Equators in the $4$-dimensional sphere. Are there infinitely many disjoint equators (centrally symmetric circles) on the surface of the $4$-dimensional sphere? There are at least two of them, namely $[0,0,x,\sqrt{1-x^2}]$ and $[x,\sqrt{1-x^2},0,0] $ , $ x\in [-1,1]$. (Note that we may have to adjust the sign of the squareroot, but that just confuses the reader.) 
I tried to take the mean of their coordinates, and rescale it to get points on he sphere, but i do not know how to show that some parametrized curve is a circle in $4$-dimensions. 
 A: Yes -- in fact the 3-sphere can be completely covered by disjoint great circles, the Hopf fibration.
(Terminology nitpick: The subset $\{ (x,y,z,w)\in\mathbb R^4\mid x^2+y^2+z^2+w^2=1\}$ is usually known as the three-sphere, counting the dimension of the surface itself rather than the space it happens to embed into).
A: Firstly, you should speak of the $3$-dimensional sphere, viewed in the $4$-dimensional euclidian space. 
Secondly, the term equator in this situation should certainly refer to the intersection in $\mathbb{R}^4$ of the sphere $S^3$ and a hyperplane, that is a $2$-dimensional object.
For your question, consider the vectors $(\cos(\phi)\cos(\theta), \sin(\phi)\cos(\theta), \cos(\phi)\sin(\theta), \sin(\phi)\sin(\theta))$. Then, when you fix $\phi$, you get a circle with the coordinate $\theta$, and if I'm not mistaken the circles corresponding to $\phi_1$ and $\phi_2$ are disjoint is $\phi_1 \neq \phi_2$.
A: Yes. Take any orthonormal basis of $\mathbb{R}^4,$ call it $v_1, v_2, v_3, v_4$ Then the circles $\cos t v_1 + \sin t v_2$ will be disjoint from $\cos s v_3 + \sin s v_4.$
