# $S:= \{{x,y} \in \mathbb R^2:x^2+y^2=1\}$ 1-manifold charts.

Define $S:= \{{x,y} \in \mathbb R^2:x^2+y^2=1\}$. This is a 1-manifold when equipped with chart $(-1,1)\to \mathbb R^2$:

$\Phi_1(x):=(x,\sqrt{1-x^2})$, $\Phi_2(x):=(x,-\sqrt{1-x^2})$, $\Phi_3(y):=(\sqrt{1-y^2}, y)$, $\Phi_4(y):=(-\sqrt{1-y^2}, y)$

It is clear $\Phi_1$ and $\Phi_3$ are going in opposite directions (as well as $\Phi_2$ and $\Phi_4$).

Formally, $V_{1,3}=S\cap(0,∞)^2$ and...

This is where I get confused.

and on $\Phi_1^{-1}(V_{1,3})=(0,1)$,

$(\Phi_3^{-1}\circ\Phi_1)(x)=\sqrt{1-x^2}$, so that

$(\Phi_3^{-1}\circ\Phi_1)'(x)=\frac{-2x}{2\sqrt{1-x^2}}<0$, so $\Phi_1$ and $\Phi_3$ have opposite orientations.

Can someone show me how these three functions are derived?

I've been trying and the ones I get don't match up.

Is $\Phi_1^{-1}(V_{1,3})=(0,1)$ because these charts only intersect on the positve axis (where $x,y\geq 0$)?

I'm not sure how to find the inverse of a chart or why it becomes one 'coordinate' in $(\Phi_3^{-1}\circ\Phi_1)(x)$ and $(\Phi_3^{-1}\circ\Phi_1)'(x)$ instead staying two like in $\Phi_i$ from the first part. (I understand the differentiation to get $(\Phi_3^{-1}\circ\Phi_1)'(x)$).

Apologies if I haven't explained that very well.

Any help greatly appreciated!

• I assume you mean $\Phi_3(y)$ and $\Phi_4(y)$ in your definitions, not $\Phi_3(x)$ and $\Phi_4(x)$. – Connor Harris Apr 6 '17 at 14:21
• Ah yes, sorry that's what it says on the example. Must be a typo. – user415105 Apr 6 '17 at 14:24

1. $V_{1,3}$ is the portion of the circle in the first quadrant, and $\Phi_1$ takes positive $x$-values to corresponding first-quadrant points, so yes, $\Phi_1^{-1}(V_{1,3})$ is the open interval $(0, 1)$. (Are you getting confused by the fact that the notation for open intervals is identical to the notation for points?)
2. The inverse of a chart is just a map from a portion of the circle back to the interval $(-1, 1)$. So $\Phi_3^{-1} \circ \Phi_1(x)$ means "given a real number $x$ between $0$ and $1$, first find the corresponding point $\Phi_1(x) = (x, \sqrt{1-x^2})$, and then find the real number $y = \sqrt{1-x^2}$ such that $\Phi_3(y)$ is the same point on the circle." Basically, the $\Phi_i$ are partially defined functions $(-1, 1) \to S$, and the $\Phi_i^{-1}$ are partially defined functions $S \to (-1, 1)$.