Spivak Calculus on Manifolds - Tangent space on a boundary point of a manifold

Question

I am an undergraduate student who is studying Spivak's calculus on manifolds. I have several questions in the pages 119 and 120 of the book, which are about the tangent space at a boundary point of a manifold.

Our definitions of vector fields, forms, and orientations can be made for manifolds-with-boundary also. If $M$ is a $k$-dimensional manifold-with-boundary and $x\in\partial M$, then $(\partial M)_x$ is a $(k-1)$-dimensional subspace of the $k$-dimensional vector space $M_x$. Thus there are exactly two unit vectors in $M_x$ which are perpendicular to $(\partial M)_x$; they can be distinguished as follows (Figure $5-8$). If $f:W\to\mathbf{R}^{n}$ is a coordinate system with $W\subset H^{k}$ and $f(0)=x$, then only one of these unit vectors is $f_*(v_0)$ for some $v_0$ with $v^{k}<0$. This unit vector is called the outward unit normal $n(x)$; it is not hard to check that this definition does not depend on the coordinate system $f$. Suppose that $\mu$ is an orientation of a $k$-dimensional manifold-with-boundary $M$. If $x\in\partial M$, choose $v_1,\ldots,v_{k-1}\in(\partial M)_x$ so that $[n(x),v_1,\ldots,v_{k-1}]=\mu _x$. If it is also true that $[n(x),w_1,\ldots,w_{k-1}]=\mu _x$, then both $[v_1,\ldots,v_{k-1}]$ and $[w_1,\ldots,w_{k-1}]$ are the same orientation for $(\partial M)_x$. This orientation is denoted by $(\partial\mu)_x$. It is easy to see that the orientations $(\partial\mu)_x$,for $x\in\partial M$, are consistent on $\partial M$. Thus if $M$ is orientable, $\partial M$ is also orientable, and an orientation $\mu$ for $M$ determines an orientation $\partial\mu$ for $\partial M$, called the induced orientation. If we apply these definitions to $\mathbf{H}^{k}$ with the usual orientation, we find that the induced orientation on $\mathbf{R}^{k-1}=\left\{x\in\mathbf{H}^{k}:x^{k}=0\right\} $ is $(-1)^{k}$ times the usual orientation. The reason for such a choice will become clear in the next section.

1. I think I understand why there are exactly two unit vectors in $M_x$ which are perpendicular to $(\partial M)_x$. This is because the dimension of the subspace of $M_x$ that is orthogonal to $(\partial M)_x$ is 1, and in a one dimensional space, there are two unit vectors which have the same size but the different directions (Is this correct?). And Spivak says in the picture that "..., then only one of these unit vectors is $f_*(v_0)$ for some $v_0$ with $v^k < 0$. Is it right that "these unit vectors" mean the two unit vectors perpendicular to $(\partial M)_x$ and that it is 'only one' because it cannot have two different directions at the same time?

2. Spivak says it is easy to see that given an orientation of a k dimensional manifold with boundary $M$, the orientations $(\partial \mu)_x$ for $x \in \partial M$ are consistent on $(\partial M)_x$. My reasoning is that following the Spivak's definition of consistent choices of orientations, consider a chart $f: W \subset H^k \rightarrow \mathbb{R}^n$ and $a, b \in W$ ($a$ is an interior point and $b$ is an interior point). Because by assumption $M$ is orientable, if $\mu_{f(a)} = [f_*((e_1)a), . . . , f_*((e_k)a)]$, then $\mu_{f(b)} = [f_*((e_1)b), . . . , f_*((e_{k - 1})b), n(b)]$ where $n(b)$ is an outward unit normal. Then, we can choose $[f_*((e_1)b), . . . , f_*((e_{k - 1})b)]$ as $(\partial \mu)_b$. In this way, by restricting $f$ to the intersection of its image and $\partial M$, we can get the same orientation for all $x \in \partial M$; thus, if $M$ is orientable, so is $\partial M$. Please check if my reasoning is okay.

3. At the bottom of the picture, Spivak says the induced orientation on ... is $(-1)^k$ times the usual orientation. I really cannot understand how this happens.

4. In the next page, he says that the vectors $n(x)$ vary continuously on $M$. Also, conversely, if a continuous family of unit normal vectors $n(x)$ is defined on all of $M$, we can determine an orientation of $M$. Could you explain why this is the case?

Thank you!

Answer 1

1. You're right.
2. No idea.
3. The best way to understand these kinds of arguments is to read the proof of Stokes' theorem. Many things are conventional, not mathematically essential.
4. Firstly you should understand correctly what orientation is. An orientation of a real vector space $V$ of dimension $n$ is simply that one fixes a basis as standard, and every other basis is positively oriented with respect to the standard basis if the change of bases matrix is of positive determinant. Now if you have a vector subspace $U\subset V$ of codimension 1, you can define an induced orientation of $U$ from $V$ as follows: You choose any vector $N$ ouside $U,$ then any basis $u_1, \ldots, u_{n-1}$ is positive if $N, u_1, u_2, \ldots, u_{n-1}$ is positive in $V.$

But the ordering $N, u_1, \ldots, u_{n-1}$ is not obligatory. Some authors may use $u_1, u_2, \ldots, u_{n-1}, N.$ It depends on the purpose. One way to understand this cleary is to read the proof (or you can prove by yourself) of Stokes' theorem.

PS: some says if you don't understand what the author writes, prove it by yourself, it will be often useful for self-study.