Holomorphic Factorization of Mappings into $ \operatorname{Sp}_{4}( \mathbb{C}) $

We prove that any null-homotopic holomorphic map from a Stein space $X$ to the symplectic group $\operatorname{Sp}_{4}(\mathbb{C})$ can be written as a finite product of elementary symplectic matrices with holomorphic entries.


Introduction
The continuous or holomorphic parameter dependence of classical linear algebra results over the fields R or C form a circle of very natural questions of general mathematical interest. For example the factorization of continuous matrices as a product of continuous elementary matrices has been studied and solved by Vaserstein [Vas88]. The corresponding holomorphic problem for the special linear group SL n has been posed by Gromov [Gro89] and finally be solved by the first two authors in [IK12]. The study of algebraic dependence is connected with famous work by Suslin [Sus77], Cohn [Co66], Bass, Milnor, Serre [BMS67] and many others.
These parameter dependence questions are a part of algebraic K-theory and the study of linear algebra over general rings. Factorization of Chevalley groups over R and C into elementary matrices is classically well known. For Chevalley groups over general rings this is much more difficult and studied a lot. For an overview see for example the paper by Vavilov and Stepanov [VS13].
Our main interest are the rings of holomorphic functions on Stein spaces. The only known holomorphic result is the existence for the special linear groups in [IK12], where Gromov's problem is solved in full generality. In the special case of an open Riemann surface the problem was solved earlier (absolutely unnoticed) by Klein and Ramspott in [KR88]. As well the authors proved the main result of this paper for any size of symplectic matrices in the special case of of an open Riemann surface in [IKL19].
In the present paper we consider the symplectic groups over rings of holomorphic function on Stein spaces. The main result is (see Section 2 for notation) Main Theorem (also Theorem 3.1). Let X be a finite dimensional reduced Stein space and f : X → Sp 4 (C) be a holomorphic mapping that is null-homotopic. Then there exist a natural number K and holomorphic mappings G 1 , . . . , G K : X → C 3 such that f (x) = M 1 (G 1 (x)) . . . M K (G K (x)).
We remind the reader that a mapping is null-homotopic if it is homotopic to a constant map. By Grauert's Oka principle it is equivalent for a holomorphic map from a Stein space into a complex Lie group to be null-homotopic via holomorphic maps or via continuous maps.
Our main tool is the Oka principle for stratified elliptic submersions, the most elaborate result in modern Oka theory. In order to apply an Oka principle one needs a topological solution which we take from our previous work on symplectic groups over rings of continuous functions on topological spaces. The Oka principle lets us homotope the topological solution to a holomorphic one. The technical details needed to prove that certain fibrations are stratified elliptic are considerable and we have so far only been able to complete these details for Sp 4 . We expect that a similar result holds for Sp 2n .
Factorization of symplectic groups over other rings (of mainly algebraic nature) has been considered before for example by Kopeiko [Kop78], Grunewald, Mennicke and Vaserstein in [GMV91]. We have especially included a section, Section 4, where we explain how our results can be formulated in algebraic/K-theoretic terms.
The paper is organized as follows. In Section 2 we recall our results on factorization of continuous matrices and prove a slight extension about the number of factors. In Section 3 we state our main results and give an overview over the proof. In Section 4 we explain how our results can be reformulated in the language used in algebraic K-theory. In Section 5 we recall the theorems from Oka theory which we use in our proof.
In Section 6 we give the proofs of Lemmata 3.3 and 3.4 where we prove that the most important fibrations in this paper, the projections of products of elementary symplectic matrices onto their last row, are surjective and we determine where they are submersive. This is done for symplectic matrices of all sizes, since we hope to be able in the future to prove that these fibrations are stratified elliptic for all sizes.
The rest of the paper is devoted to prove that our fibration (for (4×4)-matrices) is stratified elliptic in order to be able to apply Oka theory. In Section 7 we describe the stratification with respect to which we want to prove that the important fibration is stratified elliptic. This has to do with how the set of 2n algebraic equations defining a fiber in the fibration can be reduced to n equations. In the case of the Special Linear Group in [IK12] we were able to reduce to one single equation independent of the size of the matrices, which was the crucial trick to prove ellipticity by using Gromov's example of a spray, complete vector fields. This inability to reduce to less equations is the main difference between present situation of the Symplectic Group and the Special Linear Group. It causes all the hard technical work which fills the rest of the paper. In the next Section 8 we introduce our method to find complete vector fields tangent to the fibration. However not all of them are complete and we deduce that the Gromov-spray produced by them is not dominating. We determine which of them are complete. In Section 9 we explain our strategy to enlarge the set of complete vector fields so that this enlarged collection now spans the tangent space at all points and thus gives a fibre dominating spray. The realization of this strategy takes Sections 10, where we introduce useful quantities, Sections 11, 12, 13, where we prove the result for 3, 4 and 5 (elementary symplectic) factors, and finally we can give an inductive (over the number of factors) proof in Section 14. The reason for dealing with the low numbers of factors separately is that some of the fibers of our fibration are reducible in the cases of small numbers of factors, and from 5 factors on all fibers are irreducible. In the last Section 15 we end the paper with an application to the problem of product of exponentials and formulate some open questions.

Continuous factorization
Let ω = n j=1 dz j ∧ dz j+n be the symplectic form in C 2n . With respect to ω symplectic matrices are those that can be written in block form as where I n is the (n × n) identity matrix. In the special case B = C = 0 this means that D = (A T ) −1 and in the special case A = D = I n this means that B and C are symmetric and C T B = 0. Let U n denote a (n × n)-matrix satisfying U n = U T n and 0 n the (n × n) zero matrix. We call those matrices that are written in block form as I n 0 n U n I n or I n U n 0 n I n elementary symplectic matrices. Let Given a map G : X → C n(n+1)/2 let U n (G(x)) = U n (G 1 (x), . . . , G n(n+1)/2 (x)) where the G j 's are components of the map G. For odd k let M k (G(x)) = I n 0 n U n (G(x)) I n and for even k M k (G(x)) = I n U n (G(x)) 0 n I n .
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 : X → Sp 2n (C) there exist K continuous mappings G 1 , . . . , G K : X → C n(n+1)/2 such that M(x) = M 1 (G 1 (x)) . . . M K (G K (x)).
Proof. Theorem 1.3 in [IKL19] does not give a uniform bound on the number of factors depending on n and d. Suppose such a bound would not exist, i.e., for all natural numbers i there are normal topological spaces X i of dimension d and null-homotopic continuous maps f i : X i → Sp 2n (C) such that f i does not factor over a product of less than i elementary symplectic matrices. Set X = ∪ ∞ i=1 X i the disjoint union of the spaces X i and F : X → Sp 2n (C) the map that is equal to f i on X i . By Theorem 1.3. in [IKL19] F factors over a finite number of elementary symplectic matrices. Consequently all f i factor over the same number of elementary symplectic matrices which contradicts the assumption on f i .

Statement of the main result and overview of proof
We state the main result of this paper which is a holomorphic version of Theorem 2.1 for Sp 4 (C).
Theorem 3.1. There exists a natural number N(d) such that given any finite dimensional reduced Stein space X of dimension d and any null-homotopic holomorphic mapping f : X → Sp 4 (C) there exist N holomorphic mappings We have the following corollary.
Corollary 3.2. Let X be a finite dimensional reduced Stein space that is topologically contractible and f : X → Sp 4 (C) be a holomorphic mapping. Then there exist a natural number N and holomorphic mappings The strategy for proving Theorem 3.1 is as follows. Define We want to show the existence of a holomorphic map is commutative. Theorem 2.1 shows the existence of a continuous map such that the diagram above is commutative. We will prove Theorem 3.1 using the Oka-Grauert-Gromov principle for sections of holomorphic submersions over X. One candidate submersion would be to use the pull-back of Ψ K : (C 3 ) K → Sp 4 (C). It turns out that Ψ K is not a submersion at all points in (C 3 ) K . It is a surjective holomorphic submersion if one removes a certain subset from (C 3 ) K . Unfortunately the fibers of this submersion are quite difficult to analyze and we therefore elect to study where we define the projection π 4 : Sp 4 (C) → C 4 \ {0} to be the projection of a matrix to its last row: However, even the map Φ K = π 4 • Ψ K : (C 3 ) K → C 4 \ {0} is not submersive everywhere. We have the three results below (Lemma 3.3, Lemma 3.4 and Proposition 3.6) about that map which will be proved in later sections.
We introduce some notation. Projecting to the last row introduces an asymmetry between upper and lower triangulary elementary matrices and therefore we will denote by z's the variables in the lower triangular matrices and w's the variables in the upper triangular matrices. For example, the right hand side of (3.0.1) becomes  Also, when K = 2k or K = 2k + 1, let We have Lemma 3.3 that follows from a simple calculation.
is a holomorphic submersion exactly at points Z K ∈ (C 3 ) K \ S K where S K is defined by (3.0.2) above. That is, S K is the set of points where the entries in the last row of each lower triangular matrix are zero, except for the K-th matrix where no conditions are imposed, and the rank of the matrix W K , which does not involve entries from the K-th matrix, is strictly less than 2.
Remark 3.5. Lemma 3.3 and Lemma 3.4 both generalize to (2n × 2n)-matrices and the proofs are identical. In Section 6 we therefore consider the general case.
Proposition 3.6. For n = 1 and n = 2 the map is a stratified elliptic submersion.
Corollary 3.7. Let n = 1 or n = 2. Let X be a finite dimensional reduced Stein space and f : X → Sp 2n (C) be a holomorphic map. Assume that there exists a natural number K and a continuous map F : is commutative. Then there exists a holomorphic map G : X → (C n(n+1)/2 ) K \ S K , homotopic to F via continuous maps F t : X → (C n(n+1)/2 ) K \ S K , such that the diagram above is commutative for all F t .
Proof. The pull back of (3.0.3) by π 2n • f is a stratified elliptic submersion over the Stein base X. Thus by Theorem 5.6 there is a homotopy from the given continuous section to a holomorphic section. This is equivalent to the desired homotopy F t . An even better way to perform this proof is to say that the map (3.0.3) is an Oka map, see [For11, Corollary 6.14.4.(i)], which yields the desired conclusion.
Remark 3.8. The fact that the map (3.0.3) is an Oka map, see [For11, Corollary 6.14.4.(i)] yields a parametric version of Corollary 3.7. This means that the holomorphic map can be replaced by a continuous map f P : X ×P → Sp 2n (C), which is holomorphic for each fixed parameter p ∈ P and where P is a compact Hausdorff topological space.
We need the following version of the Whitehead Lemma: (3.0.4) .
Proof of Theorem 3.1. We will prove the theorem for a single map. The existence of a uniform bound N(d) follows as in the proof of Theorem 2.1. Since a finite dimensional Stein space is finite dimensional as a topological space there are K −2 continuous mappings G 1 , . . . , G K−2 : X → C 3 such that Choose a constant symmetric 2×2 matrix H with non-zero second row and replace the above factorization by This factorization by K continuous elementary symplectic matrices avoids the singularity set S K and thus we found F : Using Corollary 3.7 we know that F 0 := F is homotopic to a holomorphic map G = F 1 , via continuous maps F t , such that Since these matrices are symplectic it automatically follows that f 12,t (x) ≡ 0, f 22,t (x) ≡ 1 and f 32,t (x) ≡ 0 so that and in addition the matrix we see that f 0 = Id and thus the holomorphic map f := f 1 : X → SL 2 (C) is null-homotopic. Let ψ be the standard inclusion of Sp 2 in Sp 4 , see for example [GMV91]. By the main result from [IK12] the matrix is a product of holomorphic elementary symplectic matrices. Therefore it suffices to show that is a product of elementary symplectic matrices. In order to deduce (3.0.8) one has to use the fact that (3.0.5) is symplectic. Since  the result follows by (3.0.4). Analysing this proof and using Remark 3.8 one sees that we can actually prove a parametric version of our main theorem.
Theorem 3.9. Let X be a finite dimensional reduced Stein space, P a compact Hausdorff topological (parameter) space and f : P × X → Sp 4 (C) be a continuous mapping, holomorphic for each fixed p ∈ P , that is null-homotopic. Then there exist a natural number K and continuous mappings, holomorphic for each fixed parameter p ∈ P , G 1 , . . . , G K : In order to complete the proof of the theorem we need to establish Proposition 3.6, Lemma 3.4, and 3.3.
Remark 3.10. Proposition 3.6 is the crucial ingredient in the proof of Theorem 3.1. Its proof is by far the most difficult part of the paper. As pointed out in Remark 3.5, Lemma 3.4 holds for general n. Also if Proposition 3.6 holds for some n then Corollary 3.7 also holds for that n. Moreover the reduction of the size of the symplectic matrix from Sp 4 to Sp 2 done in the proof of Theorem 3.1 generalizes easily to a reduction from Sp 2n to Sp 2n−2 if Corollary 3.7 holds for n (see for example the proof of Lemma 4.4 in [GMV91]). Therefore if Proposition 3.6 can be proven for n = 1, . . . , m then the following holds true.
Conjecture 3.11. Let X be a finite dimensional reduced Stein space and f : X → Sp 2m (C) be a holomorphic mapping that is null-homotopic. Then there exist a natural number K and holomorphic mappings In the case of a 1-dimensional Stein space, i.e. an open Riemann surface, this Conjecture has been established in [IKL19]. The condition of null-homotopy is automatically satisfied in this case, since an open Riemann surface is homotopy equivalent to a 1-dimensional CW-complex and the group Sp 2m (C) is simply connected. The proof uses the analytic ingredient that the Bass stable rank of O(X) is 1 for an open Riemann surface and proceeds then by linear algebra arguments.

Formulation in algebraic terms
We relate our results to algebraic K-theory and reformulate them in those terms. The following is a standard notion: Since null-homotopy is an important assumption in our studies we denote the set of null-homotopic unimodular rows in U m (O(X)) by U 0 m (O(X)). This set can be seen as the path-connected component of the space of holomorphic maps from X to C m \ {0} containing the constant map (0, 0, , . . . , 0, 1) = e m . By Grauerts Oka principle C m \ {0} = GL m (C)/ GL m−1 (C) is an Oka manifold, therefore the path-connected components of continuous and holomorphic maps X → C m \ {0} are in bijection. This says that unimodular rows in U m (O(X)) are null-homotopic in the holomorphic sense iff they are null-homotopic in the continuous sense.
Algebraic K-theorists consider Chevalley groups over rings, in our example we consider the null-homotopic elements of them.
Definition 4.2. Sp 0 2n (O(X)) denotes the group of null-homotopic holomorphic maps from a Stein space X to the symplectic group Sp 2n (C), which in other words is the path-connected component of the group Sp 2n (O(X)) containing the identity.
Again by Grauert's Oka principle holomorphic maps X → Sp 2n (C) are homotopic via holomorphic maps iff they are homotopic via continuous maps.
Clearly the last row of a matrix in Sp 2n (O(X)) is unimodular, i.e., an element of U 2n (O(X)). Whether a unimodular row in U 2n (O(X)) is the last row of a matrix in Sp 2n (O(X)) is by Oka-theory a purely topological problem. Let us illustrate this by an example.
Extending a unimodular row to an invertible matrix can be reformulated as follows: Given a trivial line sub-bundle of the trivial bundle X × C n of rank n over X. Can it be complemented by a trivial bundle?
This of course is not always the case: The (non-trivial) tangent bundle T of the sphere S 2n+1 (n ≥ 4) is the complement of the trivial normal bundle N to the sphere S 2n+1 in R 2n+2 . To make this a holomorphic example consider X to be a Grauert tube around S 2n+1 , i.e., a Stein manifold which has a strong deformation retraction ρ onto its totally real maximal dimensional submanifold S 2n+1 . The bundle T is replaced by the complexified tangent bundle to the sphere pulled back onto X by the retraction ρ and equipped with its unique structure of holomorphic vector bundle (which is still not a trivial bundle). The pull-back of the complexified trivial bundle N is still a trivial line sub bundle of X × C 2n . Thus we found an example of a holomorphic row which cannot be completed to an invertible matrix in Gl 2n (O(X)) and thus not to a matrix in Sp 2n (O(X)) either.
For null-homotopic rows the situation is better.
Lemma 4.3. Every element U 0 2n (O(X)) extends to a null-homotopic matrix A ∈ Sp 0 2n (O(X)). Proof. Let F = (f 1 , . . . , f 2n ) : X → C 2n \ {0} be a null-homotopic holomorphic map, the homotopy to the constant map F 1 (x) = e 2n be denoted by F t , t ∈ [0, 1]. The map π 2n : Sp 2n (C) → C 2n \ {0} is a locally trivial holomorphic fiber bundle with typical fibre F ∼ = Sp 2n−2 (C)×C 4n−1 which is an Oka manifold. Our problem is to find a global section of the pull-back of this fibration by the map F = F 0 . Since a locally trivial bundle is a Serre fibration and the constant last row can be extended to a constant (thus null-homotopic) symplectic matrix we find a continuous section of this pull-back bundle over the whole homotopy. Thus the restriction to X × {0} is a null-homotopic continuous symplectic matrix. Since the fiber F is Oka we find a homotopy to a holomorphic symplectic matrix, which is still null-homotopic.
The notion of elementary symplectic matrices over a ring R is the same as explained in Section 2.
Let W n denote a (n × n)-matrix with entries in the ring R satisfying W n = W T n and 0 n the (n × n) zero matrix. We call those matrices that are written in block form as I n 0 n W n I n or I n W n 0 n I n elementary symplectic matrices over R. The group generated by them, the elementary symplectic group, is denoted by Ep 2n (R). We consider the group Ep 2n (O(X)) which is easily seen to be a subgroup of (multiply the symmetric matrices W n by a real number t ∈ [0, 1]) Sp 0 2n (O(X)). The meaning of Corollary 3.7 in K-theoretic terms is now the following: Proposition 4.4. Let n = 1 or n = 2. For a Stein space X the group Ep 2n (O(X)) acts transitively on the set of null-homotopic unimodular rows U 0 2n (O(X)). Proof. Let u ∈ U 0 2n (O(X)) be a null-homotopic unimodular row. By the above Lemma we can extend it to a null-homotopic symplectic matrix A ∈ Sp 0 2n (O(X)). Now we just follow the beginning of the proof of Theorem 3.1. By Theorem 2.1 we can factorize A(x) as a product of elementary symplectic matrices with continuous entries. Adding two more elementary symplectic matrices we can achieve that the factorization avoids the singularity set S K . Applying Corollary 3.7 we know that A 0 := A is homotopic to a holomorphic map G = A 1 , via continuous maps A t , such that that is, the last row of the matrices Ψ K (A t (x)) is constant. Therefore This shows that the element Ψ K (G(x)) of Ep 2n (O(X)) has the last row equal to u or equivalently moves the constant row e 2n to u.
Let ψ : SL 2 → Sp 4 be the standard embedding given by Continuing like in the proof of Theorem 3.1 we see that it gives the following "inductive step". Proposition 4.6. For a Stein space X and any n ≥ 2 the group E n (O(X)) acts transitively on the set of null-homotopic unimodular rows U 0 n (O(X)). Proposition 4.7. For a Stein space X and any n ≥ 2 holds

Stratified sprays
We will introduce the concept of a spray associated with a holomorphic submersion following [Gro89] and [FP02]. First we introduce some notation and terminology. Let h : Z → X be a holomorphic submersion of a complex manifold Z onto a complex manifold X. For any x ∈ X the fiber over x of this submersion will be denoted by Z x . At each point z ∈ Z the tangent space T z Z contains the vertical tangent space V T z Z = ker Dh. For holomorphic vector bundles p : E → Z we denote the zero element in the fiber E z by 0 z .
Remark 5.2. 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 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 ϕ t j of these vector fields V j to define s : Definition 5.3. Let X and Z be complex spaces. A holomorphic map h : Z → X is said to be a submersion if for each point z 0 ∈ Z it is locally equivalent via a fiber preserving biholomorphic map to a projection p : We will need to use stratified sprays which are defined as follows.
Definition 5.4. We say that a submersion h : Z → X admits stratified sprays if there is a descending chain of closed complex subspaces X = X m ⊃ · · · ⊃ X 0 such that each stratum Y k = X k \ X k−1 is regular and the restricted submersion h : Z| Y k → Y k admits a spray over a small neighborhood of any point x ∈ Y k .
Remark 5.5. We say that the stratification X = X m ⊃ · · · ⊃ X 0 is associated with the stratified spray.
Theorem 5.6. Let X be a Stein space with a descending chain of closed complex subspaces X = X m ⊃ · · · ⊃ X 0 such that each stratum Y k = X k \ X k−1 is regular. Assume that h : Z → X is a holomorphic submersion which admits stratified sprays then any continuous section f 0 : X → Z such that f 0 | X 0 is holomorphic can be deformed to a holomorphic section f 1 : X → Z by a homotopy that is fixed on X 0 .

Proof of Lemma 3.3 and 3.4
Lemma 3.3 and 3.4 hold for any size matrix. In this section we therefore look at (2n × 2n) matrices. Given two vectors a and b in C n (i.e. n × 1 matrices), we denote by a b the obvious vector in C 2n . We shall consider products of (2n × 2n)-matrices where Z 1 , Z 2 , · · · and W 1 , W 2 , · · · are n × n matrices of variables They are symmetric, i.e. z k,ij = z k,ji and w k,ij = w k,ji . We call the variables z k,n1 , · · · , z k,nn last row variables (this term does not apply to the w-variables). If we have K factors, there are K n(n+1) 2 variables. We will also think of the K-tuple . We will study the last row of this product, which is a map Φ K : We prefer to work with the transpose of this row, which we denote by P K , a vector in C 2n . It follows that where z = (z 1,n1 , · · · , z 1,nn ) T and e n is the last standard basis vector of C n . The set S K for K ≥ 2 is now defined as the set of K-tuples of symmetric matrices (Z 1 , W 1 , · · · ) such that in the first K − 1 matrices all the last row variables (of the Z's) are 0 and the columns of all of the W's do not span C n . (This means the totality of all W i columns of the K − 1 first factors.) Proof. We prove the result for K = 3. For K > 3, simply put W 2 = Z 3 = W 3 = · · · = 0. The proof uses an easy fact from linear algebra; given two vectors c and Pick any symmetric matrix Z 2 such that z = a − Z 2 b = 0 and let Z 1 be any symmetric matrix whose last row is z and W 1 a symmetric matrix such that Then (Z 1 , W 1 , Z 2 ) / ∈ S 3 and for this choice we have By a slight abuse of notation, we denote the Jacobian matrix of Φ K by JP K . This is a (2n × K n(n+1) 2 )-matrix whose columns are the derivatives of P K with respect to one particular variable. We denote the components of P K by P We shall look at the final part of JP 2k+1 , the part where we differentiate with respect to the new variables z k+1,11 , · · · , z k+1,n1 , z k+1,22 , · · · , z k+2,2n , · · · , z k+1,nn . This is a (2n × n(n+1) 2 )-matrix. The column where we differentiate with respect to z k+1,ij will consist of P 2k n+i in row number j and P 2k n+j in row number i. Hence the bottom half of this matrix is zero and we only look at the upper half, an (n × n(n+1) 2 )-matrix which we denote by A k+1 . If we consider just the columns which contain one particular P 2k n+i , we get a square (n × n)-matrix whose i-th row is (P 2k n+1 , · · · , P 2k 2n ), has P 2k n+i along the diagonal and is otherwise zero. The determinant of this submatrix is (P 2k n+i ) n . The situation is similar for the final part of JP 2k+2 , except now the top half is zero and the bottom half B k+1 contains P 2k+1 1 , · · · , P 2k+1 n in the same pattern as for A k+1 .
In the proof of the next lemma it will be convenient to use the following notation: if A and B are two matrices with the same column length, we let A ∪ B denote the matrix obtained by extending A with B to the right. e 2n denotes the last vector in the standard basis of C 2n . Lemma 6.2. P K is a submersion exactly on the set C K n(n+1) 2 \ S K . If K = 2k and all the last row variables are zero, then P 2k = e 2n and the span of the bottom half of the JP 2k columns equals the span of the columns of W 1 , W 2 , · · · , W k .
Proof. For N = 1 the theorem is empty. P 1 = (z 1,n1 , · · · , z 1,nn , 0, · · · , 0, 1) and where we have removed all zero columns. For N = 2 we have This implies which has full rank if and only if B 1 has full rank. Since P 1 i = z 1,ni , by the discussion preceding the lemma, B 1 has full rank if and only if at least one z 1,ni is nonzero.
If all z 1,ni are zero, then P 1 = e 2n and B 1 = 0. Hence the statement about the span is trivially true.
We now assume that the theorem is true for N = 2k. We have If at least one of the previous last row variables is nonzero, then JP 2k has full rank by the induction hypothesis and so does JP 2k+1 . If not, then P 2k = e 2n and A k+1 = I n , after removing zero columns. If JP 2k = A B , then which has full rank if and only if B has full rank. But the column span of B equals the column span of W 1 , · · · , W k . This proves the first part of the lemma for N = 2k + 1. If all the previous last row variables are zero, it also follows that P 2k+1 = (z k+1,n1 , · · · , z k+1,nn , 0, · · · , 0, 1) t .
If not, then by the above all the previous last row variables are zero and which has full rank if and only if at least one z k+1,ni is nonzero by the discussion preceding the lemma. This proves the first part of the lemma for N = 2k + 2.
If all the z k+1,ni also are zero, then P 2k+1 = e 2n and so P 2k+2 = e 2n . Also B k+1 = 0 and since the columns of W k+1 (A + Z k+1 B) are linear combinations of the columns of W k+1 , the span of the bottom half of JP 2k+2 equals the span of the columns of W 1 , · · · , W k+1 by the induction hypothesis. This completes the proof of the lemma.

The stratification
The goal in this section is to describe the stratification needed to understand that the submersion π 4 • Ψ K : Remark 7.1. We will abuse notation in the following way in the paper. A polynomial not containing a variable can be interpreted as a polynomial of that variable. More precisely, let K < L. We have the projection π : C L → C K , π(x 1 , . . . , x L , . . . , x K ) = (x 1 , . . . , x L ) and π * : We want to study the fibers a 2 , a 3 , a 4 ).
We now analyse the case when K = 2k ≥ 3 is even. Now we have a 2 ,a 3 ,a 4 ) is the solution set to the equations As in the previous case these equations simplify A similar analysis as in the previous gives us the following strata: • The stratum of generic fibers: When (a 1 , a 2 ) = (0, 0). The fibers are graphs over H K−1 (a 1 ,a 2 ) × C. Moreover the fibers are smooth. • The strata of non-generic fibers: When (a 1 , a 2 ) = (0, 0) the fibers are This set is divided into two strata as follows: -Smooth non-generic fibers: When (a 3 , a 4 ) = (0, 1) then the fibers are smooth. -Singular non-generic fibers: When (a 3 , a 4 ) = (0, 1) then the fibers are non-smooth.

Determination of complete vector fields
The description of the fibers in Section 7 leads us to study vector fields simultaneously tangent to the level sets {P = c 1 }, {Q = c 2 } of two functions P, Q : C N → C. Such fields can be constructed in the following way. Pick three variables x, y, z from the variables x 1 , . . . , x N on C N and consider the vector fields which are simultaneously tangent to the level sets. As mentioned in Section 5 we want to use a finite collection of complete vector fields spanning tangent space at every point to prove (stratified) ellipticity. It is an easy exercise to show that the collection of these vector fields over all possible triples spans the tangent space at smooth points of the variety {P = c 1 } ∩ {Q = c 2 }. It turns out that many of the vector fields we get by this method are complete but unfortunately not all of them. The complete vector fields from this collection will not span the tangent space at all points for all level sets. To overcome this difficulty and still producing dominating sprays from this collection of available complete fields is the main technical part of our paper explained in Section 9. Now we will start to describe the complete vector fields tangent to the fibers of π 4 • Ψ K = (P K 1 , P K 2 , P K 3 , P K 4 ) that we get using (8.0.1). It will be convenient to group the variables as in Section 6, and similarly for W k . Since the variable z 1 never enters in P K , we omit it from the first group Z 1 . Note that P 1 = (z 2 , z 3 , 0, 1) T . We are going to study the vector fields V K ij (x, y, z) = D xyz (P K i , P K j ). The 2 × 2 minors occuring as coefficients are denoted by C K ij (·, ·), i.e.
The description of the complete vector fields will be done inductively. We start with K = 2. We have to study G 2 (a 3 ,a 4 ) or equivalently the equations We are interested in which triples (x, y, z) of variables from the list z 2 , z 3 , w 1 , w 2 , w 3 give complete vector fields V 2 34 (x, y, z) and we denote the set of these triples by T 2 . By definition T 1 = ∅. An easy computation gives that For all the remaining noncomplete triples there is a variable such that the equation is quadratic for that variable. We are now interested in determining at every stage the triples of variables (x, y, z) such that V 2k+1 12 (x, y, z) for K = 2k + 1 odd is complete and V 2k+2 34 (x, y, z) for K = 2k + 2 even. We shall denote the set of such triples by T K . The terms occuring in P K are of degree one in the occuring variables, hence the coefficients C K ij are either of degree one or two in the occuring variables. A triple giving a coefficient which is quadratic in the integration variable (for instance if C K ij (y, z) is quadratic in the x variable) will not be complete and we shall refer to such a triple as a quadratic triple and the corresponding vector field as a quadratic vector field. The content of the next lemma is that all the remaining triples give complete vector fields. The variables that do not occur in a triple will have constant solutions and are therefore treated as such in the proof.  Proof. The result is true for T 2 . The first group is interpreted as {z 2 , z 3 } and z 3k+1 must be replaced by z 2 . The missing triplet are precisely the quadratic triples.
We shall prove (8.0.4), the proof of (8.0.5) being identical. There is a lot of symmetry in the proof and we will not repeat arguments already given in a situation symmetric to a proven statement. We first consider triples (x, y, z) not containing any variables from the new group Z k+1 , i.e. z 3k+1 , z 3k+2 and z 3k+3 . It then follows from (6.0.1) (omitting variables for shorter notation) that: A quadratic triple will still be quadratic since V 2k 34 is. For a triple in T 2k , notice that in all the first 5 terms the V 2k ij is obtained by replacing one or two of the functions P 2k 3 and P 2k 4 by P 2k 1 and/or P 2k 2 . By (6.0.2) all of the terms occuring in P 2k 1 or P 2k 2 divide a term occuring in P 2k 3 and also a term occuring in P 2k 4 . This means that all terms occuring in the 5 first vector fields above are already present in V 2k 34 and completeness is not destroyed. We also notice that for any pair x, y of previous variables, the coefficient C 2k+1 12 (x, y) will also satisfy (8.0.6). We next consider triples containing some of the new variables z 3k+1 , z 3k+2 and z 3k+3 . The Jacobian matrix is now given by (6.0.3) where If the triple contains all three variables, then 3 ) 2 ∂/∂z 3k+3 and the coefficients do not contain any of the Z k+1 variables, hence this is complete. (The solutions are just affine functions.) We now consider the case of two new variables. The first possibility is (x, z 3k+1 , z 3k+2 ). The coefficient of ∂/∂x is (P 2k 3 ) 2 . Since P 2k 3 contains all previous variables except w 3k , this is quadratic in all those variables and x = w 3k is the only possibility. The solution for w 3k is affine. The coefficient of ∂/∂z 3k+2 is now which is just a constant and the solution is again affine. Finally the coefficient of ∂/∂z 3k+1 is given by ∂P 2k ∂w 3k which is an affine function and the solution is entire. Hence this field is complete.
The precise same logic applies to the triple (x, z 3k+2 , z 3k+3 ) except now w 3k−2 is the only missing variable (now in P 2k 4 ). The final possibility of two new variables is the triple (x, z 3k+1 , z 3k+3 ). The coefficient of ∂/∂x is now P 2k 3 P 2k 4 which is of degree one in w 3k−2 and w 3k and quadratic in all other previous variables. We consider the case of x = w 3k−2 , the case x = w 3k being identical. The coefficient is an affine function of w 3k−2 , hence the solution is entire. The coefficient of ∂/∂z 3k+1 is −z 3k+1 P 2k−1 1 P 2k 4 which is just a linear function of z 3k+1 and the solution is entire. The coefficient of ∂/∂z 3k+3 is −z 3k+2 P 2k−1 2 P 2k 4 which is just a constant and the solution is affine.
We finally consider the case of one new variable and two previous variables x, y. It follows that C 2k+1 12 (x, y) satifies equation (8.0.6) hence is quadratic in z 3k+2 , so this cannot be the new variable. In order to investigate z 3k+1 and z 3k+3 we need to understand which variables are involved in the coefficients. To do this we look at each previous group of variables Z j and W j for 1 ≤ j ≤ k and see which variables are involved in the first two rows of the Jacobian with respect to these variables at level 2k + 1. For a Z j group we need to consider the matrix    and the same for a W j group. The Z 1 group only consists of z 2 and z 3 . The Z j variables do not occur in the above matrix. There is a simple formula for the above matrix which follows from (6.0.3) and (6.0.4). The matrix is the first two rows of the matrix (I = I 2 ) : and this formula makes it easy to track which variables are missing at each step, in addition to the Z j variables. We arrive at the following matrix of missing variables w 3j−3 , w 3j , z 3k+3 z 3k+3 w 3j−5 , w 3j−2 , z 3k+3 w 3j−3 , w 3j , z 3k+1 z 3k+1 w 3j−5 , w 3j−2 , z 3k+1 In the case j = 1 the missing variable matrix is We now consider a W j group. Again the W j variables do not enter. We now have to consider the two first rows of the matrix and this leads to the following missing variables matrix for j < k: For j = 1 we replace z 3j−2 by z 2 . For j = k the middle entries in the upper left and the lower right corners are replaced by z 3k+2 . We first investigate triples (x, y, z 3k+1 ), where x and y are not from Z k+1 . If x and y are from the same group, then since z 3k+1 occurs in every entry in the second row of the missing variable matrix, C 2k+1 12 (x, z 3k+1 ) and C 2k+1 12 (y, z 3k+1 ) do not depend on any of the variables x, y, z 3k+1 hence x and y are both affine functions. C 2k+1 12 (x, y) does not depend on x, y and is of degree one in z 3k+1 , hence the solution is entire. Now assume that x and y are from different groups. If x is not a missing variable in ∂P 2k+1 2 ∂y , then y is not a missing variable in ∂P 2k+1 2 ∂x . x and y are not both w 3k , let's say x. Then is quadratic in x and the field is not complete. Hence x and y must both appear in the second row of the missing variable matrix of each other. We now look at possibilities for x and y. Assume first that x is in Z j group with 1 < j ≤ k. There are now four possibilities ; x = z 3j−2 in which case y = w 3j−3 or y = w 3j or x = z 3j in which case y = w 3j−5 or y = w 3j−2 . We consider the first case. Then and from the missing variable matrix we see that this does not depend on z 3j−2 and w 3j−3 and is of degree one in z 3k+1 , hence we have an entire solution for z 3k+1 .
We also have The partial derivatives on the right hand sides do not depend on any of the variables in the triple, hence are just constants. It also follows from the missing variable matrix that P 2k 3 does not contain the product of z 3j−2 and w 3j−3 , hence the equations for these two variables is a linear system with constant coefficients. This has an entire solution. The three other cases all have similar structure and have entire solutions. In the case j = 1, we either have x = z 2 and y = w 3 or x = z 3 and y = w 1 and the discussion is the same. It also follows from the missing variable matrix that x and y cannot come from different W groups. This proves the result in the case of picking z 3k+1 from the last group. The proof in case of picking z 3k+3 from the last group is completely symmetric. This provides the final detail in the proof.
In order to produce complete fields that are also tangential to fibers of the submersion we introduce the following notation and terminology.
Definition 8.2. Let Ξ 3 = T 2 . For K ≥ 4 let We say that the triples in Ξ K are introduced on level K.
We will now use these complete fields to produce complete fields which are tangential to the fibers F K (a 1 ,a 2 ,a 3 ,a 4 ) . Here we will use triples introduced on level K to produce complete tangential fields.
First consider the case K = 2k + 1 ≥ 3 odd. If a 3 = 0 we use (7.0.1) to get Using this we define a biholomorphism a 2 ,a 3 ,a 4 ) On G 2k (a 3 ,a 4 ) × C z 3k+3 we have the complete fields ∂ 2k x 1 x 2 x 3 for x 1 , x 2 , x 3 in Ξ 2k+1 and also the complete field ∂/∂z 3k+3 . Using the biholomorphism α we get complete fields on F K   (a 1 ,a 2 ,a 3 ,a 4 ) for a 3 = 0 of the form (8.0.8) Since P 2k 3 = a 3 and P 2k 4 = a 4 on the fiber we get meromorphic fields on (C 3 ) K (8.0.10) (abusing notation) with poles on P 2k 3 = 0. Since P 2k 3 is in the kernel of these fields we can multiply the fields by (P 2k 3 ) 2 and get complete fields that are globally defined on (C 3 ) K and preserve the fibers of π 4 • Ψ K below (8.0.12) θ 2k+1 x for x 1 , x 2 , x 3 ∈ Ξ 2k+1 and the field (8.0.13) .
For the case K = 2k ≥ 3 even the analogous procedure leads to the following complete fields on (C 3 ) K tangent to the fibers of π 4 • Ψ K : Remark 8.3. It follows from the inductive formulas (7.0.1) and (7.0.2) that θ K x 1 x 2 x 3 , φ K x 1 x 2 x 3 and γ K , considered as vector fields on (C 3 ) L , are tangent to the fibers F L (a 1 ,a 2 ,a 3 a 4 ) for L ≥ K. In other words, the fields associated with triples introduced on level K are tangential to all fibers F L for L ≥ K.

Strategy of proof of stratified ellipticity
We outline the strategy for proving that the submersion is a stratified elliptic submersion. We have seen that the fibers are given by four polynomial equations. We have also seen that these four equations can be reduced to two equations. We then use the exact form of these two equations to find Ξ K so that ∂ K x 1 x 2 x 3 are complete vector fields exactly when x 1 , x 2 , x 3 ∈ Ξ K . This leads us to the globally defined complete vector fields θ K x 1 x 2 x 3 , φ K x 1 x 2 x 3 and γ K described in Section 8. Find a big (a complement of an analytic subset) "good" set on the fibers where the collection of these vector fields spans the tangent space of the fiber. For points outside the good set find a complete field V whose orbit through the point intersects the good set. At points along the orbit that are also in the good set the collection of complete vector fields above spans. Now pull back the collection of vector fields by suitable flow automorphisms of V and add these fields to the collection (see Definition 10.7). This enlarged collection of complete vector fields spans in a bigger set thus enlarging the good set. Continue this enlarging of the collection of vector fields until it spans the tangent space at every point of every fiber in the stratum. To accomplish this strategy we need the following technical results.
Lemma 9.1. Let M be a Stein manifold, N 0 ⊂ N ⊂ M analytic subvarieties. Given a finite collection θ 1 , . . . , θ k of complete holomorphic vector fields on M which span the tangent space T x M at all points x ∈ M \ N and given another complete holomorphic vector field φ on M (whose flow we denote by α t ∈ Aut hol (M), t ∈ C) with the property that the orbit through points of N \ N 0 is leaving N, i.e. {α t (x) : t ∈ C} ⊂ N ∀x ∈ N \ N 0 . Then there are finitely many times t i ∈ C i = 1, . . . , l such that In particular the finite collection {α ⋆ t i (θ m )} l,k i=1,m=1 of complete holomorphic vector fields on M is spanning the tangent space T x M at all points x ∈ M \ N 0 .
Proof. The analytic subset N has at most countably many components. Denote by B i those components which are not entirely contained in N 0 . Define a 0 to be the maximal dimension of them. Choose a point x i from each of those B i . For every i the set A i := {t ∈ C : α t (x i ) ∈ N} is discrete. Since a countable union of discrete sets is meagre in C, we find t 1 / ∈ A i ∀i. Denote byB i those components of the analytic subset N 1 := {y ∈ N : α t 1 (y) ∈ N} which are not entirely contained in N 0 and define a 1 to be the maximal dimension of them. By construction a 1 < a 0 . Choose a pointx i from each of thoseB i . For every i the set A i := {t ∈ C : α t (x i ) ∈ N} is discrete. Since a countable union of discrete sets is meagre in C, we find t 2 / ∈Ã i ∀i. Let a 2 be the maximal dimension of those components of the analytic subset N 2 := {y ∈ N : α t 1 (y) ∈ N and α t 2 (y) ∈ N} which are not entirely contained in N 0 . By construction a 2 < a 1 and continuing the construction after finitely steps we reach our conclusion.
The next Lemma is a generalized and parametrized version of the previous one. It is adapted to the stratified spray situation. Namely, we have to produce sprays not on a single fiber but in a neighborhood of the fiber in each stratum (see Definition 5.4). In fact in our case it will be on the whole stratum. The following definitions are straightforward.
Definition 9.2. Let π : X → Y is a holomorphic map between complex manifolds and denote dπ : T X → T Y the tangent map. We call a holomorphic vector field θ on X fiber preserving if dπ(θ) = 0. Proof. Let N ⊂ M be the set of points x where span{(θ 1 , . . . , θ k )} = T x M π(x) . by assumption N ∪ M y is a proper analytic subset of M y for each y ∈ Y . Since there is no invariant analytic subset different from the fibers for each x 0 ∈ N there is a field θ i whose flow starting in x 0 will leave N, i.e. go through points where (θ 1 , . . . , θ k ) span T x M π(x) Now choose (at most countably many) points, one from each component of N and as in the proof of the proceeding lemma find finitely many times t i such that enlarging the collection θ 1 , . . . , θ k by the pullbacks (α i (t i )) * (θ m )i, m = 1, . . . , k we get a new finite collection of complete fields where the set of points where this new collection does not span the tangent space of the π-fiber has smaller dimension. By finite induction on the dimension we get the desired result.

Define
for any triple x 1 , x 2 , x 3 from Z K . Removing the j-th row from M K x 1 x 2 x 3 gives us (3 × 3)-matrices which we denote by M K,j x 1 x 2 x 3 . Let R K,j x 1 x 2 x 3 = det M K,j x 1 x 2 x 3 . The significance of the functions R K,j x 1 x 2 x 3 is understood if one notices, because of (8.0.1), that and that . From (7.0.1) and (7.0.2) we get the relations x 1 x 2 x 3 R 2k,2 x 1 x 2 x 3 R 2k,3 x 1 x 2 x 3 R 2k,4 Consider the vector fields θ L x 1 x 2 x 3 and φ L x 1 x 2 x 3 where (x 1 , x 2 , x 3 ) ∈ T L and 3 ≤ L ≤ K. Rewriting (8.0.12), (8.0.14), (8.0.15) and (8.0.16) using these functions we get We see that half of the functions R K,j x 1 x 2 x 3 occur in the coefficients of the last three directions. As already observed the fields θ L x 1 x 2 x 3 and φ L x 1 x 2 x 3 for L < K have components zero along the last three directions. We have to make sure that the projection onto the last three variables of the collection of fields θ K x 1 x 2 x 3 and φ K x 1 x 2 x 3 spans a three-dimensional space. If this is true for a point we will say that the fields span all new direction in the point. In order to determine if our fields span all new directions in a point Z K ∈ F K a 1 a 2 a 3 a 4 we will use the following. Let N K = |T K | be the number of complete triples. Define the (2 × N K−1 )-matrices where (x 1 , x 2 , x 3 ) run over all triples in T K−1 . Using the formulas (10.0.8), (10.0.9), (10.0.10), (10.0.11), and remembering that a fiber F K (a 1 ,a 2 ,a 3 ,a 4 ) is called generic if (a 1 , a 2 ) = (0, 0) when K is even and if (a 3 , a 4 ) = (0, 0) when K is odd it is an exercise in linear algebra to prove the lemma below.
Lemma 10.1. If in a point Z K ∈ F K a 1 a 2 a 3 a 4 in a generic fiber Because of the formulas (10.0.6) and (10.0.7) we have the lemma below.
Lemma 10.2. Let K ≤ L and put . The importance of Lemma 10.1 and Lemma 10.2 is seen in the following corollary.
Corollary 10.3. Let L > K and Z K be a point where Rank M K K ( Z K ) = 4. Then for all points Z L contained in a generic fiber F L (a 1 ,a 2 ,a 3 ,a 4 ) such that π( Z L ) = Z K the complete fields x 3 ) ∈ T L−1 ∪ γ L span all new directions (the directions along the last three variables in (C 3 ) L ).
Proof. Two rows of the rank 4 matrix M L K ( Z L ) are linearly independent.
Corollary 10.4. Let L ≥ 3 and Z L be a point that is contained in a generic fiber F L (a 1 ,a 2 ,a 3 ,a 4 ) and such that z 2 z 3 = 0. Then In order to use this corollary we need the following lemma.
Lemma 10.5. We have the following cases for the function P = z 2 z 3 and the fibers F K : (1) P is not identically zero on F K (a 1 ,a 2 ,a 3 ,a 4 ) for K ≥ 5. For these K the fibers F K (a 1 ,a 2 ,a 3 ,a 4 ) are irreducible.

∂ 2
The singularity set of G 2 (0,1) is A 1 ∩ A 2 . Clearly P is identically zero on A 1 and not identically zero on A 2 . Observe that F 3 (a 1 ,a 2 ,0,1) are connected, their smooth part consists of the two connected components A 1 \ A 2 and A 2 \ A 1 .
The smooth generic fibers F 3 (a 1 ,a 2 ,a 3 ,a 4 ) for (a 3 , a 4 ) / ∈ {(0, 0), (0, 1)} are isomorphic to the variety G 2 (a 3 ,a 4 ) given by the two equations (10.0.15) z 2 w 1 + z 3 w 2 = a 3 and (10.0.16) z 2 w 2 + z 3 w 3 + 1 = a 4 In case z 2 = 0 these equations can be used to express w 2 and w 3 by the other variables and we get a chart isomorphic to C ⋆ z 2 × C z 3 × C w 3 . In case z 3 = 0 we can express w 2 and w 3 which gives us a similar chart. Thus G 2 (a 3 ,a 4 ) is covered by two connected charts with non-empty intersection which shows that it is connected. Thus the smooth generic fibers F 3 (a 1 ,a 2 ,a 3 ,a 4 ) are irreducible. The function P is not identically zero on both charts. The assertion (3) is completely proven.
Next we prove assertion (2). The non-generic fibers F 4 (0,0,a 3 ,a 4 ) are isomorphic to F 3 (0,0,a 3 ,a 4 ) × C 3 , where C 3 corresponds to the new variables w 4 , w 5 , w 6 . All assumptions about these fibers follow therefore from the corresponding assumptions about F 3 (0,0,a 3 ,a 4 ) . In the case of generic fibers which are known to be smooth (see Section 7) we just have to prove they are connected. For this consider (10.0.17) F 4 (a 1 ,a 2 ,a 3 ,a 4 ) = (w 4 ,w 5 ,w 6 )∈C 3 a 2 ,b 3 ,b 4 ) where b 3 = a 3 − w 4 a 1 − w 5 a 2 and b 4 = a 4 − w 5 a 1 − w 6 a 2 . In other words we consider the surjective projection ρ : F 4 (a 1 ,a 2 ,a 3 ,a 4 ) → C 3 , mapping a point to its last three coordinates (w 4 , w 5 , w 6 ) where the ρ-fibers are just fibers F 3 (a 1 ,a 2 ,b 3 ,b 4 ) . Connectedness of the ρ-fibers implies that a connected component of F 4 (a 1 ,a 2 ,a 3 ,a 4 ) has to be ρ-saturated. Since ρ is a submersion in generic points of the fiber (it is not a submersion only in singular points of an F 3 -fiber) any connected components of F 4 (a 1 ,a 2 ,a 3 ,a 4 ) is equal to ρ −1 (U), where U is some open subset of the base C 3 . Since the base is connected and ρ is surjective connectedness of F 4 (a 1 ,a 2 ,a 3 ,a 4 ) follows. The function P is not identically on any F 3 -fiber contained in F 4 (a 1 ,a 2 ,a 3 ,a 4 ) , thus not identically zero on F 4 (a 1 ,a 2 ,a 3 ,a 4 ) itself. This concludes the proof of (2). Last we prove assertion (1). The connectedness of the fibers F K (a 1 ,a 2 ,a 3 ,a 4 ) for K ≥ 5 can be proven by induction in a similar way as the connectedness of the generic F 4 -fibers is deduced from the properties of F 3 -fibers. We consider as above the surjective projection ρ : F K (a 1 ,a 2 ,a 3 ,a 4 ) → C 3 onto the last three variables who's fibers are F K−1 -fibers. Since again F K−1 -fibers are connected and ρ is a submersion in smooth points of the F K−1 -fibers, any connected component of F K (a 1 ,a 2 ,a 3 ,a 4 ) is of the form ρ −1 (U), where U is some open subset of the base C 3 .
In addition we will prove by induction that the smooth part of the singular fibers F K (a 1 ,a 2 ,a 3 ,a 4 ) \Sing F K (a 1 ,a 2 ,a 3 ,a 4 ) is connected for K ≥ 5. Together with connectedness of the fibers this implies the irreducibility of the fibers.
For even K the singular fibers are the singular F K−1 -fibers times C 3 and therefore the connectedness of the smooth part follows by induction hypothesis.
For odd K = 2k + 1 we are faced with the following situation: The singular fibre is F K (a 1 ,a 2 ,0,1) and it is fibered by F K−1 -fibers all of which are smooth except for the fibers F K−1 (0,0,0,1) . The union of those fibers forms a codimension 2 subvariety of F K (a 1 ,a 2 ,0,1) (given by the equations z 3k+2 = z 3k+3 = 0). By the argument above the complement, call it W, of this union in F K (a 1 ,a 2 ,0,1) is connected. The singular points of F K (a 1 ,a 2 ,0,1) are contained in that union and is contained in (but not equal to) the union of the singular points of the fibers F K−1 (0,0,0,1) . We want to prove that any smooth point p of F K (a 1 ,a 2 ,0,1) which is contained in a fiber F K−1 (0,0,0,1) is contained in the in the connected component containing W . Since the complement of W has codimension 2 in F K (a 1 ,a 2 ,0,1) an open neighborhood of p in F K (a 1 ,a 2 ,0,1) has to intersect W , which gives the desired conclusion.
As in the proof of (2) the function P cannot be identical zero on any fiber F K (a 1 ,a 2 ,a 3 ,a 4 ) since this fiber contains F K−1 -fibers on which by induction hypothesis P is not identical zero.
Remark 10.6. The fact that after a certain number of factors the fibers of the fibration become all irreducible is very general. It was proven by J. Draisma as an outcome of an interesting discussion with the second author. The irreducubility statement in our lemma is just an example of a much more general property. We refer the interested reader to [Dr20]. The exact number from which on irreducibility of the fibers holds (in our case 5) is not known in general, although Draisma gives a bound. Obviously Γ(A) consists of complete vector fields and we call it the collection generated by A.
Definition 10.8. Let L ≥ 3. We define At each step of the induction we will prove the following Proposition, which plays a crucial role in the inductive proof of Proposition 3.6.
Proposition 10.9. For each L ≥ 4 holds: There are finitely many (complete) fields from Q L which span the tangent space T x F L at each smooth point of any generic fiber F L . For L = 3 there are finitely many (complete) fields from Q 3 which span the tangent space T x F 3 at each point of any smooth generic fiber F 3 .
Remark 10.10. For L = 3 singular generic fibers F 3 (a 1 ,a 2 ,0,1) have two irreducible components and we can prove the statement about smooth points on generic fibers only for one of those components. It is false for the other component.
Notice that z 2 = z 3 = 0 is contained in F 3 (z 5 ,z 6 ,0,1) and therefore z 2 and z 3 is never simultaneuously zero on any fiber in this stratum. It is enough to show that G 2 (a 3 ,a 4 ) is elliptic. We see from the table that the fields ∂ 2 z 3 w 1 w 3 , ∂ 2 z 2 w 1 w 3 , ∂ 2 w 1 w 2 w 3 span the tangent space T Z 2 G 2 (a 3 ,a 4 ) for all points Z 2 where z 2 z 3 = 0. The complement of this good set is the disjoint union of the analytic subsets A = Z 2 ; z 2 = 0 and B = Z 2 ; z 3 = 0 . From the table we see that ∂ 2 z 2 w 2 w 3 (z 2 ) = z 2 3 which is nowhere zero on A. Also ∂ 2 z 3 w 1 w 2 (z 3 ) = z 2 2 which is nowhere zero on B. By Lemma 9.1 there exist finitely many complete fields from Γ ∂ 2 x 1 x 2 x 3 ; (x 1 , x 2 , x 3 ) ∈ T 2 that span the tangent space T Z 2 G 2 (a 3 ,a 4 ) for all points in the stratum. Therefore G 2 (a 3 ,a 4 ) is elliptic. It follows that there are finitely many complete fields from Γ θ 3 a 2 ,a 3 ,a 4 ) for all points in the stratum. Now we consider the stratum of non-smooth generic fibers (a 3 , a 4 ) = (0, 1). The two equations defining G 2 (0,1) can be written in matrix form as Recall that G 2 (0,1) has two irreducible components. The components are given by (see (10.0.13) and (10.0.14)) (11.0.2) A 1 = {z 2 = z 3 = 0} ∼ = C 3 w 1 w 2 w 3 and (11.0.3) A 2 = w 1 w 2 w 2 w 3 z 2 z 3 = 0 0 and det w 1 w 2 w 2 w 3 = 0 .
The singularity set of G 2 (0,1) is A 1 ∩ A 2 . We have to show that the smooth part of G 2 (0,1) , that is the disjoint union of A 1 \A 2 and A 2 \A 1 , is elliptic. In the proof for the smooth generic case it is shown that on the set where z 2 and z 3 are not both zero then there exists a collection of complete spanning vector fields. Since A 2 \ A 1 is contained in that set we need only consider A 1 \A 2 . The set A 1 \A 2 is biholomorphic to ∂ ∂w 2 are complete on C 3 \{w 1 w 3 −w 2 2 = 0} and span the tangent space in all points outside the analytic set A ′ = {w 1 = w 3 = 0}∩(A 1 \A 2 ). Since w 2 is nowhere zero on A ′ any of the four complete fields points out of A ′ . By Lemma 9.1 the proof is complete. Observe that we also have proved Propsition 10.9 for L = 3. Notice that the fields 2w 2 ∂ ∂w 1 + w 3 ∂ ∂w 2 , 2w 2 ∂ ∂w 3 + w 1 ∂ ∂w 2 are not in Q 3 and this explains the difference between L = 3 and L ≥ 4 in Proposition 10.9. See Remark 10.10.
The stratum of non-generic fibers is a locally trivial bundle with fibers C 5 ( ∼ = F 3 (a 1 ,a 2 ,0,0) ) which is an elliptic submersion.

Proof of Proposition 3.6: 4 matrix factors
We begin the proof by studying the stratum of generic fibers, (a 1 , a 2 ) = (0, 0). We write a 2 ,b 3 ,b 4 ) where b 3 = a 3 − w 4 a 1 − w 5 a 2 and b 4 = a 4 − w 5 a 1 − w 6 a 2 . We need to find finitely many complete vector fields spanning T Z 4 F 4 (a 1 ,a 2 ,a 3 ,a 4 ) for points Z 4 in the stratum of generic fibers. Because of (12.0.1) there are b 3 and b 4 so that Z 3 ∈ F 3 (a 1 ,a 2 ,b 3 ,b 4 ) and Z 4 = ( Z 3 , w 4 , w 5 , w 6 ). We consider first the set of points in these fibers having the property that (b 3 , b 4 ) = (0, 0) or (0, 1). Under these assumptions Z 4 lies in a generic smooth fiber F 3 (a 1 ,a 2 ,b 3 ,b 4 ) and we know from Section 11 that there is a finite collection of fields from Q 3 which spans a 2 ,a 3 ,a 4 ) . Corollary 10.4 together with Lemma 10.5(3) shows that for the set defined by z 2 z 3 = 0 (which is a Zariski open and dense set of points of the generic fiber F 4 (a 1 ,a 2 ,a 3 ,a 4 ) ) the fields θ 4 x 1 x 2 x 3 ; (x 1 , x 2 , x 3 ) ∈ T 3 ∪ γ 4 span the new directions w 4 , w 5 , w 6 . Since these new directions are complementary to a 2 ,a 3 ,a 4 ) we have found finitely many complete fields spanning T Z 4 F 4 (a 1 ,a 2 ,a 3 ,a 4 ) for points in a Zariski open dense set in all smooth generic fiber F 3 (a 1 ,a 2 ,b 3 ,b 4 ) . Using Lemma 9.1 we get finitely many complete fields spanning T Z 4 F 4 (a 1 ,a 2 ,a 3 ,a 4 ) for all points in all generic fibers F 4 (a 1 ,a 2 ,a 3 ,a 4 ) with the property that (b 3 , b 4 ) = (0, 0) or (0, 1). Next we consider points Z 4 where (b 3 , b 4 ) = (0, 1), i.e, Z 4 ∈ F 3 (a 1 ,a 2 ,0,1) ⊂ F 4 (a 1 ,a 2 ,a 3 ,a 4 ) . Remember that (12.0.2) F 3 (a 1 ,a 2 ,0,1) = A 1 ∪ A 2 = A 1∪ (A 2 \ A 1 ) (see (10.0.13) and (10.0.14)) where A 1 and A 2 are irreducible components. In the proof for K = 3 we saw that there is a finite collection from Q 3 which spans all tangent spaces a 2 ,a 3 ,a 4 ) for all points in A 2 \ A 1 . Lemma 10.5(3) gives that z 2 z 3 is not identically zero on A 2 \ A 1 and as above appealing to Lemma 9.1 we get spanning fields for the fiber F 4 (a 1 ,a 2 ,a 3 ,a 4 ) in all points of A 2 \ A 1 . Our aim is to exclude the existence of a subset of the fiber invariant under the flows of fields from Q 4 . By the reasoning above such a subset must be contained in A 1 or the set of points Z 4 where (b 3 , b 4 ) = (0, 0). Next we show that such a subset is disjoint from A 1 . A calculation shows that Therefore the complete fields θ 4 z 2 z 3 z 6 = a 2 1 ∂ 3 z 2 z 3 z 6 + . . . and φ 4 z 2 z 3 z 6 = a 2 2 ∂ 3 z 2 z 3 z 6 + . . . moves points out of A 1 (into the big orbit) unless in addition to z 2 = z 3 = 0 also (12.0.3) 1 + z 4 w 1 + z 5 w 2 = z 4 w 2 + z 5 w 3 = 0.
We now study the stratum of singular generic fibers. Here (a 3 , a 4 ) = (0, 1). Again notice that when (b 1 , b 2 ) = (0, 0) then is a generic smooth fiber for Φ 4 and Proposition 10.9 (for L = 4), Corollary 10.4 and Lemma 10.5 shows that for these points we have spanning fields as above.
The fact that D 1 = D 2 = D 3 = 0 means that the rank drops when we remove the third column from these matrices. This implies that the third column is non-zero and the other columns are multiples of a non-zero vector v which moreover is linearly independent of the third column. Now we use the field γ 3 (see (8.0.13)) to show that the set I = C \ Sing(F 5 (0,0,0,1) ) ∩ {D 1 = D 2 = D 3 = 0} does not contain an invariant subset under fields from Q 5 . In the points that we are considering γ 3 = ∂ ∂z 4 . We consider two cases. Case 1: (w 5 , w 6 ) = (0, 0) In this case det w 4 w 5 w 5 w 6 = 0.
14. Proof of Proposition 3.6: Induction steps Recall the description of the stratification for the submersion Φ M = π 4 • Ψ M given in Section 7. When M is odd we have the following strata: • The strata of generic fibers: When (a 3 , a 4 ) = (0, 0). The fibers are graphs over G M −1 (a 3 ,a 4 ) × C. This set is divided into two strata as follows: -Smooth generic fibers: When (a 3 , a 4 ) = (0, 1) then the fibers are smooth. -Singular generic fibers: When (a 3 , a 4 ) = (0, 1) then the fibers are non-smooth.
14.1. Even number of factors. We begin by showing that the stratified submersion is elliptic when the number of matrix factors is even. This case is easier than the case when the number of factors is odd which we will deal with in subsection 14.2. Assume that K = 2k − 1 ≥ 5 and that the submersions Φ L = π 4 • Ψ L are stratified elliptic submersions when 3 ≤ L ≤ K and that Propostion 10.9 is true when 3 ≤ L ≤ K. Study (14.1.1) F K+1 (a 1 ,a 2 ,a 3 ,a 4 ) = F 2k (a 1 ,a 2 ,a 3 ,a 4 ) = (w 3k−2 ,w 3k−1 ,w 3k )∈C 3 a 2 ,b 3 ,b 4 ) where b 3 = a 3 − w 3k−2 a 1 − w 3k−1 a 2 and b 4 = a 4 − w 3k−1 a 1 − w 3k a 2 . That is we use the new group of variables w 3k−2 , w 3k−1 and w 3k to present F 2k (a 1 ,a 2 ,a 3 ,a 4 ) as a fibration over C 3 with fibers F 2k−1 .

Product of exponentials and open questions
For a Stein space X, a complex Lie group G and its exponential map exp : g → G we say that a holomorphic map f : X → G is a product of k exponentials if there are holomorphic maps f 1 , . . . , f k : X → g such that f = exp(f 1 ) · · · exp(f k ).
It is easy to see that any map f which is a product of exponentials (for some sufficiently large k) is null-homotopic. In the case where G is the special linear group SL n (C) the converse follows from [IK12] as explained in [DK19]. In the same way we prove: Theorem 15.1. For a Stein space X there is a number N depending on the dimension of X such that any null-homotopic holomorphic map f : X → Sp 4 (C) can be factorized as f (x) = exp(G 1 (x)) . . . exp(G K (x)).
where G i : X → sp 4 (C) are holomorphic maps.
Proof. By Theorem 3.1 we find K elementary symplectic matrices A i (x) ∈ Sp 4 (O(X)), i = 1, 2, . . . K, such that f (x) = A 1 (x) . . . A K (x). Now remark that the logarithmic series ln(Id + B) = 1 n B n is finite for the nilpotent matrices B i = A i − Id.
Open Problem 15.2. Determine the optimal number K in Theorem 15.1.
Open Problem 15.3. Determine the optimal numbers of factors in Theorem 3.1.

The smooth fibers
F K (a 1 ,a 2 ,a 3 ,a 4 ) = (π 4 • Ψ K ) −1 (a 1 , a 2 , a 3 , a 4 ). of the fibration projecting the product of K elementary symplectic matrices to its last row are smooth affine algebraic varieties. They are new examples of Oka manifolds, since we prove as a by-product of Proposition 3.6 that they are holomorphically flexible (for definition see [AFKKZ]). Our proof does not give the algebraic flexibility of them. Even if our initial complete fields obtained in Section 8 are algebraic. The problem is that their flows are not always algebraic (not all of them are locally nilpotent). Therefore the pull-backs by their flows are merely holomorphic vector fields.
Open Problem 15.4. Which other (stronger) flexibility properties like algebraic flexibility, algebraic (volume) density property, or (volume) density property do the fibers F K (a 1 ,a 2 ,a 3 ,a 4 ) admit? For the definition of these flexibility properties we refer to the overview article [Ku14].
Let us remark that the fibers of the fibration for 5 elementary factors in [IK12] have been thoroughly studied in [KaKu11] and [KaKu16] Section 7. They were the starting point for the introduction of the class of generalized Gizatullin surfaces whose final classification was achieved in [KKL20]. The topology of these fibers for any number of elementary factors has been studied in [DeVi20] where it was also proven that they admit the algebraic volume density property. Such studies are interesting since the possible topological types of Oka manifolds or manifolds with the density property are not understood at the moment.
Open Problem 15.5. Determine the homology groups of the fibers F K (a 1 ,a 2 ,a 3 ,a 4 ) . And finally: Open Problem 15.6. Prove Conjecture 3.7.