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?

Thanks in advance!


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 $0$.


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