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The question is as follows:

Show that the 2-dimensional distribution in $\mathbb{R}^3$ defined by the vector fields

$$X_1 = \frac{\partial}{\partial x}, ~~~~ ~~~~ ~~~ X_2 = e^{-x}\frac{\partial}{\partial y} + \frac{\partial}{\partial z}$$ has no integral manifolds.

$\textbf{My solution:}$

By using Frobenius theorem we have:

\begin{align}[X_1 , X_2] = [\frac{\partial}{\partial x}, e^{-x}\frac{\partial}{\partial y} + \frac{\partial}{\partial z}] &= [\frac{\partial}{\partial x}, e^{-x}\frac{\partial}{\partial y}]+ [\frac{\partial}{\partial x}, \frac{\partial}{\partial z}] \\&= [\frac{\partial}{\partial x}, e^{-x}\frac{\partial}{\partial y}]+ 0 \\&= [\frac{\partial}{\partial x}, e^{-x}\frac{\partial}{\partial y}] \\& =\frac{\partial}{\partial x}(e^{-x}\frac{\partial}{\partial y}) - e^{-x}\frac{\partial}{\partial y}(\frac{\partial}{\partial x}) \\&=-e^{-x} \frac{\partial}{\partial y}-0 \\&= -e^{-x} \frac{\partial}{\partial y} \end{align} And this will be in $D = \left< X_1 , X_2 \right>$ iff $-e^{-x}=0$. But we know that $-e^{-x} < 0$. So it is not involutive and so has no integral manifolds.

Okay, now as @Jack Lee point it out, "Not involutive" does not imply "no integral manifolds." A non-involutive distribution might have isolated integral manifolds; it just won't have one through every point.

So lets try to find it.

As in link to this result, let $\gamma(t) = (\gamma_1(t), \gamma_2(t), \gamma_3(t)$ be an integral curve with initial point $\gamma(0) = (x_0, y_0, z_0)$ of vector field $X_1$. Which means that $u$ at the point $\gamma(t)$ is equal to $\gamma'(t)$. Let's write this out.

$$X_1(\gamma(t)) = \partial_1 $$ and $$\gamma'(t) = (\gamma_1'(t), \gamma_2'(t), \gamma_3'(t)) = \gamma_1' \partial_1 + \gamma_2'\partial_2+\gamma_3' \partial_3.$$ Setting these equal to each other and equating coefficients gives us a system of ODEs to solve:

$$\gamma_1' = 1$$ $$\gamma_2' = 0$$ $$\gamma_3' = 0$$ And by doing the same for vector field $X_2$, we get

$$\gamma_1' = 0$$ $$\gamma_2' = e^{-y_1}$$ $$\gamma_3' = 0$$ But as we see they are not compatible with each other. And maybe it can show that we do not have an integral manifold for this distribution?

Can you please let me know if I am wrong?

Thanks!

user513784
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    "Not involutive" does not imply "no integral manifolds." A non-involutive distribution might have isolated integral manifolds; it just won't have one through every point. – Jack Lee Dec 18 '17 at 15:41
  • In fact, you wrote down an example of such a distribution here! – Jack Lee Dec 18 '17 at 15:46
  • @JackLee Many thanks! So how can I show it has no integral manifolds? – user513784 Dec 18 '17 at 17:14
  • Note that the commutator must again be a vector field, and you're not correctly understanding the computation. You should have $-e^{-x}\frac{\partial}{\partial y}$ for the answer. – Ted Shifrin Dec 18 '17 at 17:20
  • @TedShifrin Many thanks! Can you please let me know how can I compute it correctly? – user513784 Dec 18 '17 at 17:34
  • @TedShifrin Thanks! I understood how to compute it and corrected it. Now can you please let me know how to show that it has no integral manifold? – user513784 Dec 18 '17 at 17:48
  • You really should write $\dfrac{\partial}{\partial y}(e^{-x})\dfrac{\partial}{\partial x}$, not what you wrote. ... If there were an integral surface for $D$, it would have to be a cylinder with rulings parallel to the $x$-axis. Is this possible? – Ted Shifrin Dec 18 '17 at 18:04
  • @TedShifrin Thanks! Can you please explain more? And I cannot understand $\dfrac{\partial}{\partial y}(e^{-x})\dfrac{\partial}{\partial x}$? And why it it would have to be a cylinder with rulings parallel to the $x$-axis? – user513784 Dec 18 '17 at 18:08
  • Why can you not understand that? You have to end up with a vector field, so the final expression cannot have second-order derivatives in it. Go back through the basics of why $[X,Y]$ is a vector field when $X$ and $Y$ are. ... Since $X_1$ is everywhere tangent to your surface, the surface contains integral curves of that vector field. What are those? – Ted Shifrin Dec 18 '17 at 18:19
  • @TedShifrin Many thanks! Yes I understood that we cannot have second order derivative in the final expression and I corrected it. But I really do not understand why did you write $\dfrac{\partial}{\partial y}(e^{-x})\dfrac{\partial}{\partial x}$? Since we do not have such expression in our vector fields? I can see that we can have $-e^{-x}\frac{\partial}{\partial y}$, but not what you wrote? At least I cannot see it! Can you please let me know where it comes from? – user513784 Dec 18 '17 at 18:29
  • Oh, you're right. My error. But your term is canceled by $e^{-x}\dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial y}\right)$ from the first term. It's not that it's zero to start with. – Ted Shifrin Dec 18 '17 at 18:34
  • @TedShifrin Many thanks! Can you please let me know that why our distribution here has no integral manifolds? – user513784 Dec 18 '17 at 18:37
  • What did you get when you followed up on my comment about rulings? Think about the tangent planes to a cylinder (as opposed to those of a general ruled surface). – Ted Shifrin Dec 18 '17 at 18:51
  • @TedShifrin Many thanks! But can you please let me know how did you get the cylinder with rulings parallel to the $x$-axis? I added some calculation that shows integral curves for the vector fields $X_1$ and $X_2$ are not compatible? – user513784 Dec 19 '17 at 22:43
  • Your additional computation is confusing you because you're trying to find two different integral curves, not one. As I already said, the integral curves of $X_1$ are all (portions of) lines parallel to the $x$-axis. That means any surface containing them would have to be a (generalized) cylinder with rulings parallel to the $x$-axis. What is true about the tangent planes to a cylinder? – Ted Shifrin Dec 19 '17 at 23:06
  • @TedShifrin They have to pass trough those parallel lines to the $x$-axis? – user513784 Dec 19 '17 at 23:18
  • @TedShifrin Thanks! Did you mean that? Or I just said the trivial? – user513784 Dec 19 '17 at 23:57
  • I'm going to let you think about this on your own. Yes, the tangent plane at each point contains the actual line. But cylinders (and cones) are different from a general ruled surface. – Ted Shifrin Dec 19 '17 at 23:58
  • @TedShifrin Many thanks! Sorry, can you please let me know why if there were an integral surface for $D$, it would have to be a cylinder with rulings parallel to the $x$-axis? – user513784 Dec 28 '17 at 22:16
  • I answered this already. What are integral curves of $X_1$? – Ted Shifrin Dec 28 '17 at 22:19
  • @TedShifrin I am so sorry, but you said it would be a cylinder with rulings parallel to the $x$-axis. But now let us suppose that is the case. But why our whole distribution has no integral manifolds? Thanks! – user513784 Dec 28 '17 at 22:23
  • Is the plane spanned by $X_1$ and $X_2$ independent of $x$? – Ted Shifrin Dec 28 '17 at 22:26
  • @TedShifrin Yes it is! (I am not sure!) – user513784 Dec 28 '17 at 22:42
  • @TedShifrin Many thanks! But I think it is not independent of $x$, because we have $e^{-x}$? Can you please explain a bit more? – user513784 Dec 29 '17 at 11:21

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