Verify that this is not orientable.

Question

Verify that this is not orientable.

Möbius transformation:

$$U=\{(t,\theta) \mid \frac{-1}{2}\lt t\lt \frac{1}{2}, 0\lt \theta \lt 2\pi \}$$ $\sigma (t, \theta)=<((1-t\sin (\theta/2))\cos (\theta), (1-t\sin (\theta/2)\sin \theta, t \cos (\theta /2)>$

$\tilde{U}=\{(t,\theta) \mid \frac{-1}{2}\lt t\lt \frac{1}{2}, -\pi\lt \theta \lt \pi \}$ $$\tilde{\sigma}(t,\theta)=\sigma(t, \theta) $$

$\{\sigma, \tilde{\sigma}\}$ forms an atlas for möbius band

$$N_{\sigma}=\lambda (t,\theta)N_{\tilde{\sigma}}$$

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I have written all I know. Please help me to verify that this is not orientable.

And also I calculated the following:

$\sigma_{\theta}=(-\sin \theta, \cos \theta, 0)$

and $\sigma_t=(-\sin (\theta /2)\cos \theta, -\sin (\theta /2)\sin \theta, \cos (\theta /2)$

and $\sigma_t \times \sigma_{\theta}=(-\cos \theta \cos (\theta /2), -\sin \theta cos(\theta /2), -\sin (\theta/2))$

$N_{\sigma}=\sigma_t \times \sigma_{\theta}$

But I dont understand why the möbius band is not orientable.

Answer 1

Hint: A hypersurface of $\mathbb{R}^n$ is orientable if, and only if there is a global, smooth, unit length normal vector field.