Homotopy and Framed Cobordisms
3.2 The Product Neighborhood Theorem 3.3 The Hopf Degree Theorem3.3The Hopf Degree Theorem
35
Chapter 4
Intersection Theory
The purpose of the present chapter is to extend the degree theory developed in Chapters 1 and 2 to smooth maps between manifolds of different dimen-sions, with the dimension of the source being smaller than the dimension of the target. The relevant transversality theory is the subject of Section 4.1, orientation and intersection numbers are introduced in Section 4.2, self-intersection numbers are discussed in Section 4.3, and Section 4.4 examines the Lefschetz number of a smooth map from a compact manifold to itself and establishes the Lefschetz–Hopf theorem and the Lefschetz fixed point theorem.
4.1 Transversality
This section introduces the notion of transversality of a smooth map to a submanifold of the target space.
Definition 4.1.1 (Transversality). Let m, n, k be nonnegative integers such that k≤n, let M be a smooth m-manifold, let N be a smooth n-manifold, and let Q⊂N be a smooth submanifold of dimension n−k. The number k is called thecodimension of Q and is denoted by
codim(Q) := dim(N)−dim(Q).
Let f :M →N be a smooth map and letp∈f−1(Q). The map f is said to be transverse to Q at p if
Tf(p)N = im (df(p)) +Tf(p)Q (4.1.1) It is calledtransverse toQif it is transverse toQat everyp∈f−1(Q). The notation f −
tQ signifies that the map f is transverse to the submanifoldQ.
37
f
Q M
Figure 4.1: Transverse and nontransverse intersections.
Example 4.1.2. (i)IfQ=N, then every smooth mapf :M →N is trans-verse toQ.
(ii) If Q={q} is a single point in N, then a smooth map f :M →N is transverse toQif and only if q is a regular value off.
(iii)Iff :M →N is an embedding, then its imageP :=f(M) is a smooth submanifold ofN (see [21, Theorem 2.3.4]). In this situationf is transverse toQif and only if
TqN =TqP+TqQ for allq ∈P ∩Q. (4.1.2) If (4.1.2) holds we say thatP is transverse to Qand write P−
tQ.
(iv) Assume ∂M =∅, let T M ={(p, v)|p∈M, v∈TpM} be the tangent bundle, and letZ ={(p, v)∈T M|v= 0} be the zero section inT M. Iden-tify a vector field X ∈Vect(M) with the map M →T M :p7→(p, X(p)).
This map is transverse to the zero section if and only if the vector field X has only nondegenerate zeros. (Exercise: Prove this).
(v) Assume ∂M =∅. Then the graph of a smooth map f :M →M is transverse to the diagonal ∆ ={(p, p)|p∈M} ⊂M×M if and only if ev-ery fixed point p = f(p) ∈ M is nondegenerate, i.e. det(1l−df(p))6= 0.
(Exercise: Prove this).
The next lemma generalizes the observation that the preimage of a regu-lar value is a smooth submanifold (see Lemma 1.3.7 and [21, Thm 2.2.17]).
Lemma 4.1.3. Let M be an m-manifold with boundary, let N be an n-manifold without boundary, and letQ⊂N be a codimension-k submanifold without boundary. Assumef andf|∂M are transverse to Q. Then the set
P :=f−1(Q) =
p∈M
f(p)∈Q
is a codimension-k submanifold of M with boundary ∂P =P ∩∂M and its tangent space atp∈P is the linear subspace
TpP =
v∈TpM
df(p)v∈Tf(p)Q .
4.1. TRANSVERSALITY 39 Proof. Let p0 ∈P =f−1(Q) and define q0 :=f(p0)∈Q. Then it follows from [21, Theorem 2.3.4] that there exists an open neighborhood V ⊂N of q0 and a smooth mapg:V →Rk such that the origin 0∈Rk is a regular value of g and V ∩Q=g−1(0). We prove the following.
Claim: Zero is a regular value of the map g◦f :U :=f−1(V)→Rk and also of the map g◦f|U∩∂M :U ∩∂M →Rk.
To see this, fix an elementp∈U such thatg(f(p)) = 0 and letη ∈Rk. Then q:=f(p)∈V ∩Q, g(q) = 0.
Since zero is a regular value of g, there exists a vectorw∈TqN such that dg(q)w=η.
Since f is transverse toQ, there exists a vector v∈TpM such that w−df(p)v ∈TqQ.
Since TqQ= kerdg(q), this implies
d(g◦f)(p)v =dg(q)df(p)v=dg(q)w=η.
Thus zero is a regular value ofg◦f :U →Rk, and the same argument shows that zero is also a regular value of the restriction of g◦f toU ∩∂M.
By Lemma 1.3.7 it follows from the claim that the set P∩U =f−1(Q)∩U = (g◦f)−1(0)
is a smooth (m−k)-dimensional submanifold of M with boundary
∂(P∩U) =P∩U∩∂M and the tangent spaces
TpP = kerd(g◦f)(p)
= kerdg(q)df(p)
=
v∈TpM
df(p)∈kerdg(q) =TqQ forp∈U withq :=f(p)∈Q. This proves Lemma 4.1.3.
The next goal is to show that, given a compact submanifold Q⊂N without boundary, every smooth map f :M →N is smoothly homotopic to a map that is transverse to Q. This is in contrast to Sard’s theorem in Chapter 1 which asserts, in the case where Q = {q} is a singleton, that almost every element q∈N is a regular value of f. Instead, the results of the present section imply that, given an element q ∈N, every smooth map f :M →N is homotopic to one that hasq as a regular value.
Thom–Smale Transversality
Assume throughout thatM is a smoothm-manifold with boundary, thatN is a smoothn-manifold without boundary, and thatQ⊂N is a codimension-ksubmanifold without boundary that is closed as a subset of N.
Definition 4.1.4 (Relative Homotopy). Let A ⊂M be any subset and let f, g : M → N be smooth maps such that f(p) = g(p) for all p ∈ A.
A smooth map F : [0,1]×M →N is called a homotopy from f to g relative to A if
F(0, p) =f(p), F(1, p) =g(p) for allp∈M (4.1.3) and
F(t, p) =f(p) =g(p) for all t∈[0,1]and all p∈A. (4.1.4) The maps f and g are called homotopic relative to A if there exists a smooth homotopy fromf to g relative to A. We write
f ∼A g
to mean that f is homotopic to g relative to A. That relative homotopy is an equivalence relation is shown as in Section 1.5.
Theorem 4.1.5(Local Transversality). Letf :M →N be a smooth map and letU ⊂M be an open set with compact closure such that
f U \U
∩Q=∅.
Then the following holds.
(i) There exists a smooth map g:M →N such that g is homotopic to f relative to M\U and bothg|U andg|U∩∂M are transverse to Q.
(ii)Iff|U∩∂M is transverse toQ, then there exists a smooth mapg:M →N such thatgis homotopic tof relative to∂M∪(M \U)andg|U∩∂M is trans-verse toQ.
Proof. See page 44.
Corollary 4.1.6 (Global Transversality). Assume M is compact. Then every smooth map f :M →N is homotopic to a smooth map g:M →N such that both g and g|∂M are transverse to Q, and the homotopy can be chosen relative to the boundary whenever the restriction off to the boundary is transverse to Q.
Proof. Theorem 4.1.5 withU =M.
The proof of Theorem 4.1.5 relies on the following lemma.
4.1. TRANSVERSALITY 41 Lemma 4.1.7. Let N be an n-manifold without boundary, let Q⊂N be a closed set, let K ⊂N be a compact set, and let V ⊂N be an open neighbor-hood ofK∩Q with compact closure. Then there exists an integer`≥0and a smooth map G:R`×N →N such that, for all λ∈R` and all q ∈N, Moreover, if W ⊂ N is an open neighborhood of V, then G can be chosen such that G(λ, q) =q for allλ∈R` and all q∈N\W.
Proof. The proof has three steps.
Step 1. Let W ⊂N be an open neighborhood of V with compact closure.
Then there are vector fields X1, . . . , X`∈Vect(N) such that supp(Xi)⊂W for all iand TqN = span{X1(q), . . . , X`(q)} for allq ∈V.
Assume without loss of generality that N ⊂R` is a smooth submanifold of the Euclidean spaceR` for some integer`and thatN is a closed subset ofR` (see Theorem A.3.1). By Theorem A.2.2 there exists a partition of unity subordinate to the open cover M =W ∪(M\V) and hence there exists a constant ε >0 such that the following holds.
(I) If q ∈V and t∈R` satisfies maxi|ti|< ε, then
The vector fieldsXi have compact support and hence are complete. Thus the mapψ:R`×N →N is well defined. It satisfies
ψ(0, q) =q, ∂ψ
∂ti
(0, q) =Xi(q)
for allq∈N and all i∈ {1, . . . , `}. Hence (4.1.8) holds for t= 0 by Step 1 and so assertion (I) follows from the fact that V is compact and the set of all pairs (t, q)∈R`×N that satisfy (4.1.8) is open.
To prove (II) we argue by contradition and assume that (II) is wrong for every constant ε >0. Then there exist sequences tν ∈R` and qν ∈K\V such that limν→∞tν = 0 and ψ(tν, qν)∈Q for all ν. Since K is compact, there exists a subsequence (still denoted by qν) that converges to an el-ement q ∈K. Moreover, since G is continuous and Q is a closed subset of N, we have q=ψ(0, q) = limν→∞ψ(tν, qν)∈Q. Thus q∈K∩Q⊂V. SinceV is an open subset ofN, this implies qν ∈V forν sufficiently large, a contradiction. Thus (II) must hold for someε >0 and this proves Step 2.
Step 3. We prove Lemma 4.1.7.
Letψ be as in Step 2 and define the mapG:R`×N →N by G(λ1, . . . , λ`, q) :=ψ
ελ1
pε2+λ21, . . . , ελ`
q ε2+λ2`
, q
(4.1.9)
for λi ∈R and q∈N. Then G(0, q) =q for all q∈N and so G satis-fies (4.1.5). Moreover,G satisfies (4.1.6) by (II) and satisfies (4.1.7) by (I).
This proves Lemma 4.1.7.
Remark 4.1.8. The assertion of Lemma 4.1.7 holds with `≤2n. To see this, suppose that the vector fields X1, . . . , X` satisfy the requirements of Step 1 in the proof of Lemma 4.1.7 with ` >2n. Choose a Riemannian metric onN and define the mapf :T N →R` by
f(q, w) := hw, X1(q)i, . . . ,hw, X`(q)i
forq ∈N and w∈TqN.
This map has a regular value ξ = (ξ1, . . . , ξ`) ∈ R` by Sard’s theorem.
Since ` >2n= dim(T N), we have ξ /∈ f(T N) and, in particular, ξ 6= 0.
Assume without loss of generality thatξ`6= 0 and defineYi ∈Vect(N) by Yi(q) :=Xi(q)− ξi
ξ`X`(q) forq∈N and i= 1, . . . `−1.
Then, sinceξ /∈f(T N), it follows that TqN = span{Y1(q), . . . , Y`−1(q)} for allq ∈K. (Exercise: Verify the details.)
4.1. TRANSVERSALITY 43 We also need the following lemma. LetQ⊂N be a codimension-k sub-manifold without boundary and letF :R`×M →N be a smooth map such that both F and F|
R`×∂M are transverse toQ. Then Lemma 4.1.3 asserts that the set
Now it follows from (A) that there exists a tangent vector v0 ∈TpM such that (bλ, v0)∈T(λ,p)M and so dF(λ, p)(bλ, v0)∈TqQ. This implies
w−dFλ(p)(v−v0) =w−dF(λ, p)(bλ, v)−dF(λ, p)(bλ, v0)∈TqQ and so (B) holds. This shows that (A) is equivalent to (B) and this proves (i).
The proof of (ii) is analogous and this proves Lemma 4.1.9.
Proof of Theorem 4.1.5. We prove part (i). Since U is compact, so is K :=f(U)⊂N.
Moreover,f(U \U)∩Q=∅ and this impliesK∩Q⊂N \f(U\U). Since the setN\f(U \U) is open, Lemma A.1.2 asserts that there exists an open setV ⊂N with compact closure such that
K∩Q⊂V ⊂V ⊂N \f(U\U).
Hencef(U \U)∩V =∅ and so the set
B :=U∩f−1(V) =U ∩f−1(V)
is compact. Hence there exists a smooth functionβ :M →[0,1] such that
supp(β)⊂U, β|B= 1. (4.1.10)
(See Theorem A.2.2.) Choose a map G:R`×N →N as in Lemma 4.1.7 and defineF :R`×M →N by
F(λ, p) :=Fλ(p) :=G(β(p)λ, f(p)) for (λ, p)∈R`×M. (4.1.11) Then
F0=f, Fλ|M\U =f|M\U for allλby (4.1.5) in Lemma 4.1.7. We prove thatF|
R`×U andF|
R`×(U∩∂M)
are transverse to Q. Fix an element (λ, p)∈R`×U with F(λ, p)∈Q.
Then G(β(p)λ, f(p)) =F(λ, p)∈Q by definition of F, and so it follows from (4.1.6) with q :=f(p)∈K and λ replaced by β(p)λ that f(p)∈V. This impliesp∈U∩f−1(V) =B, and hence the vectors
∂F
∂λi(λ, p) =β(p)∂G
∂λi(β(p)λ, f(p)) = ∂G
∂λi(λ, f(p))
span the tangent space TF(λ,p)N by (4.1.7) in Lemma 4.1.7. This shows that F|
R`×U and F|
R`×(U∩∂M) are transverse to Q as claimed. Hence, by Lemma 4.1.3, the set
M := (R`×U)∩F−1(Q)
is a smooth submanifold of R`×U with boundary ∂M =R`×(U ∩∂M).
By Sard’s theorem there exists a common regular valueλ∈R` of the pro-jectionπ:M →R` and of π|∂M :∂M →R`. Hence, by Lemma 4.1.9, the homotopyft(p) :=F(tλ, p) satisfies the requirements of part (i).
4.1. TRANSVERSALITY 45 We prove part (ii). Thus assume thatf|U∩∂M is transverse toQ. As in the proof of (i), define the compact set
K:=f(U)⊂N,
choose an open neighborhood V ⊂N of K∩Q with compact closure such that
f(U\U)∩V =∅, and define the compact set B⊂M by
B :=U ∩f−1(V).
We prove that there exists a smooth function β:M →Rsuch that
supp(β)⊂U, β|U∩∂M = 0, βB\∂M >0. (4.1.12) To see this choose a smooth function β1:M →[0,1] with
supp(β1)⊂U, β1|B = 1
as in (4.1.10). Choose an atlas {Uα, φα}α∈A on M and let ρα:M →[0,1]
be a partition of unity subordinate to the cover, i.e. each point in M has an open neighborhood on which only finitely many of theραdo not vanish and
supp(ρα)⊂Uα, X
α
ρα= 1.
(See Theorem A.2.2.) For α∈Adefine βα:Uα→Rby βα◦φ−1α (x) :=xm
for x∈φα(Uα)⊂Hm. Then the function ραβα:Uα→R extends uniquely to a smooth function on M that vanishes onM\Uα, the function
β0 :=X
α
ραβα :M →R
vanishes on the boundary and is positive in the interior, and so the product functionβ :=β0β1 satisfies (4.1.12).
With this understood, the proof of part (ii) proceeds exactly as the proof of (i). The key observation is that the functionF :R`×M →N in (4.1.11) still has the property that F|
R`×U and F|
R`×(U∩∂M) are transverse to Q, because F(λ,·)|∂M =f|∂M for all λ∈R` and f|U∩∂M is transverse to Qby assumption. This proves Theorem 4.1.5.