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$SO(3,\mathbb{R})$ is a 3-dimensional manifold.

However, when we can represent a fixed angle using the sphere coordinate, there are only $2$ parameters. Why?

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2 Answers

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Here's an informal argument. You can think of a matrix $M \in SO(3,\mathbb{R})$ as a choice of three orthonormal vectors. Choosing the first one, we have a choice of $\mathbb{S}^2$ to pick the first vector, and this is a 2D manifold. Choosing the second, we must pick a vector orthogonal to the first, so we have the choice from $\mathbb{S}^1$ sitting inside the 2-sphere, i.e. the circle orthogonal to our original choice. The circle is 1D, so we now have $2+1 = 3$ dimensions. For the final choice we just have the choice of a sign, $\pm$, for the choice of the final vector, and we pick whichever one makes the matrix $M$ have determinant $1$. Thus $SO(3,\mathbb{R})$ is three dimensional.

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From wikipedia:

"Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation."

You can pick an axis by picking a point on a sphere centered at the origin. A line passing through the origin passes through two points on the sphere so you really only need half of the sphere. It takes two degrees of freedom to pick the axis (it takes two degrees of freedom to specify a point on a sphere). After you pick the axis, you need an additional degree of freedom to specify how much you're rotating on the axis.

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