# Submanifold of $\mathbb{R}^3$ with tangent space spanned by given vector fields

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 $$N$$.

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

$$\theta'^{p}(t)=X_{\theta^{p}(t)},$$ 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.