Question on the proof of the Poincare-Hopf theorem

Question

OP is reading Milnor's celebrated Topology from the Differentiable Viewpoint's 6th chapter, where he deals with indices of vector fields and in particular the Poincare-Hopf theorem. A lemma he used is stated and proved as following:

For any (smooth) vector field $v$ on (smooth, compact, boundaryless) manifold $M$ With only non degenerate zeros the index sum $\sum \iota$ is equal to the degree of the Gauss mapping: $g:\partial N_{\epsilon}\to S^{k-1}$. In particular this sum does not depend on the choice of vector field.(OP:$N_\epsilon$ is the $\epsilon$-thickening of $M$, which in the first place is embedded into $\mathbb R ^k$.)

Proof

For $x\in N_{\epsilon}$ let $r(x)$ denote the closest point on $M$. ...... $x-r(x)$ will be perpendicular to $TM_x$. For sufficiently small $\epsilon$, $r(x)$ is well defined and smooth.

Consider the squared distance function $\phi(x)=||x-r(x)||^2$. An EASYcomputation shows $\operatorname {grad} \phi (x) = 2(x-r(x))$. Hence, for each point on the level surface $\partial N_\epsilon = \phi ^{-1} (\epsilon^2)$ the outward normal vector is given by $g(x) = \frac {\operatorname {grad} \phi (x) }{||\operatorname {grad} \phi (x) ||}=(x-r(x))/\epsilon$. Extend $v$ to a vector field $w$ on the neighbourhood $N_\epsilon$ by setting $w(x)=x-r(x)+v(r(x))$. Then $w$ points outward along the boundary, since the inner product $w(x)\bullet g(x)$ is equal to $\epsilon >0$. Note that $w$ can vanish only at the zeros of $v$ in $M$; this is CLEAR since the summands $x-r(x)$ and $v(r(x))$ are mutually orthogonal.(Unfinished, but the OP is unable to proceed)

There is lack of clarity in the proof. The two places I used bold letters are my problems.

First, how is it easily computed that $\operatorname{grad} \phi(x)=2(x-r(x))$, without the derivative of $r(x)$ with respect to $x$ appearing?

Second and the most severe one that makes me doubt my sanity: how can $x-r(x)$ be generally perpendicular to $v(r(x))$? I don't think the assumption that vector field $v(x)$ is a tangent vector field is ever made in this chapter, so we have freedom of choosing the direction of $v(x)$ at a certain point, yet Milnor claims it must be tangent?

(Milnor's book is OP's first exposure to differential topology and therefore I entrusted the introduction to him skinny book. So the OP has no other knowledge on diff-top (e.g. transversality, cobordism, etc..) other than what's in this booklet.)

Answer 1

The proof applies to a tangent vector field. Milnor is proving that the sum of the indices of a tangent vector field is the Euler Characteristic of the manifold.This is not true for vector fields in other vector bundles.

The vector x - r(x) is perpendicular to the tangent plane to the manifold. So it is lineally independent of the vector field on the manifold.

The derivative of r-r(x) is zero along any level surface in the tubular neighborhood of the manifold and points perpendicular to these level surfaces. In the normal direction its derivative is the unit vector in the normal direction since r(x) remains constant in this direction.

• Note that the level surfaces are well defined n-1 dimensional manifolds in the tubular neighborhood. Outside of this neighborhood that may not be true since the rays normal to the manifold might converge. The Tubular Neighborhood Theorem guarantees that convergence will not occur in a small enough tubular neighborhood.