Timeline for Is $\mathrm{End}-\{0\}=\mathrm{Aut}$ for derivation Lie algebra?
Current License: CC BY-SA 4.0
23 events
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| S Aug 11, 2020 at 10:03 | history | bounty ended | CommunityBot | ||
| S Aug 11, 2020 at 10:03 | history | notice removed | CommunityBot | ||
| Aug 3, 2020 at 10:25 | comment | added | solver6 | We also need $A$ to be of finite trancendence degree. If $D$ is locally nilpotent then for each $a\in A$ and each $U\in\text{Der}(A)$, $(\text{ad}D)^n(U)(a) = \ldots+D^iUD^{n-i}(a)+\ldots = 0$ for $n$ large. So since we know that $A$ has finite trancendence degree so $(\text{ad}D)^n(U)=0$ for large $n$ for each $U$. | |
| Aug 3, 2020 at 10:17 | history | edited | YCor | CC BY-SA 4.0 |
fixed English
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| Aug 3, 2020 at 10:16 | comment | added | solver6 | For each commutative algebra $A$ we have that ad of $D\in\text{Der}(A)$ is locally nilpotent if and only if $D$ is locally nilpotent derivation of $A$. It is because for each $a\in A$ we have $(\text{ad}D)(aD) = [D, aD] = (Da)D$ and so $(\text{ad}D)^n(aD) = (D^na)D$ so if for some $n$ it is known that $(\text{ad}D)^n(aD) = 0$ then $(D^na)D = 0$ and $D^na=0$. | |
| Aug 3, 2020 at 10:11 | comment | added | YCor | 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. | |
| Aug 3, 2020 at 10:05 | history | edited | solver6 | CC BY-SA 4.0 |
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| Aug 3, 2020 at 9:59 | comment | added | solver6 | For $n=1$ it is obviously true since the only nilpotent elements of derivation algebra are $\mathbb{C}\partial_x$. | |
| Aug 3, 2020 at 9:58 | history | edited | solver6 | CC BY-SA 4.0 |
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| Aug 3, 2020 at 9:04 | comment | added | YCor | Could you elaborate on why does a positive answer imply the Jacobian conjecture? Also, do you know the answer for $n=1$? | |
| S Aug 3, 2020 at 8:21 | history | bounty started | solver6 | ||
| S Aug 3, 2020 at 8:21 | history | notice added | solver6 | Draw attention | |
| Aug 3, 2020 at 8:20 | history | edited | solver6 | CC BY-SA 4.0 |
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| Jul 25, 2020 at 11:27 | comment | added | solver6 | As I know Jacobian conjecture is a consequence of this question. So is it equivalent to Jacobian conjecture for $\mathbb{A}^n$? | |
| Jul 24, 2020 at 16:37 | comment | added | abx | Oh sure, I am stupid again. Thanks. | |
| Jul 24, 2020 at 15:48 | comment | added | user108998 | @abx, that's a derivation of the lie algebra. U can exponentiate it (bc it's loc nilpotent) to an endom, but ofc this will be an aut | |
| Jul 24, 2020 at 14:48 | comment | added | abx | But isn't $\operatorname{ad}(\partial_{x_1}) $ an endomorphism which kills the $\partial _{x_i}$ for $i>1$? | |
| Jul 24, 2020 at 8:50 | comment | added | abx | Oh, OK, that was stupid, sorry. | |
| Jul 24, 2020 at 8:01 | history | edited | YCor | CC BY-SA 4.0 |
fixed title, added tags
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| Jul 24, 2020 at 7:59 | comment | added | YCor | @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 $[T,U]Q=TUQ-UTQ$. | |
| Jul 24, 2020 at 7:59 | comment | added | solver6 | There is no elements $x_i$ in this Lie algebra. For $a, b$, $[a, b] = ab - ba$ | |
| Jul 24, 2020 at 7:30 | comment | added | abx | What is the Lie algebra structure? For instance, what is $[x_1,\partial _{x_1}]$? | |
| Jul 24, 2020 at 7:07 | history | asked | solver6 | CC BY-SA 4.0 |