📂Geometry

Coordinates of a Three-Dimensional Unit Sphere

Formulas

3D space’s unit sphere can be represented by the following six coordinate patch mappings. For $(u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\}$,

$$ \begin{align*} \mathbf{x} _{(0,0,1)}(u, v) &= \left( u, v , \sqrt{1- u^{2} -v^{2} } \right) \\ \mathbf{x}_{(0,0,-1)}(u, v) &= \left( u, v , -\sqrt{1- u^{2} -v^{2} } \right) \\ \mathbf{x}_{(0,1,0)}(u, v) &= \left( u, \sqrt{1- u^{2} -v^{2}}, v \right) \\ \mathbf{x}_{(0,-1,0)}(u, v) &= \left( u, -\sqrt{1- u^{2} -v^{2}}, v \right) \\ \mathbf{x}_{(1,0,0)}(u, v) &= \left( \sqrt{1- u^{2} -v^{2}}, u, v \right) \\ \mathbf{x}_{(-1,0,0)}(u, v) &= \left( -\sqrt{1- u^{2} -v^{2}}, u, v \right) \end{align*} $$

Description ¹

A coordinate patch mapping is a mathematical expression of what we know as a surface concept. Let’s look at a specific example to see how a surface is actually represented by a coordinate patch mapping. Consider the unit sphere in 3D space. Then, consider the following coordinate patch mapping defined as $\mathbf{x} _{(0,0,1)} : \R^{2} \to \R^{3}$.

$$ \mathbf{x} _{(0,0,1)}(u, v) = \left( u, v , \sqrt{1- u^{2} -v^{2} } \right) ,\quad (u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} $$

This coordinate patch represents the part of the sphere defined by $z \gt 0$. Moreover, the following coordinate patch

$$ \mathbf{x}_{(0,0,-1)}(u, v) = \left( u, v , -\sqrt{1- u^{2} -v^{2} } \right),\quad (u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} $$

represents the sphere defined by $z \lt 0$. Therefore, except for the equator, the part of the sphere that intersects with the plane represented by a red line $xy-$, i.e.,

Now let’s bring in more coordinate patch mappings to cover the equator as well.

$$ \mathbf{x}_{(0,1,0)}(u, v) = \left( u, \sqrt{1- u^{2} -v^{2}}, v \right),\quad (u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} \\ \mathbf{x}_{(0,-1,0)}(u, v) = \left( u, -\sqrt{1- u^{2} -v^{2}}, v \right),\quad (u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} $$

The side of the sphere can be represented as the left picture below with the above two coordinate patches.

At first glance, it might seem that the sphere can be completely represented with these four coordinate patches, but that’s not the case. The two points marked in the right picture, $(1,0,0)$ and $(-1, 0, 0)$, are not represented by any of the four coordinate patches mentioned above. Hence, the following two coordinate patches are necessary.

$$ \mathbf{x}_{(1,0,0)}(u, v) = \left( \sqrt{1- u^{2} -v^{2}}, u, v \right),\quad (u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} \\ \mathbf{x}_{(-1,0,0)}(u, v) = \left( -\sqrt{1- u^{2} -v^{2}}, u, v \right),\quad (u,v) \in U = \left\{ (u,v) : u^{2} + v^{2} \lt 1 \right\} $$

Now, with the six coordinate patches defined above, the 3D unit sphere can be represented, and every point on the sphere is represented by at least one coordinate patch.