# Representation of 3-sphere

Question: In $S^2$we have latitude-longitude representation, though a sphere does not have the same topology as a plane. Do we have something similar to represent the $S^3$space (i.e. 3-sphere space)?

• The four Euclidean coordinates for S3 are redundant since they are subject to the condition that $x_0^2 + x_1^2 + x_2^2 + x_3^2 = 1$.

• As a 3-dimensional manifold one should be able to parameterize S3 by three coordinates, just as one can parameterize the 2-sphere using two coordinates (such as latitude and longitude).

• Due to the nontrivial topology of S3 it is impossible to find a single set of coordinates that cover the entire space.

• Just as on the 2-sphere, one must use at least two coordinate charts.

• Some different choices of coordinates are given below.

• Hyperspherical coordinates

• Hopf coordinates

• Stereographic coordinates

References: Coordinate systems on the 3-sphere space.

Some quick notes followed:

• Stereographic projection can map $S^2 \setminus N$to $\mathbb{R}^2$, where N is the North Pole. This is one coordinate chart. If we pick another point (anyone but not North Pole) on S2 and project again, we can obtain another coordinate chart. With these two charts, we can cover the entire S2 space.

• $SO(3)$ and $S^2 \times S^1$ are not the same space (not homeomorphic). See this question at StackExchange.

• The full space in 4D space is $R^4 \times S^3 \times S^2 \times S^1$.

• The space of unit quaternion$S^3$(which is a 3D surface embedded in 4D space) double covers the space of rotation SO(3). See this note for more information.

• Basically the unit quaternion $q = (cos(\theta/2), x sin(\theta/2), y sin (\theta/2), z sin (\theta/2))$can represents a counterclockwise rotation by the angle $\theta$ around the normalized axis $\mathbb{n} = (x, y, z)^\top$. If we set the rotation to be $\theta + 2\pi$, the corresponding quaternion representation will become $-q$; if we set the rotation to be $\theta + 4\pi$, the corresponding quaternion representation will remain $q$. That said, for each rotation we will have exactly two representations in quaternion.

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