I'm a beginner at differential geometry and I'm having some trouble with the following problem:
Let $M \subset \mathbb{R}^n$ be a $k$-dimensional smooth manifold (smoothly embedded in $\mathbb{R}^n$), where $k<n$. Define
$$N = \{ (q, v) \left. \right| q \in M, v \in \mathbb{R}^n \text{ is perpendicular to } M \text{ at } q \} \subset M \times \mathbb{R}^n.$$
Show that $N$ is an $n$-dimensional smooth manifold.
My thoughts so far:
Let $(q,v) \in N$. Since $M$ is a $k$-dim. manifold we can choose an open neighbourhood $U \subset \mathbb{R}^n$ of $q$ and an open subset $V \subset \mathbb{R}^k$ and a diffeomorphism
$\varphi: M \cap U \rightarrow V$.
Consider (for fixed $q$) the set
$$A := \{ (q,w) \left. \right| w \in \mathbb{R}^n \text{ perpendicular to } M \text{ at } q \}.$$
Since the tangent space $T_qM$ is a $k$-dim subspace of $\mathbb{R}^n$ and $A$ is a vector space itself with $\mathbb{R}^n = T_qM \oplus A$, we know that $A$ is in fact an $(n-k)$-dimensional subspace of $\mathbb{R}^n$.
Therefore I suspect that we could just consider the open neighbourhood $U \times \mathbb{R}^n \subset \mathbb{R}^{2n}$ of the point $(q,v)$. If we intersect this with $N$ we should get something like
$$(U \cap M) \times ((n-k)\text{-dim. subspace of } \mathbb{R}^n)$$
which we should be able to map diffeomorphically onto an open subset of $\mathbb{R}^n$ of the form $V \times \mathbb{R}^{n-k}$ via the map $\varphi \times id$.
Is my argumentation somewhat correct? And how could one write this down in a simple way?
Thaks in advance for any help!
Best Answer
My guess is that the proof will be easier if you also use the fact that $M$ is locally the regular level set of a function. Namely, there exists a function $f : U \to \mathbb{R}^{n-k}$ with the following properties:
Now you have enough information to write down the function $T(U \cap M) \to V \times \mathbb{R}^{n-k}$ which takes $(q,w) \in T(U \cap M)$ to $(\phi(q),Df_q(w))$. Of course you still have to put the pieces together to show that this function is a diffeomorphism.