Smooth function, manifold-tangent vector fields $X=(0,1,0) \qquad Y=(1,0,y)$

Question

I'm looking for a smooth function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ with a non-vanishing gradient, such that the vector fields: $$X=(0,1,0) \qquad Y=(1,0,y)$$ are tangent to each 2-dim submanifold given by $f_c=\{x\in\mathbb{R}^n; f(x)=c\}$.

My attempt:

$df=df_c$ and additionally: $\langle X(\textbf x),df (\textbf x)\rangle=0 \quad\langle Y(\textbf x),df (\textbf x)\rangle=0 $. This gives: $$\frac{\partial}{\partial y} f=0 \qquad(1)$$

$$\text{and}:$$ $$\frac{\partial}{\partial x}f+y\frac{\partial}{\partial z}f=0 \qquad(2)$$

Conclusion:If $(2)$ needs to hold for all $y$, then there is no such function, because $f$ must be independent of $y$ according to $(1)$.

My question:

Is my conclusion correct? I understand that there is an explanation using Lie brackets and the Fröbenious theorem, can someone help me with that?

Answer 1

Your conclusion looks correct to me.

For the alternative approach, you just compute the Lie bracket $[X,Y]=(0,0,1)$. This is not at each point in the space spanned by $X$ and $Y$. Hence Frobenius tells us that the spaces spanned by $X$ and $Y$ are not the tangent spaces of the leaves of a foliation. The family of submanifolds $f_c$ would have been such a foliation of the region where $f$ is regular, i.e. everywhere since you want $f$ to have non-vanishing gradient. So $f$ doesn’t exist.

Answer 2

I commented above that the Frobenius criterion is a (very) enhanced version of the integrability criterion that mixed partial derivatives (of a $C^2$ function) are equal. We can apply that directly here.

Observe that if you differentiate the second equation with respect to $y$, then you get $$0=f_{xy}+ f_z + yf_{zy} = f_{yx} + f_z + yf_{zy} = f_z + (f_y)_x + y(f_y)_z = f_z,$$ since the first equation gives us $f_y = 0$. Now you end up with all three partial derivatives everywhere $0$, so $f$ must be constant.

REMARK: As you can find in numerous other posts of mine, a powerful way of applying Frobenius is to use the version stated in terms of differential forms. In your case, the distribution is defined by $\omega = 0$, where $\omega = dz-y\,dx$. (This is a very famous $1$-form, e.g., because it defines a contact structure on $\Bbb R^3$.) Note that $d\omega = dx\wedge dy$ is not in the ideal generated by $\omega$. Therefore, the distribution cannot be integrable.