Claim: Show that the circle $S^1$ is a one-dimensional manifold.
Proof:
$(x,y)$ lies on the upper semi-circle, i.e., $y>0$.
The upper semi-circle, $V$ is open in $S^1$ containing $(x,y)$, and $U = (-1,1)$ is an open interval in $\mathbb{R}^1$. Now we give a parametrization of $V$, i.e., a diffeomorphism $f: (-1,1) \to V$ defined as $$ f(x) = \left(x,\sqrt{1-x^2}\right) $$
$f$ is one-one: $f(x_1)=f(x_2) \implies \left(x_1,\sqrt{1-x_1^2}\right) = \left(x_2,\sqrt{1-x_2^2}\right) \implies x_1=x_2$
$f$ is onto: For any $(x_0,y_0) \in S^1$ with $y_0>0$ we have $x_0 \in (-1,1)$ such that $\phi(x_0) = (x_0,y_0)$
$f$ is smooth: $$ f'(x) =\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{\left(x+h \ ,\ \sqrt{1-(x+h)^2}\right)-\left(x \ , \ \sqrt{1- x^2}\right)}{h} = \lim_{h \to 0} \frac{\left(h \ , \ \sqrt{1-(x+h)^2}-\sqrt{1-x^2}\right)}{h} = \left(1,\frac{-x}{\sqrt{1-x^2}}\right) $$
$$ f''(x) = \lim_{h \to 0} \frac{f'(x+h)-f'(x)}{h} = \lim_{h \to 0} \frac{\left(1,\frac{-x+h}{\sqrt{1-(x+h)^2}}\right)-\left(1,\frac{-x}{\sqrt{1-x^2}}\right)}{h} = \left(0,\lim_{h\to 0}\frac{\frac{-x+h}{\sqrt{1-(x+h)^2}}-\frac{-x}{\sqrt{1-x^2}}}{h}\right) = \left(0,\lim_{h \to 0}\frac{\frac{\sqrt{1-(x+h)^2}-(-x+h)\frac{-2(x+h)}{2\sqrt{1-(x+h)^2}}}{1-(x+h)^2}}{1}\right) = \left(0,\frac{1}{1-x^2}\right) $$
- $f^{-1}$ is smooth:
(x,y) lies on the lower semi-circle, i.e., $y<0$.
My Questions:
- Is there a simpler and a better way to show that $n^{th}$ order partial derivative is continuous? As finding them using above method is becoming very difficult.
