Resently I’m reading Bott-Tu’s

differential forms in algebraic topology, and comparing it with some differential topology textbook.While proving the Poincare-Hopf theorem, it defines the intersection number I(M,N), where M, N are submanifold of another smooth manifold K with condition dimM+dimN=dimK, to be I(M,N)=∫KηM∧ηN. Here ηM is the Poincare dual of M, so ηM∧ηN is a top form in K, assuming the dimension condition.

My question is why this definition is coincide with the definition appeared in differential topology, which using the orientable intersection number that “sum up” ±1 locally according to the orientation.

If it’s possible, I perfer an answer without using much algebraic geometry. Any links to webpage or other reference are welcomed as well. Great Thanks!

**Answer**

Imagine nonnegative function f:R→R that is supported in the interval [−ϵ,ϵ] with ∫Rfdx=1. If γ:[0,1]→R is any path with γ(0) and γ(1) outside of [−ϵ,ϵ] then ∫[0,1]γ∗(fdx) is equal to the algebraic intersection number relative to its boundary, of γ with 0.

We can play the same game in Rn, defining an n-form ω supported in an ϵ-ball of →0, so that

∫Rnω=1. Once again if γ:B→Rn is a smooth map of the ball with γ(∂B) outside the ϵ-ball where ω is supported, then

∫Bγ∗(ω)

will be the intersection number of B relative to its boundary, or alternately the winding number of ∂B about →0.

The reason for this is the full change of variables formula for integrating n-forms of compact support on an oriented n-manifold. If f:X→Y is a smooth, proper map of oriented n-manifolds, with Y connected, define the degree of f as follows. Let y∈Y be a regular value of f. Since f is proper f−1(y) is compact. Since it is a zero manifold, it is a finite set of points. At each x∈f−1(y) define the sign of x to be ±1 depending on whether dfx:TxX→TyY is orientation preserving or orientation reversing. The degree of f, denoted deg(f) is

∑x∈f−1(y)sgn(x). Standard elementary arguments show that deg(f) is independent of the point y, and unchanged by homotopies of f through smooth proper maps.

Change of variables says that if ω is an n-form with compact support on Y, then f∗ω will have compact support. Both integrals are well defined and

∫Xf∗ω=deg(f)∫yω.

Here is a slightly flawed but intuitive proof. The set of singular points of f has measure zero by Sard’s theorem. Remove the singular points of f, S(f)⊂Y from Y and f−1(S(f)) from X. You now have a map

f:X−f−1(S(f))→Y−S(f) that is a local diffeomorphism.

Since you removed a set of measure zero,

∫Y−S(f)ω=∫Yω.

Using the stack of records theorem,

∫X−f−1(S(f))ω=deg(f)∫Y−S(f)ω.

Since the pull back of an n-form at points where df is singular is zero, we have

∫Xf∗ω=∫X−f−1(S(f))f∗ω. Completing the proof.

If M⊂N is a compact regular submanifold, you can cover M with coordinate patches (Uα,ϕα) of N where M∩Uα is a slice for all α. That is in local coordinates ϕα, M is the set of points with xm+1=xm+2…=xn=0. We call (xm+1,…,xn) the complementary coordinates to M.

Complete this to a cover of N by adding N−M.

Choose a partition of unity subordinate to the open cover of N so that ρα:M→R≥0 has support inside Uα. Choose ϵ>0 so that inside the support of each ρα if the complementary coordinates M have norm less than ϵ are contained in Uα. Let ωα be an n−m form with support in the points in Uα whose complementary coordinates have norm less than ϵ and ∫Rn−mωα=1. Finally let

ω=∑αραωα. The form is smooth.

With a little more care you can make sure its closed.

The point is if K is a regular manifold that is transverse to M of dimension n−m, then

the pull back under inclusion i:K→N,

i∗(ω) has its support near the points where K intersects M, and if ϵ is small enough, the integral near each point is ±1 depending on the sign of the intersection of K and M at that point.

That is it.

**Attribution***Source : Link , Question Author : Ricanry , Answer Author : Charlie Frohman*