Suppose you have two smooth functions $F,G:\mathbb{R}^n\rightarrow \mathbb{R}$ with the property that none of the partial derivatives vanish anywhere. I wonder if it's possible to find functions $P:\mathbb{R}^n\leftrightarrow \mathbb{R}^n$ and $Q:\mathbb{R}\rightarrow \mathbb{R}$ such that

$$G = Q\circ F\circ P.$$

I feel like it should be possible, using Jacobians or something as a change of coordinates, but I'm not sure how to begin or if I need further restrictions. If this result is true, it would imply that in some sense all real functions of this form with non-vanishing partial derivatives are equivalent through a kind of change of coordinates.

Here's a solution for a special case. If the dimension is $n=1$, then $F$ and $G$ are smooth monotone functions and are therefore invertible. Let $P$ be the identity function, and let $Q=G\circ F^{-1}$. Of course, if $F$ is not also surjective, then $Q$ is not well-defined everywhere; perhaps it is possible to extend $Q$ to all of $\mathbb{R}$, though I'm not sure how smoothly.

And if the dimension $n>1$, then here is a sketch of what I think shows the result:

  • Intuitively, there ought to be (integral?) curves $\gamma_1,\gamma_2:\mathbb{R}^1\rightarrow \mathbb{R}^n$ through the domains of $F$ and $G$ such that $F\circ \gamma_1$ is a monotonic function passing through all possible values of $F$, and similarly for $G$. These curves define a fixed transformation between the codomains of $F$ and $G$, by $Q\equiv (G\circ \gamma_2) \circ (F\circ \gamma_1)^{-1}$.
  • Intuitively, it should be possible to parametrize the graphs of $F$ and $G$ using $n$ coordinates, where the first coordinate tells you the level and the remaining coordinates uniquely specify a point in the domain with that level. Because of the conditions on $F$ and $G$, I expect these can be expressed as smooth functions $\mathbb{R}^n\rightarrow \mathbb{R}^n$.
  • In another way of looking at it, there should be smooth functions $T_1, T_2 : \mathbb{R}^{n-1} \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ which carry points in $\mathbb{R}^n$ smoothly and invertibly through all other points in $\mathbb{R}^n$ with the same $F$ (respectively $G$) level. We think of the second argument as being the point $x$, and the first argument as being a "displacement" within the level surface at $F(x)$. (I am a little concerned about level sets shaped like $n$-spheres, in which case this construction doesn't work invertibly— but perhaps such level sets are not possible if $F$, $G$ have everywhere nonzero partials. In that case, the level sets will be sort of $\mathbb{R}^{n-1}$-shaped, always.)

  • By combining the flow functions $T_i$ and the canonical curves $\gamma_i$, we obtain functions $H_i \equiv T_i(s, \gamma_i(t))$ which send $\mathbb{R}^{n-1} \times \mathbb{R}$ smoothly into all of $\mathbb{R}^n$.

  • Intuitively, there should be a smooth isomorphism $P:\mathbb{R}^n\rightarrow \mathbb{R}^n$ which sends the level sets of $G$ to the level sets of $F$. In particular, we should be able to pick one which sends the canonical points $\gamma_2(\mathbb{R})$ to the canonical points $\gamma_1(\mathbb{R})$ so that $P\circ \gamma_2 = \gamma_1$.

    We can do so concretely by defining $P(x) = H_1\circ H_2^{-1}$

  • Hence our solution is to use $Q\equiv (G\circ \gamma_2) \circ (F\circ \gamma_1)^{-1}$, sending levels of $F$ monotonically to levels of $G$, and $P\equiv H_1\circ H_2^{-1}$, sending level sets of $G$ to level sets of $F$ in a monotonic-like way.

    (specifically, if $G(x) < G(y)$, then $F\circ P(x) < F\circ P(y)$ or something like that).

    (Here, $H_2^{-1}$ looks up the unique way of expressing point $x$ as "distance" traveled along a $G$-level curve from some canonical other point with the same $G$-level. The result is a flow amount $s$; $H_1$ looks up the point with the same $F$-level as $x$ and applies the flow amount $s$ to it, resulting in a uniquely defined new point $x^\prime$, with certain special properties.)

Does this kind of construction work? I think if $n>1$ then we can construct $H_i$ as follows: because the partials of $F$ and $G$ don't vanish, we should be able to find unit vector fields $\mathbb{R}^n\rightarrow \mathbb{R}^n$ which are smooth and perpendicular to the gradients of $F$ and $G$, respectively; then the flow functions $H_i$ can simply translate along the integral surfaces of these vector fields. I think the only obstacle I see is if the integral surfaces are of incompatible shape— for example, if one is a line and the other a circle. But maybe the level surfaces can't be things like circles, because then the functions $F$ and $G$ wouldn't have everywhere nonvanishing partials (?).

Here's a concrete example of this construction for if $F$ is the sum function and $G$ is the multiplication function (ignoring the fact that $G$'s level sets have many branches):

  • We can pick canonical curves $\gamma_1(t) = \langle \vec{0}, t \rangle$ and $\gamma_2(t) = \langle \vec{1}, t\rangle$ so that $F\circ \gamma_1$ and $G\circ \gamma_2$ pass through all possible values.

  • These curves define the level-transforming function $Q \equiv (G\circ \gamma_2) \circ (F \circ \gamma_{1})^{-1}$; in this case, you can find that $Q$ is just the identity.

  • We can pick flow functions to send points to all other points on the same level surface : $T_1(s,x) = x + \bar{s}$, where $\bar{s} \equiv \langle s_1, \ldots, s_{n-1}, \sum_i s_i\rangle$.

    $T_2(s, x) \equiv x \cdot \bar{s}$, where $\bar{s} \equiv \langle e^{s_1}, \ldots, e^{s_{n-1}}, e^{\sum_i s_i}\rangle.$

  • In combination with the canonical curves $\gamma_i$, these flow functions give us $H_i(s,t) = T_i(s, \gamma_i(t))$.

    Here, $H_1(s,t) = \langle \vec{s}, t - \sum_i s_i\rangle$ and $H_2(s,t) = \langle e^\vec{s}, t\cdot e^{-\sum_i s_i} \rangle$.

  • Hence we define $P \equiv H_1 \circ H_{2}^{-1}$.

    To be clear, $H_2^{-1}(y) = \langle \log{y_1}, \ldots, \log{y_{n-1}}, \prod_i y_i \rangle$. In which case,

    $$P(x_1,\ldots, x_n) = \langle \log{x_1}, \ldots, \log{x_{n-1}}, \prod_i x_i - \sum_{i=1}^{n-1} \log{x_i}\rangle$$

  • And we find that $Q\circ F \circ P = F \circ P = \prod_i x_i = G$ (if you add up all the entries in $P$, the log terms all cancel and you're left with the product.)

  • There are of course other $P$ and $Q$ with this same effect; it all depends on the choice of canonical curve $\gamma_1, \gamma_2$, for instance.

  • $\begingroup$ The crux of this question seems to be showing that the level sets of a smooth function $\mathbb{R}^n\to\mathbb{R}$ with no critical points are diffeormorphic to $\mathbb{R}^{n-1}$. This is "obviously true" but I'm struggling to prove it... $\endgroup$
    – user7530
    Oct 27 '17 at 18:36

An attempt:

The image of $F$ is an open interval, so composing with a diffeomorphism from that interval to $\mathbb{R}$, we may assume that the image of $F$ is $\mathbb{R}$.

Consider the vector field $X(x) = \frac{\nabla F(x)}{\|\nabla F(x)\|^2}$. If $x(t)$ is an integral curve of $X$ then $\frac{d}{dt}F(x(t))\equiv 1$. Therefore, if $\phi_t$ is the $1$-parameter group of diffeomorphisms corresponding to $X$ then we have $$F(\phi_t(x)) = F(x) + t$$ Assume that $\phi_{\cdot}$ is defined on $\mathbb{R}\times \mathbb{R}^n$. Let $V_0=F^{-1}(0)$. We have a diffeomorphism $V_0 \times \mathbb{R} \to \mathbb{R}^n$. Under this diffeomorphism, $F$ corresponds to the projection on the last component in $V_0 \times \mathbb{R}$.

$V_0$ should be diffeomorphic to $\mathbb{R}^{n-1}$.


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