How to prove given set is a smooth manifold

Question

I want to prove that the following set is a smooth manifold: $$M=\{(\vec x,\vec y)\in\Bbb{R}^n\times\Bbb{R}^n:\|\vec x-\vec y\|=1\}$$

My idea was to define $F:\Bbb{R}^n\times\Bbb{R}^n\rightarrow\Bbb{R} $ by $F(\vec x,\vec y)=\|\vec x-\vec y\|-1$, and so $M=F^{-1}(\{0\})$. One can easily check that for every $a=(\vec x,\vec y)\in M, RankJ_F(a)=1$, but I don't know how to take it from here.

Is my approach correct? what am I missing?

Answer 1

Let $L: \mathbb R^n \times \mathbb R^n \to \mathbb R^n \times \mathbb R^n$, $L(\vec x, \vec y) = (\vec x, \vec y - \vec x)$. $L$ is a diffeomorphism with inverse $(\vec x, \vec y) \mapsto (\vec x, \vec y + \vec x)$. Then

\begin{align} L(M) &= \{(\vec x,\vec y-\vec x)\in\Bbb{R}^n\times\Bbb{R}^n:\|\vec x-\vec y\|=1\}\\ &= \{(\vec x,\vec z)\in\Bbb{R}^n\times\Bbb{R}^n:\|\vec z\|=1\}\\ &= \mathbb R^n \times \mathbb S^{n-1} \end{align}

is clearly a smooth manifold. Thus $M = L^{-1} (L(M))$ is also a smooth manifold.