Timeline for "Jacobian Conjecture" for $k[x_1,\ldots,x_n,x_1^{-1},\ldots,x_n^{-1}]$?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2018 at 10:29 | comment | added | Armando j18eos | I know: it's only a typo. :) | |
| Apr 22, 2018 at 21:42 | comment | added | YCor | @Armandoj18eos yes you're right! fortunately it does not matter! | |
| Apr 22, 2018 at 19:00 | comment | added | Armando j18eos | @YCor The determinant Jacobian of your example is $2$ and not $1$. ;) | |
| Apr 14, 2017 at 0:11 | comment | added | user237522 | (Thanks!) Concluding that regardless of the Jacobian of $\{f_1,\ldots,f_n\}$, there exist infinitely many endomorphisms which are not automorphisms, according to the matrix of exponents, as YCor explained in a comment above. In the above answer, $h_1=x_1^{1}x_2^{-1}$ and $h_2=x_2^2$, the corresponding matrix is not invertible in $M_n(\mathbb{Z})$, so $h$ is not an automorphism. This answers my question (I wonder if my edit is worthless or not). | |
| Apr 13, 2017 at 23:55 | comment | added | YCor | More precisely, a nonzero scalar multiple of a monomial (in ACL's comment it's understood that a scalar multiple of a monomial is a monomial). | |
| Apr 13, 2017 at 23:41 | comment | added | user237522 | Namely, for an arbitrary $k$-endomorphism $h$, $h_i=h(x_i)$ must be invertible (since $x_i$ is), so it is a monomial $x_1^{c_1}\cdots x_n^{c_n}$, $c_j \in \mathbb{Z}$ (an invertible element in $k[x_1,\ldots, x_n ,x_1^{-1},\ldots, x_n^{-1}]$ is necessarily a monomial $x_1^{c_1}\cdots x_n^{c_n}$, if I am not wrong). | |
| Apr 13, 2017 at 23:27 | comment | added | YCor | In spirit yes. The MathSE question only considered automorphisms and the point is that they can easily be described. It does not consider endomorphisms, but it works in the same way. | |
| Apr 13, 2017 at 23:13 | comment | added | user237522 | Thank you very much YCor and ACL! very interesting discussion. I guess that YCor's last comment is actually the answer in math.stackexchange.com/questions/2232869/… | |
| Apr 13, 2017 at 23:07 | comment | added | YCor | @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 $\mathrm{GL}_n(\mathbf{Z})$) | |
| Apr 13, 2017 at 22:58 | comment | added | ACL | Consider a morphism $f\colon\mathbf G_m^n\to\mathbf G_m^n$. It is given by monomials $f_1,\dots,f_n$. Write $f_i(x)=\prod x_j^{a_{ij}}$. Then $Jf=\det(\partial f_i/\partial x_j)=\det( a_{ij} f_i / x_j ) = f_1\dots f_n / x_1\dots x_n \det (a_{ij})$. So either $Jf=0$ (and the image of $f$ is a subtorus), or $Jf$ is a nonzero monomial, and then $f$ is an isogeny. | |
| Apr 13, 2017 at 22:57 | comment | added | ACL | 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. | |
| Apr 13, 2017 at 22:51 | comment | added | YCor | Thanks; I was thinking of composition, but it's indeed irrelevant. | |
| Apr 13, 2017 at 22:50 | comment | added | ACL | Well, it is invertible. In the sense that it is a non vanishing function on the torus. | |
| Apr 13, 2017 at 22:47 | comment | added | YCor | @ACL Indeed the Jacobian is not even invertible in this case. | |
| Apr 13, 2017 at 22:45 | comment | added | ACL | 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. | |
| Apr 13, 2017 at 22:44 | comment | added | YCor | 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. | |
| Apr 13, 2017 at 22:28 | comment | added | user237522 | Or perhaps the right condition should be as follows: The Jacobian of $\{f_1,f_2\}$, call it $J$, is in $\mathbb{C}-\{0\}$ and the Jacobian of $\{f_1^{-1},f_2^{-1}\}$, call it $\hat{J}$, is in $\mathbb{C}-\{0\}$. But it seems a little too restrictive. So perhaps the right condition should be that $J \hat{J}$ is in $\mathbb{C}-\{0\}$? (this allows non-invertible $J$'s). | |
| Apr 13, 2017 at 22:16 | comment | added | user237522 | Perhaps it is enough to assume injectivity of $f$ instead of invertibility of the Jacobian? or in addition to invertibility of the Jacobian? (In my original question there was no restriction on $f$, but then @YCor suggested $x_1 \mapsto 1$, so I commented that perhaps I should add either invertibility of the Jacobian or injectivity). | |
| Apr 13, 2017 at 22:08 | comment | added | user237522 | Thanks! Firstly, as you know, I did not impose the condition that the Jacobian is invertible. Do you think that there exists a similar (or additional) condition which will guarantee that $f$ is an automorphism? | |
| Apr 13, 2017 at 22:03 | vote | accept | user237522 | ||
| Apr 13, 2017 at 22:02 | history | answered | YCor | CC BY-SA 3.0 |