Consider the map: \begin{equation} x: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}: (u,v) \rightarrow \bigg(\bigg(1+v\cos\frac{u}{2}\bigg)\cos(u), \bigg(1+v\cos\frac{u}{2}\bigg)\sin u, v\sin\frac{u}{2}\bigg) \end{equation} There is given that , for $\epsilon$ sufficient small, $M_{\epsilon}=x(\mathbb{R}\times(-\epsilon,\epsilon))$ is a surface.
My question: Is $M$ orientable and how does $M$ look like?
Approach: I have to check that there is an differentiable unit normal vectorfield that is defined on whole $M$. I first started calculating $\dfrac{x_u\times x_v}{\|x_u\times x_v\|}$, but the result was very complicated. Can someone help me? Thanks in advance!
First of all, you can check via Implicit Function Theorem that the image $M$ of $x$ is a differentiable surface in $\mathbb{R}^3$, and therefore it is locally orientable!
After, you can note that $M$ is the Möbius strip (link); in particular, $M$ is diffeomorphic to the algebraic surface \begin{equation} \left\{(x,y,u,v)\in\mathbb{R}^4\mid\begin{cases} x^2+y^2=1\\ (u^2-v^2)y=2uvx \end{cases}\right\} \end{equation} e via Jacobian matrix, you can prove that $M$ is not orientable!