Let $M$ be a subvariety of $\mathbb{R}^n$. Show that the canonical projection $\pi: \nu M \rightarrow M$ is a submersion. Where $\nu M=\{(x,v)\in \mathbb{R}^n \times\mathbb{R}^n;x\in M \text{ and } v\perp T_{x}M\}$.

My initial idea was to show that restriction of differentiable is differentiable, but I can not write. It's been some time since I studied geometry.


We can only answer whether $\pi$ is differentiable (submersion) if some differentiable structure on $\nu M$ is given. We are considering the canonical one, given as follows:

Let $(x,v)\in \nu M$. Take a chart $\phi: U\subset M\to V\subset \Bbb R^n$, with $x\in U$ (this exists, since $M$ is a differentiable manifold). Then define $\nu\phi: \nu U\to V\times \Bbb R^n$ by $\nu \phi(y,u)=(\phi(y),u)$. The collection of all such $\nu\phi$'s is an atlas for $\nu M$. Indeed, the transition maps between $\nu\phi$ and $\nu\psi$ are given by $(q,u)\mapsto (\psi\phi^{-1}q,u)$ and $(q,u)\mapsto (\phi\psi^{-1}q,u)$ which are differentiable because its coordinates are differentiable.

Let's show that, with this structure, $\pi:\nu M \to M$ is differentiable.

Let $(x,v)$ be in $\nu M$ and $\nu \phi$ be just as above. Then, let us look to the expression of $\pi$ in the coordinates given by $\nu \phi$ about $(x,v)\in \nu M$ and $\phi$ about $\pi(x,v)=x\in M$. Avoiding to write down the open sets (domains and codomains) involved, the expression will be $(q,u)\in \Bbb R^n\times \Bbb R^n\overset{(\nu \phi)^{-1}}{\longmapsto} (\phi^{-1}(q),u) \in \nu M\overset{\pi}{\longmapsto} \phi^{-1}(q)\in M\overset{\phi}{\longmapsto}q\in \Bbb R^n$, i.e. $(q,u)\mapsto q$, which is differentiable in the usual ($\Bbb R^k$) sense. Therefore, $\pi:\nu M \to M$ is differentiable.

Now, we show that $\pi:\nu M\to M$ is a submersion.

Consider $(x,v)\in \nu M$. We have to show that $d\pi_{(x,v)}:T_{(x,v)}\nu M\to T_xM$ is surjective. So, consider a $v\in T_xM$ and a directive curve for it, i.e. some differentiable curve $\alpha:(-\epsilon,\epsilon)\to M$ such that $\alpha(0)=x$ and $\alpha'(0)=v$. Consider any normal local vector field $X:U\to \Bbb R^n$, $x\in U\subset M$, with $X(x)=v$. Then we can define $\beta:I\to \nu M$ by $\beta(t)=(\alpha(t),X(\alpha(t)))$. We have $\beta(0)=(x,v)$. Of course $\beta$ is differentiable. Denote $\beta'(0)=:w\in T_{(x,v)}\nu M$. We claim that $d\pi_{(x,v)}\,w=v$. In fact, \begin{align*} d_{(x,v)}\pi\,w:=\left.\frac{d}{dt}\right|_{t=0}(\pi\circ \beta(t))=\left.\frac{d}{dt}\right|_{t=0}\alpha(t)=\alpha'(0)=v. \end{align*} This concludes the proof.

Note that choosing a normal vector field $X$ was important to guarantee that $X(\alpha(t))\in (T_{\alpha(t)}M)^\perp$, i.e. to guarantee that $\beta:I\to \nu M$ is well-defined (I mean, we could not just take any curve in $\Bbb R^n$ going through $v$: we really need this normal vector field, although it disappears in the computation of $d\pi_{(x,v)}w$).


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