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

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?

• You need to finish your argument. There certainly are such functions, namely constants. Why no others? Aug 6, 2020 at 16:30
• @TedShifrin *Apart from the trivial constant functions of course. Well, I thought my argument is finished. For instance we take $y=0$, then $(2)$ gives $\frac{\partial}{\partial x}f=0$. Plugging this back into $(2)$ we get that $\frac{\partial}{\partial z}f=0$. Meaning: $\frac{\partial}{\partial x}f=0,\frac{\partial}{\partial y}f=0, \frac{\partial}{\partial z}f=0$, which is a constant function.
– Luka
Aug 6, 2020 at 16:58
• Noooo ... Way too sloppy. You only know $\partial f/\partial x=0$ at the points with $y=0$. Please write out a careful proof. You'll find that you need to use second-order partial derivatives. Aug 6, 2020 at 17:24
• Philosophical comment: The most basic integrability criterion (of which Frobenius is a vast strengthening) is the equality of mixed partial derivatives. Aug 6, 2020 at 17:33
• You don't need Frobenius at all. That was my main point. Just observe that if you differentiate the second equation with respect to $y$, you get $$0=f_{xy} + f_z + yf_{zy} = f_{yx} + f_z + yf_{yz} = f_z,$$ since $f_y = 0$. Now you end up with all three partial derivatives $0$ everywhere. Aug 7, 2020 at 22:22

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.

• An excellent point, thanks! Aug 6, 2020 at 17:30
• I think, I understand, what you mean.
– Luka
Aug 7, 2020 at 22:17
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.