Remarks on the Milnor fibration at infinity

Laurenţiu Păunescu, Alexandru Zaharia

1 Introduction

For a ∈ (0, ∞), we denote: D_{a} := { w ∈ C | |w| ≤ a }, B_{a} := { z ∈ C^{n} | ∥z∥ ≤ a },

S_{a}^{1} := ∂D_{a}, S_{a} := ∂B_{a} and E_{a} := C^{n}\ B^{∘}_{a}.

Let f : C^{n} ⟶ C be a polynomial function such that f(0) = 0. In [6] J. Milnor showed that if R > 0 is sufficiently small, then the map

Φ : S_{R} \ f^{−1}(0) ⟶ S^{1} , Φ(z) := \frac{f(z)}{|f(z)|}  (1)

is the projection map of a smooth fibre bundle. This fibration, called the Milnor fibration of f at 0, is an important tool in the study of singular points of complex analytic functions. Its properties have been intensively studied, starting with [6], especially in the case of isolated singular points.

Consider now a polynomial f, without the condition f(0) = 0. In general, for R sufficiently large, the map (1) is not a fibration. (Here and in the sequel, “for R sufficiently large” will mean that “there exists R_{0} ∈ (0, ∞) such that if R ≥ R_{0}”.) In [10] it was introduced a class of polynomials, the so-called semitame polynomials, for which the map (1) is the projection map of a smooth fibre bundle when R is sufficiently large. This fibration is called the Milnor fibration at infinity, and its fibres, F_{θ} := Φ^{−1}(e^{iθ}), are called the Milnor fibres at infinity. In [10], the topological description of the fibre F_{θ} was done only for a subclass of semitame polynomials, namely for the so-called M-tame polynomials.

The aim of this paper is to continue the study started in [10] and to give a topological description of the Milnor fibre at infinity. We give a formula for the Euler-Poincaré characteristic of F_{θ} and we describe a class of polynomials f for which the Milnor fibre at infinity has the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1}, and we compute the number of spheres S^{n−1} in this bouquet.

As an application, we show that the links at infinity of some hypersurfaces diffeomorphic to affine spaces C^{k}, given in [3], are knotted spheres. These results and their proofs are given in the next two Sections.

In [11] we investigated how some metric properties of polynomials are affected by the group of algebraic automorphisms (preserving the origin) of C^{n}. For example, we found that, when n ≥ 3, the property of satisfying Malgrange condition (see [12] for the definition and some properties) of a linear form may disappear under the action. In the last Section of this paper we give examples which show that the property of being M-tame also depends on the algebraic coordinate system of C^{n}, when n ≥ 4. In particular, when n ≥ 4, our definition of M-tameness is not equivalent to the definition given in [8]. This is rather surprising and contrasts deeply to the case n = 2, where, by [16], being M-tame does not depend on the coordinate system.

Aknowledgements. The first author was supported by the ARC Grant A69530267. The second author was supported by a grant of Professors E. Bierstone, A. Khovanskii, P. Milman, and M. Spivakovsky of the University of Toronto.

Dedication. To Professor Takuo Fukuda at his sixtieth birthday anniversary.

2 Main results

Let f : C^{n} ⟶ C be a polynomial and let denote by B_{f} the minimal set B ⊂ C such that the restriction f : C^{n} \ f^{−1}(B) ⟶ C \ B is a locally trivial fibration. It is well-known, see for instance [15], [13], that B_{f} is finite. The elements of B_{f} are called atypical values of f. A generic fibre of f is a fibre f^{−1}(a) for some a ∈ C \ B_{f}. Let δ ∈ (0, ∞) be such that δ > |y| for all y ∈ B_{f}. The global monodromy fibration of f is the restriction

f : f^{−1}(S_{δ}^{1}) ⟶ S_{δ}^{1} .  (2)

The total Milnor number of f, denoted by μ(f), is the sum of the local Milnor numbers in all the singular points of f. If f has only isolated singularities, then

μ(f) = dim_{C} \frac{C[x_{1}, …, x_{n}]}{(\frac{∂f}{∂x_{1}} (x), …, \frac{∂f}{∂x_{n}} (x))} .

The gradient of f : C^{n} ⟶ C is defined by

grad f(x) := (\overline{\frac{∂f}{∂x_{1}} (x)}, …, \overline{\frac{∂f}{∂x_{n}} (x)}) ,

where x = (x_{1}, ⋯, x_{n}) and the bar denotes the conjugation. We put:

M(f) := { x ∈ C^{n} | ∃λ ∈ C such that grad f(x) = λ · x }  (3)

Geometrically, a point x ∈ M(f) \ {0} if and only if either x is a critical point of f, or x is not a critical point of f, but the hypersurface f^{−1}(f(x)) does not intersect transversally the sphere S_{∥x∥} (at x) .

For a sequence {y^{k}} ⊆ M(f) we consider the conditions

lim_{k→∞} ∥y^{k}∥ = ∞ and lim_{k→∞} f(y^{k}) = c.  (4)

We denote by Σ_{f} the set of critical values of f and we put:

S_{f} := { c ∈ C | there exists a sequence {y^{k}} ⊆ M(f) such that (4) is fulfilled }.

In general Σ_{f} ⊆ B_{f} and it may happen that B_{f} ≠ Σ_{f}. In [9] it is shown that B_{f} ⊆ Σ_{f} ∪ S_{f}. When n = 2, it follows from from [16] that B_{f} = Σ_{f} ∪ S_{f}. In the last Section we give an example, with n = 4, which shows that in general B_{f} ≠ Σ_{f} ∪ S_{f}.

The polynomial f : C^{n} ⟶ C is called semitame if S_{f} ⊆ {0}. The polynomial f is called M-tame if S_{f} = ∅. For examples and some counterexamples, we refer to [9], [11].

Let f : C^{n} ⟶ C be a semitame polynomial function. From the definition, it follows easily:

(i) If z is a non-isolated critical point of f, then f(z) = 0. If f is M-tame, then f has only isolated singularities.

(ii) Let α > 0. For R sufficiently large, we have: M(f)∩E_{R}∩f^{−1}(C\D_{α}) = ∅. We recall from [10] some properties of semitame and M-tame polynomials. (iii) For R sufficiently large, the Milnor map at infinity

Φ = \frac{f}{|f|} : S_{R} \ f^{−1}(0) ⟶ S^{1}  (5)

is the projection map of a smooth fibre bundle with fibres F_{θ} := Φ^{−1}(e^{iθ}). Assume that δ ∈ (0, ∞) is such that δ > |y| for all y ∈ Σ_{f}. (iv) For R sufficiently large, the fibrations (2) and

Φ : S_{R} \ f^{−1}(D_{δ}) ⟶ S^{1}  (6)

are diffeomorphic. Moreover, if f is M-tame, then the fibrations (2) and (5) are diffeomorphic and the Milnor fibre at infinity has the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1}, the number of spheres S^{n−1} being equal to μ(f).

(v) Let D′, D′′ ⊆ C be closed discs centered in 0, such that D′ ⊆ D′′ and D′ ≠ D′′. For R sufficiently large, and if z ∈ S_{R} \ f^{−1}(0) satisfies grad log f(z) = λz, then we have: either f(z) ∈ D′, or f(z) ∉ D′′. In the first case, arg(−λ) ∈ (\frac{−π}{4}, \frac{π}{4} ); in the second case, arg(λ) ∈ (\frac{−π}{4}, \frac{π}{4} ).

By [6], Corollary 2.6, the real algebraic sets M(f) and f^{−1}(0) can be expressed as finite disjoint unions, M(f) = M_{1} ∪ … ∪ M_{p} and f^{−1}(0) = M_{p+1} ∪ … ∪ M_{q}, where each M_{j} is a smooth manifold with only finitely many components. By [6], Corollary 2.8, we can suppose that

(vi) For R sufficiently large, the sphere S_{r} is transverse to M_{j}, for all j and for all r ≥ R.

Our first result describes the fibration Φ : S_{R}∩f^{−1}(D^{∘}_{δ} \{0}) ⟶ S^{1}. Together with (iv), this give us a better picture of the Milnor fibration at infinity.

Theorem 1. Let f : C^{n} ⟶ C be a semitame polynomial function and let δ be as above. Let R > 0 be a large number such that above conditions (ii) … (vi) hold, with α = δ/2 in (ii) and D′ = D_{δ} in (iv). Let ε ∈ (0, δ) \ Σ_{f} be such that the sphere S_{R} is transverse to f^{−1}(c) for all c ∈ D_{ε} (transversality to f^{−1}(0) in a stratified sense, as in (vi) ). Then the fibration

Φ : S_{R} ∩ f^{−1}(D^{∘}_{δ} \{0}) ⟶ S^{1}  (7)

is diffeomorphic to

f : f^{−1}(S_{ε}^{1})∩ E^{∘}_{R}⟶ S_{ε}^{1} .  (8)

Corollary 2. With the hypothesis of Theorem 1, we have:

χ(F_{θ}) = χ(f^{−1}(ε)) + χ(f^{−1}(ε) ∩ E_{R}) .  (9)

If 0 is a regular value of f, then

χ(F_{θ}) = 2χ(f^{−1}(ε)) − χ(f^{−1}(0)) .  (10)

In particular, if 0 is a regular value of f and χ(f^{−1}(ε)) ≠ χ(f^{−1}(0)), then F_{θ} is not diffeomorphic to f^{−1}(ε).

Let f : C^{n} ⟶ C be a polynomial and let B_{f} = {b_{1}, …, b_{s}} denote its atypical values. Let Δ_{i} be a small closed disc centered at b_{i}. Assume that these discs are pairwise disjoint, and let a_{i} ∈ Δ \ {b_{i}}. In [2], (proof of) Theorem 1.1, Broughton shows that for R sufficiently large and for a ∈ C \ B_{f},

H_{n−1}(f^{−1}(a)) ≅ Z^{μ(f)} ⊕ ⨁_{i=1}^{s} H_{n}(f^{−1}(Δ_{i}) ∩ E_{R}, f^{−1}(a_{i}) ∩ E_{R}) .

We denote by λ(f) the rank of ⨁_{i=1}^{s} H_{n}(f^{−1}(Δ_{i}) ∩ E_{R}, f^{−1}(a_{i}) ∩ E_{R}). With this notation, we have the following:

Theorem 3. Let f : C^{n} ⟶ C be a semitame polynomial. Suppose that f has only isolated singularities and that its generic fibre has the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1} with μ(f) + λ(f) spheres. Then the Milnor fibre at infinity has also the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1} with μ(f) + 2λ(f) spheres.

3 Proofs

Proof of Theorem 1. We define ψ : C^{n} ⟶ R^{2} by: ψ(z) := (|f(z)|, ∥z∥). By semitameness, the set B := M(f) ∩ f^{−1}(\overline{D_{δ} \ D_{ε}}) is compact. Therefore, the set Y := ψ (B) is also compact. By the choice of δ, we have: Y ∩ ({δ} × [R, ∞)) = ∅. Therefore, there exists a line L ⊂ R^{2}, with slope equal to −b_{0}, for some b_{0} ∈ (0, ∞), such that L ∩ Y = ∅, Y is situated under L and (δ, R) ∈ L. Let U be a neighborhood of B such that ψ(U) is situated under L and such that L ∩ ψ(U) = ∅.

The set A := E_{R} ∩ B is compact, hence, by (i), there exists σ > 0 such that for all z ∈ A, if λ satisfies grad log f(z) = λz, then |λ| ≥ σ. Therefore, for a suitable large b′ > b_{0}, we have:

sup \{ Re\frac{−λ}{|λ|^{2}} | z ∈ A and grad log f(z) = λz \} < b′ .

Let b > b′ and let R_{1} := R + b(δ − ε). Then the segment Δ, which joins the points (ε, R_{1}) and (δ, R) in R^{2}, is above L, and the fibrations (8) and

f :B^{∘}_{R_{1}} ∩f^{−1}(S_{ε}^{1})∩ E^{∘}_{R}⟶ S_{ε}^{1}  (11)

are diffeomorphic.

We will construct a vector field v in a neighborhood X of E_{R} ∩ f^{−1}(\overline{D_{δ} \ D_{ε}}), satisfying the following properties:

(∗) For all z ∈ X, ⟨v(z), grad log f(z)⟩ = −1.

(∗∗) For all z ∈ X \ U, ⟨v(z), z⟩ = b.

(∗ ∗ ∗) For all z ∈ X ∩ U, Re⟨v(z), z⟩ ∈ (0, b].

We take X = E_{r} ∩ f^{−1}(D_{δ′} \ D_{ε′}), for some r < R close to R, some δ′ > δ close to δ and some ε′ < ε close to ε. For any z ∈ X \ M(f), the vectors z and grad log f(z) are linearly independent, hence there exists a neighborhood W_{z} ⊆ X \ M(f) of z such that for any u ∈ W_{z}, the equations ⟨w, grad log f(u)⟩ = −1 and ⟨w, u⟩ = b have a common solution, w = w_{z}(u), which depends smoothly on u.

If z ∈ X ∩ M(f), the vectors z and grad log f(z) are linearly dependent, hence there exists λ ∈ C such that grad log f(z) = λz. By hypothesis, we have arg(−λ) ∈ (\frac{−π}{4}, \frac{π}{4} ). Thus, the vector w = \frac{−λz}{∥λz∥^{2}} satisfy ⟨w, grad log f(z)⟩ = −1 and Re⟨w, z⟩ = Re\frac{−λ}{|λ|^{2}} ∈ (0, b′). We can extend the vector w to a a vector field w_{z}, defined in a small ball U_{z} centered at z, such that conditions ⟨w_{z}(u), grad log f(u)⟩ = −1 and Re⟨w_{z}(u), u⟩ ∈ (0, b′) are satisfied for all u ∈ U_{z}. Moreover, we can assume that the radius of U_{z} is small enough, such that U_{z} ⊆ U.

Next, using a partition of the unity, we glue together the vector fields w_{z}, for all z ∈ X, and we obtain the desired vector field v.

Let p(s) be an integral curve of the vector field v. Property (∗) tells us that the argument of f(p(s)) is constant and that |f(p(s))| is strictly decreasing as a function of s. Properties (∗∗) and (∗ ∗ ∗) guarantee that ∥p(s)∥ is a strictly increasing function of s. Thus, starting at any point z ∈ F_{θ}∩f^{−1}(D) and following the trajectory through z, we travel in X, “away” from the origin, in a direction of decreasing |f|, until we reach a point z′ ∈ f^{−1}(εe^{iθ}) ∩ E_{R}. Note also that starting at a point z ∈ f^{−1}(δe^{iθ}) ∩ S_{R} and following the trajectory of the vector field v, the point ψ(p(s)) stays in Δ, hence we rich a point in f^{−1}(εe^{iθ}) ∩ S_{R_{1}}. Therefore, the correspondence z ⟼ z′ gives us that the fibrations (11) and

Φ : S_{R} ∩ f^{−1}(D^{∘}_{δ} \D_{ε}) ⟶ S^{1}  (12)

are diffeomorphic. To complete the proof, we note that the choice of ε guarantees that the fibrations (12) and (7) are diffeomorphic.

♢

Proof of Corollary 2. Let δ′ > δ be close to δ. Relation (9) follows from the Mayer-Vietoris sequence associated to F_{θ} = (F_{θ} ∩ f^{−1}(D_{δ′}))∪(F_{θ} ∩ f^{−1}(C \ D_{δ})) and from the fact that χ(f^{−1}(δ) ∩ S_{R}) = 0.

When 0 is a regular value of f, relation (10) follows from (9), since f^{−1}(0) is isomorphic to f^{−1}(ε)∩ B^{∘}_{R} and χ(f^{−1}(0) ∩ S_{R}) = 0.

♢

Proof of Theorem 3. By [10], Lemma 15, the closure \overline{F_{θ}} is embedded in S_{R} in such a way as to have the same homotopy type as its complement S_{R} \ \overline{F_{θ}}. By Alexander duality, the reduced homology group H̃_{i}(F_{θ}) = H̃_{i}(S_{R} \ \overline{F_{θ}}) is isomorphic to the reduced cohomology group H̃^{2n−i}(\overline{F_{θ}}). By [10], Corollary 14, F_{θ} is obtained from a generic fibre f^{−1}(a) by attaching cells of dimension at least n − 1. Since f^{−1}(a) has the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1}, it follows that F_{θ} is (n − 2)-connected. And now, the argument in [6], page 58, shows that F_{θ} has the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1}. Assume that the number of spheres in this bouquet is μ(f) + λ(f) + ν. We have to show that ν = λ(f). We will take ε > 0 very small. Relation (9) implies that

(−1)^{n−1}ν = χ(f^{−1}(ε) ∩ E_{R}) .  (13)

The exact sequence of the pair (f^{−1}(D_{ε}) ∩ E_{R} , f^{−1}(ε) ∩ E_{R}) implies that

χ(f^{−1}(ε) ∩ E_{R}) = χ(f^{−1}(D_{ε}) ∩ E_{R}) − χ(f^{−1}(D_{ε}) ∩ E_{R} , f^{−1}(ε) ∩ E_{R}) . (14)

If ε is sufficiently small, we have (see for instance [1], page 320)

χ(f^{−1}(D_{ε})) = χ(f^{−1}(0)) .

This implies, for R sufficiently large, that χ(f^{−1}(D_{ε}) ∩ E_{R}) = 0. Thus, the relations (13) and (14) show that the equality ν = λ(f) is equivalent to

χ(f^{−1}(D_{ε}) ∩ E_{R} , f^{−1}(ε) ∩ E_{R}) = (−1)^{n}λ(f) .

And now, this equality follows easily since the generic fiber of f has the homotopy type of a bouquet S^{n−1} ⋁ … ⋁ S^{n−1}, by using semitameness and [2], relation (1.2) and pages 232-233.

♢

Remark. The monodromy of the fibration (2) was studied in some papers, see for instance [7], [5], [14] and the references quoted there. When f is a semitame polynomial, one can compute the zeta-function of the monodromy associated with the Milnor fibration at infinity of f by combining Theorems 1 and 4 in [5] with our Theorem 1.

Example. As in [3], Theorem 5, let m, d and a be integers such that m ≥ 2, 0 < a < d − 1 and (a, d) = (a, d − 1) = 1. Let X_{d,a} ⊂ C^{2m} be the hypersurface defined by the equation f(x) = 0, where

f(x_{0}, ...x_{2m−1}) = x_{0}^{a}x_{1}^{d−a} + x_{1}x_{2}^{d−1} + ... + x_{2m−3}x_{2m−2}^{d−1} + x_{2m−2} + x_{2m−1}^{d} .

The polynomial f is quasihomogeneous (with integer weights), hence, by [10], f is semitame. Moreover, μ(f) = 0. In [3], page 174, it is shown that the link at infinity of X_{d,a} is diffeomorphic to a sphere S^{2m−3}. Moreover, Theorem 5 and its proof in [3], Theorem 5.2 in [2] and the argument in [6], pag. 58, show that the generic fibre of f has the homotopy type of a bouquet S^{2m−1} ⋁ … ⋁ S^{2m−1} and that, for a > 1, we have λ(f) > 0. Thus, when a > 1, the Milnor fibre at infinity is not contractible. Now, the argument in [6], page 61, shows that, for a > 1, the link at infinity of X_{d,a} is a knotted sphere.

Note also that when a = 1, the polynomial f is M-tame.

4 M-tameness and the coordinate system of C^{n}

The problem of describing the atypical values of a polynomial f : C^{n} ⟶ C was considered by many authors, see for instance [16], [9], [12], [4] and the references quoted there. One approach to this problem was made by considering the behavior of the gradient of f near infinity. In [9] it is shown that B_{f} ⊆ Σ_{f} ∪ S_{f}. When n = 2, it follows from from [16] that B_{f} = Σ_{f} ∪ S_{f}. We give an example, with n = 4, which shows that in general B_{f} ≠ Σ_{f} ∪ S_{f}.

Theorem 4. Let f : C^{4} ⟶ C be defined by

f(x, y, z, u) = x + y − 2x^{2}y^{3} + x^{3}y^{6} + zy^{3} − z^{2}y^{5} + uy^{5} .

Then we have:

(1) There exists a polynomial automorphism φ : C^{4} ⟶ C^{4} such that φ = (f, φ_{2}, φ_{3}, φ_{4}). In particular, B_{f} = ∅.

(2) The polynomial f is not M-tame. More precisely, S_{f} = {0}, hence f is semitame.

Remark. By Theorem 3, the Milnor fibre at infinity of the above polynomial is still contractible, despite the fact that it is obtained by adding cells of dimension at least 3 to the generic fibre (which is diffeomorphic to C^{3}).

Corollary 5. Let f : C^{4} ⟶ C be as in Theorem 4, let g : C ⟶ C be any polynomial and let h : C^{5} ⟶ C be defined by h(x, y, z, u, v) = f(x, y, z, u)+g(v). Then h is a component of a polynomial automorphism of C^{5} and has the property that S_{h} = Σ_{g}. In particular, if g has at least two critical values, then h is not semitame.

Proof of Theorem 4. The coordinate change U = u−z^{2} gives us f(x, y, z, U) = x + y − 2x^{2}y^{3} + x^{3}y^{6} + zy^{3} + Uy^{5}. Next, the coordinate change Z = z − 2x^{2} + x^{3}y^{3} + Uy^{2} gives us f(x, y, Z, U) = x + y + Zy^{3}, and now the first point follows.

We prove now that f is semi–tame. Using Curve Selection Lemma at infinity, see for instance [10], we can replace the sequences in the definition of S_{f} by functions, p : (0, ϵ) ⟶ C^{4} and λ : (0, ϵ) ⟶ C, given by Laurent power series, such that the following conditions hold:

(a) lim_{s→0} ∥p(s)∥ = ∞.

(b) lim_{s→0} f(p(s)) ∈ C.

(c) The relation grad f(p(s)) = λ(s) · p(s) is satisfied for all s ∈ (0, ϵ).

We have to show that that the limit in (b) is equal to 0. Let us write p(s) = (x(s), y(s), z(s), u(s)) and let us denote by f_{x}, f_{y}, f_{z} and f_{u} the partial derivative of f with respect to x, y, z and respectively u. Let ord (λ) be the order of the Laurent power series λ. If φ : C^{4} ⟶ C is a polynomial, ord (φ) will denote the order of φ(p(s)).

First we note that the condition (b) implies that lim_{s→0} grad f(p(s)) = 0. Considering f_{u}, it follows that lim_{s→0} y(s) = 0. Next, considering f_{x}, we obtain that

lim_{s→0} x(s)y(s)^{3} is a root of the equation 1 − 4T + 3T^{2} = 0, hence it can be equal to

\frac{1}{3} or 1. Thus, lim_{s→0} ∥x(s)∥ = ∞ and ord (x) = −3ord (y) < 0.

Suppose now that lim_{s→0} x(s)y(s)^{3} = \frac{1}{3}.

Since 5f −yf_{y} = x(5+xy^{3})(1−xy^{3})+2y^{3}z+4y tends to a finite value along the curve p(s), when s tends to 0, and since ord (x(5 + xy^{3})(1 − xy^{3})) = ord (x) < 0, we must have ord (y^{3}z) = ord (x), hence ord (z) = −6ord (y). This implies that ord (f_{z}) = −ord (y) < 0, hence lim_{s→0} f_{z}(p(s) ≠ 0, a contradiction.

It remains that lim_{s→0} x(s)y(s)^{3} = 1. Let us put a = ord (1 − xy^{3}). Then 0 <

a = ord (f_{x}) = ord (\overline{λx}), hence ord (λ) = a + 3ord (y). Now it follows that ord (u) = 2ord (y) − a and ord (f_{y}) = a + 4ord (y).

Let us put b = ord (z). Then ord (f_{z}) = ord (\overline{λz}) = a + 3ord (y) + b, hence ord (1 − 2zy^{2}) = a + b. Since f_{z} = y^{3} − 2zy^{5} tends to 0 along p(s), it follows that b > −5ord (y).

Suppose that a + b = 0. Then a = −b = 2ord (y) (since ord (1 − 2zy^{2}) = a + b = 0). It follows that ord (5f − yf_{y}) = −ord (y) < 0, in contradiction to the fact that 5f − yf_{y} tends to a finite value along p(s).

Suppose now that 0 > a + b. Using the inequality b > −5ord (y), it follows that 5ord (y) > a, hence ord (y^{4}u) = 6ord (y) − a > 0. Note also that 0 > a + b = ord (1 − 2zy^{2}) = ord (zy^{2}) = b + 2ord (y), hence a = 2ord (y) and ord (y^{2}z(3 − 5y^{2}z)) = 2b + 4ord (y) < 0.

We consider now f_{y} = −6x^{2}y^{2}(1 − xy^{3}) − y^{2}z(3 − 5y^{2}z) + 5y^{4}u + 1. Since f_{y} tends to 0 along p(s), we must have ord (x^{2}y^{2}(1 − xy^{3})) = 2b + 4ord (y), hence −2ord (y) = 2b + 4ord (y), i.e. b = −3ord (y). And now it follows that ord (5f − yf_{y}) = −ord (y) < 0, a contradiction.

It remains to consider the case when 0 < a + b. It follows that 0 < a + b = ord (1 − 2zy^{2}), hence b = ord (z) = −2ord (y) and y^{2}z tends to \frac{1}{2} along the curve p(s). And now, since ord (x^{2}y^{2}(1−xy^{3})) = a−4ord (y) and ord (y^{4}u) = 6ord (y)− a, the condition that f_{y} tends to infinity along p(s) implies that a = 4ord (y) or a = 6ord (y). Both these values for a implies that 5f − yf_{y} tends to 0 along p(s), hence S_{f} ⊆ {0}.

To prove that f is not M-tame, we show that there exist differentiable functions p : (0, ϵ) ⟶ C^{4} and λ : (0, ϵ) ⟶ C such that the above conditions (a), (b) and (c) hold. We take

\overline{λ(s)} = s^{7} · \frac{ϱ(s) · (2 + 3s^{4}ϱ(s))}{1 + s^{4}\overline{ϱ(s)}}

and p(s) = (x(s), y(s), z(s), u(s)), where

x(s) = s^{−3} + s · ϱ(s) , y(s) = s , z(s) = \frac{1}{2}s^{−2} − η(s)

and

\overline{u(s)} = s^{−2} · \frac{1 + s^{4}\overline{ϱ(s)}}{ϱ(s) · (2 + 3s^{4}ϱ(s))} .

We will choose the functions ϱ, η : [0, ϵ) ⟶ C such that conditions (a), (b) and (c) hold.

First, we note that for any continuous functions ϱ, η : [0, ϵ) ⟶ C, the conditions (a) and (b) hold, if ϱ(0) ≠ 0.

Next, we observe that

\frac{∂f}{∂x}(p(s)) = s^{4}ϱ(s)(2 + 3s^{4}ϱ(s)) = \overline{λ(s) · x(s)}

and

\frac{∂f}{∂u}(p(s)) = s^{5} = \overline{λ(s) · u(s)} .

Thus, condition (c) will be satisfied if

\frac{∂f}{∂y} (p(s)) = \overline{λ(s) · y(s)}  (15)

and

\frac{∂f}{∂z} (p(s)) = \overline{λ(s) · z(s)} .  (16)

Relation (16) implies

2η(s) = \frac{ϱ(s)(2 + 3s^{4}ϱ(s))}{1 + s^{4}\overline{ϱ(s)}} · (\frac{1}{2} − s^{2}\overline{η(s)})  (17)

and relation (15) implies

\frac{5}{4} + 6ϱ(s) · (1 + s^{4}ϱ(s))^{2} + s^{2} [2η(s) + \frac{5(1 + s^{4}ϱ(s))}{\overline{ϱ}(s)(2 + 3s^{4}\overline{ϱ}(s))}] − 5s^{4}(η(s))^{2} =

= s^{8} · \frac{ϱ(s) · (2 + 3s^{4}ϱ(s))}{1 + s^{4}\overline{ϱ(s)}}  (18)

We consider now in R × C^{2}, with coordinates (s, ϱ, η), the real algebraic set X described by the equations which corresponds to (17) and (18), namely:

2η = \frac{ϱ(2 + 3s^{4}ϱ)}{1 + s^{4}\overline{ϱ}} · (\frac{1}{2} − s^{2}\overline{η})

and

\frac{5}{4} + 6ϱ · (1 + s^{4}ϱ)^{2} + s^{2} [2η + \frac{5(1 + s^{4}ϱ)}{\overline{ϱ}(2 + 3s^{4}\overline{ϱ})}] − 5s^{4}η^{2} = s^{8} · \frac{ϱ · (2 + 3s^{4}ϱ)}{1 + s^{4}\overline{ϱ}}

It is easy to see that the point P := (0, \frac{−5}{24}, \frac{−5}{48}) ∈ X and that X is smooth in P. Moreover, X is transversal to the hyperplane { s = 0 } in P. Therefore there exists a C^{∞} parametrization of X near P, hence there exists C^{∞} functions

ϱ, η : [0, ϵ) ⟶ C such that (15) and (16) hold and ϱ(0) = \frac{−5}{24}.

♢

Proof of Corollary 5. Follows easily from the definitions.

♢

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