# Problem understanding Milnor Proof: (Theorem 1, Vector Fields chapter)

(THIS QUESTION HAS BEEN EDITED)

I am reading Milnor's ''Topology From a Differentiable Viewpoint'' and in the chapter about vector fields, page 38, there is a theorem that states that:

Given any vector field $$v$$ on $$M \subset \mathbb{R^n}$$, ($$M$$ an m-dimensional, compact boundaryless manifold) with only nondegenerate zeros, then the index sum of $$v$$ is equal to the degree of the Gauss mapping.

So, I am going to avoid writing the proof (if anyone needs it to give a better answer I will write it, just prefer to avoid this if not necessary).

You have:

1. $$N_{\epsilon}$$ a closed $$\epsilon$$-neighborhood of $$M$$

2. $$r$$: $$N_{\epsilon} \to M$$, a differentiable map that maps $$x$$ to the point in $$M$$, closest to $$x$$. (Making $$\epsilon$$ small enough this works well)

3. $$w$$ a vector field in $$N_{\epsilon}$$ that extends $$v$$, given by: $$w(x)= (x-r(x))+v(r(x))$$

So.. What I think Milnor is trying to do, is to use Hopf's lemma (Lemma 3 in the book),you can see that $$w$$ points outwards along the boundary and if it has a zero, it must be a zero of $$v$$ (since $$x-r(x)$$ and $$v(r(x))$$ are orthogonal), so all its zeros are isolated, then you are in condition to apply Hopf's lemma.

Now he states that: For any $$z \in M$$

$$d_zw(h)= d_zv(h)$$ in $$T_zM$$

and

$$d_zw(h)=h$$ in $$T_zM$$'s orthogonal complement.

So... maybe it is easy to see, but I cannot see why this is. Trying to calculate $$\frac{\partial w_i}{\partial x_j}$$ for arbitrary $$i,j$$ seems usless, since I do not know how to calculate the derivative of $$r(x)$$. This is my first doubt.

My second doubt.. Supposing that the previous statement is true, do you have that $$d_zw$$ and $$d_zv$$ have the same determinant at any $$z$$, zero of $$w$$, hence the same index. Now Milnor uses Hopf's Lemma, to prove that the index sum of $$w$$ is equal the Gauss mapping. To finish this off I need to see that $$v$$ has the same zeros than $$w$$, why is this true? (You know that a zero of $$w$$ is a zero of $$v$$, why is the reciprocal true?)

I am really stuck with this, any help would be appreciated, thanks in advance.

(FROM HERE ON I EDITED IT)

So... I am going to answer my second doubt: If $$z$$ is a zero of $$v$$, then $$z \in M$$ so $$r(z)=z$$ and thus $$w(z)=v(z)=0$$

And for my first doubt, I was thinking that I never used that $$v$$ has non-degenerate zeros (Or so I think), so maybe that is a hint to anwering my first doubt, cannot seem to see it though. If I had used that $$v$$ has only non-degenerate zeros, please tell me where. Thanks in advance.