How these charts are written?

Question

The spherical coordinate map$$σ(u, v) = (\cos u \cos v, \cos u \sin v,\sin u), −π/2 < u < π/2, −π < v < π,$$ and its variation $$σ˜(u, v) = (\cos u \cos v,\sin u, \cos u \sin v), −π/2 < u < π/2, 0 < v < 2π,$$ are charts on the unit sphere. I know that a unit sphere can be covered by using 6 charts according to this question .why not just 2 charts to make atlas for sphere?

.But how the above written charts are made$\sigma (u,v)$?

I am reading charts and atlases from here Manifolds in euclidean space

Answer 1

You get a parametrization of a rotational surface $S\subset{\mathbb R}^3$ by first drawing its meridian curve $\mu$ in the $(\rho\geq0,z)$-halfplane $M$. Let $u$ be the parameter used for $\mu$. Then we have $$\mu:\quad u\mapsto\bigl(\rho(u),z(u)\bigr)\qquad(a\leq u\leq b)\ .$$ Now we rotate the meridian halfplane $M$ around the $z$-axis. In this way each point $(\rho,z)\in M$ generates a free floating circle $\gamma$ having its center on the $z$-axis: $$\gamma:\quad v\mapsto(\rho\cos v,\rho\sin v,z)\qquad(-\pi\leq v\leq \pi)\ .$$ Here $v$ is the usual polar angle in the $(x,y)$-plane. Applying this to each point $\bigl(\rho(u),z(u)\bigr)\in\mu$ we obtain the following parametrization of $S$ in terms of the parameters $u$ and $v$: $$S:\quad(u,v)\mapsto\bigl(\rho(u)\cos v,\rho(u)\sin v,z(u)\bigr)\ ,$$ with $a\leq u\leq b$, $\ -\pi\leq v\leq\pi$.

In the case of the unit sphere the meridian curve $\mu$ is a semicircle, and is given by $$\rho(u)=\cos u,\quad z(u)=\sin u\qquad\left(-{\pi\over2}\leq u\leq{\pi\over2}\right)\ .$$ Now put it all together.

Answer 2

You absolutely can use only two charts to cover a sphere. The way that I think of it is via stereographic projection. The question is as to how complicated you want your transition functions to be...

Here is one way to cover it with two charts, using a little bit of complex analysis. Let us consider $\mathbb{CP}^1$ to be the quotient of $\mathbb{C}^2 \setminus \{(0,0)\}$ via the diagonal action of $\mathbb{C}^\times$; that is, we say that $$ (a, b) \sim (\lambda a, \lambda b) $$ for every $\lambda \in \mathbb{C}^\times$. We can show that this is homeomorphic to $\mathbb{C} \cup \{\infty\} \cong S^2$ by considering the two sets (which also provides a covering by two charts at the same time!) $$ U = \{[a, b] \mid a \neq 0\} \qquad V = \{[a, b] \mid b \neq 0\} $$ Each of these is homeomorphic to $\mathbb{C}$, the first via the map $$ [a, b] \mapsto b/a \in \mathbb{C} $$ and the second via $$ [a, b] \mapsto a/b. $$ Note that the overlap of these two sets is $$ U \cap V = \{[a, b] \mid a, b \neq 0\} \cong \mathbb{C}^\times $$ and so the only point that is missing from $U$ is $[0, 1]$. In particular, the quotient we are describing is homeomorphic to $\mathbb{C} \cup \{[0, 1]\}$ which is homeomorphic to $S^2$.

What about the transition maps? Well, the intersection is just a copy of $\mathbb{C}^\times$. If we look at the maps to $\mathbb{C}$ above, we see that the transition map is just $z \mapsto z^{-1}$, which in the end is not so bad!

Answer 3

To get any parameterization of a manifold (of course locally), think of how you would give directions to someone sitting on the manifold in an unambiguous way.

For the sphere, start at, say point, $z$ (but a generic one, avoid poles and stuff for now), to describe a displacement to another point (nearby for the time being), first I look at the horizontal circle that contains $z$. and I think of going along this circle $\theta$ much, and I measure this not in terms of the distance I walk, but in terms of the angle from my position to the center of this circle. Then I say, from my new position, I'll move in on the shortest path towards the North pole. And again I measure this in terms of the change of angle, call $\phi$ from my position to the center of the sphere. Notice that my latter movement is occurring on the great circle containing the poles.

Now, I if you are on the sphere, and I ask you move $(30,15)$ degrees, you know what to do, right?! So, there we are, a local parameterization of a patch on the sphere. Note that we only need two parameters to find directions on the ball, which agrees with it being 2-dimensional.

If I set my "origin", that is $(\theta,\phi)=(0,0)$ to be point $(x,y,z)=(1,0,0)$, then where will the point $(\theta,\phi)$ be in $(x,y,z)$ coordinates? That is, what is the relationship between our introduced parameters and the physical position of a point?

Ok, draw a line perpendicular from $z=(x,y,z)$ to the $x$-$y$ plane. The shadow of $z$ makes an angle of $\theta$ with the $x$-axis. What is the distance $r$ from it to $(0,0)$? To find $r$, look at the right triangle made by $(o,o,o)$, $z$, and its shadow. The angle at origin is $\phi$. So, z, the height of $z$, is $$z=1*sin(\phi)$$ (sphere is radius 1.)

And, $$r=1*cos(\phi).$$

Now that $r$ is known, on the x-y plane project the shadow of $z$ to axes x and y separately to find $x$ and $y$ as $$x=r*cos(\theta)=cos(\phi)*cos(\theta)$$ $$y=r*sin(\theta)=cos(\phi)*sin(\theta)$$

Check which one is u, and which is v in your notation.

Now, it remains to ask: how far can $\theta$ and $\phi$ be stretched without running into trouble?

Answer: $\theta$ can run from $0$ all the way to $2\pi$ but has to stop just before that, because then we are back at where we started. And we don't want two addresses for one point.

$\phi$ will span all of sphere nicely, except for poles maybe, if it runs from $-\pi$, where it approaches south pole, to $\pi$ when it reaches the north pole.

Thus, only excluding the north pole, we could cover the whole manifold with one coordinate chart.