# Velocity vectors on $S^{3}$

Consider $S^{3}$ as the unit sphere in $C^{2}$ under the usual identification $C^{2}\leftrightarrow R^{4}$. For each $z=(z^{1},z^{2})\in S^{3}$, define a curve $\gamma _{z}:R\rightarrow S^{3}$ by $\gamma _{z}(t)(e^{it}z^{1},e^{it}z^{2})$. Show that $\gamma_{z}$ is a smooth curve whose velocity is never zero.

This is a problem from Lee "Introduction to smooth manifolds". On what I am strugling is velocity vector calculation. I am given a velocity vector formula $${\gamma }'(t_{0})=\frac{d\gamma^{k}}{dt}(t_{0})\frac{\partial }{\partial x^{k}}\mid _{\gamma(t_{0})}$$ where $x^{k}$ are coordinate functions under smooth chart $(U,\varphi)$ and $\gamma(t_{0})\in U$.

My solution would be: since $\frac{d\gamma^{k}}{dt}(t_{0})=ie^{it}z^{k}$ is not zero averywhere (i.e. on the $S^{3}$), velocity vector is non-zero as well.

Is it enough and it just seems easy, or am I missing something?

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Yes, it's that easy. – Neal Jan 21 '13 at 19:18
@Neal Thanks for confirmations. I am self-studing differential geometry and do not have any advisor to turn for advise. So it is great to have a possibility to ask in Mathematics. – Tomas Jan 21 '13 at 19:23
that's wrong. You need to use coordinate representations. – TKM Oct 9 '14 at 3:12
@TKM So, could you elaborate more on why this is wrong and what is the solution you have in mind? – Tomas Oct 9 '14 at 6:47