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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.

user023049
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3 Answers3

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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.

Arthur
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  • Sorry, I think the last part of my post obfuscated the original question. I know of this vector field, but can it be modified to be transitive? I'm almost certain the answer is yes, just adding in an irrational number somewhere, but I don't have the intuition for exactly how and how to prove it is transitive. – user023049 May 15 '17 at 16:07
  • This can be seen as decomposing $\mathbb{R}^{2n}$ into a product of $n$ planes and rotating infinitesimally in every plane. Love this construction! – lisyarus May 15 '17 at 16:08
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    @lisyarus There's another natural way to describe this (which is essentially the same as yours): Regard $\Bbb R^{2n}$ as $\Bbb C^n$. Then, $\Bbb S^1$ acts by simultaneous rotation of each factor of $\Bbb C$, as $\zeta \cdot {\bf z} = \zeta \cdot (z_1, \ldots, z_n) = (\zeta z_1, \ldots, \zeta z_n)$. Since this action clearly preserves lengths, it restricts to an action on $\Bbb S^{2 n - 1}$, and $v$ is the infinitesimal generator of the flow this defines, that is, $$v_{\bf z} = \left.\frac{d}{dt}\right\vert_0 (e^{it} {\bf z}) = i {\bf z}.$$ – Travis Willse May 15 '17 at 16:16
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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 $l_{z,w}:(-\varepsilon,\varepsilon)\to S^3$, $l_{z,w}(t) = (ze^{it},we^{it})$.

JJR
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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.

Travis Willse
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