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Is it true that every nonzero endomorphism of Lie $\mathbb{C}$-algebra $\mathbb{C}[x_1,\ldots, x_n]\partial_{x_1}\oplus\ldots\oplus\mathbb{C}[x_1,\ldots, x_n]\partial_{x_n}$ is an automorphism?

As I know a positive answer to question implies the Jacobian conjecture for $\mathbb{A}_n$.

So is it equivalent or a stronger result than JC?

Idea of proof that (positive answer to Question) $\Rightarrow$ JC

If $(F_1,\ldots, F_n)$ is a set of polynomials with constant nonzero Jacobian, then $\mathbb{C}[F_1,\ldots, F_n]\subseteq\mathbb{C}[x_1,\ldots, x_n]$ is subalgebra. It can be shown that the constant Jacobian means that each derivation from $\mathbb{C}[F_1,\ldots, F_n]$ can be uniquely continued to a derivation of $\mathbb{C}[x_1,\ldots, x_n]$. So we have the inclusion $\text{Der}(\mathbb{C}[F_1,\ldots, F_n])\to\text{Der}(\mathbb{C}[x_1,\ldots, x_n])$. If the main question has a positive answer, then this inclusion is surjective and there are derivations $D_1,\ldots, D_n$ of $\mathbb{C}[x_1,\ldots, x_n]$ with $D_i|_{\mathbb{C}[F_1,\ldots, F_n]}=\partial_{F_i}$. We know that $F_i$ are slices of $D_i$, so it is enough to check that $D_i$ are locally nilpotent commuting derivations. Since $D_i$ are locally nilpotent on $\mathbb{C}[F_1,\ldots, F_n]$ one can prove that $\text{ad}D_i|_{\text{Der}(\mathbb{C}[F_1,\ldots, F_n])}$ are locally nilpotent derivations of $\text{Der}(\mathbb{C}[F_1,\ldots, F_n])$, so (from Question statement) $\text{ad}D_i$ is locally nilpotent on $\text{Der}(\mathbb{C}[x_1,\ldots,x_n])$. Thus one can prove that $D_i$ is locally nilpotent on $\mathbb{C}[x_1,\ldots, x_n]$ and the $D_i$ are locally nilpotent with slices $F_i$ which commute with each other, so we have JC.

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  • $\begingroup$ What is the Lie algebra structure? For instance, what is $[x_1,\partial _{x_1}]$? $\endgroup$
    – abx
    Commented Jul 24, 2020 at 7:30
  • $\begingroup$ There is no elements $x_i$ in this Lie algebra. For $a, b$, $[a, b] = ab - ba$ $\endgroup$
    – solver6
    Commented Jul 24, 2020 at 7:59
  • $\begingroup$ @abx $x_1$ is not in the Lie algebra. The Lie algebra structure is the commutator: $P(x)\partial $ is the operator $Q(x)\mapsto P(x)\partial x$, and the bracket is the commutator of operators $[T,U]Q=TUQ-UTQ$. $\endgroup$
    – YCor
    Commented Jul 24, 2020 at 7:59
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    $\begingroup$ Could you elaborate on why does a positive answer imply the Jacobian conjecture? Also, do you know the answer for $n=1$? $\endgroup$
    – YCor
    Commented Aug 3, 2020 at 9:04
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    $\begingroup$ Actually I did the "exercise" for $n=1$. I had to be careful because the image of an ad-locally-nilpotent element by an (non-surjective) endomorphism is possibly not ad-locally-nilpotent, but in the present case it can be checked by hand, and eventually $\partial_x$ is mapped to a nonzero multiple of itself, which easily implies that the subspace of elements of degree $\le n$ is stable for all $n$, and hence we can run the "injective implies surjective" argument. $\endgroup$
    – YCor
    Commented Aug 3, 2020 at 10:11

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