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For an orientable 2-manifold in $\mathbb{R}^4$, it's somewhat obvious that if the manifold has a normal vector field then it has a pair of normal vector fields.

I am trying to understand why it is not true for the Klein bottle embedded in $\mathbb{R}^4$. It has a normal vector field, but I can't visualize how adding a second normal vector will cause a problem.

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    $\begingroup$ This is somehow equivalent to the other answer, but you can note for $\omega$ a volume form on $\Bbb R^4$ if $X_1$ and $X_2$ are linearly independent normals to an embedding of $K$ then $(i_{X_1}(i_{X_2}(\omega))$ restricted to $TK$ is a volume form hence $K$ would be orientable. $i$ denotes the contraction operator. $\endgroup$
    – PVAL-inactive
    May 16, 2016 at 15:19

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By a couple of normal vector fields, I assume you mean two linearly independent normal vector fields.

Let $i : K \to \mathbb{R}^4$ be an embedding. On $K$ we have an exact sequence of vector bundles

$$0 \to TK \to i^*T\mathbb{R}^4 \to N \to 0$$

where $N$ is the normal bundle; note that $\operatorname{rank}N = \dim\mathbb{R}^4 - \dim K = 4 - 2 = 2$. If $K$ had two linearly independent normal vector fields, $N$ would be trivial. Applying the first Stiefel-Whitney class, we see that $w_1(i^*T\mathbb{R}^4) = w_1(TK) + w_1(N)$, but as $T\mathbb{R}^4$ and $N$ are both trivial, $w_1(i^*T\mathbb{R}^4) = i^*w_1(T\mathbb{R}^4) = i^*0 = 0$ and $w_1(N) = 0$, so $w_1(TK) = 0$. But this is impossible as the first Stiefel-Whitney class of the tangent bundle vanishes if and only if the manifold is orientable, which $K$ is not.

In general, a closed manifold $M$ admits an embedding into $\mathbb{R}^n$ with trivial normal bundle if and only if the manifold is stably parallelisable, i.e. $TM\oplus\varepsilon^k$ is trivial for some $k$. Therefore, if such an embedding exists, all the Stiefel-Whitney classes of $M$ must vanish; in particular, we see that $M$ must be orientable.

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  • $\begingroup$ "i.e. $TM\oplus\varepsilon^1$ is trivial". I'm not sure if I believe this. Can't $TM$ be stably trivial (in other words for some $k$, $TM\oplus\varepsilon^k$ is trivial) but $TM\oplus\varepsilon^1$ be non-trivial? $\endgroup$
    – PVAL-inactive
    May 16, 2016 at 14:56
  • $\begingroup$ @PVAL: The two conditions are apparently equivalent, though I must admit, I have not worked out why. See page $69$ of this. $\endgroup$
    – Michael Albanese
    May 16, 2016 at 14:59
  • $\begingroup$ @PVAL: I have edited to avoid confusion. $\endgroup$
    – Michael Albanese
    May 16, 2016 at 17:10
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Came up with a bit simpler solution. Let $e_1(x)$ be a tangent vector field on the Klein bottle $K$. Suppose that $n_1(x)$ and $n_2(x)$ - pair of linearly indenpendent normal vector fields. Then we can choose $e_2(x)$ in the $TK_x$ so that four vectors $(e_1(x), e_2(x), n_1(x), n_2(x))$ form a positively oriented basis in $\mathbb{R}^4$. This would give us orientation on the Klein bottle.

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    $\begingroup$ No this doesn't work. In fact $S^2$ embeds in $\Bbb R^4$ with two linearly independent normal vectors, but of course there is no non-vanishing tangent vector field on $S^2$. $\endgroup$
    – PVAL-inactive
    May 30, 2016 at 4:08
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    $\begingroup$ But on $K$ there is as its Euler characteristic is zero. $\endgroup$
    – Stan
    May 30, 2016 at 4:41

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