For manifolds with boundary , for each point in boundary, there are exactly two unit normal vectors to tangent space of boundary

Question

Let $X$ be a smooth manifold with boundary in $\mathbb R^n$ . Let $\partial X$ denote the manifold boundary of $X$ . Let $\newcommand{\del}{\partial}x \in \del X$ , then $\dim T_x (X) - \dim T_x(\del X)=1$ . How to show that there are exactly two unit vectors in $T_x(X)$ that are peroendicular to $T_x (\del X)$ ? (If needed, assume $X$ is connected )

Answer 1

You said

I can see that any basis of unit vectors of $T_x(\partial X)$ can be extended to a basis of $T_x(X)$ by adding one unit vector ... I am not able to see anything else , one main problem being to figure out what is the innerproduct on $T_x(X)$

From the comments above, I've told you what is probably (in my opinion) the inner product on $T_x(X)$ (namely the one inherited form the inclusion $X\subseteq \Bbb R^n$), so you can take that one extra basis vector that you know and Gram-Schmidt it* to get a non-zero normal vector. Divide it by its own length (as dictated by the inner product) to get a unit normal vector $v$. Take its negative to get the other.

As for why there are only two, let's take an arbitrary unit normal vector (among all possible ones) and call it $u$. It is orthogonal to the whole of $T_x(\partial X)$, so we can Gram-Schmidt $u$ with $v$ to get a $u'$ that is orthogonal to both $v$ and $T_x(\partial X)$ (a linear combination of vectors normal to $T_x(\partial X)$, which is what $u'$ is, is still normal to $T_x(\partial X)$).

That means that it is either $0$ or linearly independent of $(T_x(\partial X), v)$. Since it's an element of $T_x(X)$, and $\dim T_x (X) - \dim T_x(\del X)=1$, the latter cannot be the case, so it must be equal to $0$. The only vectors that become $0$ when Gram-Schmidt-ed with $v$ are multiples of $v$. Therefore $u$ was a multiple of $v$ and therefore it is one of the two we had already found.

*Gram-Schmidt is really something you do to all vectors in a basis, one at a time. Here I just mean to use one step on your original new basis vector, subtracting from it its orthogonal projection onto the subspace $T_x(\partial X)$ without really caring about orthogonalizing the other basis vectors with respect to one another.