# Connection between Euler characteristic and degree of the Gauss map

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

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.