I am reading John Lee's Introduction to smooth manifold, second edition, proof of Theorem 16.11 and stuck at understanding some statement.
Theorem 16.11 (Stokes's Theorem). Let $M$ be an oriented smooth $n$-manifold with boundary, and let $w$ be a compactly supported smooth $(n-1)$-form on $M$. Then
$$\int_Mdw=\int_{\partial M}w.$$
We accept this theorem for $M=\mathbb{H}^n$, the upper half-space. In fifth paragraph of the proof, he argues as follows :
Now let $M$ be an arbitrary smooth manifold with boundary, but consider an $(n-1)$-form $w$ that is compactly supported in the domain of a single positively or negatively oriented smooth chart $(U,\varphi)$. Assuming that $\varphi$ is a positively oriented boundary chart, the definition yeilds
$$ \int_Mdw := \int_{\mathbb{H}^{n}}(\varphi^{-1})^{*}dw = \int_{\mathbb{H}^{n}}d((\varphi^{-1})^{*}w) = \int_{\partial\mathbb{H}^{n}}(\varphi^{-1})^{*}w,$$
where $\partial\mathbb{H}^{n}$ is given the induced orientation. Since $d\varphi$ takes outward-pointing vectors on $\partial M$ to outward-pointing vectors on $\mathbb{H}^{n}$ ( by Proposition 5.41 ), it follows that $\varphi |_{U\cap \partial M}$ is an orientation-preserving diffeomorphism onto $\varphi(U) \cap \partial \mathbb{H}^{n}$, and thus
$$\int_{\partial\mathbb{H}^{n}}(\varphi^{-1})^{*}w = \int_{\varphi(U) \cap \partial \mathbb{H}^{n}}(\varphi^{-1})^{*}w = \int_{U \cap \partial M}w = \int_{\partial M} w.$$
Q. And why the bold statement ( $\varphi |_{U \cap \partial M}^{\varphi(U)\cap \partial \mathbb{H}^n}$ is orientation preserving ) is true? Intuitively it seems true, but I can't make a formal proof.
Here, the defintion of 'orientation-preserving' map is as follows :
Definition. A local diffeomorphism $F: M \to N$ between oriented smooth manifolds with or without boundary of positive dimension is orientation-preserving if for each $p\in M$, the isomorphism $dF_p$ takes oriented bases of $T_pM$ to oriented bases of $T_{F(p)}N$.
Can anyone helps?
Best Answer
It seems like the piece you're missing is the definition of the induced orientation on the boundary $\partial M$. This definition is buried in Propositions 15.21 and 15.24, and can be summarised as follows:
Now let's answer your question. We want to show that $\varphi|_{U \cap \partial M} : U \cap \partial M \to \varphi(U) \cap \partial \mathbb H^n$ is orientation preserving. This amounts to showing that $(d\varphi)_p$ maps a positively oriented basis of $T_p(\partial M)$ to a positively oriented basis of $T_{\varphi(p)}(\partial \mathbb H^n)$, for any $p \in U \cap \partial M$
So pick a $p \in U \cap \partial M$. Suppose $(E_1, \dots, E_{n-1})$ is a positively oriented basis for $T_p(\partial M)$. This means that for some outward-pointing vector $N \in T_p(M)$, $(N, E_1, \dots, E_{n-1})$ is a positively oriented basis for $T_p(M)$. (In fact, $(N, E_1, \dots, E_{n-1})$ is a positively oriented basis for all outward-pointing vectors $N \in T_p(M)$.)
Since $\varphi$ is a positively-oriented chart, the diffeomorphism $\varphi : U \to \varphi(U)$ is orientation preserving. So $((d\varphi)_p(N), (d\varphi)_p(E_1), \dots, (d\varphi)_p(E_{n-1}))$ is a positively-oriented basis for $T_{\varphi(p)}(\mathbb H^n)$.
$(d\varphi)_p(N)$ is an outward pointing vector in $T_{\varphi(p)}(\mathbb H^n)$, since $N$ is by assumption an outward pointing vector in $T_p(U)$. (The property of being an outward-pointing vector is preserved under pushforward by a diffeomorphism.)
So $((d\varphi)_p(E_1), \dots, (d\varphi)_p(E_{n-1}))$ is a positively oriented basis for $T_{\varphi(p)}(\partial \mathbb H^n)$, as required.
Edit: By the way, if it was me writing this proof, I would explain it differently. I even wrote notes for myself on this section of Lee to help me understand the accountancy of the plus and minus signs in my own way.
Let $(U, \varphi)$ be a boundary chart on $M$. Consider a $(n-1)$-form of the form $$ \omega := g(x^1, \dots, x^{n}) dx^1 \wedge \dots \wedge dx^{n-1},$$ where $g(x^1, \dots, x^{n})$ is a smooth function compactly supported within $U$, and where $x^1 \dots x^n$ are the local coordinates associated with the chart $(U, \varphi)$.
[There are other $(n-1)$-forms to consider, but those are less interesting from the perspective of fixing plus or minus signs because they integrate to zero on both sides of the Stoke's theorem.]
Taking the exterior derivative of $\omega$, we get $$ d\omega = (-1)^{n-1} \frac{\partial g}{\partial x^n}(x^1, \dots, x^{n}) dx^1 \wedge \dots \wedge dx^{n-1} \wedge dx^n. $$
Integrating this on $M$, we get $$ \int_M d \omega = \begin{cases} + (-1)^{n-1} \int_{-\infty}^{\infty} dx^1 \dots \int_{-\infty}^{\infty} dx^{n-1} \int_{0}^{\infty} dx^n \frac{\partial g}{\partial x^n} (x^1, \dots, x^{n}) & \text{if }\varphi \text{ positive oriented} \\ - (-1)^{n-1}\int_{-\infty}^{\infty} dx^1 \dots \int_{-\infty}^{\infty} dx^{n-1} \int_{0}^{\infty} dx^n \frac{\partial g}{\partial x^n} (x^1, \dots, x^{n}) & \text{if }\varphi \text{ negatively oriented} \end{cases}$$
By the fundamental theorem of calculus, this simplifies to $$ \int_M d \omega = \begin{cases} + (-1)^{n} \int_{-\infty}^{\infty} dx^1 \dots \int_{-\infty}^{\infty} dx^{n-1} g(x^1, \dots, x^{n-1}, 0) & \text{if }\varphi \text{ positive oriented} \\ - (-1)^{n}\int_{-\infty}^{\infty} dx^1 \dots \int_{-\infty}^{\infty} dx^{n-1} g (x^1, \dots, x^{n-1}, 0) & \text{if }\varphi \text{ negatively oriented} \end{cases}$$
The chart $(U, \varphi)$ on $M$ naturally induces a chart $(U \cap \partial M, \varphi|_{U \cap \partial M})$ on $\partial M$. The local coordinates associated with the chart $(U \cap \partial M, \varphi|_{U \cap \partial M})$ are $x^1, \dots x^{n-1}$, i.e. they are the first $n-1$ of the local coordinates associated with the chart $(U, \varphi)$. Relative to these two charts, the inclusion map $\iota: \partial M\to M$ is given by $(x^1,\dots, x^{n-1})\mapsto(x^1,\dots,x^{n-1},0)$
We have $$ \int_{\partial M} \iota^\star \omega = \begin{cases} + \int_{-\infty}^{\infty} dx^1 \dots \int_{-\infty}^{\infty} dx^{n-1} g(x^1, \dots, x^{n-1}, 0) & \text{if }\varphi|_{U \cap \partial M} \text{ positive oriented} \\ -\int_{-\infty}^{\infty} dx^1 \dots \int_{-\infty}^{\infty} dx^{n-1} g(x^1, \dots, x^{n-1}, 0) & \text{if }\varphi|_{U \cap \partial M} \text{ negatively oriented} \end{cases} $$
All that remains is to relate the orientation of $\varphi|_{U \cap \partial M}$ to the orientation of $\varphi$.
You can easily verify my claim using Lee's definition of the orientation on $\partial M$. I suggest working with the basis $(\frac{\partial}{\partial x^1}, \dots, \frac{\partial}{\partial x^{n-1}})$ for $T_p(\partial M)$, using $N = -\frac{\partial}{\partial x^n} \in T_p(M)$ as the outward pointing vector. Or check my claim against your multivariable calculus intuition in low dimensions.
Comparing our expressions for $\int_M d \omega$ and $\int_{\partial M} \iota^\star \omega$, we see that $$ \int_M d \omega = + \int_{\partial M} \iota^\star \omega,$$ regardless of whether the original chart $(U, \varphi)$ is positively or negatively oriented, and regardless of whether $n$ is odd or even.