Consider two vector-valued functions $f$ and $g$ of a vector $x$. If the jacobians of these two functions with respect to $x$ are exactly transposes of each other, what can we say directly about the relationship between $f$ and $g$? Specifically:

$$ x \in \mathbb{R}^n,\ \ \ f: \mathbb{R}^n \to \mathbb{R}^n,\ \ \ g: \mathbb{R}^n \to \mathbb{R}^n$$

$$ \nabla f := \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \dots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} & \dots & \frac{\partial f_n}{\partial x_n}\end{bmatrix},\ \ \ \nabla g := \begin{bmatrix} \frac{\partial g_1}{\partial x_1} & \dots & \frac{\partial g_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial g_n}{\partial x_1} & \dots & \frac{\partial g_n}{\partial x_n}\end{bmatrix}$$

$$\text{Suppose:}\ \ \nabla f = (\nabla g)^T\ \ \forall x\ \ $$

Using this relationship between the derivatives of $f$ and $g$, can we express any relation between $f$ and $g$ themselves?

Worded another way: given $f(x)$ and its jacobian $\nabla f$, can you find a function $g(x)$ with jacobian equal to $(\nabla f)^T$? Presumably you could find it in terms of $f(x)$? At first I thought it would just be some simple rearrangement of variables, perhaps $f$ of a permutation of $x$, but the actual relation is proving to be rather complicated.


\begin{align} f(x) = \begin{bmatrix} x_1 x_2 \\ x_1^2 + x_2^3 \end{bmatrix}\ \ \ &\implies\ \ \ \nabla f = \begin{bmatrix} x_2 & x_1 \\ 2x_1 & 3x_2 \end{bmatrix} \\ g(x) = \begin{bmatrix} g_1(x) \\ g_2(x) \end{bmatrix}\ \ \ &\implies\ \ \ \nabla g = \begin{bmatrix} x_2 & 2x_1 \\ x_1 & 3x_2 \end{bmatrix} \end{align}

Is there a general form for all of these special function-pairs $(f, g)$ which have transposed jacobians? Even simple examples are, in general, complicated PDE's to actually solve, but I'm hoping there is some clever way to think about what having the transpose of a given jacobian would mean, perhaps geometrically.

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    $\begingroup$ There's going to be some pretty strong condition on $f$: in order for $(\nabla f)^T$ to be a Jacobian, its rows must be gradient vector fields. This imposes a constraint on the derivative of $f$ - if I'm not mistaken, $f$ must have constant curl. Chase this idea and see how far you can take it - for example if you can in fact conclude that $f$ has zero curl then you will have proven that the only examples are where $f = g + \rm constant$. $\endgroup$ – Anthony Carapetis Feb 13 '17 at 10:43
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    $\begingroup$ The example is wrong. $\frac{\partial g_1}{\partial x_1} = x_2$ and $\frac{\partial g_1}{\partial x_2} = 2x_1$ lead to $1 = \frac{\partial^2 g_1}{\partial x_1 \partial x_2} = 2$, which is a contradiction. $\endgroup$ – Michał Miśkiewicz Feb 13 '17 at 11:26
  • $\begingroup$ Interesting point that the example I gave leads to a contradiction. It exemplifies that the transpose of some given jacobian is not necessarily the jacobian of any other function; in many cases no (continuous) solution $g$ even exists. Thanks! $\endgroup$ – jnez71 Feb 13 '17 at 20:27

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