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Question: Consider $\mathbb{S}^{3}$ as the unit sphere in $\mathbb{C}^{2}$ under the usual identification $\mathbb{C}^{2} \leftrightarrow \mathbb{R}^{4}$. For each $z = (z^1, z^2) \in \mathbb{S}^{3}$, define a curve $\gamma_z : \mathbb{R} \to \mathbb{S}^{3}$ 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.

Given some $t_0 \in \mathbb{R}$ we define the velocity of $\gamma_z$ at $t_0$, denoted by $\gamma_z'(t_0)$, to be the vector: $$\gamma_z'(t_0) = d \gamma_z \left( \dfrac{d}{dt} \bigg|_{t_0} \right) \in T_{\gamma_z(t_0)}\mathbb{S}^{3}$$ where $\left( \dfrac{d}{dt} \bigg|_{t_0} \right)$ is the standard coordinate basis vector in $T_{t_0}\mathbb{R}$. Therefore in our case we have that: $$d\gamma_z^1 \dfrac{d}{dt} \bigg|_{t_0} = (ie^{it}z^{1}, e^{it}z^2) \hspace{1cm} \text{and} \hspace{1cm} d\gamma_z^2 \dfrac{d}{dt} \bigg|_{t_0} = (e^{it}z^1, ie^{it}z^2)$$ both of which are nonzero in both component spaces for every $t_0 \in \mathbb{R}$.

As far as showing that $\gamma_z$ is smooth, since we can identify the tangent bundle of the embedded submanifold with a subbundle of the ambient manifold. But not sure how to explicitly show that.

Also, I'm not sure if I showed the velocity is never zero correctly.

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    $\gamma_z(t)$ is a function of $t$ with derivative $||\gamma'_z(t)||=||(ie^{it}z^1, ie^{it}z^2)||\neq 0$ for all $z=(z^1,z^2)\in\mathbb{S}^3$. For smoothness: If you calculate higher order derivatives you can see what the n-th derivative is. Then argue that it is nth order differentiable with n-th derivative continuous. – 2017-02-11
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    Its enough to say that $n$th order derivative given by: $\gamma_z^{(n)}(t) = (i^{n}e^{it}z^1, i^{n}e^{it}z^2)$ is continuous for every $(z^1,z^2) \in \mathbb{S}^3$ and every $n \in \mathbb{N}$, so that $\gamma_z$ must be smooth? – 2017-02-12
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    In this way you say that the nth derivative exists and is continuous for all n, so it is smooth – 2017-02-12

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