# Degree of Gauss map equal to half the Euler characteristic and Poincaré-Hopf

The Poincaré-Hopf theorem states that for a smooth compact $m$-manifold $M$ without boundary and a vector field $X\in\operatorname{Vect}(M)$ of $M$ with only isolated zeroes we have the equality $$\sum_{\substack{p\in M\\X(p)=0}}\iota(p,X)=\chi(M)$$ where $\iota(p,X)$ denotes the index of $X$ at $p$ and $\chi(M)$ denotes the Euler characteristic of $M$.

Let $m$ be even and $M\subset\mathbb{R}^{m+1}$ be a $m$-dimensional smooth compact submanifold without boundary and denote by $\nu:M\to S^m$ its Gauss map. How can I deduce from the Poincaré-Hopf theorem that the Brouwer degree of $\nu$ is equal to half the Euler characteristic of $M$ i.e. $$\deg(\nu)=\frac{1}{2}\chi(M)?$$

After many repeated (unsuccessful) tries I was hoping hat someone else might shed some light onto this...

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I recommend that you read Chapter 6 of Milnor's Topology from the differentiable viewpoint and then read up the missing details in his Morse theory. –  t.b. Aug 25 '11 at 9:31
Milnor's "Topology from the differentiable viewpoint" is precisely where I learned about the Poincaré-Hopf theorem. Unfortunately he doesn't address the degree of the Gauss map being half of the Euler characteristic anywhere in this book. Are you saying that in his "Morse theory" there is indeed an exposition of the argument I'm after? –  Martin Worsek Aug 25 '11 at 11:21
No I'm not saying that, that is, I don't know. I was convinced that it was in Milnor and I remembered that he referred to Morse theory for some details in the proof of Poincaré-Hopf. –  t.b. Aug 25 '11 at 12:32
This is not true for odd $m$. For example if $M=\mathbb{S}^{1}$ then $\chi(M)=0$, while the degree of the Gauss map equals 1. –  symboleon Aug 25 '11 at 19:17
Indeed, I'm going to edit my original post to include the assumption of an even dimension. –  Martin Worsek Aug 26 '11 at 17:28

One way to go about this is to start with your manifold $M \subset \mathbb R^{m+1}$ and consider a height function $f : \mathbb R^{m+1} \to \mathbb R$ (orthogonal projection onto a vector) restricted to $M$. So there is some fixed vector $v \in S^m$ such that $f(x)=\langle x,v\rangle$ for al $x \in \mathbb R^{m+1}$.

Generically, this is a Morse function so its gradient (the orthogonal projection of $v$ to $TM$) is a vector field which is transverse to the $0$-section of $TM$. So Poincare-Hopf tells you how you can compute the Euler characteristic from this.

Now how is that related to the Gauss map? If you were to compute the degree of the Gauss map $\nu : M \to S^m$ by computing its intersection number with $v \in S^m$ you would have a very similar looking sum to compute! But do notice that the orthogonal projection of $v$ to $TM$ can be zero at both $\nu^{-1}(v)$ and $\nu^{-1}(-v)$. When you work out the details this ultimately explains why there's the $1/2$ and why it only works in even dimensions.

I hope that gives you the idea without giving too much away.

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@Sam: I think perhaps you've misunderstood the content of my post. –  Ryan Budney Aug 26 '11 at 18:52
@Sam: "Aren't you saying..." My answer is yes. "And that sum..." this question is too vague, I'm not saying anything that vague. "But if we can only..." this seems confused. The point is that solutions to $\nabla (f_{|M}) = 0$ are divided into two sorts -- the ones in $\nu^{-1}(v)$ and the ones in $\nu^{-1}(-v)$. –  Ryan Budney Aug 26 '11 at 19:27
This answer definitely helps –  Martin Worsek Aug 26 '11 at 19:53