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Let $M\subset \mathbb{R}^n$ be a closed compact smooth oriantable manifold. Let $E\to M$ be the normal bundle of the embedding and $N\subset \mathbb{R}^n$ a (closed) tubular neighbourhood. $N$ is an $n$-manifold with boundary $\partial N$, an $(n-1)$-manifold. The Gauss map $G:\partial N\to S^{n-1}$ sends an element $v$ of $\partial N$ to the normalized vector pointing from $v$ outwards (image).

Let a vector field with isolated zeros on $M$ be given. The Poincare-Hopf theorem of Milnor's book ''Topology from the differentiable viewpoint'' states that the Euler characteristic of $M$ equals the sum of the indices at the zeros of the vector field. The latter number equals by Theorem 1, page 38 of the same book the degree $deg(G)$ of the Gauss map $G$, if I read this correctly.

But some questions and answers on this site (e.g. this or that) state that $$ \chi(M)=2 deg(G). $$

What do I mix up here? Are there different definitions of degree or the Gauss map involved?

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In Lemma 3, $X$ is $N$, not $M$. –  user641 Jan 22 '13 at 13:31
    
Thanks, I meant Theorem 1 on page 38, sorry. –  user59218 Jan 22 '13 at 13:49

1 Answer 1

up vote 3 down vote accepted

I think you are just mixing up two definitions of Gauss map. One is intrinsic to the manifold (and is equal half of the Euler characteristic) while the other depends on the embedding into a Euclidean space and a tubular neighborhood and this will be twice bigger, since the boundary of this neighborhood has two components (i.e. inside and outside).

As a very simple illustration of this, take $M=S^2$. The Gauss map is just identity, and so its degree is $1 = {1 \over 2} \chi(S^2)$. But for the tubular neighborhood the Gauss map is essentially a projection from the inside and outside component of the neighborhood to the sphere, which means its degree is 2.

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Thank for this great and illustrative answer! –  user59218 Jan 22 '13 at 14:47

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