There is a formula of to calculate the angle between two normalized vectors:
$$\alpha=\arccos \frac {\vec{a} \cdot\ \vec{b}} {|\vec {a}||\vec {b}|}.$$
The formula of 3D solid angle between three normalized vectors is (borrowed from wikipedia):
$$\alpha = 2\arctan \frac{|\vec {a}\ \vec {b}\ \vec {c}|}{|\vec{a}| |\vec{b}| |\vec{c}| + (\vec{a} \cdot \ \vec{b})|\vec{c}| + (\vec{a} \cdot \vec {c})|\vec{b}| + (\vec{b} \cdot \vec{c})|\vec{a}|}.$$
How to figure out a formula of solid angle between $n$ normalized vectors in $n$-dimensional space?
UPDATE
OK, I've found out the following formula for the angle between two vectors in 2d space through the $\arctan$:
$$\alpha= 2\arctan \frac {|\vec{a} \wedge \vec {b}|} {|\vec{a}| |\vec{b}| + \vec{a}\ \cdot\ \vec{b}}$$
How can I apply this $\arctan$ formula for the 4D space, for example?
For more detail, I explain my assumptions.
The solid angle of orthogonal basis in 4D space must be $\alpha = \frac {2 \cdot \pi ^ 2 \cdot R^3} {2 ^ 4} = \frac {\pi ^ 2 R^3} {8} $. Here we can see a $\pi^2$ factor. Does this mean that the 4D solid angle formula contains multiplication of two $\arctan$ there is one $\arctan$ in this formula? I think yes, but still have some difficulties with such formula inference.
Best Answer
See the paper by Ribando, "Measuring Solid Angles Beyond Dimension Three", published in Discrete & Computational Geometry 2006. An electronic version may be found here: https://link.springer.com/content/pdf/10.1007%2Fs00454-006-1253-4.pdf
It seems there is no closed formula for solid angle in dimension > 3, but a multi-variable Taylor series is given in Theorem 2.2 there. Later in this paper, its radius of convergence is discussed.