Some properties of polynomial maps in terms of newton polyhedrons

c as s→ 0; (b7) gj(ϕ(s)) = 0 for all j = 1, . . . , p, and all s ∈ (0, ϵ); (b8) ∇f(ϕ(s)) +∑pj=1 λj(s)∇gj(ϕ(s)) = λp+1(s)ϕ(s) for s ∈ (0, ϵ). If λp+1 ≡ 0, then it is clear that c ∈ K∞(f, S) and there is nothing to prove. So we may assume that λp+1 is not identically zero. It follows from (b7) and (b8) that 0 ̸≡ 1 2 d∥ϕ(s)∥2 ds = ℜ 〈 dϕ(s) ds , ϕ(s) 〉 = ℜ 〈 dϕ(s) ds , 1 λp+1(s) [ ∇f(ϕ(s)) + p∑ j=1 λj(s)∇gj(ϕ(s)) ]〉 = ℜ 1 λp+1(s) [ d ds (f ◦ ϕ)(s) + p∑ j=1 λj(s) d ds (gj ◦ ϕ)(s) ] = ℜ 1 λp+1(s) [ d ds (f ◦ ϕ)(s) ] . In particular, f ◦ ϕ ̸≡ c. On the other hand, we may write ∥ϕ(s)∥ = asα + (higher-order terms), f(ϕ(s)) = c+ bsβ + (higher-order terms), 17 where a ̸= 0, b ̸= 0 and α, β ∈ Q. By the conditions (b5) and (b6) respectively, then α 0. Furthermore, a simple computation shows that ords|λp+1(s)| ≥ β − 2α. It turns out from (b8) that ords ( ∥ϕ(s)∥∥∇f(ϕ(s)) + p∑ j=1 λj(s)∇gj(ϕ(s))∥ ) = ords|λp+1(s)|+ 2α ≥ β. Since β > 0, we get lim s→0+ ( ∥ϕ(s)∥∥∇f(ϕ(s)) + p∑ j=1 λj(s)∇gj(ϕ(s))∥ ) = 0. Therefore, c ∈ K∞(f, S), and so the inclusion (2.2) holds. For the proof of the inclusion (2.1) we fix c∗ ∈ C \ (K0(f |S) ∪ T∞(f |S)) and D a small open disc centered at c∗, with the closure D ⊂ C \ (K0(f |S) ∪ T∞(f |S)). Then it is not hard to see that there exists a real number R0 > 0 such that for all c ∈ D and all R ⩾ R0, the fiber (f |S)−1(c) is non-singular and intersects transversally with the sphere S2n−1R (this is possible if D is small enough). By continuity, there exists an open neighbourhood U of (f |S)−1(D)∩{x ∈ Cn | ∥x∥ ≥ R0} in Cn such that the vectors ∇f(x),∇g1(x), . . . ,∇gp(x), and x are C-linearly independent for all x ∈ U. Therefore, we can find a smooth vector field v1 on U satisfying the following conditions • ⟨v1(x),∇f(x)⟩ = 1; • ⟨v1(x),∇gj(x)⟩ = 0 for j = 1, . . . , p; • ⟨v1(x), x⟩ = 0. (We can construct such a vector field locally, then extend it over U by a smooth partition of unity.) We now fix ϵ > 0. Since D ∩K0(f, S) = ∅, the vectors ∇f(x),∇g1(x), . . . ,∇gp(x) are C-linearly independent for all x belonging to some open neighbourhood V of (f |S)−1(D)∩ 18 {x ∈ Cn | ∥x∥ ≤ R0 + ϵ} in Cn. Consequently, there exists a smooth vector field v2 on V such that the following conditions hold • ⟨v2(x),∇f(x)⟩ = 1; • ⟨v2(x),∇gj(x)⟩ = 0 for j = 1, . . . , p. (We can construct such a vector field locally, then extend it over V by a smooth partition of unity.) Next, we fix a partition of unity θ1 and θ2 subordinated to the covering{ x ∈ U | ∥x∥ > R0 + ϵ 3 } and { x ∈ V | ∥x∥ < R0 + 2ϵ 3 } of (f |S)−1(D), and define the smooth vector field v on (f |S)−1(D) by v := θ1v1 + θ2v2. Then we can see that the following conditions hold: • ⟨v(x),∇f(x)⟩ = 1; • ⟨v(x),∇gj(x)⟩ = 0 for j = 1, . . . , p; • ⟨v(x), x⟩ = 0 provided that ∥x∥ ≥ R0 + ϵ. Finally, integrating the vector field v we have that the restriction f : (f |S)−1(D) → D is a trivial C∞-fibration, which means that c∗ ̸∈ B(f |S). Remark 2.4. (i) The inclusion (2.1) provides an extension to algebraic sets of Theorem 1 in (Ne´methi & Zaharia, 1990), where the case S = Cn was studied. (ii) The inclusions (2.1) and (2.2) may be strict in general, see (Pa˘unescu & Zaharia, 1997, 2000) and (Ha` & Pha.m, 2008). 19 (iii) The proof of Theorem 2.3 also implies the following inclusion, which was proved in (Rabier, 1997; Jelonek, 2004; Jelonek & Kurdyka, 2005), B(f |S) ⊂ K0(f |S) ∪K∞(f |S). (iv) A straightforward modification shows that Lemma 3.1 and Theorem 2.3 still hold in the case where S does not have the explicit form as it was assumed; in fact, it suffices to suppose that S is a non-singular constructive subset of Cn. As we shall not use this “improve” statement, we leave the proof as an exercise. Under the non-degeneracy condition of Definition 1.18, we obtain the following bound of tangency values at infinity of f |S in terms of critical values of certain polynomial functions. Theorem 2.5. Assume that the restriction f |S of f on S is Newton non-degenerate at infinity. Then T∞(f |S) ⊂ Σ∞(f |S) ∪K0(f |S) ∪ {0}. Moreover, if the polynomial f : Cn → C is convenient, then T∞(f |S) = ∅. Proof. For convenience we will write g0 instead of f. Take arbitrarily c ∈ T∞(g0|S)\(K0(g0|S)∪{0}).We will show that c ∈ Σ∞(f |S). Indeed, by definition, there exist sequences {x(k)}k∈N ⊂ Cn and {λ(k)j }k∈N ⊂ C, j = 1, . . . , p+ 1, such that (c1) ∥x(k)∥ → ∞ as k →∞; (c2) g0(x (k))→ c as k →∞; (c3) gj(x (k)) = 0 for all j = 1, . . . , p, and all k ∈ N; (c4) ∇g0(x(k)) + ∑p j=1 λ (k) j ∇gj(x(k)) = λ(k)p+1x(k) for all k ∈ N. 20 By the Curve Selection Lemma at infinity (see Theorem 1.14), there exist analytic curves ϕ : (0, ϵ)→ Cn and λj : (0, ϵ)→ C, j = 1, . . . , p+ 1, such that (c5) ∥ϕ(s)∥ → ∞ as s→ 0; (c6) g0(ϕ(s))→ c as s→ 0; (c7) gj(ϕ(s)) = 0 for all j = 1, . . . , p, and all s ∈ (0, ϵ); (c8) ∇g0(ϕ(s)) + ∑p j=1 λj(s)∇gj(ϕ(s)) = λp+1(s)ϕ(s) for s ∈ (0, ϵ). Put I := {i | ϕi ̸≡ 0}. By the condition (c5), I ̸= ∅. For i ∈ I, we can write the curve ϕi in terms of parameter, say ϕi(s) = x 0 i s qi + (higher-order terms), where x0i ̸= 0 and qi ∈ Q. We have mini∈I qi < 0, because of the condition (c5). If λp+1 ≡ 0, then it follows from the conditions (c7) and (c8) that d ds (g0 ◦ ϕ)(s) = 〈 dϕ(s) ds ,∇g0(ϕ(s)) 〉 = − p∑ j=1 λj(s) 〈 dϕ(s) ds ,∇gj(ϕ(s)) 〉 = − p∑ j=1 λj(s) d ds (gj ◦ ϕ)(s) = 0. Consequently, g0(ϕ(s)) = c for s ∈ (0, ϵ), and so c ∈ K0(g0|S) by (c8), which is a contradiction. Therefore, λp+1 ̸≡ 0. Put J := {j ∈ {1, . . . , p} | λj ̸≡ 0}. For j ∈ J ∪ {p+ 1}, we can write λj(s) = cjs mj + (higher-order terms), where cj ̸= 0 and mj ∈ Q. Put J1 := {j ∈ {0} ∪ J | gj|CI ̸≡ 0}. The condition (c6) and the assumption that c ̸= 0 together imply that 0 ∈ J1, and so J1 ̸= ∅. For each j ∈ J1, let dj be the minimal 21 value of the linear function ∑ i∈I αiqi on RI ∩ Γ(gj) and ∆j be the face of RI ∩ Γ(gj), where this linear function takes its minimum value, respectively. A simple calculation shows that gj(ϕ(s)) = gj,∆j(x 0)sdj + (higher-order terms), where x0 := (x01, . . . , x 0 n) can be understood as x 0 i = 1 for i ̸∈ I because the face function gj,∆j , which is associated with gj and ∆j, does not depend on the variable xi. The condition (c6) and the assumption that c ̸= 0 together imply that d0 ⩽ 0 and d0g0,∆0(x0) = 0. (2.3) Otherwise, d0 < 0 and g0(ϕ(s))→ +∞, which contradicts (c6). Furthermore, it follows from the condition (c7) that gj,∆j(x 0) = 0 for all j ∈ J1 \ {0}. (2.4) On the other hand, we have for all i ∈ I and all j ∈ J1, ∂gj(ϕ(s)) ∂xi = ∂gj,∆j ∂xi (x0)sdj−qi + (higher-order terms). (2.5) Put ℓ := min{mj + dj | j ∈ J1}, J2 := {j ∈ J1 | ℓ = mj + dj}. Then the condition (c8) and the equality (2.5) together imply that for all i ∈ I,(∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) sℓ−qi + (higher-order terms) = cp+1x0i s mp+1+qi + (higher-order terms), where c0 := 1 and m0 := 0. Consequently, ℓ− qi ⩽ mp+1 + qi for all i ∈ I. Therefore, ℓ−mp+1 ⩽ 2min i∈I qi < 0. (2.6) 22 Let I1 := {i ∈ I | ℓ− qi = mp+1 + qi}. We show that the set I1 is empty. To see this, we observe that ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) =  cp+1x0i if i ∈ I1, 0 if i ∈ I \ I1, 0 if i ̸∈ I, where the last equality holds because for all i ̸∈ I and all j ∈ J2, the polynomial gj,∆j does not depend on the variable xi. Consequently, n∑ i=1 (∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) x0i qi = ∑ i∈I1 (∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) x0i qi = ∑ i∈I1 cp+1|x0i |2 ℓ−mp+1 2 . The left side of the above equality can be written as n∑ i=1 (∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) x0i qi = ∑ j∈J2 cj ( n∑ i=1 ∂gj,∆j ∂xi (x0)x0i qi ) = ∑ j∈J2 cjdjgj,∆j(x 0), where the second equality follows from the Euler relation: n∑ i=1 ∂gj,∆j ∂xi (x0)x0i qi = djgj,∆j(x 0) for j ∈ J2. Therefore, ∑ i∈I1 cp+1|x0i |2 ℓ−mp+1 2 = ∑ j∈J2 cjdjgj,∆j(x 0). As the left side of this equality is nonzero if I1 ̸= ∅, we get from (2.3) and (2.4) that I1 = ∅. Thus, ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0 for all i = 1, . . . , n. Since the restriction of g0 on S is Newton non-degenerate at infinity, the case where d0 < 0 and g0,∆0 = 0 does not occur. Thus d0 = 0 and c = g0,∆0(x 0) ∈ Σ∞(g0|S). 23 Finally, assume that the polynomial f : Cn → C is convenient. Then d0 < 0, which is a contradiction. Hence T∞(f |S) = ∅. For S = Cn, the next statement was shown in (Ne´methi & Zaharia, 1990, Theorem 2). Corollary 2.6. Under the assumption of Theorem 2.5, we have B(f |S) ⊂ Σ∞(f |S) ∪K0(f |S) ∪ {0}. Moreover, if the polynomial f : Cn → C is convenient, then B(f |S) = K0(f |S). Proof. This is an immediate consequence of Theorems 2.3 and 2.5 and the fact that the bifurcation set B(f |S) contains the set K0(f |S) of critical values of the restriction f |S. 2.2 The stability of global monodromies By the inspiration of Theorem 0.3 , we begin our works about the stability result. Recall that the (non-singular) algebraic set S is given by S := {x ∈ Cn | g1(x) = 0, . . . , gp(x) = 0}. In what follows, let f : C× Cn → C, (t, x) 7→ f(t, x), be a polynomial function. We will write ft(x) := f(t, x) and assume that for each t ∈ [0, 1], the restriction ft|S : S → C is dominant (i.e., the image set ft(S) is dense in C). Theorem 2.7. Let the following conditions are satisfied: (i) The Newton polyhedron of ft is independent of t; (ii) For each t ∈ [0, 1], the restriction ft|S is Newton non-degenerate at infinity. Then the global monodromies of the ft|S are all isomorphic. 24 The proof of Theorem 2.7 will be divided into several steps, which, for convenience, will be called lemmas. Lemma 2.8 (Boundedness of affine singularities). There exists a real number r > 0 such that K0(ft|S) ⊂ Dr for all t ∈ [0, 1]. Proof. Suppose the lemma were false. Then by the Curve Selection Lemma at infinity (see Theorem 1.14), there exist analytic curves ϕ : (0, ϵ)→ Cn, t : (0, ϵ)→ [0, 1], and λj : (0, ϵ)→ C, j = 1, . . . , p, such that (d1) ∥ϕ(s)∥ → ∞ as s→ 0; (d2) t(s)→ t0 ∈ [0, 1] as s→ 0; (d3) ft(s)(ϕ(s))→∞ as s→ 0; (d4) gj(ϕ(s)) = 0 for all j = 1, . . . , p, and all s ∈ (0, ϵ); (d5) ∇ft(s)(ϕ(s)) + ∑p j=1 λj(s)∇gj(ϕ(s)) = 0 for s ∈ (0, ϵ). Put I := {i | ϕi ̸≡ 0}. By the condition (d1), I ̸= ∅. For i ∈ I, we can write the curve ϕi in terms of parameter, say ϕi(s) = x 0 i s qi + (higher-order terms), where x0i ̸= 0 and qi ∈ Q. Observe that mini∈I qi < 0 because of the condition (d1). Recall from our assumptions that the Newton polyhedron Γ(ft) of ft does not depend on t. By the condition (d3), RI ∩ Γ(ft) ̸= ∅. Let d0 be the minimal value of the linear 25 function ∑ i∈I αiqi on RI ∩ Γ(ft) and ∆0 be the face of RI ∩ Γ(ft) where this linear function takes its minimum value. We can write ft(s)(ϕ(s)) = ft0,∆0(x 0)sd0 + (higher-order terms), ∂ft(s) ∂xi (ϕ(s)) = ∂ft0,∆0 ∂xi (x0)sd0−qi + (higher-order terms) for all i ∈ I, where x0 := (x01, . . . , x 0 n) with x 0 i = 1 for i ̸∈ I and ft0,∆0 denotes the face function of ft0 with respect to the face ∆0. By the condition (d3), d0 < 0. Furthermore, for i /∈ I, the function ft0,∆0 does not depend on the variable xi, and so ∂ft0,∆0 ∂xi (x0) = 0 for all i ̸∈ I. (2.7) Put J := {j ∈ {1, . . . , p} | λj ̸≡ 0}. If J = ∅, then from the condition (d5) we deduce for all i ∈ I that ∂ft(s) ∂xi (ϕ(s)) = 0, and hence that ∂ft0,∆0 ∂xi (x0) = 0. It turns out from (2.7), the Euler relation, and the inequality d0 < 0 that ft0,∆0(x 0) = 0, which contradicts the non-degeneracy condition. Therefore, J ̸= ∅. For j ∈ J , we can write λj(s) = cjs mj + (higher-order terms), where cj ̸= 0 and mj ∈ Q. Put J1 := {j ∈ J | gj|CI ̸≡ 0}. If J1 = ∅, then ∂gj ∂xi (ϕ(s)) ≡ 0 for all i ∈ I and all j ∈ J. We deduce from the condition (d5) that ∂ft(s) ∂xi (ϕ(s)) ≡ 0 for all i ∈ I. Consequently, ∂ft0,∆0 ∂xi (x0) = 0 for all i ∈ I. It follows from (2.7), the Euler relation, and the inequality d0 < 0 that ft0,∆0(x 0) = 0, which contradicts the non-degeneracy condition. Hence J1 ̸= ∅. For each j ∈ J1, let dj 26 be the minimal value of the linear function ∑ i∈I αiqi on RI ∩ Γ(gj) and ∆j be the face of RI ∩ Γ(gj) where this linear function takes its minimum value. We can write gj(ϕ(s)) = gj,∆j(x 0)sdj + (higher-order terms), where gj,∆j is the face function of gj with respect to the face ∆j. By the condition (d4), then gj,∆j(x 0) = 0 for all j ∈ J1. (2.8) On the other hand, for i ∈ I and j ∈ J1, ∂gj ∂xi (ϕ(s)) = ∂gj,∆j ∂xi (x0)sdj−qi + (higher-order terms). For i /∈ I and j ∈ J1, the function gj,∆j does not depend on the variable xi, and hence, ∂gj,∆j ∂xi (x0) = 0. (2.9) Let ℓ := min j∈J1 (mj + dj), J2 := {j ∈ J1 | ℓ = mj + dj}. The condition (d5) implies that for all i ∈ I, ∂ft0,∆0 ∂xi (x0)sd0−qi + (higher-order terms) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0)sℓ−qi + (higher-order terms) = 0.(2.10) There are three cases to be considered. Case 1: ℓ = d0 We deduce from (2.7), (2.9) and (2.10) that ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0 for i = 1, . . . , n. 27 Consequently, 0 = n∑ i=1 qix 0 i ∂ft0,∆0 ∂xi (x0) + n∑ i=1 ∑ j∈J2 cjqix 0 i ∂gj,∆j ∂xi (x0) = n∑ i=1 qix 0 i ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj n∑ i=1 qix 0 i ∂gj,∆j ∂xi (x0) = d0ft0,∆0(x 0) + ∑ j∈J2 cjdjgj,∆j(x 0) = d0ft0,∆0(x 0), where the last equality follows from (2.8). Since d0 < 0, we get ft0,∆0(x 0) = 0, which contradicts the non-degeneracy condition. Case 2: ℓ > d0 By (2.7) and (2.10), we have ∂ft0,∆0 ∂xi (x0) = 0 for i = 1, . . . , n. Then by a similar argument to that in Case 1, we also get a contradiction. Case 3: ℓ < d0 By (2.9) and (2.10), we obtain∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0 for i = 1, . . . , n. This fact and (2.8) combined give a contradiction with the non-degeneracy condition. Lemma 2.9 (Boundedness of singularities at infinity). There exists a real number r > 0 such that Σ∞(ft|S) ⊂ Dr for all t ∈ [0, 1]. 28 Proof. Suppose the assertion of the lemma is false. By the Curve Selection Lemma at infinity (see Theorem 1.14), we can find a nonempty subset I of {1, . . . , n} with ft|CI ̸≡ 0, a (possibly empty) subset J of {j ∈ {1, . . . , p} | gj|CI ̸≡ 0}, faces ∆0 of Γ(ft|CI ) and ∆j of Γ(gj|CI ) for j ∈ J, and analytic curves ϕ : (0, ϵ)→ (C∗)I , t : (0, ϵ)→ [0, 1], and λj : (0, ϵ)→ C, j ∈ J, such that the following conditions hold (e1) ∥ϕ(s)∥ → ∞ as s→ 0; (e2) t(s)→ t0 ∈ [0, 1] as s→ 0; (e3) ft(s),∆0(ϕ(s))→∞ as s→ 0; (e4) gj,∆j(ϕ(s)) = 0 for all j ∈ J and all s ∈ (0, ϵ); (e5) ∇ft(s),∆0(ϕ(s)) + ∑ j∈J λj(s)∇gj,∆j(ϕ(s)) = 0 for all s ∈ (0, ϵ). Then by a similar argument to that given in the proof of Lemma 2.8, with ft and gj replaced by ft,∆0 and gj,∆j respectively, we also reach a contradiction. The details are left to the reader. Lemma 2.10 (Transversality in the neighbourhood of infinity). Let r be a positive real number such that the conclusions of Lemmas 2.8 and 2.9 are fulfilled. Then there exists a real number R0 > 0 such that for all t ∈ [0, 1], all R ⩾ R0 and all c ∈ S1r, we have the fiber (ft|S)−1(c) intersects transversally with the sphere S2n−1R . Proof. If the assertion is not true, then by the Curve Selection Lemma at infinity (see Theorem 1.14), there exist t0 ∈ [0, 1], c ∈ S1r and analytic curves ϕ : (0, ϵ)→ Cn, t : (0, ϵ)→ [0, 1], and λj : (0, ϵ)→ C, j = 0, 1, . . . , p+ 1, satisfying the following conditions 29 (f1) ∥ϕ(s)∥ → ∞ as s→ 0; (f2) t(s)→ t0 as s→ 0; (f3) ft(s)(ϕ(s))→ c as s→ 0; (f4) gj(ϕ(s)) = 0 for all j = 1, . . . , p, and all s ∈ (0, ϵ); (f5) λ0(s)∇ft(s)(ϕ(s)) + ∑p j=1 λj(s)∇gj(ϕ(s)) = λp+1(s)ϕ(s) for all s ∈ (0, ϵ). Put I := {i | ϕi ̸≡ 0}. By the condition (f1), I ̸= ∅. For i ∈ I, we can write the curve ϕi in terms of parameter, say ϕi(s) = x 0 i s qi + (higher-order terms), where x0i ̸= 0 and qi ∈ Q. Observe that mini∈I qi < 0, because of the condition (f1). By the condition (f3) and the fact that |c| = r > 0, we have RI ∩ Γ(ft) ̸= ∅. Let d0 be the minimal value of the linear function ∑ i∈I αiqi on RI ∩ Γ(ft) and ∆0 be the face of RI ∩ Γ(ft) where this linear function takes its minimum value. As the Newton polyhedron Γ(ft) of ft does not depend on t, we can write ft(s)(ϕ(s)) = ft0,∆0(x 0)sd0 + (higher-order terms), ∂ft(s) ∂xi (ϕ(s)) = ∂ft0,∆0 ∂xi (x0)sd0−qi + (higher-order terms) for i ∈ I, where x0 := (x01, . . . , x 0 n) with x 0 i = 1 for i ̸∈ I. The condition (f3) and the fact that |c| = r > 0 together imply that d0 ⩽ 0 and d0ft0,∆0(x0) = 0. (2.11) Furthermore, for i /∈ I, the function ft0,∆0 does not depend on the variable xi, and so ∂ft0,∆0 ∂xi (x0) = 0 for all i ̸∈ I. (2.12) On the other hand, we deduce from the condition (f5), Lemmas 3.1 and 2.8 that λ0 ̸≡ 0 and λp+1 ̸≡ 0 (perhaps after reducing ϵ). Replacing λj by λjλ0 if necessary, we 30 may assume that λ0 ≡ 1. Put J := {j ∈ {1, . . . , p} | λj ̸≡ 0}. For j ∈ J ∪ {p + 1}, we can write λj(s) = cjs mj + (higher-order terms), where cj ̸= 0 and mj ∈ Q. Put J1 := {j ∈ J | gj|CI ̸≡ 0}. We only consider the case J1 ̸= ∅ because the case J1 = ∅ is handled similarly. For each j ∈ J1, let dj be the minimal value of the linear function ∑ i∈I αiqi on RI ∩ Γ(gj) and ∆j be the face of RI ∩ Γ(gj) where this linear function takes its minimum value. We can write gj(ϕ(s)) = gj,∆j(x 0)sdj + (higher-order terms). By the condition (f4), then gj,∆j(x 0) = 0 for all j ∈ J1, (2.13) On the other hand, for i ∈ I and j ∈ J1, ∂gj ∂xi (ϕ(s)) = ∂gj,∆j ∂xi (x0)sdj−qi + (higher-order terms). For i /∈ I and j ∈ J1, the function gj,∆j does not depend on the variable xi, and hence, ∂gj,∆j ∂xi (x0) = 0. (2.14) Let ℓ := minj∈J1(mj + dj) and J2 := {j ∈ J1 | mj + dj = ℓ}. From the condition (f5), for i ∈ I we have ∂ft0,∆0 ∂xi (x0)sd0−qi + (higher-order terms) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0)sℓ−qi + (higher-order terms) = cp+1x0i s mp+1+qi + (higher-order terms). (2.15) There are three cases to be considered. 31 Case 1: ℓ = d0 From (2.15) we have d0 − qi ⩽ mp+1 + qi for all i ∈ I. Therefore d0 −mp+1 ⩽ 2min i∈I qi < 0. Put I1 := {i ∈ I | d0 − qi = mp+1 + qi}. Hence, i ∈ I \ I1 if, and only if, ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0, and in this case d0 − qi < mp+1 + qi. If I1 = ∅, then ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0 for all i = 1, . . . , n. Hence, the non-degeneracy condition, (2.11) and (2.13) together imply that d0 = 0. Consequently, by the condition (f2), c = ft0,∆0(x 0) ∈ Σ∞(ft0 |S), which contradicts our assumption. If I1 ̸= ∅, then from (2.15) we have for all i ∈ I1, ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = cp+1x0i , d0 −mp+1 = 2qi. This, together with the Euler relation, (2.11), (2.12), (2.13) and (2.14), yield 0 = d0ft0,∆0(x 0) + ∑ j∈J2 cjdjgj,∆j(x 0) = n∑ i=1 qix 0 i ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 n∑ i=1 cjqix 0 i ∂gj,∆j ∂xi (x0) = n∑ i=1 qix 0 i ( ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) = ∑ i∈I1 qix 0 i ( ∂ft0,∆0 ∂xi (x0) + ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) = ∑ i∈I1 |x0|2i d0 −mp+1 2 cp+1 ̸= 0, 32 which is impossible. Case 2: ℓ > d0 The same argument as in Case 1 yields a contradiction. Case 3: ℓ < d0 From (2.15) we have ℓ− qi ⩽ mp+1 + qi for all i ∈ I. Therefore ℓ−mp+1 ⩽ 2min i∈I qi < 0. Put I2 := {i ∈ I | ℓ− qi = mp+1 + qi}. Hence, i ∈ I \ I2 if, and only if,∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0, and in this case ℓ− qi < mp+1 + qi. If I2 = ∅, then ∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = 0 for all i = 1, . . . , n, which, together with (2.13), leads to a contradiction with the non-degeneracy condition. If I2 ̸= ∅, then from (2.15) we have for all i ∈ I2,∑ j∈J2 cj ∂gj,∆j ∂xi (x0) = cp+1x0i , ℓ−mp+1 = 2qi. 33 This, together with the Euler relation and (2.13), yields 0 = ∑ j∈J2 cjdjgj,∆j(x 0) = ∑ j∈J2 n∑ i=1 cjqix 0 i ∂gj,∆j ∂xi (x0) = n∑ i=1 ∑ j∈J2 cjqix 0 i ∂gj,∆j ∂xi (x0) = ∑ i∈I2 qix 0 i (∑ j∈J2 cj ∂gj,∆j ∂xi (x0) ) = ∑ i∈I2 |x0|2i ℓ−mp+1 2 cp+1 ̸= 0, which is impossible. We now can complete the proof of Theorem 2.7. Proof of Theorem 2.7. Let r and R0 be the positive real numbers such that the conclu- sions of Lemmas 3.1, 2.8, 2.9 and 2.10 are fulfilled. Put X := {(t, x) ∈ [0, 1]× S | f(t, x) ∈ S1r}. By Corollary 2.6, then B(ft|S) ⊂ Dr for all t ∈ [0, 1]. Furthermore, for all (t, x) ∈ X ∩ {∥x∥ ⩾ R0}, the vectors ∇ft(x),∇g1(x), . . . ,∇gp(x), and x are C-linearly independent. Therefore, we can find a smooth map v1 : X ∩ {∥x∥ ⩾ R0} → Cn, (t, x) 7→ v1(t, x), satisfying the following conditions • ⟨v1(t, x),∇ft(x)⟩ = −∂ft∂t (x); • ⟨v1(t, x),∇gj(x)⟩ = 0 for j = 1, . . . , p; • ⟨v1(t, x), x⟩ = 0. We take arbitrary (but fixed) ϵ > 0. Since S1r∩K0(ft|S) = ∅ for all t ∈ [0, 1], the vectors ∇ft(x),∇g1(x), . . . ,∇gp(x) are C-linearly independent for all (t, x) ∈ X∩{∥x∥ ⩽ R0+ϵ}. 34 Consequently, there exists a smooth map v2 : X∩{∥x∥ ⩽ R0+ϵ} → Cn, (t, x) 7→ v2(t, x), such that the following conditions hold • ⟨v2(t, x),∇ft(x)⟩ = −∂ft∂t (x); • ⟨v2(t, x),∇gj(x)⟩ = 0 for j = 1, . . . , p. Next, we fix a partition of unity θ1 and θ2 subordinated to the covering{ (t, x) ∈ X | ∥x∥ > R0 + ϵ 3 } and { (t, x) ∈ X | ∥x∥ < R0 + 2ϵ 3 } of X, and define the smooth map v : X → Cn, (t, x) 7→ v(t, x), by v := θ1v1 + θ2v2. Then we can see that the following conditions hold: • ⟨v(t, x),∇ft(x)⟩ = −∂ft∂t (x); • ⟨v(t, x),∇gj(x)⟩ = 0 for j = 1, . . . , p; • ⟨v(t, x), x⟩ = 0 provided that ∥x∥ ⩾ R0 + ϵ. Finally, we define the smooth vector field w on X by w(t, x) := ∂ ∂t + v1(t, x) ∂ ∂x1 + · · ·+ vn(t, x) ∂ ∂xn , where vi are the coordinates of the map v, and then integrating this vector field we can see that for each t ∈ [0, 1], there exists a C∞-diffeomorphism Φt : f −1 0 (S1r) ∩ S → f−1t (S1r) ∩ S, which makes the following diagram commutes f−10 (S1r) ∩ S Φt−−−→ f−1t (S1r) ∩ S f0 y fty S1r id−−−→ S1r where id denotes the identity map. The proof is completed. 35 In summary, this chapter establishes the following main results: • The bifurcation set and the monodromy of a complex polynomial function f re- stricting to a non-singular algebraic set S in terms of its Newton polyhedron where f |S is Neton non-degenerate at infinity. • The global monodromy of a family polynomial {ft} restricting on an algebraic set in which the Newton polyhedrons of {ft} are independent from t and satisfy the non-degeneracy condition. 36 Chapter 3 Compactness criteria for real algebraic set and Newton polyhedron The following chapter bases on the result [BP-1] in List of Author’s Related Papers. Let f : Rn → R be a polynomial and Z(f) its zero set. In this chapter, in terms of the so-called Newton polyhedron of f, we present a necessary criterion and a sufficient condition for the compactness of Z(f). From this we derive necessary and sufficient criteria for the stable compactness of Z(f). 3.1 The compactness of an algebraic set. From now on let f : Rn → R be a nonconstant polynomial in n ≥ 2 variables and let Z(f) be its zero set: Z(f) := {x ∈ Rn | f(x) = 0}. 37 The results of Theorem 0.4 and 0.5 are the roots of the following works which focus on the case n ≥ 2. Lemma 3.1. If Z(f) is compact, then f is bounded either from below or from above. Proof. Suppose the assertion of the lemma is false. We have lim R→+∞ min ∥x∥=R f(x) = −∞ and lim R→+∞ max ∥x∥=R f(x) = +∞. Then, for R sufficiently large, there exist a, b ∈ Rn with ∥a∥ = ∥b∥ = R such that f(a) < 0 < f(b). Furthermore, since Z(f) is compact, we may assume that Z(f) ⊂ {x ∈ Rn | ∥x∥ < R}, after perhaps increasing R. On the other hand, the sphere Sn−1R := {x ∈ Rn | ∥x∥ = R} is path-connected (note that n ≥ 2). Hence, there is a continuous ϕ : [0, 1]→ Sn−1R , t 7→ ϕ(t), such that ϕ(0) = a and ϕ(1) = b. Consequently, the composition function f◦ϕ : [0, 1]→ R is continuous and satisfies (f ◦ ϕ)(0)× (f ◦ ϕ)(1) = f(a)× f(b) < 0. Thanks to the mean value theorem, we can find t0 ∈ (0, 1) such that (f ◦ ϕ)(t0) = 0, a contradiction. Lemma 3.2. Assume that f is bounded from below and its zero set Z(f) is compact. Then, the sub-level set {x ∈ Rn | f(x) ≤ 0} is compact. Proof. In order to obtain a contradiction, suppose that {x ∈ Rn | f(x) ≤ 0} is not compact. Then there exists a sequence {ak}k≥1 ⊂ Rn such that lim k→∞ ∥ak∥ = +∞ and f(ak) ≤ 0 for all k. 38 Let bk be an optimal solution of the problem max ∥x∥ = ∥ak∥ f(x). Since f is bounded from below, it cannot be bounded from above. In particular, lim k→∞ f(bk) = +∞. Therefore, for all k sufficiently large, f(ak)× f(bk) ≤ 0. As in the proof of Lemma 3.1, we can find ck ∈ Rn with ∥ck∥ = ∥ak∥ = ∥bk∥ such that f(ck) = 0, which contradicts the compactness of Z(f). The following is a generalization for the necessary criterion for compactness of real algebraic sets Theorem 3.3. Suppose that Z(f) is compact. The following assertions hold true: (i) f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n}, (ii) One of the following statements holds (ii1) f is bounded from below and f∆ ≥ 0 on Rn for all ∆ ∈ Γ∞(f). (ii2) f is bounded from above and f∆ ≤ 0 on Rn for all ∆ ∈ Γ∞(f). Proof. (i) This is obvious. (ii) By Lemma 3.1, f is bounded either from below or from above. Assume that f is bounded from below; the case f is bounded from above is treated similarly. Take any ∆ ∈ Γ∞(f). We will show that f∆ ≥ 0 on Rn. In fact, since f is continuous, it suffices to prove that f∆ ≥ 0 on (R \ {0})n. Suppose to the contrary that 39 there is a point x0 ∈ (R\{0})n such that f∆(x0) < 0. By definition, there exists a vector q ∈ Rn with minj=1,...,n qj < 0 such that ∆ = ∆(q,Γ(f)). Define the monomial curve ϕ : (0,+∞)→ Rn, t 7→ (x01tq1 , . . . , x0ntqn). Then ∥ϕ(t)∥ → +∞ as t → 0+. Furthermore, a simple calculation shows that for all t > 0 small enough we have f(ϕ(t)) = f∆(x 0)td + higher-order terms, where d := d(q,Γ(f)). Since f∆(x 0) 0 sufficiently small. Hence, the sub-level set {x ∈ Rn | f(x) ≤ 0} is not compact, which contradicts Lemma 3.2. The following remark shows that the converse of Theorem 3.3 does not hold. Remark 3.4. Let n = 2 and consider the polynomial f(x1, x2) := (x1 − x2)2. By definition, the Newton polyhedron Γ(f) is a segment joining the two points (2, 0) and (0, 2), and so the Newton boundary Γ∞(f) is the union of the faces: ∆1 := {(2, 0)}, ∆2 := {(0, 2)}, and ∆3 := {(1− t)(2, 0) + t(0, 2) | 0 ≤ t ≤ 1}. Clearly, the polynomials f∆1(x1, x2) = x 2 1, f∆2(x1, x2) = x 2 2, and f∆3(x1, x2) = (x1− x2)2 are all non-negative on Rn. However, Z(f) = {(x1, x2) ∈ R2 | x1 = x2} is not compact. On the other hand, we have the following statement, which provides a sufficient condition for compactness of real algebraic sets. Theorem 3.5. Suppose the following conditions: (i) f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n}, (ii) One of the following statements holds 40 (ii1) f∆ > 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). (ii2) f∆ < 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). Then Z(f) is compact. Proof. Assume the assertion of the theorem is false, i.e., there exists a sequence of points {ak}k≥1 ⊂ Z(f) such that limk→∞ ∥ak∥ = +∞. By the Curve Selection Lemma at infinity (see Theorem 1.14), there exists an analytic curve ϕ : (0, ϵ)→ Rn, t 7→ (ϕ1(t), . . . , ϕn(t)), such that (a) ∥ϕ(t)∥ → +∞ as t→ 0+, (b) f(ϕ(t)) = 0 for t ∈ (0, ϵ). Let J := {j | ϕj ̸≡ 0} ⊆ {1, . . . , n}. Recall that RJ := {α ∈ Rn | αj = 0 for j /∈ J}. By Condition (a), J ̸= ∅. For j ∈ J , we can expand the function ϕj in terms of parameter, say ϕj(t) = x 0 j t qj + higher-order terms, where x0j ̸= 0 and qj ∈ Q. By Condition (a) again, we obtain minj∈J qj < 0. Note that the curve ϕ lies in RJ∩Z(f). Hence, by the assumption (i), the restriction of f on RJ is not constant; in particular, the polyhedron Γ(f |RJ ) is nonempty and different from {0}. Let d be the minimal value of the linear function ∑j∈J qjαj on Γ(f |RJ ) and let ∆ be the maximal face of Γ(f |RJ ) (maximal with respect to the inclusion of faces) where the linear function takes this value, i.e., d := d(q,Γ(f |RJ )) and ∆ := ∆(q,Γ(f |RJ )) (here we put qj := 0 for j ̸∈ J.) Then ∆ ∈ Γ∞(f) because mini∈J qj < 0. Furthermore, we have asymptotically as t→ 0+, f(ϕ(t)) = f∆(x 0)td + higher-order terms, 41 where x0 := (x01, . . . , x 0 n) with x 0 j := 1 for j /∈ J (note that the polynomial f∆ does not depend on xj for j ̸∈ J). Combining this with Condition (b) gives f∆(x0) = 0, which contradicts the assumption (ii). 3.2 The stability of compactness of an algebraic set. In the rest of this chapter we study the stable compactness of real algebraic sets, which is easier to check than compactness. Definition 3.6. The set Z(f) is called stably compact if there is ϵ > 0 such that Z(f+g) is compact for all polynomials g : Rn → R with Γ(g) ⊆ Γ(f) and |g| < ϵ. By definition, the set Z(f) is stably compact if, and only if, remains compact for all sufficiently small perturbations of the “Newton” coefficients of the polynomial f. Lemma 3.7. The following conditions are equivalent: (i) f∆ ̸= 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). (ii) One of the following statements holds (ii1) f∆ > 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). (ii2) f∆ < 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). Proof. It suffices to show the implication (i) ⇒ (ii). Assume this is not the case, which means that there exist faces ∆1,∆2 ∈ Γ∞(f) such that f∆1 > 0 > f∆2 on (R \ {0})n. We may assume further that these faces are adjacent, i.e., ∆ := ∆1 ∩ ∆2 ̸= ∅. Then ∆ ∈ Γ∞(f). By assumption, f∆ ̸= 0 on (R \ {0})n. Fix x0 := (x01, . . . , x0n) ∈ (R \ {0})n, and without loss of generality, we may assume that f∆(x 0) > 0. By definition, there exists a vector q with minj=1,...,n qj < 0 such that ∆ = ∆(q,Γ(f)). A simple calculation shows that f∆2(t q1x01, . . . , t qnx0n) = t df∆(x 0) + higher-order terms, 42 where d := d(q,Γ(f)). Since f∆(x 0) > 0, this implies that f∆2(t q1x01, . . . , t qnx0n) > 0 for all t > 0 small enough, which contradicts the fact that f∆2 < 0 on (R \ {0})n. In what follows, let P(x) := ∑ α∈Γ(f)∩Zn+ |xα| and for each face ∆ of the polyhedron Γ(f), set P∆(x) := ∑ α∈∆∩Zn+ |xα|. By definition, the functions P and P∆ are positive on (R \ {0})n. Remark 3.8. Let P˜(x) := ∑ α |xα|, where the sum is taken over all the vertices of Γ(f). Then there exist positive constants c1, c2, and R such that c1P(x) ≤ P˜(x) ≤ c2P(x) for all x ∈ Rn. Indeed, the right-hand inequality clearly holds with c2 := 1. To see the left-hand in- equality, let v1, . . . , vs be the vertices of the polyhedron Γ(f). Then, for each α ∈ Γ(f), there exist non-negative real numbers λ1, . . . , λs, with λ1 + · · ·+ λs = 1, such that α = λ1v 1 + · · ·+ λsvs. Consequently, for all x ∈ Rn we have |xα| = |xλ1v1+···+λsvs| = (|xv1|)λ1 · · · (|xvs|)λs ≤ λ1|xv1|+ · · ·+ λs|xvs| ≤ |xv1 |+ · · ·+ |xvk | = P˜(x). Hence 1 #(Γ(f)∩Zn+)P(x) ≤ P˜(x), which completes the proof. The following lemma is a version at infinity of (Bui & Pham, 2016, Theorem 3.2). In the lemma, the equivalent of the statements (i) and (ii) was proved in (Gindikin, 1974; Mikhailov, 1967); for the sake of completeness we give a proof, which is different from the ones in these papers. Lemma 3.9. The following conditions are equivalent (i) f∆ > 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). 43 (ii) There exist positive constants c1, c2, and R such that c1P(x) ≤ f(x) ≤ c2P(x) for all ∥x∥ > R. (3.1) (iii) f is Newton non-degenerate at infinity and there exists R > 0 such that f(x) ≥ 0 for all ∥x∥ > R. Proof. (i) ⇒ (ii) Suppose that f is written as f =∑α aαxα. We have for all x ∈ Rn, f(x) ≤ ∑ α |aα||xα| ≤ max α |aα| ∑ α |xα| ≤ max α |aα|P(x), and so the right-hand inequality in (3.1) holds with c2 := maxα |aα| > 0. Suppose the left-hand inequality in (3.1) was false. By the Curve Selection Lemma at infinity (see Theorem 1.14), then we could find analytic curves ϕ : (0, ϵ) → Rn, t 7→ (ϕ1(t), . . . , ϕn(t)), and c : (0, ϵ)→ R such that the following assertions hold: (a) ∥ϕ(t)∥ → +∞ as t→ 0+; (b) c(t) > 0 for t ∈ (0, ϵ), c(t)→ 0 as t→ 0+; (c) c(t)P(ϕ(t)) > f(ϕ(t)) for t ∈ (0, ϵ). Let J := {j | ϕj ̸≡ 0} ⊂ {1, . . . , n}. By Condition (a), J ̸= ∅. We can expand the functions c(t) and ϕj(t) for j ∈ J, in terms of the parameter, say c(t) = c0t p + higher-order terms ϕj(t) = x 0 j t qj + higher-order terms, where c0 ̸= 0, x0j ̸= 0 and p, qj ∈ Q. By conditions (a) and (b), c0 > 0 and p > 0 > minj∈J qj. If RJ ∩Γ(f) = ∅, then for each α ∈ Γ(f), there exists an index j /∈ J such that αj > 0. Consequently, P(ϕ(t)) ≡ ∑ α∈Γ(f)∩Zn+ |ϕ(t)α| ≡ ∑ α∈Γ(f)∩Zn+ ∏ j∈J |ϕj(t)αj | ∏ j /∈J |ϕj(t)αj |  ≡ 0. 44 Similarly, we also have f(ϕ(t)) ≡ 0, which contradicts Condition (c). Therefore, RJ ∩Γ(f) ̸= ∅. Let d be the minimal value of the linear function∑j∈J αjqj on RJ ∩ Γ(f) and let ∆ be the maximal face of Γ(f) where this linear function takes its minimum value. Then ∆ ∈ Γ∞(f) since minj∈J qj < 0. Furthermore, we have asymptot- ically as t→ 0+, c(t)P(ϕ(t)) = c0P∆(x 0)td+p + higher-order terms, f(ϕ(t)) = f∆(x 0)td + higher-order terms, where x0 := (x01, . . . , x 0 n) with x 0 j := 1 for j /∈ J. Note that P∆(x0) > 0 and f∆(x0) > 0. Therefore, by Condition (c), we get d+ p ≤ d, which contradicts the fact that p > 0. (ii) ⇒ (iii) The left-hand inequality in (3.1) shows that f(x) ≥ 0 for all ∥x∥ > R. Take any x0 ∈ (R \ {0})n and ∆ ∈ Γ∞(f). By definition, there exists a vector q ∈ Rn with minj=1,...,n qj < 0 such that ∆ = ∆(q,Γ(f)). Consider the monomial curve ϕ : (0,+∞)→ Rn, t 7→ (x01tq1 , . . . , x0ntqn). Clearly, ∥ϕ(t)∥ → +∞ as t→ 0+. Furthermore, we have asymptotically as t→ 0+, P(ϕ(t)) = P∆(x 0)td + higher-order terms, f(ϕ(t)) = f∆(x 0)td + higher-order terms, where d := d(q,Γ(f)). Since P∆(x 0) > 0, it follows from (3.1) that f∆(x 0) > 0. In particular, f is Newton non-degenerate at infinity. (iii) ⇒ (i) Take any ∆ ∈ Γ∞(f). We first show that f∆ ≥ 0 on (R \ {0})n. On the contrary, suppose that f∆(x 0) < 0 for some x0 ∈ (R \ {0})n. By definition, there exists a vector q ∈ Rn with minj=1,...,n qj < 0 such that ∆ = ∆(q,Γ(f)). Consider the monomial curve ϕ : (0,+∞)→ Rn, t 7→ (x01tq1 , . . . , x0ntqn). 45 Clearly, ∥ϕ(t)∥ → +∞ as t→ 0+. Furthermore, we have asymptotically as t→ 0+, f(ϕ(t)) = f∆(x 0)td + higher-order terms, where d := d(q,Γ(f)). Since f∆(x 0) < 0, it follows that f < 0 on the curve ϕ, which contradicts our assumption. Therefore, f∆ ≥ 0 on (R \ {0})n, and by continuity, we have f∆ ≥ 0 on Rn. We next show that f∆ > 0 on (R \ {0})n. By contradiction, suppose that f∆(x0) = 0 for some x0 ∈ (R \ {0})n. Since f∆ ≥ 0 on Rn, it follows that x0 is a global minimizer of f∆ on Rn, and so x0 is a critical point of f∆. Therefore, f∆(x 0) = ∂f∆(x 0) ∂x1 = · · · = ∂f∆(x 0) ∂xn = 0, which contradicts the non-degeneracy condition of f. The following result presents necessary and sufficient conditions for the stable com- pactness in terms of the Newton polyhedron of the defining polynomial. Theorem 3.10 (Compare Theorem 0.6). The following conditions are equivalent: (i) Z(f) is stably compact. (ii) f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n} and f∆ ̸= 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). (iii) f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n} and one of the following statements holds (iii1) f∆ > 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). (iii2) f∆ < 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). (iv) f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n} and there exist σ ∈ {−1, 1} and constants c1 > 0, c2 > 0, and R > 0 such that c1P(x) ≤ σf(x) ≤ c2P(x) for all ∥x∥ > R. (3.2) 46 (v) f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n}, f is Newton non-degenerate at infinity, and there exist σ ∈ {−1, 1} and R > 0 such that σf(x) ≥ 0 for all ∥x∥ > R. Proof. The equivalences (ii)⇔ (iii)⇔ (iv)⇔ (v) follow immediately from Lemmas 3.7 and 3.9. Hence, it suffices to show (i) ⇒ (iii) and (iv) ⇒ (i). (i)⇒ (iii) By assumption, the set Z(f) is compact. Thanks to Theorem 3.3, f |RJ ̸≡ 0 for all J ⊂ {1, . . . , n}. Replacing f by −f if necessary, we may assume that f is bounded from below and f∆ ≥ 0 on (R \ {0})n for all ∆ ∈ Γ∞(f). We will show that (iii1) holds. On the contrary, suppose that there exist x0 ∈ (R \ {0})n and ∆ ∈ Γ∞(f) such that f∆(x 0) = 0. This implies that ∆ contains at least two vertices, say ∆1 and ∆2. Note that all the coordinates of the vertices ∆1 and ∆2 are even integer numbers because f is bounded from below. This implies easily that f∆1(x 0) > 0 and f∆2(x 0) > 0. Now, for each ϵ > 0 consider the polynomial gϵ(x) := −ϵx∆1 . Clearly, gϵ(x) = −ϵf∆1(x), Γ(gϵ) ⊂ Γ(f), and Γ∞(f + gϵ) = Γ∞(f) for all ϵ > 0 small enough. Fur- thermore, we have (f + gϵ)∆(x 0) = f∆(x 0) + gϵ,∆(x 0) = −ϵf∆1(x0) < 0, (f + gϵ)∆2(x 0) = f∆2(x 0) > 0. By Theorem 3.3, Z(f + gϵ) is not compact, a contradiction. (iv) ⇒ (i) Without loss of generality, we may assume that (iv) holds with σ = 1. Let f be written as f = ∑ α aαx α and set ϵ := min {c1 2 ,min α |aα| } > 0, where the second minimum is taken over all the vertices of Γ(f). Take any polynomial g : Rn → R with Γ(g) ⊆ Γ(f) and |g| < ϵ. By definition, Γ∞(f + g) = Γ∞(f). Furthermore, for all x ∈ Rn we have |g(x)| ≤ |g| ∑ α∈Γ(f) |xα| ≤ c1 2 P(x). 47 It follows from (3.2) that c1 2 P(x) ≤ (f + g)(x) ≤ (c1 2 + c2 ) P(x) for ∥x∥ > R. (3.3) Consequently, for all J ⊂ {1, . . . , n} we have (f + g)|RJ ̸≡ 0 since otherwise P|RJ ≡ 0, and hence f |RJ ≡ 0 by (3.2), a contradiction. Furthermore, from (3.3) and Lemma 3.9, we deduce that (f + g)∆ > 0 on (R \ {0})n for all ∆ ∈ Γ∞(f + g). Therefore, in view of Theorem 3.5, the set Z(f + g) is compact. Let us make some final comments. Remark 3.11. (i) The criteria presented in this paper can be easily extended to examine the (stable) compactness of basic closed semi-algebraic sets. To see this, let X be a basic closed semi-algebraic set defined by X := {x ∈ Rn | g1(x) = 0, . . . , gl(x) = 0, h1(x) ≥ 0, . . . , hm(x) ≥ 0}, where g1, . . . , gl, h1, . . . , hm are polynomial functions on Rn. It is easy to see that X is compact if, and only if, the set Y := {(x, y) ∈ Rn×Rm | g1(x) = 0, . . . , gl(x) = 0, h1(x)− y21 = 0, . . . , hm(x)− y2m = 0}, is compact. Then the statement follows because Y is the zero set of the polynomial function Rn × Rm → R, (x, y) 7→ [g1(x)]2 + · · ·+ [gl(x)]2 + [h1(x)− y21]2 + · · ·+ [hm(x)− y2m]2. (ii) In the two-dimensional case, the criteria remain valid if one checks only the codi- mension one faces. Unfortunately, this is not true in the general case as can be seen with the polynomial f(x, y, z) := x2 + (y − z)2. (iii) Finally, we would like to mention that the established criteria can be checked, at least in principle. Indeed, given a polynomial function f : Rn → R, there are algorithms 48 with polynomial time complexity that can generate all faces of the Newton polyhedron Γ(f) (see (Fukuda, 2004; Fukuda & Rostal, 1994; K. Fukuda & Margot, 1997; Murty & Chung, 1995)). Moreover, for each face ∆ ∈ Γ∞(f), it is not hard to see that the problem of checking positivity (or non-negativity) of the polynomial f∆ (corresponding to the face ∆) on (R \ {0})n can be reduced to the problem of minimizing polynomial functions over semi-algebraic sets; on the other hand, for each polynomial optimization problem, by using standard results about the existence of sums of squares certificates (i.e., Positivstellensa¨tze), we can construct an appropriate sequence of computationally feasible semidefinite programming relaxations (Lasserre’s hierarchy), whose optimal val- ues converge monotonically, increasing to the optimal value of the original problem. For more detailed information on the subject, see the surveys (Laurentl, 2009; Scheiderer, 2009) and the monographs (Ha` & Pha.m, 2017; Lasserrel, 2009; Marshall, 2008), as well as references therein. In summary, this chapter establishes the following main results: • Giving a necessary condition and a sufficient condition for the compactness of an algebraic set Z(f) which is defined by a real polynomial function which is bounded either from above or from below. • The necessary and sufficient criteria for the stable compactness of Z(f). 49 Conclusions The main goals of this thesis are to study properties of a class of functions satisfying non-degenerate conditions. Singularity Theory and Semi-algebraic Geometry are main tools for our study. Our main results include: - Investigating the global monodromy of a family polynomial {ft} restricting on an algebraic set in which the Newton polyhedrons of {ft} are independent from t and satisfy the non-degenerated condition. (see Theorem 2.7). - Giving a necessary condition and a sufficient condition for the compactness of an algebraic set Z(f) which is defined by a real polynomial function which is bounded either from above or from below. 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