Assume I am given a smooth function $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ with jacobian matrix $\mathrm{J}$ (i.e. $\mathrm{J}(x) = \frac{\partial f_i}{\partial x_j}$). Can we say anything about existence / practical computation of a function $g:\mathbb{R}^n \rightarrow \mathbb{R}^n$ such that is jacobian is exctly the transpose of $\mathrm{J}$ ?

In other words, does there exist a function $g:\mathbb{R}^d \rightarrow \mathbb{R}^n$ such that : $\forall x \in \mathbb{R}^n$, $\forall\ 1 \le i,j \le n$,

$$ \frac{\partial f_i}{\partial x_j}(x) = \frac{\partial g_j}{\partial x_i}(x)$$

My guess is that such a function probably does not always exist, but I couldn't find provable counter-examples, not relevant litterature.

  • $\begingroup$ When you set ${{\partial f_i}\over {\partial x_j} }={{\partial g_j}\over {\partial x_i} }$, the domain of $f_i$ is $R^n$ and the domain of $g_j$ is $R^d$ the equality is possible only if $n=d$. $\endgroup$ – Tsemo Aristide Aug 5 '16 at 13:18
  • $\begingroup$ Indeed, I meant to write only one letter. I'll fix the question. Thanks ;-) $\endgroup$ – G. Fougeron Aug 5 '16 at 13:20

You need to find $n$ functions $g_i, i=1,...,n$ such that ${{\partial f_i}\over {\partial x_j} }={{\partial g_j}\over {\partial x_i} }$.

Consider the $1$-form $\alpha_j=\sum_i{{\partial f_i}\over{\partial x_j}}dx_i$, this form is closed if and only if ${{\partial^2 f_i}\over{\partial x_i\partial x_l}}={{\partial^2 f_l}\over{\partial x_i\partial x_l}}$.

So if ${{\partial^2 f_i}\over{\partial x_i\partial x_l}}={{\partial^2 f_l}\over{\partial x_i\partial x_l}}$ you can apply the Poincare lemma to find a function $g_j$ whose differential is $\alpha_j$, in this case $g=(g_1,...,g_n)$ answer your question.

To find counterexamples, try to construct functions for which the condition above is not verified.

  • $\begingroup$ Interesting approach. however consider $f_x(x,y) = x + y$, $f_y(x,y) = xy$. Then $\frac{\partial^2 f_x}{\partial x \partial y} = 0 \neq \frac{\partial^2 f_y}{\partial x \partial y} = 1$, but nevertheless $g_x(x,y) = \frac{1}{2} y^2$ and $g_y(x,y) = x + y$ work perfectly fine. $\endgroup$ – G. Fougeron Aug 5 '16 at 14:17
  • $\begingroup$ The condition is sufficient, but not necessary $\endgroup$ – Tsemo Aristide Aug 5 '16 at 14:24
  • $\begingroup$ You have $f_x(x,y)=x+y, f_y(x,y)=xy$ The Jacobian is $\pmatrix{1 & 1\cr y & x}$, you want a function whose Jacobian is $\pmatrix{1 & y\cr 1 & x}$, set $g_x(x,y)={1\over 2}y^2$ and $g_y(x,y)=x+y$, the Jacobian of $g$ is $\pmatrix{0 & y\cr 1 &1}$ $\endgroup$ – Tsemo Aristide Aug 5 '16 at 14:36
  • $\begingroup$ Indeed, how silly of me ! Thanks fo the remark. $\endgroup$ – G. Fougeron Aug 5 '16 at 15:10

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