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In this question, first we fix an isomorphism between $TS^{3}$ and $S^{3}\times \mathbb{R}^{3}$.(To be more precise we consider the global trivialization of $TS^{3}$ with help of $3$ global sections $\{ix,jx,kx\}$, $x\in S^{3}$, with quaternioun multiplications). So every map $S^{3}\to S^{2}\subset \mathbb{R}^{3}$ is counted as a unique (unit speed) vector field on $S^{3}$. In particular the Hopf map

$$p:S^{3}\to S^{2}\; \text{with} \;p(z,w)= (\parallel z\parallel^{2}-\parallel w\parallel^{2},\;2z\bar{w})$$.

defines a unique vector field on $S^{3}$, a non singular foliation of $S^{3}$. We denote this vector field on $S^{3}$ by $\tilde{X}$.

What is known about the dynamics of this vector field(Foliation)? Existence of periodic orbit? Invariant torus? (A possible full dynamical description?) According to the following notation what type of vector field $X$ on $S^{2}$ can be lifted to $\tilde{X}$? Is there a reference which investigated this particular foliation of $S^{3}$?

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Ali Taghavi
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  • I'm not sure I understand the definition of the map $p$... You use complex coordinates on $S^{3} \subseteq \mathbb{C}^2$, but what coordinates do you use on $S^{2}$? Can you write the definition of $p$ in terms of the frame ${ix, jx, kx}$, please? – Oldřich Spáčil Oct 25 '14 at 20:43
  • @OldřichSpáčil I got that formula of hopf fibration from http://en.wikipedia.org/wiki/Hopf_fibration. The coordinate of $S^{2}$ is the usual one:$S^{2}={(x,z)\in \mathbb{R}\times \mathbb{C}\mid |x|^{2}+\parallel z\parallel^{2}=1}$. But the hopf map is a map defined on $S^{3}$ so how you expect to define it in term of fram which are elements of $TS^{3}$ not $S^{3}$? Please let me know if I should more clarify the question. – Ali Taghavi Oct 26 '14 at 06:22
  • This is what I meant: as you pointed out, the moving frame ${ix, jx, kx}$ trivializes the tangent bundle $TS^{3}$. But it also trivializes the unit tangent bundle because the frame is orthonormal, hence $STS^{3} \cong S^{3}\times S^{2}$. So, as you said, any unit vector field on $S^{3}$ is identified using the moving frame with a smooth map $S^{3} \to S^{2}$. The map $p$ then describes the coordinates in the frame ${ix, jx, kx}$ of the corresponding unit vector field and I thought it would help to write down the vector field as a linear combination of $ix, jx, kx$. – Oldřich Spáčil Oct 27 '14 at 12:07
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    Is not there still freedom left in choosing identifications between $S^2$ and the set of unit vectors in each tangent space? For example, does anything obstruct you in choosing them in such a way that the field will be just the field of unit tangents to the fibers of the Hopf map? – მამუკა ჯიბლაძე Nov 08 '14 at 18:29
  • @მამუკაჯიბლაძე very interesting point. I am thinking about it... – Ali Taghavi Nov 08 '14 at 19:01
  • @მამუკაჯიბლაძე .. and are you concluding that the answer to my main question is independent of a trivialization of $TS^{3}$, up to topological equivalent? So all thing is the hopf fibration, up to topological equivalent? So we do not get some new foliation of $S^{3}$? – Ali Taghavi Nov 08 '14 at 19:14
  • On the contrary, I think the answer depends on the trivialization. For example I believe you also can choose trivialization in such way that the field becomes a constant field (of unit vectors parallel in the sense of some embedding of $S^3$ into $R^6$). – მამუკა ჯიბლაძე Nov 08 '14 at 19:22

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