The setup is expressed quite nicely in this question which I quote below:
It is a classical fact that if $(x_1,\ldots,x_n)$ is a random vector uniformly distributed on the sphere $S^{n-1} \subseteq \mathbb{R}^n$, then the random vector $(x_1,\ldots,x_{n-2})$ is uniformly distributed in the unit ball $B_{n-2} = \{ (y_1,\ldots,y_{n-2}) \mid \sum_{i=1}^{n-2} y_i^2 \le 1\} \subseteq \mathbb{R}^{n-2}$.
In the $n=4$ case, dropping two co-ordinates, say $x_3 $ and $x_4$, from $(x_1, x_2, x_3,x_4)$ leaves us with $\bar r_1=(x_1,x_2)$ which is a uniformly random vector inside a circle. $\bar r_2=(x_3,x_4)$ is also uniformly random inside a circle.
Question: Are $r_1$ and $r_2$ independent?
No. There is a nonzero chance that $r_1^2>0.9$. Similarly, there is a nonzero chance that $r_2^2>0.9$. However, these two events cannot happen simultaneously, since that would imply $$ x_1^2+x_2^3+x_3^2+x_4^2=r_1^2+r_2^2>0.9+0.9=1.8, $$ contradicting $x_1^2+x_2^2+x_3^2+x_4^2=1$. Therefore, $r_1$ cannot be independent of $r_2$.