Is there a simple example of a transitive vector field on the three sphere? I was just thinking about the transitive field on the torus and the possibility of defining a vector field without singularities on $S^3$ and this idea popped up.
Edit: I think my question was misunderstood. I am specifically asking about transitivity, or the existence of a dense orbit (flow). I know about the existence of a vector field without singularities (which is what every answer is mentioning), I just worded my initial post poorly. What I want to know is if it is possible to define a vector field on $S^3$ that has a dense orbit.
 A: You can define a nowhere-zero vector field on any odd-dimension sphere. For instance, one is given by embedding the $(2n-1)$-sphere as the unit sphere in $\Bbb R^{2n}$, and at the point $x = (x_1, x_2, x_3,\ldots,x_{2n})\in S^{2n-1}$ define the tangent vector
$$
v_x = (x_2, -x_1, x_4, -x_3, \ldots, x_{2n}, -x_{2n-1})
$$
This is readily seen to be orthogonal to $x$, and therefore tangent to the unit sphere at $x$. At the same time, it is never zero, because the origin of $\Bbb R^{2n}$ is not part of our sphere.
On an even-dimension sphere, it can never be done, for the same reasons that it cannot be done on $S^2$ in particular. For instance, my proof here (stolen from theorem 2.28 in Hatcher, where my above example for odd-dimensional spheres can also be found) covers all of those cases simultaneously.
A: This is not straight forward. You can take equivalence classes of paths through any point $(z,w)\in S^3$, where $S^3=\{(z,w)\in \mathbb{C}^2\mid |z|^2+|w|^2=1\}$, with paths given by $l_{z,w}:(-\varepsilon,\varepsilon)\to S^3$, $l_{z,w}(t) = (ze^{it},we^{it})$.
A: If one knows a little Lie theory this is immediate: We may identify $S^3$ topologically with the Lie group $\textrm{SU}(2) \cong \textrm{Spin}(3)$ of unit quaternions, and hence as the unit sphere in the space $\Bbb H$ of quaternions.
In particular, all of its left-invariant vector fields, which are parameterized by $T_1 \textrm{SU}(2)$---and which we may in turn identify with the space $\textrm{Im} \, \Bbb H$ of imaginary unit quaternions---vanish nowhere (with the exception of the usual vector field): Given any $r \in \textrm{Im} \,\Bbb H$, the vector field $X$ along $S^3$ defined by $$X_q := rq$$ is tangent to $\Bbb S^3$ and (again, unless $r = 0$) vanishes nowhere. (Here, we are implicitly using the canonical identifications $T_q \Bbb H \cong \Bbb H$). The choice $r = i$ gives the example in Arthur's answer in the special case $n = 2$, which also coincides with the example in JJR's answer.
