I was thinking about this after I read about Jacobian conjecture. But I can't see what I did wrong? Maybe you can help me.

Let $F: \mathbb{C}^n \to \mathbb{C}^n$ be of the form $F(x_1, \dots, x_n)= (F_1(x_1, \dots, x_n), \dots, F_n( \dots , x_1, \dots, x_n) )$ for some $F_1,\dots , F_n \in \mathbb{C}[X_1, \dots, X_n ]$, which I will call polynomial, with constant Jacobian determinant $\det DF$.

Then by Implicit Function Theorem $F$ has a local inverse $F'$ around $0$. Since $F$ is analytical $F'$ is also analytical. We also know that $ DF' = (DF)^{-1}= \frac{1}{\det DF} \text{adj}(DF) $ but the adjugate is computed in terms of the minors of $DF$, $\det DF$ is constant and thus $DF'$ is given by polynomial functions in its components, in particular $D^k F =0$ for some $k\in \mathbb{N}$. Thus $F'$ is polynomial because of its Taylor expansion. Since $F'$ is polynomial it is defined on the whole space $\mathbb{C}^n$.

Now we have $F' \circ F\mid_U= \text{id}$ for some open neighborhood $U$ of $0$, but this implies $F' \circ F= \text{id}$ since $F' \circ F$ is a polynomial.

  • $\begingroup$ Hi user60589. There were some ‘$’ signs missing, hence the edit. $\endgroup$ – Haskell Curry Feb 1 '13 at 20:13
  • $\begingroup$ Why do you presume the adjugate is polynomial? Computing cofactors involves determinants of sub-matrices. $\endgroup$ – copper.hat Feb 1 '13 at 20:27
  • $\begingroup$ They are sub-matrices of $DF$ which consist of the derivatives of $F_i$, so the determinant of it is again a polynomial? $\endgroup$ – user60589 Feb 1 '13 at 20:31
  • $\begingroup$ I think copper.hat is right. The minors involve divisions, and it could happen that you were dividing polynomials with no common factors in the process, which would not yield a polynomial in the end. $\endgroup$ – busman Feb 2 '13 at 13:17
  • $\begingroup$ A minor is just a determinant? Why should they involve divisions? Or am I wrong? $\endgroup$ – user60589 Feb 2 '13 at 13:28

You have used the wrong formula $$df^{-1}(y)=(df)^{-1}(y)$$ instead of the correct formula $$df^{-1}(y)=(df)^{-1}\bigl(f^{-1}(y)\bigr)\ .$$


Look what you did wrong: $$Id=D(F\circ F^{-1})=DF\circ F^{-1} \cdot DF^{-1}\Rightarrow DF^{-1}=(DF\circ F^{-1})^{-1},$$ thus, $DF^{-1} \neq (DF)^{-1}$, which seems to be your crucial assumption.


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.