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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?

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    $\begingroup$ You need to finish your argument. There certainly are such functions, namely constants. Why no others? $\endgroup$ Commented Aug 6, 2020 at 16:30
  • $\begingroup$ @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. $\endgroup$
    – Luka
    Commented Aug 6, 2020 at 16:58
  • $\begingroup$ 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. $\endgroup$ Commented Aug 6, 2020 at 17:24
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    $\begingroup$ Philosophical comment: The most basic integrability criterion (of which Frobenius is a vast strengthening) is the equality of mixed partial derivatives. $\endgroup$ Commented Aug 6, 2020 at 17:33
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    $\begingroup$ 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. $\endgroup$ Commented Aug 7, 2020 at 22:22

2 Answers 2

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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.

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  • $\begingroup$ An excellent point, thanks! $\endgroup$
    – Knaus
    Commented Aug 6, 2020 at 17:30
  • $\begingroup$ I think, I understand, what you mean. $\endgroup$
    – Luka
    Commented Aug 7, 2020 at 22:17
  • $\begingroup$ Many thanks for your answer! $\endgroup$
    – Luka
    Commented Aug 7, 2020 at 22:18
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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.

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  • $\begingroup$ Thank you! I will use this while studying for my exam, when I get to the part with differential forms, that is. $\endgroup$
    – Luka
    Commented Aug 8, 2020 at 21:44

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