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The question is as follows:

In $M = \mathbb{R}^3$, for the distribution $D = \left<X1,X2\right>$ generated by the vector fields: $$X_1 = \frac{\partial}{\partial x} + z^2 \frac{\partial}{\partial y}, ~~~~ ~~~~ ~~~ X_2 = \frac{\partial}{\partial y} + z^2 \frac{\partial}{\partial z}$$ and such that every vector field $X \in \varkappa(D)$ (The set of vector fields tangent to $D$) is a linear combination $aX_1 +bX_2$, where the smooth functions $a = a(x, y)$ and $b = b(x, y)$ are uniquely determined.

Show that the only points through which there exist integral manifolds are the points in the plane $z = 0$.

$\textbf{Some effort:}$

We know that the integral curves of $X_1$ are $(x_0+t , y_0 e^{t^2}, z_0)$ and those of $X_2$ are $(x_0 , y_0 + t, z_0 e^{t^2})$. So that the respective local flows are $$\varphi_t(x,y,z) = (x + t , y e^{t^2}, z), ~~~~ ~~~~ ~~~ \psi_s(x,y,z) =(x , y + s , z e^{s^2}) $$ And the map $$(s,t) \in \mathbb{R}^2 \mapsto (\psi_s \circ \varphi_t)(x_0, y_0 , z_0) = \psi_s (x_0 + t , y_0 e^{t^2}, z_0)= (x_0 + t , y_0 e^{t^2} + s , z_0 e^{s^2})$$ will be the integral surface through $(x_0, y_0, z_0)$.

After this I do not know what to do?!

Can someone help me in getting to to result?

Thanks!

user513784
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1 Answers1

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You can solve this with the Frobenius theorem, but otherwise you can check directly that the surface you define $\varphi: (s,t) \mapsto (x_{0} + t, y_{0}e^{t^{2}} + s, z_{0}e^{s^{2}})$ is an integral surface only when $z_{0}$ vanishes. The tangent space to this surface in $\mathbb{R}^{3}$ is spanned by the vectors

\begin{align*} &\frac{\partial \varphi}{\partial t} = (1, 2ty_{0}e^{t^{2}}, 0),\\ &\frac{\partial \varphi}{\partial s} = (0, 1, 2sz_{0}e^{s^{2}}). \end{align*}

Then you can show that $X_{2} = \frac{\partial}{\partial y} + z^{2}\frac{\partial}{\partial z} = (0, 1, z_{0}^{2}e^{2s^{2}}) \in \left\langle \frac{\partial \varphi}{\partial t}, \frac{\partial \varphi}{\partial s} \right\rangle$ if and only if $z_{0} = 0$ (and similarly for the other vector field $X_{1}$). You can and should verify this by direct calculation.

  • Many thanks! How can we use Frobenius theorem for this question? – user513784 Dec 18 '17 at 08:55
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    The Frobenius theorem gives conditions for when there exists an integral manifold tangent to your distribution. It happens precisely when the Lie bracket $[X_{1}, X_{2}]$ belongs to your distribution $D$, so you can compute the Lie bracket directly to see whether or not you have integral manifolds. –  Dec 18 '17 at 09:00
  • Many thanks! I found it. We have $[X_1 , X_2] = -2z^3 \frac{\partial}{\partial y}$. For this to be in our distribution $D$, we should have $z=0$. Am I right? – user513784 Dec 18 '17 at 18:04
  • I haven't checked your computation but that sounds right :) –  Dec 18 '17 at 18:33