{\partial}{\partial{{z}_{3,11}}}\cdots\frac{\partial}{\partial{{z}_{3,(n-1)(n-1)}}}\det(Z_{3}e_{1},...,Z_{3}e_{n-1},e_{n})=\det(e_{1},...,e_{n})=1.$

By Lemma 3.34 and the previous step, we conclude the existence of a finite
collection $A\subset\mathcal{Q}_{4}$ such that

$\mathcal{F}^{4}_{y}\cap\\{\vec{Z}_{4}\in\mathcal{N}_{1}:P_{f}^{1}(\vec{Z}_{4})=0,\partial_{x}^{3}(z_{1,nn})=0\\}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\tilde{N})$

and

$\mathcal{F}^{4}_{y}\cap\tilde{N}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\\{\vec{Z}_{4}:Q_{f}^{1}\neq 0,P_{s}^{2}\neq
0\\})$

for each generic fiber $\mathcal{F}^{4}_{y}$. In particular, this implies

$\mathcal{F}^{4}_{y}\cap X_{3}\subset C_{A}(\mathcal{F}^{4}_{y}\cap
X_{2})\subset C_{A}(\mathcal{F}^{4}_{y}\cap\mathcal{U}_{4})$

for each generic fiber $\mathcal{F}^{4}_{y}$.

For the final step, observe that $\mathcal{N}_{1}=X_{1}\cup X_{2}\cup X_{3}$.
We put all of the involved vector fields above together and obtain a finite
collection $A\subset\mathcal{Q}_{4}$ satisfying

$\mathcal{F}^{4}_{y}\cap\mathcal{N}_{1}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\mathcal{U}_{4})$

for each generic fiber $\mathcal{F}^{4}_{y}$. This finishes the proof. ∎

######  Lemma 3.45.

Define the set

$\mathcal{N}_{2}:=\\{\vec{Z}_{4}:P_{s}^{2}(\vec{Z}_{4})=0\\}.$

Then there is a finite collection $A\subset\mathcal{Q}_{4}$ such that

$\mathcal{F}^{4}_{y}\cap\mathcal{N}_{2}\subset
C_{A}(\mathcal{F}^{4}_{y}\cap\mathcal{U}_{4})$

for each generic fiber $\mathcal{F}^{4}_{y}$.

In order to prove this lemma, the symplectic nature of the elementary matrices
$M_{i}(Z)$ comes into play. The following result, also called complementary
bases theorem, is proved by Dopico and Johnson [DJ06].

######  Theorem 3.46 (Complementary bases theorem).

Let

$M=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\in\SpC$

be a symplectic $2n\times 2n$-matrix and let $k:=\mathrm{rank}(B)$, i.e. there
are $k$ indices $j_{1},...,j_{k}$ such that the vectors
$Be_{j_{1}},...,Be_{j_{k}}$ form a basis of the image $\mathrm{Im}(B)$. Let
$i_{1},...,i_{n-k}$ denote the complementary indices in $\\{1,...,n\\}$, that
is, $\\{1,...,n\\}=\\{i_{1},...,i_{n-k}\\}\dot{\cup}\\{j_{1},...,j_{k}\\}.$

Then the $n\times n$ matrix

$X=\begin{pmatrix}Ae_{i_{1}}&\cdots&Ae_{i_{n-k}}&Be_{j_{1}}&\cdots&Be_{j_{k}}\end{pmatrix}$

is regular.

######  Example 3.47.

Consider the symplectic elementary matrix

$\begin{pmatrix}I_{n}&Z\\\ 0&I_{n}\end{pmatrix},$

and assume the symmetric matrix $Z\in\mathbb{C}^{\frac{n(n+1)}{2}}$ to be of
rank $k$. We find complementary indices as in the theorem and obtain that the
matrix

$X=\begin{pmatrix}e_{i_{1}}&\cdots&e_{i_{n-k}}&Ze_{j_{1}}&\cdots&Ze_{j_{k}}\end{pmatrix}$

is regular. By applying some Gauss-elimination to $X$, we conclude that $Z$
has a non-vanishing principal minor of order $k$. More precisely, let
$(Z)_{i_{1},...,i_{n-k}}$ be the symmetric $k\times k$-matrix, obtained by
removing columns and rows $i_{1},...,i_{n-k}$ from $Z$. Then
$\det((Z)_{i_{1},...,i_{n-k}})=\pm\det(X)\neq 0$.

###### Proof of Lemma 3.45.

Let’s stratify the variety $\mathcal{N}_{2}$ in the following way. For $0\leq
k\leq n$, define

$\mathcal{A}_{k}:=\\{\vec{Z}_{4}\in\mathcal{N}_{2}:\mathrm{rank}(Z_{3})\leq
k\\}$

and as a convention let $\mathcal{A}_{-1}:=\emptyset$. Then each stratum

$\mathcal{B}_{k}:=\mathcal{A}_{k}\setminus\mathcal{A}_{k-1},\quad 0\leq k\leq
n,$

consists of those points $\vec{Z}_{4}\in\mathcal{N}_{2}$ with
$\mathrm{rank}(Z_{3})=k$.