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 $T_zM$


$d_zw(h)=h$ 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.


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.


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