Set : Consider $$S^3:=\{(z_1,z_2)\in \mathbb{C}^2 | |z_1|^2+|z_2|^2=1\} $$ So if $S^3$ has a Riemannian metric induced from $\mathbb{R}^4$, we have an isometric $S^1$-action : $$ g\cdot (z_1,z_2)=(gz_1,gz_2) $$ for $g\in S^1\subset \mathbb{C}$. Hence $$ \pi : S^3\rightarrow S^3/S^1=S^2(r)$$ where quotient space is a two dimensional sphere of radius $r=\frac{1}{2}$.
(1) Note that each orbit is a great circle.
(2) Topologically, $S^3$ is a union of two solid torus $T_i$. And $\partial T_1=\partial T_2$ is a totally geodesic flat torus.
Question : Can we find some geodesic curve in $S^3$ that is mapped onto some not geodesic curve ?
Thank you in advance.
Yes, we can find such a curve.
Consider the curve on $S^3$: $\gamma(t) = \left(\cos t, \frac{1}{\sqrt{2}}\sin t , \frac{1}{\sqrt{2}}\sin t, 0\right)$. Note that $\gamma$ has unit length and is given by the intersection of $S^3$ with the plane spanned by $(1,0,0,0)$ and $(0,1,1,0)$, so $\gamma(t)$ is a geodesic. (Alternatively, $\gamma''(t) = -\gamma(t)$, obviously, so $\gamma''(t)$ points to the center of $S^3$. In otherwords, the accleration vector projects to $0$).
According to Wikipedia (but using $S^2(1/2)$), the Hopf map sends $\gamma(t)$ to $\left( \cos t \sin t, \frac{1}{\sqrt{2}}\sin^2 t, \frac{1}{\sqrt{2}}\cos^2(t)\right)$.
This is not a geodesic on $S^2$. One way to see it is to note that the $z$-coordinate is always non-negative, so the image lies in a hemisphere, but also goes through the pole of that hemisphere. No geodesic on $S^2$ does this.