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This is part of another old qualifying exam problem, so any help is appreciated! We are given two vector fields $X$ and $Y$ on $\mathbb{R}^3$ described by

$$X = \frac{\partial}{\partial y} + z\frac{\partial}{\partial x}\qquad Y = \frac{\partial}{\partial z} + y\frac{\partial}{\partial x}$$

and we are asked to give a parameterization $(x(s,t), y(s,t), z(s,t))$ for the (unique) surface passing therough the point $p = (1,0,0)$ and tangent to the vector fields $X$ and $Y$ at each point, and to then give an equation of the form $F(x,y,z) = 0$ for this surface.

I was able to determine that $[X, Y] = 0$, and that $X$ and $Y$ were linearly independent at $p$, so I computed the flows $\theta_t$ of $X$ and $\phi_s$ of $Y$ to be $$\begin{align*} \theta_t(x,y,z) &= (xe^{zt}, y + t, z)\\ \phi_s(x,y,z) &= (xe^{ys}, y, z + s) \end{align*}$$ which gives us $$\theta_t\circ \phi_s = \phi_s\circ \theta_t = (xe^{zt + ys + st}, y + t, z + s),$$ but it is at this point that I get stuck. To solve this, I have tried to use the following theorem from Lee's textbook:

Theroem 9.46: Let $M$ be a smooth $n$ manifold, and let $(V_1, \ldots, V_k)$ be a linearly independent $k$-tuple of smooth commuting vector fields on an open subset $W\subseteq M$. For each $p\in W$, there exists a smooth coordinate chart $(U, (s^i))$ centered at $p$ such that $V_i = \frac{\partial}{\partial s^i}$ for $i = 1,\ldots,k$. If $S\subseteq W$ is an embedded codimension-$k$ submanifold and $p$ is a point of $s$ such that $T_pS$ is complementary to the span of $(V_1\vert_p,\ldots,V_k\vert_p)$, then the coordinates can also be chosen such that $S\cap U$ is the slice defined by $s^1=\cdots=s^k=0$.

At $p = (1,0,0)$ we have that $\text{span}\left\{X\vert_p,Y\vert_p\right\} = \text{span}\left\{\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\}$, so if we take $S$ to be the $x$-axis, then we can apply the theorem. However, since $S$ is not just a point, I am having trouble finding the explicit parameterization $(x(s,t), y(s,t), z(s,t))$. Agian, any help on how I need to go about this is greatly appreciated. Thank you!

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One can do this without directly thinking about or computing flows.

Hint Taking a general $1$-form $$\omega_0 := p(x, y, z) \,dx + q(x, y, z) \,dy + r(x, y, z) \,dz$$ on $\Bbb R^3$, imposing $\omega(X) = \omega(Y) = 0$, and solving for the coefficient functions $p, q, r$, shows that the annihilator of $X$ and $Y$ at each point is spanned by $$\omega_0 := dx - z \,dy - y \,dz,$$ but this form is exact: $$\omega_0 = d(x - y z) .$$ (Involutivity of the distribution $D := \operatorname{span}\{X, Y\}$ implies that any annihilator $\omega$ of $D$ satisfies $d\omega = \theta \wedge \omega$ for some $1$-form $\theta$, and since $\Bbb R^3$ is simply connected, if $\omega$ vanishes nowhere there is a function $f$ such that $f \omega$ is exact.)

So, by construction, $D$ is tangent everywhere to the level sets $G^{-1}(c)$, where $G(x, y, z) := x - y z$. In particular, $G(p) = G(1, 0, 0) = 1$, so the (maximal) integral submanifold $S$ of $D$ through $p$ is $S = G^{-1}(1) = \{x - y z = 1\}$. Rearranging the defining equation gives $x - y z - 1 = 0$, so we can take $$F(x, y, z) = x - y z - 1 .$$ Rearranging again to solve for $x$ gives $x = y z + 1$, giving the global parameterization $$(y, z) \mapsto (y z + 1, y, z)$$ of $S$.

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    $\begingroup$ That’s so cool! Thank you so much for sharing! $\endgroup$ – peabody Aug 14 at 21:22
  • $\begingroup$ You're welcome, I'm glad you found it useful. NB the statement about the exactness of the annihilating $1$-form originally wasn't quite stated correctly---please see the edits. $\endgroup$ – Travis Willse Aug 14 at 22:26
  • $\begingroup$ Also, since the source of this is problem is on an old qualifying exam, I feel obligated to point out that while finding a closed annihilator was easy in this case---it happened to be the simplest nonvanishing annihilator to write down, a is often the case for such problems--- in general after you have found an annihilator $\omega$, solving $d(f \omega) = 0$ to find a closed annihilator $f\omega$ amounts to solving a p.d.e. for one function in three variables. This may or may not be easier than computing the flows. $\endgroup$ – Travis Willse Aug 14 at 22:45
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With $g (x,y,z)=x-yz-1$, I think that it works. (To find it, I considered the equation that $Ker (g (x,y,z)) $ must satisfy, to find a candidate for $dg $ and then some $g $ that could work, and then check that it indeeds work). Considering that $[X,Y]=0$ over $\mathbb {R}^3$, we have the unicity from the Frobenius theorem, but I'm not sure about that last point, I'm not really on point around that theorem.

Tell me if you think something there isn't right!

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  • $\begingroup$ Yeah, I think that does it. Thank you! I'm still working out the explicit parameterization that I need, though. $\endgroup$ – peabody Aug 13 at 23:23
  • $\begingroup$ $\lbrace (x,y,z) \in \mathbb {R}^3 | g (x,y,z)=0 \rbrace$ can be rewritten as $\lbrace (yz+1,y,z) | (y,z) \in \mathbb {R}^2 \rbrace $. You can then check that this the defintion of a submanifold as the graph of a smooth function! $\endgroup$ – Paul Aug 14 at 10:21
  • $\begingroup$ Oh, duh, I was way overthinking it. Thank you so much! $\endgroup$ – peabody Aug 14 at 14:07
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Ok, I finally figured out what I was doing wrong: I messed up the computations for the flows of the vector fields. The flows $\theta_t$ of $X$ and $\phi_t$ of $Y$ should be \begin{align*} \theta_t(x,y,z) &= (x + zt, y + t, z)\\ \phi_s(x,y,z) &= (x + ys, y, z + s). \end{align*}

This would then give us that $$\theta_t\circ \phi_s = \phi_s \circ \theta_t = (x + ys + zt + st, y + t, z + s).$$

If we take $S$ to then be the $x$-axis parameterized by $S(r) = (r, 0, 0)$, we can then use the given theorem to construct the function $$\Psi_r(t,s) = \theta_t\circ \phi_s(r,0,0) = (r + st, t, s) = (x(s,t),y(s,t),z(s,t))$$ and it is fairly easy to check that

\begin{align*} \frac{\partial}{\partial t}\Psi_r(t,s) &= s\frac{\partial}{\partial x} + \frac{\partial}{\partial y} = z(s,t)\frac{\partial}{\partial x} + \frac{\partial}{\partial y} = X\\ \frac{\partial}{\partial s}\Psi_r(t,s) &= t\frac{\partial}{\partial x} + \frac{\partial}{\partial z} = y(s,t)\frac{\partial}{\partial x} + \frac{\partial}{\partial z} = Y. \end{align*}

Then if we want to construct the zero set $F(x,y,z) = 0$ that coincides with this surface passing through $(1,0,0)$, we simply pick $r = 1$ and solve the rest to get that $F(x,y,z) = x - yz - 1$.

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