2

The question is as follows:

In the 3-sphere $S^3 \subset \mathbb{R}^4 $ consider the 1-dimensional distribution defined by $$X = -y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y } - \omega \frac{\partial}{\partial z } + z \frac{\partial}{\partial \omega }$$ Determine the foliation $F$ integrating this distribution.

$\textbf{Some attempt:}$

Let $\gamma (t) = (x(t ), y(t), z(t), \omega (t))$, with initial point $\gamma (0) = (x_0, y_0, z_0 , \omega_0)$. Hence, $\gamma'(t)=(x'(t ), y'(t), z'(t), \omega' (t))$, where $$ \begin{cases} \frac{dx}{dt} = -y \\ \frac{dy}{dt} = x \end{cases} ~~~~ \text{and}~~~~ \begin{cases} \frac{dz}{dt} = -\omega \\ \frac{d\omega}{dt} = z \end{cases}$$ From where this suggests we try $x(t) = A \cos t$ and plugging this into the second equation gives $y(t) = A \sin t$, and it's easy to check that this choice of $x(t)$ and $y(t)$ solves the second equation in the first system. And the same for second system, we have $z(t) = B \cos t$ and $\omega (t) = B \sin t$.

So we have $\gamma (t) = (A \cos t , A \sin t , B \cos t, B \sin t)$ with $\gamma (0) = (A, 0 , B, 0)$. So we get $A = x_0$ and $B=z_0$.

Can some one please let me know if I am wrong?

Thanks!

user513784
  • 333
  • This is fine. By the way, a 1 -dimensional distribution is usually called a vector field. Now all you need to do is understand what the resulting foliation looks like. – Danu Dec 23 '17 at 15:39
  • 1
    @Danu: Not quite — a $1$-dimensional distribution is a line field, not a vector field. – Ted Shifrin Dec 23 '17 at 18:00
  • @Danu Many thanks! Can you please let me know how to know the resulting foliation looks like what? – user513784 Dec 23 '17 at 18:27
  • @Danu Many thanks! Can you please let me know what can be the resulting foliation? – user513784 Dec 28 '17 at 22:12

1 Answers1

3

So the question seems rather to be: how do I understand the shape of the curve $\gamma(t)$ inside $\mathbb S^3$? There are a number of ways.

The easiest may be the following: choose a point $N$ such that: $\gamma(t)\ne N$ for any $t$. Now you can use the fact that $\mathbb S^3\setminus N$ is diffeomorphic to $\mathbb R^3$ (via stereographic projection from $N$). Composing $\gamma$ with the stereographic projection gives you a (closed) curve in $\mathbb R^3$ that you can visualize easily.

The previous one was not much specific of your curve and can be done for every curve in $\mathbb S^3$. For something more tailored on this $\gamma$ you should consider that your curve lies on the two hyperplanes $x_1/A-x_3/B=0$ and $x_2/A-x_4/B=0$ (here I'm assuming $A\ne0\ne B$), and both hyperplanes are through the origin. Therefore your curve sits inside a $2$-plane through the center intersecting the $3$-sphere and it should be then easy to convince you that this is nothing but a great circle on $\mathbb S^3$.

Then try and understand what happens when changing to various choices of $A$ and $B$...

N. Ciccoli
  • 749
  • 3
  • 5
  • Many thanks for your knowledgeable answer! But can you please let me know why our curve lies on the two hyperplanes $x_1/A-x_3/B=0$ and $x_2/A-x_4/B=0$? Is there any observation to see that? – user513784 Dec 29 '17 at 12:09
  • 1
    It was just easy to see that such equations are satisfied by looking at the explicit form of $\gamma$. Of course it also does belong to generalized cylinders like $(x_1/A)^2+(x_2/A)^2=1$ and the like for $x_3,x_4$... – N. Ciccoli Dec 29 '17 at 13:00
  • Many thanks! I understood that. You talked about cylinders, can you also please let me know why there is no integral manifolds in the example attached to this comment [linked to the result][1] [1]:https://math.stackexchange.com/questions/2571733/show-that-the-2-dimensional-distribution-in-mathbbr3-has-no-integral-manif?noredirect=1#comment5332986_2571733 – user513784 Dec 29 '17 at 13:29