Is there exist a similar conjecture to the famous Jacobian Conjecture with $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$ instead of $\mathbb{C}[x_1,\ldots,x_n]$?

Namely, let $f$ be $\mathbb{C}$-algebra endomorphism of $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, denote $f_i:= f(x_i)$, and further assume that the Jacobian of $\{f_1,\ldots,f_n\}$ is in $\mathbb{C}-\{0\}$. Is such $f$ an automorphism?

(I guess that first one should be familiar with the group of automorphisms of $\mathbb{C}[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$, see this question).

Edit: I also wonder if there exists any nice connection between the Jacobian Conjecture and my above conjecture (which is not exactly phrased yet); for example, are the two conjectures equivalent?


Counterexample: the endomorphism of the product of two punctured lines (complement of the curve of equation $xy=0$ in the plane) given by $$(x,y)\mapsto f(x,y)=\left(\frac{x}{y},y^2\right)$$

We have $$\begin{pmatrix}\partial_1f_1 & \partial_2f_1\\ \partial_1f_2 & \partial_2f_2\end{pmatrix}=\begin{pmatrix}\frac{1}{y} & -\frac{x}{y^2}\\ 0 & 2y\end{pmatrix},$$ so the Jacobian is constant equal to 1.

It's not an automorphism, since both $(1,1)$ and $(-1,-1)$ map to $(1,1)$.

| cite | improve this answer | |
  • 2
    $\begingroup$ About injectivity, see: en.wikipedia.org/wiki/Ax%E2%80%93Grothendieck_theorem (I haven't checked it works in this context, although it sounds likely at first sight). But it's clearly of a different spirit, because the Jacobian condition is some immediately checkable condition, while the injectivity is not. $\endgroup$ – YCor Apr 13 '17 at 22:44
  • 1
    $\begingroup$ Note that constancy of Jacobian is not necessary, even in 1-d: take $x\mapsto 1/x$, for example. A necessary condition is that the Jacobian be a monomial. $\endgroup$ – ACL Apr 13 '17 at 22:45
  • 1
    $\begingroup$ Well, it is invertible. In the sense that it is a non vanishing function on the torus. $\endgroup$ – ACL Apr 13 '17 at 22:50
  • 3
    $\begingroup$ More generally, any isogeny of algebraic tori: $f\colon \mathbf G_m^n\to\mathbf G_m^n$ has an invertible Jacobian. (Proof: either by computation, or by remarking that there is an “inverse” isogeny, whose composition is $(x_1,\dots,x_n)\to (x_1^d,\dots,x_n^d)$, where $d$ is the degree of $f$.) On the other hand, the first proof gives the clue to the initial question. $\endgroup$ – ACL Apr 13 '17 at 22:57
  • 1
    $\begingroup$ @user237522: the point in ACL's last comments is that the endomorphism semigroup of the torus is considerably smaller than that of the affine $n$-space, so checking whether it's an automorphism is immediate (namely, you need that the matrix of exponents, which is in $\mathrm{M}_n(\mathbf{Z})$, belongs to $\mathrm{GL}_n(\mathbf{Z})$) $\endgroup$ – YCor Apr 13 '17 at 23:07

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.