Queen’s University - Math 844
Winter 2016
Problem Set #10
Posted: Thursday, 7/4/2016
Due: Thursday, 14/4/2016
• Recall that we proved in class that if f : S kP→ S k is a C ∞ map, then deg(f ), which defines the vector space homomorphism
f ? : H k (S k ) → H k (S k ), is equal to the sum p∈f −1 (q) p , where q is any regular value of f , and for p ∈ f −1 (q), p = 1 if Tp f is
orientation preserving, and −1 otherwise; in particular any point NOT in the image of f is a regular value for f , and since the
preimage of such a point is the empty set, it follows immediately that if f is not surjective, then deg(f ) = 0.
1. Let S n = {p ∈ Rn+1 | kpk = 1} (the unit sphere of Rn+1 ), and let f, g : S n → S n C ∞ maps. Show that if f (p) 6= −g(p) ∀p ∈ S n
then deg(f ) = deg(g). (Hint: Show that if f (p) 6= −g(p) ∀p ∈ S n , then (1 − t)f (p) + tg(p) 6= 0 ∀(p, t) ∈ S n × [0, 1], and use this
to construct a C ∞ homotopy between f and g.)
2. Let S 1 = {z ∈ C | |z| = 1} (i.e. we identify the unit circle S 1 with the set of complex numbers of unit modulus). Let n ∈ Z, and
consider the map fn : S 1 → S 1 defined by fn (z) = z n . Compute deg(fn ).
3. Consider the unit sphere S 2 of R3 , and let pN = (0, 0, 1) ∈ S 2 (the “North Pole”). We can identify S 2 \ {pN } with R2 ' C
via stereographic projection, i.e. we have a homeomorphism φN : S 2 \ {pN } → R2 which is nothing other than stereographic
projection with respect to the North Pole of S 2 . Let now n ∈ N, n ≥ 1, and consider the mapping z 7→ z n on C; via the usual
identification of C with R2 , this mapping induces a C ∞ mapping on gn : R2 → R2 . We now define the mapping fn : S 2 → S 2
2
∞
by fn (q) = φ−1
mapping. Compute deg(fn ).
N ◦ gn ◦ φN (q) ∀q ∈ S \ {pN }, and fn (pN ) = pN . It is easily verified that fn is a C
4. (The Fundamental Theorem of Algebra – first proof via degree theory)
P
Pn−1
j
j
Let P (z) = z n + n−1
j=0 aj z be a polynomial of degree n ≥ 1 in the complex variable z. Let Q(z) =
j=0 aj z ; we then have
P (z) = z n + Q(z). We wish to prove that ∃z ∈ C such that P (z) = 0. We shall assume that ∀z ∈ C P (z) 6= 0, and we shall
derive a contradiction. In what follows, we shall identify the unit circle S 1 with the subset {z ∈ C | |z| = 1} of C.
Assume then that P (z) 6= 0, ∀z ∈ C. and define ∀r ∈ R, hr : S 1 → S 1 by:
(
hr (z) =
P (re−1/r z)
,
|P (re−1/r z)|
P (0)
, r≤
|P (0)|
r > 0,
0
It is easily verified that the mapping (r, z) 7→ hr (z) is C ∞ on R × S 1 .
(a) Show that the Z−valued mapping r 7→ deg(hr ) is constant on R. (Hint: Use the expression for deg(hr ) in terms of an
orientation form on S n ).
(b) Deduce from (a) that deg(hr ) = 0 ∀r ∈ R.
(c) Show that for r > 0 large enough, (re−1/r z)n + tQ(re−1/r z) 6= 0, ∀z ∈ S 1 , ∀t ∈ [0, 1].
(d) For r > 0 large enough (as in (c)), consider the C ∞ map Fr : S 1 × [0, 1] → S 1 defined by
Fr (z, t) =
(re−1/r z)n + tQ(re−1/r z)
;
|(re−1/r z)n + tQ(re−1/r z)|
Use this map to deduce a contradiction.
5. (The Fundamental Theorem of Algebra – second proof via degree theory)
P
j
n
Let P (z) = z n + n−1
j=0 aj z = z +Q(z) be a polynomial of degree n ≥ 1 on C. Using the construction (and notation) in Problem
(3), we can extend the polynomial P to a C ∞ mapping from S 2 to S 2 , denoted PS 2 : S 2 → S 2 , and satisfying PS 2 (pN ) = pN .
Consider now the C ∞ map F : S 2 × [0, 1] → S 2 defined by F (q, t) = φ−1 ((φ(q))n + tQ(φ(q))) for q 6= pN , and F (pN , t) = pN
∀t ∈ [0, 1].
(a) Using F , show that deg(PS 2 ) = deg(fn ) (with fn defined as in Problem (3).
(b) Assume now P (z) 6= 0, ∀z ∈ C; using (a), derive a contradiction.