# Is $\mathrm{End}-\{0\}=\mathrm{Aut}$ for derivation Lie algebra?

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.

• What is the Lie algebra structure? For instance, what is $[x_1,\partial _{x_1}]$?
– abx
Jul 24, 2020 at 7:30
• There is no elements $x_i$ in this Lie algebra. For $a, b$, $[a, b] = ab - ba$ Jul 24, 2020 at 7:59
• @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 $[T,U]Q=TUQ-UTQ$.
– YCor
Jul 24, 2020 at 7:59
• Could you elaborate on why does a positive answer imply the Jacobian conjecture? Also, do you know the answer for $n=1$?
– YCor
Aug 3, 2020 at 9:04
• 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.
– YCor
Aug 3, 2020 at 10:11