INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES
WARSZAWA 1996
AN ALGEBRAIC METHOD FOR
CALCULATING THE TOPOLOGICAL DEGREE
A N D R Z E J L E ¸ C K I
Institute of Mathematics, Gda´ nsk University Wita Stwosza 57, 80-952 Gda´ nsk, Poland
E-mail: matal@halina.univ.gda.pl
Z B I G N I E W S Z A F R A N I E C Institute of Mathematics, Gda´ nsk University
Wita Stwosza 57, 80-952 Gda´ nsk, Poland E-mail: matzs@halina.univ.gda.pl
1. Introduction. Effective methods for calculating the topological degree for a con- tinuous mapping are very useful. In this paper we present an algebraic method which applies to polynomial mappings. We shall show that in this case the topological degree can be expressed in terms of signatures of some effectively defined bilinear forms (see Theorem 4.1).
The method may be derived from the theory of bilinear forms on finite intersection algebras given by Scheja & Storch [10], Eisenbud & Levine [5], Khimshiashvili [8], Kunz [7] and Cardinal [4]. All facts needed for the proof of Theorem 4.1 are presented in [2].
The complete proof requires some advanced facts concerning complete intersection algebras. In this paper we explain the method for polynomial mappings having only non-degenerate roots. This way we may avoid difficult details and make the main idea of the method to be more clear.
In the case of the local topological degree there is a similar formula (so called Eisenbud
& Levine formula). One can find its proof in [1], [2], [5], [8]. In [9] one may find a description of an algorithm which has been used to create a computer program which can calculate the local topological degree.
1991 Mathematics Subject Classification: Primary 55M25; Secondary 14P05.
Both autors supported by grant KBN 610/P3/96/07.
The paper is in final form and no version of it will be published elsewhere.
[73]
2. Preliminaries. In this section we shall collect some useful facts concerning bilinear forms and polynomial algebras.
Let R (resp. C) denote the field of real (resp. complex) numbers. Let V be a finite dimensional real vector space and let Φ : V × V → R be a bilinear symmetric form. Let V
+(resp. V
−) denote a maximal subspace of V on which Φ is positive (resp. negative) definite, i.e. if x ∈ V
+− {0} (resp. x ∈ V
−− {0}) then Φ(x, x) > 0 (resp. Φ(x, x) < 0).
We define
signature Φ = dim V
+− dim V
−.
We shall say that Φ is non-degenerate if its matrix is non-singular.
Lemma 2.1. Let ϕ : R → R be an R-linear functional and let Φ : R × R → R be the bilinear form given by Φ(x, y) = ϕ(xy).Then signature Φ = sign ϕ(1). Moreover Φ is non-degenarate if and only if ϕ(1) 6= 0.
P r o o f. Since ϕ is R-linear then for every x ∈ R − {0} we have Φ(x, x) = ϕ(x
2) = ϕ(x
2· 1) = x
2ϕ(1). Because x
2> 0 then signature Φ = sign ϕ(1).
Lemma 2.2. Let ϕ : C → R be an R-linear functional and let Φ : C × C → R be the bilinear form given by Φ(z, w) = ϕ(zw). Then signature Φ = 0.
P r o o f. Let V
+⊂ C denote a maximal R-subspace on which Φ is positive definite, i.e. Φ(z, z) = ϕ(z
2) > 0 for every z ∈ V
+− {0}. Then √
−1 V
+is an R-subspace of C and if w = √
−1 z ∈ √
−1 V
+− {0} then Φ(w, w) = ϕ(w
2) = ϕ(−z
2) = −ϕ(z
2) < 0.
Hence dim V
−≥ dim √
−1 V
+= dim V
+.
By similar arguments dim V
+≥ dim V
−. Hence dim V
+= dim V
−and signature Φ = 0.
Let
B = R
⊕· · ·
⊕R
⊕C
⊕· · ·
⊕C =
m⊕
1
R ⊕
r1
C.
Then B is a finite dimensional R-algebra. Let ϕ : B → R be an R-linear functional.
Denote
s
1= ϕ(1
⊕0
⊕· · ·
⊕0), .. .
s
m= ϕ(0
⊕· · ·
⊕1
⊕· · ·
⊕0).
From previous lemmas we get
Proposition 2.3. Let Φ : B × B → R be the bilinear form given by Φ(f, g) = ϕ(f g).
Then
signature Φ = #{1 ≤ i ≤ m : s
i> 0} − #{1 ≤ i ≤ m : s
i< 0}.
Moreover if Φ is non-degenerate then s
16= 0, . . . , s
m6= 0.
Let f
1, . . . , f
n∈ R[x
1, . . . , x
n], let F
R= (f
1, . . . , f
n) : R
n→ R
nand let F
C: C
n→ C
nbe its complexification. Let
J = ∂(f
1, . . . , f
n)
∂(x
1, . . . , x
n)
denote the determinant of the Jacobian matrix. Let A = R[x
1, . . . , x
n] / I, where I is the ideal in R[x
1, . . . , x
n] generated by polynomials f
1, . . . , f
n. Then A is an R-algebra.
From now on we shall assume that d = dim A < ∞ and that F
Chas only non- degenerate complex roots, i.e. if z ∈ F
C−1(0) then J (z) 6= 0.
The next two facts generalize the Fundamental Theorem of Algebra. They follow immediately from Corollary 1 in [6], p.57.
Theorem 2.4. #{z ∈ C
n: F
C(z) = 0} = dim A = d.
So there are d complex roots for F
Cand we may assume that F
C−1(0) = {p
1, . . . , p
m, q
1, ¯ q
1, . . . , q
r, ¯ q
r},
where p
1, . . . , p
m∈ R
n, q
1, . . . , q
r∈ C
n− R
nand ¯ q
iis the complex conjugate of q
i. Clearly m + 2r = d.
If f ∈ I then f = 0 on F
C−1(0). Then there is an R-homomorphism of algebras Ψ : A → B =
m⊕
1
R ⊕
r1
C
given by Ψ(f ) = f (p
1)
⊕· · ·
⊕f (p
m)
⊕f (q
1)
⊕· · ·
⊕f (q
r). It is easy to see that dim B = m + 2r = d = dim A.
Theorem 2.5. If f = 0 on F
C−1(0) then f ∈ I. Hence Ψ : A → B is an isomorphism of R-algebras. Thus g = h in A if and only if g(p
i) = h(p
i) for 1 ≤ i ≤ m and g(q
j) = h(q
j) for 1 ≤ j ≤ r.
3. The construction of bilinear forms. Denote x = (x
1, . . . , x
n), y = (y
1, . . . , y
n).
Define A
2= R[x, y] / I
2, where I
2is the ideal in R[x, y] generated by f
1(x), . . . , f
n(x), f
1(y), . . . , f
n(y). One may check that A
2is isomorphic to A ⊗ A.
For 1 ≤ i, j ≤ n define
T
ij(x, y) = f
i(y
1, . . . , y
j−1, x
j, . . . , x
n) − f
i(y
1, . . . , y
j, x
j+1, . . . , x
n) x
j− y
jIt is easy to see that each T
ijextends to a polynomial, thus we may assume that T
ij∈ R[x, y]. Define
T (x, y) = det [T
ij(x, y)].
It is easy to see that J (x) = T (x, x).
Theorem 3.1. For any polynomial q(x) we have
q(x)T (x, y) = q(y)T (x, y) in A
2. P r o o f. Note B
jthe j-th column of [ T
ij(x, y)]. Then
(x
j− y
j)B
j=
f
1(y
1, . . . , y
j−1, x
j, . . . , x
n) − f
1(y
1, . . . , y
j, x
j+1, . . . , x
n) .. .
f
n(y
1, . . . , y
j−1, x
j, . . . , x
n) − f
n(y
1, . . . , y
j, x
j+1, . . . , x
n)
We do not change the determinant if we add to this column a linear combination of the form
X
k6=j
(x
k− y
k)B
k.
The j-th column then becomes
n
X
k=1
(x
k− y
k)B
k=
f
1(x
1, . . . , x
n) − f
1(y
1, . . . , y
n) .. .
f
n(x
1, . . . , x
n) − f
n(y
1, . . . , y
n)
Developing this determinant relatively to the j-th column we get an element of the ideal I
2. Hence
(x
j− y
j)T (x, y) = 0 in A
2, and then x
jT (x, y) = y
jT (x, y) in A
2. Hence
x
kx
jT (x, y) = x
ky
jT (x, y) = y
ky
jT (x, y) in A
2and by induction
x
a11· · · x
annT (x, y) = y
a11· · · y
nanT (x, y) in A
2.
So the theorem is true if q(x) is a monomial. One gets the general case by linearity.
Proposition 3.2. Suppose that p, q ∈ F
C−1(0). If p = q then T (p, q) = T (p, p) = J (p), if p 6= q then T (p, q) = 0.
P r o o f. We have already proved that T (p, p) = J (p). Suppose that p 6= q. There is a polynomial Q(x) ∈ C[x] such that Q(p) 6= 0 and Q(q) = 0. Applying the same arguments as in the proof of the previous theorem one can see that there are h
1, . . . , h
n, g
1, . . . , g
n∈ C[x, y] such that
Q(x)T (x, y) = Q(y)T (x, y) +
n
X
i=1
h
i(x, y)f
i(x) +
n
X
j=1
g
j(x, y)f
j(y).
Since f
1(p) = . . . = f
n(p) = f
1(q) = . . . = f
n(q) = 0 then Q(p)T (p, q) = Q(q)T (p, q) = 0, and then T (p, q) = 0.
Suppose that e
1(x), . . . , e
d(x) form a basis in A. Since A
2is isomorphic to A⊗A then e
i(x)e
j(y) for 1 ≤ i, j ≤ d form a basis in A
2. Hence there are t
ij∈ R such that
T (x, y) =
d
X
i,j=1
t
ije
i(x)e
j(y) =
d
X
i=1
e
i(x)ˆ e
i(y) in A
2,
where ˆ e
i=
d
P
j=1
t
ije
j.
Theorem 3.3. ˆ e
1, . . . , ˆ e
dform a basis in A.
P r o o f. According to Theorem 2.5, A is isomorphic to the product B = ⊕
m1
R ⊕
r1
C.
Let E
1, . . . , E
dbe the basis given by
E
1= 1
⊕0
⊕· · ·
⊕0, E
2= 0
⊕1
⊕· · ·
⊕0, . . . , E
m+1= 0
⊕· · ·
⊕1
⊕· · ·
⊕0, E
m+2= 0
⊕· · ·
⊕√ −1
⊕· · ·
⊕0, . . . , E
d−1= 0
⊕· · ·
⊕0
⊕1, E
d= 0
⊕· · ·
⊕0
⊕√ −1.
Using Proposition 3.2 it is easy to see that elements ˆ E
1, . . . , ˆ E
dconstructed as above form a basis. Moreover, since e
1, . . . , e
dare non-singular combinations of E
1, . . . , E
dthen ˆ
e
1, . . . , ˆ e
dare non-singular combinations of E ˆ
1, . . . , ˆ E
d, and then they form a basis.
Then there are a
1, . . . , a
d∈ R such that 1 = a
1ˆ e
1+ · · · + a
de ˆ
din A. Hence if p ∈ F
C−1(0) then
a
1e ˆ
1(p) + · · · + a
dˆ e
d(p) = 1.
Definition. Let ϕ : A → R be the linear functional given by ϕ(f ) = a
1b
1+ · · · + a
db
d,
for f = b
1e
1+ · · · + b
de
d∈ A.
Lemma 3.4. If p
i∈ F
R−1(0) f or 1 ≤ i ≤ m and T
i(x) = T (x, p
i) ∈ A then ϕ(T
i) = 1.
P r o o f. Since T (x, y) =
d
P
j=1
e
j(x)ˆ e
j(y) in A
2then there are h
k, g
k∈ R[x, y] such that
T (x, y) =
d
X
j=1
e
j(x)ˆ e
j(y) +
n
X
k=1
(h
k(x, y)f
k(x) + g
k(x, y)f
k(y)).
Because f
1(p
i) = . . . = f
n(p
i) = 0 then T
i(x) =
d
X
j=1
e
j(x)ˆ e
j(p
i) +
n
X
k=1
h
k(x, p
i)f
k(x),
and then T
i= ˆ e
1(p
i)e
1(x)+· · ·+ ˆ e
d(p
i)e
d(x) in A. So ϕ(T
i) = a
1e ˆ
1(p
i)+· · ·+a
dˆ e
d(p
i) = 1.
Take p
i∈ F
R−1(0). We have assumed that J (p) 6= 0 for every p ∈ F
C−1(0), so J (p
i) 6= 0.
Let t
i= T
i/ J (p
i) ∈ A. From Proposition 3.2, t
i(p
i) = 1 and t
i(q) = 0 for every q ∈ F
C−1(0) , q 6= p
i. Let Ψ : A → B be the isomorphism of algebras defined before.
Then Ψ(t
i) = 0
⊕· · ·
⊕1
⊕· · ·
⊕0, where 1 is in the i-th factor.
Let Φ : A × A → R be the bilinear form given by Φ(f, g) = ϕ(f g).
Lemma 3.5. signature Φ =
m
P
i=1
sign J (p
i).
P r o o f. From Lemma 3.4, ϕ(t
i) = ϕ(T
i/ J (p
i)) = J (p
i)
−1ϕ(T
i) = J (p
i)
−1for 1 ≤ i ≤ m. Then sign ϕ(t
i) = sign J (p
i). Now it is enough to apply Proposition 2.3.
Let M : R
n→ R be a polynomial, let ϕ
M: A → R be the linear functional given by ϕ
M(f ) = ϕ(M f ), let Φ
M: A × A → R be the bilinear form given by Φ
M(f, g) = ϕ
M(f g) = ϕ(M f g).
Lemma 3.6. signature Φ
M=
m
P
i=1
sign M (p
i)J (p
i). If Φ
Mis non-degenerate then
M (p
i) 6= 0 for every 1 ≤ i ≤ m.
P r o o f. Using the same arguments as in the proof of the previous lemma one can show that ϕ
M(t
i) = M (p
i) / J (p
i). From Proposition 2.3,
signature Φ
M=
m
X
i=1
sign M (p
i)J (p
i).
Moreover, if Φ
Mis non-degenerate then 0 6= ϕ
M(t
i) = M (p
i) / J (p
i).
4. A formula for the topological degree. Let F
R= (f
1, . . . , f
n) : R
n→ R
nbe a polynomial mapping, let M : R
n→ R be a polynomial and let B = { x ∈ R
n: M (x) >
0 }. If B is bounded and ∂B ∩ F
R−1(0) = ∅ then deg(F
R, B, 0) will denote the topological degree of F
Rwith respect to B and 0 ∈ R
n.
Let A = R[x
1, . . . , x
n] / I, where I is the ideal in R[x
1, . . . , x
n] generated by f
1, . . . , f
n. If dim A < ∞ then one may define bilinear forms Φ and Φ
M: A × A → R the same way as in Section 3.
Theorem 4.1 (A formula for the topological degree). If Φ
Mis non-degenerate then
∂B ∩ F
R−1(0) = ∅. So if B is bounded then deg(F
R, B, 0) is defined and deg(F
R, B, 0) = 1
2 (signature Φ + signature Φ
M).
In this paper we shall give the proof under the additional assumption that all complex roots are non-degenerate, i.e. if p ∈ F
C−1(0) then J (p) 6= 0. We want to point out that this assumption is not necessary.
P r o o f. From Lemma 3.6, M
−1(0) ∩ F
R−1(0) = ∅. Since ∂B ⊂ M
−1(0) then ∂B ∩ F
R−1(0) = ∅. According to Theorem 2.4, F
R−1(0) is finite. In that case
deg(F
R, B, 0) = X
i∈P