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Question: This is part of a problem from an old qualifying exam that I am having a little trouble with, so any help is greatly appreciated! In the problem, we are given two vector fields on $\mathbb{R}^3$ expressed by $$X = x(2y + \cos(y))\frac{\partial}{\partial x} - \frac{\partial}{\partial z},\qquad Y = \frac{\partial}{\partial y} + \frac{\partial}{\partial z} $$ and we need to write down a submanifold $N$ of $\mathbb{R}^3$ containing the point $p = (0,1,0)$ such that there is an open neightborhood $U_p$ around $p$ such that for every point $q\in U_p$ we have $T_qN = \text{span}\{X\vert_q, Y\vert_q\}$, but I can't figure out a good way of going about this.

My Attempt: My first thought was to check and see if $X$ and $Y$ commute: they are obviously linearly independent at $p = (0,1,0)$, so if they commute, there is some coordinate representation of $N = (s^1,s^2,s^3)$ such that for a chart $(V_p, (s^i))$ of $N$ centered at $p$, we have that $X = \frac{\partial}{\partial s^1}$ and $Y = \frac{\partial}{\partial s^2}$ [see Theorem 9.46 of Lee's Intro to Smooth Manifolds]. However, when I compute the Lie bracket $[X, Y]$, I get $$\begin{align*} [X,Y] &= X(1)\frac{\partial}{\partial y} + X(1)\frac{\partial}{\partial z} - Y(x(2y + \cos(y)))\frac{\partial}{\partial x} + Y(1)\frac{\partial}{\partial z}\\ &=0 + 0 - (2x-x\sin(y))\frac{\partial}{\partial x}+0\\ &=(x\sin(y) - 2x)\frac{\partial}{\partial x}\\ &\not= 0, \end{align*}$$ so the vector fields don't commute. At this point I got stuck. I did compute that the flow $\theta_t$ of $X$ is given by $$\theta_t(x,y,z) = (xe^{2yt + t\cos(y)}, y, z - t),$$ and that the flow $\psi_t$ of $Y$ is given by $$\psi_t(x,y,z) = (x, y + t, z + t),$$ but I don't really know how to use this information to describe $N$.

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  • $\begingroup$ I guess you mean for every $q \in U_p \cap N$ the condition about the tangential space should hold? $\endgroup$ Commented Aug 8, 2019 at 18:29
  • $\begingroup$ Yes, that's what I mean, sorry. $\endgroup$
    – peabody
    Commented Aug 8, 2019 at 18:33

1 Answer 1

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$\theta'^{p}(t)=X_{\theta^{p}(t)},$ and $\psi'^{p}(t)=Y_{\psi^{p}(t)},$ so using your calculation, i.e. $\theta_t(0,1,0) = (0,1,-t)$ and $\psi_t(0,1,0) = (0, 1 + t, t),$ we get a vector normal to a desired submanifold by taking the cross-product: $X_{\theta_t(0,1,0)}\times Y_{\psi_t(0,1,0)}=(1,0,0)$, and it follows now that $N$ is the $yz$-plane.

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  • $\begingroup$ Wonderful, thank you so much! That also helps me out with a later part of the question. $\endgroup$
    – peabody
    Commented Aug 9, 2019 at 2:56

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