Cross product of the gradient of two functions I am having a bit of a confusion with some claims I keep finding on a book of Fluid Dynamics. Let's say we have two functions in 3-D space, $f(\mathbf{x})$ and $g(\mathbf{x})$, with the following constraint:
$\nabla f \times \nabla g = 0$
From my understanding of mathematics (I'm not mathematician) is that the level sets of both functions coincide. What I don't understand is why can I make the claim that I can write one function in terms of the other, that is:
$f = f(g)$
Thanks in advance
 A: I think I've understood your question, correct me If I'm wrong. You have two scalar fields $f,g$ and know that
$$\nabla f \times\nabla g = \mathbf 0$$
and you want to show that there is some function $h$ such that $f(\mathbf x) = h(g(\mathbf x))$ for all $\mathbf x.$
The first thing is that the condition on the gradients gives you that
$$\nabla f = \lambda \nabla g,$$
where $\lambda$ is some scalar field. In fact if $\lambda(\mathbf x) = h'(g(\mathbf x))$ then we are done.
I think however, that there are some extra conditions you have missed as your assertion is false in general.
Here is a counterexample: Define scalar fields $f,g$ as:
\begin{align}
f(x,y,z)&=x+y+z\\
g(x,y,z)&=(x+y+z)^2
\end{align}
Then we can find
\begin{align}
\nabla f &=\begin{pmatrix}1\\1\\1\end{pmatrix}\\
\nabla g &=2(x+y+z)\begin{pmatrix}1\\1\\1\end{pmatrix}
\end{align}
Then certainly $\nabla f \times\nabla g=\mathbf 0$ but we clearly can't have $f$ as a function of $g$ as then we would have $f(1,1,1) = f(-1,-1,-1)$ as $g(1,1,1)=g(-1,-1,-1)$ which clearly isn't true.
A: The equation defines a 1D curve. It's actually 3 equations but the third is not independent. Each equation defines a 2D surface and where they cross is a 1D curve.
At each point on the curve is a value for $f$ and $g$. So we could write the curve as a function $h(f,g)=0$ for some function $h$ which gives the relation between $f$ and $g$ on the curve.
BTW. Which book did you get this out of I would like to read it?
