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Consider $S^3 := \{(z,w) \in \mathbb{C}^2:|z|^2 + |w|^2 = 1\}$ be the unit $3$-sphere with equivalence relation

$$(z,w) \sim (z',w') \iff z' = e^{i \theta }z, w' = e^{i\theta} w$$ for some $\theta \in \mathbb{R}$.

My definition of foliation:

A rank $k$-foliation of a manifold $M$ is a collection $\{L_\alpha\}_{\alpha\in A}$ of connected immersed submanifolds of $M$ such that

(1) $M= \coprod_{\alpha \in A} L_\alpha$

(2) For every point $p \in M$, there is a chart $(U, \phi=(x^1, \dots, x^m))$ with $p \in U$ such that for every leaf $L_\alpha$ we have that $U \cap L_\alpha$ is empty or the countable union of slices of the form $\{x^{k+1}= constant, \dots, x^m = constant\}$.

Now, I see that

$$[(z,w)] = \{e^{i \theta}(z,w): \theta \in \mathbb{R}\}$$

but I cannot proceed after that. I guess we need to view this as immersed submanifold somehow? Also, I will have to work with charts on spheres with seems rather painful. Any help will be appreciated!

  • Notice that this foliation can be seen as the orbits of a Lie group action, namely the natural action of $S^1$ on $S^3$. You can show that the orbit at $x$ can be identified with the Lie group modulo the isotropy at $x$. What is the isotropy at $x$? And therefore, what do the orbits look like? – Jonas Jan 02 '20 at 13:48
  • I have not heard of isotropy before. Also this came up before we covered Lie groups and orbits. –  Jan 02 '20 at 13:49
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    Can you see that $[(z,w)]$ is just an embedded circle, and how that makes (1) trivial? For (2), I would write everything in real coordinates and use projections as my charts. – Steve D Jan 07 '20 at 14:38
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    As Jonas alluded to, there is a smooth fiber bundle $S^1 \to S^3 \xrightarrow{\pi} S^2$ where the fibers are precisely the orbits of the action you specify. Being a fiber bundle means that for each $x \in S^2$, there is an open set $U$ with $x \in U$ such that $\pi^{-1}(U)$ is diffeomorphic to $U\times S^1$. Using such diffeomorphisms, one can construct the desired charts in $(2)$. In general, if $F \to E \xrightarrow{\pi} B$ is a smooth fiber bundle, then $E$ is foliated by the fibers of $\pi$ which are all diffeomorphic to $F$. – Michael Albanese Jan 10 '20 at 12:28
  • @user661541: Would an answer along these lines be acceptable? – Michael Albanese Jan 10 '20 at 19:42
  • @MichaelAlbanese I'm not too familiar with fiber bundles (I didn't see much more than the definition of this), but as long as it's elementary, it is fine! –  Jan 10 '20 at 19:53
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    The 1 line answer is that on the open set $z \ne 0$ the map $(z,w) \mapsto \frac{w}{z}$ is a submersion each of whose point inverses is contained in a leaf, and similarly on the open set where $w \ne 0$; now apply the implicit function theorem. – Lee Mosher Jan 11 '20 at 00:23
  • @Lee Mosher. Kindly expand on this in an answer. Thanks! –  Jan 11 '20 at 00:25
  • I think this question will also help you out. – ImHackingXD Oct 24 '23 at 17:39

1 Answers1

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As I mentioned in my comment above, this is a special case of a more general result: if $F \to E \xrightarrow{p} B$ is a smooth fiber bundle, then the fibers of $\pi$ define a foliation of $E$ by leaves which are diffeomorphic to $F$. Let me first give the steps of how one proves this result.

  1. The fibers of $p$ are embedded submanifolds of $E$ by the regular value theorem.
  2. By local triviality, we can find a collection of charts $\{(U_i, \eta_i)\}$ which cover $B$, and diffeomorphisms $\psi_i : p^{-1}(U_i) \to F\times U_i$ which respect the two natural projections to $U_i$.
  3. As $F$ is a manifold, we can find a collection of charts $\{(V_j, \chi_j)\}$ which cover $F$.
  4. The desired charts on $E$ are $\{(\psi_i^{-1}(V_j\times U_i), \phi_{i,j})\}$ where $\phi_{i,j} = (\chi_j, \eta_i)\circ\psi_i$.

I will carry out these steps (and verify the claims made) in the explicit case you asked about, namely the Hopf fibration.

Consider the map $p : S^3 \to \mathbb{CP}^1$ given by $p(z, w) = [z, w]$; note that the preimages under this map are precisely the equivalence classes: $p^{-1}(p(z, w)) = \{e^{i\theta}(z, w) \mid \theta \in \mathbb{R}\} = [(z, w)]$. The map $p$ is a surjective submersion, so every point in $\mathbb{CP}^1$ is a regular value and hence every preimage is an embedded submanifold of $S^3$. This establishes property $(1)$.

Consider the open set $U_1 = \mathbb{CP}^1\setminus\{[1, 0]\}$ and the map $\psi_1 : p^{-1}(U_1) \to S^1\times U_1$ given by $(z, w) \mapsto (\frac{w}{|w|}, [z, w])$. This is a smooth map with smooth inverse $(e^{i\theta}, [a, b]) \mapsto \frac{|b|e^{i\theta}}{b\sqrt{|a|^2+|b|^2}}(a, b)$. That is, $\psi_1$ is a diffeomorphism. Moreover, if we denote the natural projection $S^1\times U_1 \to U_1$ by $\operatorname{pr}_2$, then we have

$$\operatorname{pr}_2(\Psi((z, w))) = \operatorname{pr}_2((\tfrac{w}{|w|}, [z, w])) = [z, w] = p((z, w)).$$

It follows that $[(z, w)]$ is mapped diffeomorphically to $S^1\times\{[z, w]\}$.

Note that the points $(i, [z, w]) \in S^1\times U_1$ are precisely those for which $\operatorname{arg}(w) = \frac{\pi}{2}$, so $\psi_1^{-1}((S^1\setminus\{i\})\times U_1) = p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \frac{\pi}{2}\}$. Each of the factors of $(S^1\setminus\{i\})\times U_1$ are diffeomorphic to Euclidean spaces. Explicitly, we have diffeomorphisms $\eta_1 : U_1 \to \mathbb{R}^2$ given by $([z, w]) \mapsto (\operatorname{Re}(\frac{z}{w}), \operatorname{Im}(\frac{z}{w}))$ and $\chi_1 : S^1\setminus\{i\} \to \mathbb{R}$ given by $\lambda \mapsto \frac{\operatorname{Re}(\lambda)}{1-\operatorname{Im}(\lambda)}$; note that $\chi_1$ is precisely stereographic projection from the 'north pole'.

Putting what we have together, the map

$$\phi_{1,1} : p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \tfrac{\pi}{2}\} \to \mathbb{R}^3$$

given by $\phi_{1,1} := (\chi_1, \eta_1)\circ\psi_1$ is a diffeomorphism and hence viewed as a coordinate chart. Moreover, if $L_{\alpha} = [(z, w)]$, then $(p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \frac{\pi}{2}\})\cap L_{\alpha}$ is mapped diffeomorphically to $\mathbb{R}\times\{(\operatorname{Re}(\frac{z}{w}), \operatorname{Im}(\frac{z}{w})\}$. That is, if $\varphi_{1,1} = (x^1, x^2, x^3)$, the intersection of the leaf with the coordinate chart is given by the equations $x^2 = \operatorname{Re}(\frac{z}{w}), x^3 = \operatorname{Im}(\frac{z}{w})$.

Replacing $\chi_1 : S^1\setminus\{i\} \to \mathbb{R}$ with $\chi_2 : S^1\setminus\{-i\}$ given by $\lambda \mapsto \frac{\operatorname{Re}(\lambda)}{1+\operatorname{Im}(\lambda)}$ (stereographic projection from the 'south pole'), we obtain a similar chart

$$\phi_{1,2} = (\chi_2, \eta_1)\circ\psi_1 : p^{-1}(U_1)\setminus\{\operatorname{arg}(w) = \tfrac{3\pi}{2}\} \to \mathbb{R}^3.$$

Now let $U_2 = \mathbb{CP}^1\setminus\{[0, 1]\}$, then there is a diffeomorphism $\phi_2 : p^{-1}(U_2) \to S^1\times U_2$ respecting the two projections to $U_2$ given by $(z, w) \mapsto (\frac{z}{|z|}, [z, w])$. Replacing $\eta_1 : U_1 \to \mathbb{R}^2$ with $\eta_2 : U_2 \to \mathbb{R}^2$ given by $[z, w] \mapsto (\operatorname{Re}(\frac{w}{z}), \operatorname{Im}(\frac{w}{z}))$, we obtain two more charts

$$\phi_{2,1} = (\chi_1, \eta_2)\circ\psi_2 : p^{-1}(U_2)\setminus\{\operatorname{arg}(w) = \tfrac{\pi}{2}\} \to \mathbb{R}^3$$

and

$$\phi_{2,2} = (\chi_2, \eta_2)\circ\psi_2 : p^{-1}(U_2)\setminus\{\operatorname{arg}(w) = \tfrac{3\pi}{2}\} \to \mathbb{R}^3.$$

The domains of the four charts $\phi_{1,1}, \phi_{1, 2}, \phi_{2, 1}, \phi_{2,2}$ cover $S^3$ and satisfy the requirements of $(2)$.

Michael Albanese
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  • Thanks for the answer. Is there an easy way to see that $p$ is a submersion? –  Jan 17 '20 at 19:29
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    If $\gamma(t) = [\gamma_1(t), \gamma_2(t)]$ is a curve in $\mathbb{CP}^1$, then $\hat{\gamma}(t) = \frac{1}{\sqrt{|\gamma_1(t)|^2+|\gamma_2(t)|^2}}(\gamma_1(t), \gamma_2(t))$ is a curve in $S^3$ such that $(p\circ\hat{\gamma})(t) = \gamma(t)$. It follows that $dp(\hat{\gamma}'(0)) = \gamma'(0)$. As $\gamma'(0)$ is arbitrary, we see that $dp$ is surjective (at every point), and hence $p$ is a submersion. – Michael Albanese Jan 18 '20 at 03:51