# Is this map 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 and $\partial N$ an $(n-1)$-manifold.

The Gauss map $G:\partial N\to S^{n-1}$ sends an element $v$ of $\partial N$ to the normalized normal vector pointing from $v$ outwards (image).

How can I define this map more rigorously? I.e., how can I make ''pointing from $v$ outwards'' more precise?

Now $\partial N$ is homotopic to the total space $E'$ of the sphere bundle $E'\to M$ of $E$.

Let $T$ be the tangent bundle of $M$ and $E\oplus T\cong \mathbb{R}^n\times M$ the trivial bundle on $M$. The inclusion $E\subset E\oplus T\cong \mathbb{R}^n\times M$ of vector bundles on $M$ restricts to sphere bundles and composing with the projection is $$G':\partial N\cong E'\to (S^{n-1}\times M)\to S^{n-1}$$

How can I see that $G'$ is homotopic to the Gauss map $G$?

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Pointing outwards is easy: a vector $w$ in $\mathbb{R}^n$ is said to point outwards in $\nu \in \partial N$ if $\exists \epsilon > 0$ such that for every $s\in (0,\epsilon)$, $\nu + s w \not\in N$. The second requires some thought. – Willie Wong Jan 22 '13 at 13:17
@WillieWong: Yes, I see, thanks for the answer to the first question. – user59218 Jan 22 '13 at 13:20
For the second question, I sketch what I think should be a method. By definition of the tubular neighborhood you have that $N \setminus M$ is diffeomorphic to $E' \times (0,1]$. For $s\in (0,1]$ let $N_s$ be the image of $E'\times \{s\}$. Let $G_s$ be the Gauss map associated to $N_s$. You directly compute that as $s\to 0$ $G_s \to G'$ continuously.ă€€This establishes $G_s$ as the homotopy. – Willie Wong Jan 22 '13 at 13:26