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Let $M$ a submanifold of $\mathbb R^n$, for all $x$ in $M$, let $\pi_x:\mathbb R^n\rightarrow T_xM$ the orthogonal projection onto the tangent space $T_xM$ of $M$ at $x$.
How could you show that for all $x$ in $M$, it exists an open neighborhood $U$ of $x$, such as for all $y$ in $U\cap M$, $\pi_y$ restricted on $U$ is a diffeomorphism from $U$ onto $\pi_y(U)$?

Thank you.

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    The neighborhood isn't being taken with respect $M$; as the problem states, it's an open subset of $\mathbb{R}^n$ around $x\in M.$ Then the intersection with $M$ is taken.2012-06-03

1 Answers 1

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Change the coordinates such as $x=0$ and $\mathbb R^n=T_xM\oplus V=\{(t,w)\}$, so $T_xM:w=0$.
By the local inverse function theorem there is an open neighborhood of $x$ such as $\pi_x:U\rightarrow\pi_x(U)$ is a diffeomorphism.
Denote by $f:V=\pi_x(U)\rightarrow U$ the inverse function. So $f$ is a local parameterization of $M$ at $x$ by a neighborhood of $0$ in $T_xM$. We have $f(0)=0$ and $d_xf=0$ (since $T_xM:w=0$).

Let $y=(t,f(t))\in U\cap M$, then $T_yM=\{(t+h,f(t)+d_tf(h))\}$.
It exists $\varepsilon>0$ such as $\|d_tf\|<\varepsilon$ for all $(t,f(t))\in U$, so $p_y$ is injective on $U$.
Let $z\in U$, we check that $d_z(p_y):T_zU\rightarrow T_yU=T_yM$ coincides with the orthogonal projection $T_zU\rightarrow T_yM$ : $p_y:U\hookrightarrow R^N\rightarrow T_yM$ so $d_z(p_y)=T_zU\hookrightarrow R^N\rightarrow T_yM$.So $d_z(p_y)$ is invertible.
So $p_y$ is a diffeomorphism on $U$.