8
$\begingroup$

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...

  • 0
    I$n$deed, I'm going to edit my original post to include the assumption of an even dimension.2011-08-26

2 Answers 2

4

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.

  • 0
    This answer definitely helps2011-08-26
0

Here is a characteristic classes way of doing it.

  • One uses the naturally of the Euler class for oriented vector bundles. This means that the Euler class of the induced bundle with induced orientation under a mapping is the pull back of the Euler class, f*(E)

The Euler class of the unit sphere is 1/(2pi)time its volume element. The pull back of the Euler class of the sphere under the Gauss map is the Euler class of the induced bundle - which in this case is the tangent bundle to the surface.

On the other hand the degree of the Gauss map is the integral of the pullback of 1(4pi)volume element of the sphere since the total volume of the sphere is 4pi.