0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.