# Proof of that if $M$ is a $n$-manifold oriented, then $\partial M$ is a $(n-1)$-manifold oriented

I'm trying understand the result described on the title of the topic by the book "Differential forms and Applications" by do Carmo. The proof given by author can be found below

$$\textbf{Proposition 2.}$$ The boundary $$\partial$$ of an $$n$$-differentiable manifold $$M$$ with a boundary is an $$(n-1)$$-differentiable manifold. Furthermore, if $$M$$ is orientable, an orientation for $$M$$ induces an orientation for $$\partial M$$.

$$\textbf{Proof.}$$ Let $$p \in M$$ be a point of the boundary and let $$f_{\alpha}: U_{\alpha} \subset H^n \longrightarrow M^n$$ be a parametrization around $$p$$, then $$f_{\alpha}^{-1}(p) = q = (0,x_2, \cdots, x_n) \in U_{\alpha}$$. Let

$$\overline{U}_{\alpha} = U_{\alpha} \cap \{ (x_1, \cdots, x_n) \in \mathbb{R}^n \ ; \ x_1 = 0\}.$$

By identifying the set $$\{ (x_1, \cdots, x_n) \in \mathbb{R}^n \ ; \ x_1 = 0\}$$ with $$\mathbb{R}^{n-1}$$, we see that $$\overline{U}_{\alpha}$$ is an open set in $$\mathbb{R}^{n-1}$$. By denoting by $$\overline{f}_{\alpha}$$ the restriction of $$f_{\alpha}$$ to $$\overline{U}_{\alpha}$$, we see, by lemma $$3$$, that $$\overline{f}_{\alpha}(\overline{U}_{\alpha}) \subset \partial M$$. Finally, by letting $$p$$ run of points of $$\partial M$$, we easily check that the family $$\{ (\overline{U}_{\alpha},\overline{f}_{\alpha}) \}$$ is a differentiable structure for $$\partial M$$. This proves the first part of the Proposition.

To prove the second part, assume that $$M$$ is orientable and choose an orientation of $$M$$, i.e., a differentiable structure $$\{(U_{\alpha},f_{\alpha}) \}$$ such that the change of the coordinates has positive jacobian. Consider the elements of the family that satisfy the condition $$f_{\alpha}(U_{\alpha}) \cap \partial M \neq \emptyset$$. Then the family $$\{ (\overline{U}_{\alpha},\overline{f}_{\alpha}) \}$$ described in the first part is a differentiable structure for $$\partial M$$. We want to show that if $$\overline{f}_{\alpha}(\overline{U}_{\alpha}) \cap \overline{f}_{\beta}(\overline{U}_{\beta}) \neq \emptyset$$, the change of coordinates has positive jacobian, i.e., that

$$\det (d(\overline{f}_{\alpha}^{-1} \circ \overline{f}_{\beta})_q) > 0,$$

for all $$q$$ whose image, by some parametrization, is in the boundary.

Observe that the chagne of the coordinates $$f_{\alpha} \circ f_{\beta}^{-1}$$ takes a point of the form $$(0,x_2^{\beta}, \cdots, x_n^{\beta})$$ into a point of the form $$(0,x_2^{\alpha}, \cdots, x_n^{\alpha})$$. Thus, for a point $$q$$ whose the image is in the boundary,

$$\det (d({f}_{\alpha}^{-1} \circ {f}_{\beta})) = \frac{\partial x_1^{\alpha}}{\partial x_1^{\beta}} \det (d(\overline{f}_{\alpha}^{-1} \circ \overline{f}_{\beta})),$$

but $$\frac{\partial x_1^{\alpha}}{\partial x_1^{\beta}} > 0$$, because $$x_1^{\alpha} = 0$$ in $$q = (0,x_2^{\alpha}, \cdots, x_n^{\alpha})$$ and both $$x_1^{\alpha}$$ and $$x_1^{\beta}$$ are negative in a neighborhood of $$p$$. Since $$\det (d({f}_{\alpha}^{-1} \circ {f}_{\beta})) > 0$$ by hypothesis, we conclude that $$\det (d(\overline{f}_{\alpha}^{-1} \circ \overline{f}_{\beta})) > 0$$ as we wished. $$\square$$

My doubts exactly are

• How obtain the relation with the determinants?
• Why $$x_1^{\alpha} = 0$$ is important to conclude that $$\frac{\partial x_1^{\alpha}}{\partial x_1^{\beta}} > 0$$?
• This would not be $$\frac{\partial x_1^{\alpha}}{\partial x_1^{\beta}} \geq 0$$? Because I can have $$x_1^{\alpha} = 0$$ or can I not have this?

For points $p$ in the common domain of $f_\alpha$ and $f_\beta$, $x_1^\alpha(p) = 0$ if and only $p$ is on the boundary, which is true if and only $x_1^\beta(p) = 0$. So if at a point on the boundary, we hold $x_1^\beta$ constant (at $0$) while measuring the response of $x_1^\alpha$ to variances in any of the other $\beta$ coordinates, $x_1^\alpha$ is forced to remain at $0$ as well. Therefore $$\frac{\partial x_1^\alpha}{\partial x_j^\beta} = 0, \quad j > 1$$
If we use Cramer's rule to expand the Jacobian determinant on that column, we get that $\det(d(f_\alpha \circ f_\beta^{-1}))$ is the single (potentially) non-zero entry $\frac{\partial x_1^\alpha}{\partial x_1^\beta}$ times the minor determinant, which is $\det(d(\overline{f_\alpha} \circ \overline{f_\beta}^{-1}))$. I'll leave it you to figure how $\det(d(f_\alpha \circ f_\beta^{-1}))$ and $\det(d(f_\alpha^{-1}\circ f_\beta))$ are related.
As for the derivative question, since he says $x_1^\alpha$ and $x_1^\beta$ are negative near the border point $p$, apparently he defines $H^n := \{(x_1, ..., x_n) \in \Bbb R^n\mid x_1 \le 0\}$ (I am used to it being the other side, $x_1 \ge 0$, but either way gives the same result.) If we take the derivative at the boundary point (where both coordinates are $0$), holding the other $\beta$ coordinates constant so that $x_1^\alpha$ varies with $x_1^\beta$ only, the derivative is given by $$\frac{\partial x_1^\alpha}{\partial x_j^\beta} = \lim_{x_1^\beta \to 0-} \frac{x_1^\alpha(x_1^\beta) - 0}{x_1^\beta - 0}$$ Since both $x_1^\beta < 0$ and the corresponding $x_1^\alpha < 0$, the fraction is always positive, which means the limit is $\ge 0$.
But if the partial derivative is $0$, then by the determinant formula we would also have $\det(d(f_\alpha \circ f_\beta^{-1})) = 0$, which would imply $d(f_\alpha \circ f_\beta^{-1})$ is singular, which is not allowed for two coordinate systems from the same atlas. So $\frac{\partial x_1^\alpha}{\partial x_j^\beta}$ cannot be $0$.