2
$\begingroup$

I have asked several mathematicians about the following questions,but all of them think they are good questions,but can not give a complete answer.Now I have to come here to ask mathematicians all over the world,and hope that I can get a complete answer or some constructive suggestions.

Given $$\frac{P_1(x_1,x_2,\dots,x_n)}{P_2(x_1,x_2,\dots,x_n)}\text{, and }\frac{P_3(y_1,y_2,\dots,y_n)}{P_4(y_1,y_2,\dots,y_n)}$$ where $P_i$ is polynomial(that is $P_1(x_1,x_2,\dots,x_n), P_2(x_1,x_2,\dots,x_n), P_3(y_1,y_2,\dots,y_n), P_4(y_1,y_2,\dots,y_n)$ are polynomial) whose coefficients are over $\mathbb{Q}$,let $J$ be Jacobian matrix, when does $f(x_1,x_2,\dots,x_n)$ exist,such that $$\frac{P_1(x_1,x_2,\dots,x_n)}{P_2(x_1,x_2,\dots,x_n)}=\frac{P_3(f(x_1,x_2,\dots,x_n))}{P_4(f(x_1,x_2,\dots,x_n))}|\det (J(f(x_1,x_2,\dots,x_n)))|\text{ ,}$$ and $f(x_1,x_2,\dots,x_n)$ is bijective ?

When $f(x_1,x_2,\dots,x_n)$ is restricted to differential polynomials, rational functions, algebraic functions,or just all differential function including transcendental functions over $\mathbb{C}$ , $\mathbb{R}$, what is the condition or non-condition for the existence of $f(x_1,x_2,\dots,x_n)$ ?

$\endgroup$
10
  • $\begingroup$ Related: mathoverflow.net/questions/180876/… $\endgroup$ Commented Sep 24, 2014 at 5:48
  • $\begingroup$ @NoahS,yes,it is related,but some people think it is not clearly expressed,so it has been closed.Now I rephrase them and having asked mathematicians around me without any complete answer and constructive suggestion,I have to post it here. $\endgroup$ Commented Sep 24, 2014 at 9:27
  • 3
    $\begingroup$ Your question is far too general for there to be any hope of a comprehensive solution. This is why you aren't getting any answers. I suggest that you try to solve the case $n=1$, i.e., give necessary and sufficient conditions for a bijective solution in that case for general polynomials in one variable. If you are able to find computable and interesting criteria in this case, try $n=2$. If you can't do that, then asking the question for general $n$ is probably not worthwhile. By the way, it seems highly unlikely to me that such an approach will shed any light on the Jacobian Conjecture. $\endgroup$ Commented Sep 24, 2014 at 11:01
  • 1
    $\begingroup$ @XL_at_China: For example, consider the question of whether there is a bijective solution $f:\mathbb{R}\to\mathbb{R}$ to the equation $$\frac1{x^2+1} = \frac{a_k}{f(x)^{2k}+1}f'(x)$$ when $k>1$ is an integer for some constant $a_k$ and how you would determine that constant $a_k$. That's only one sample of the kind of thing you'd have to be able to answer, even in the case $n=1$. $\endgroup$ Commented Sep 24, 2014 at 15:31
  • 1
    $\begingroup$ @XL_at_China: I agree that it is not literally the Jacobian Conjecture, but it does seem that, in order to answer your question, you will have to have a constructive method for finding a candidate mapping $f$ and deciding whether or not it is polynomial. If you could do that, it seems highly likely to me that you would have to be able to solve the Jacobian conjecture along the way, so I think it's actually stronger. For example, in the case, $n=1$, the Jacobian Conjecture is trivial, but your problem is definitely not. $\endgroup$ Commented Sep 26, 2014 at 8:46

0

You must log in to answer this question.