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Let $\mathfrak g_1,\mathfrak g_2$ be finite-dimensional real or complex Lie algebras such that ${\rm Der}(\mathfrak g_1)$ and ${\rm Der}(\mathfrak g_2)$ are isomorphic as Lie algebras, where ${\rm Der}(\mathfrak h)$ denotes the algebra of derivations of the Lie algebra $\mathfrak h$.

In that case, is it true that $\mathfrak g_1$ is isomorphic to $\mathfrak g_2$?

I tried to find some reference dealing with that question, but couldn't. Since it's a very simple question to ask, I believe this probably means the answer is 'not necessarily'. However, I was not able to find, or to produce, a counter-example for it either.

If the answer is 'yes', can you sketch the argument or point out some reference for that? If the answer is 'not necessarily', can you describe a counter-example?

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  • $\begingroup$ I just found a rather antique paper that had escaped my first attempts to look for an answer, in which it is claimed that such an example exists: $\endgroup$ – Dry Bones Jul 25 '20 at 22:28
  • $\begingroup$ On the Derivation Algebras of Lie Algebras, by Shigeaki Tôgô. $\endgroup$ – Dry Bones Jul 25 '20 at 22:29
  • $\begingroup$ Nevertheless, I leave the question unanswered because someone might know a better characterization of such examples. The one in the above paper is a somewhat trivial one. $\endgroup$ – Dry Bones Jul 26 '20 at 18:35
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The question is very natural and has an earlier counterpart in group theory. If we have two groups $G$ and $H$ such that $\operatorname{Aut}(G)\cong \operatorname{Aut}(H)$, does it follow that $G\cong H$? The answer is negative both for Lie algebras and groups. There are many counterexamples, and it depends on your taste if you find them "somewhat trivial". In any case, the claim is not true.

For groups, a standard example is $$ \operatorname{Aut}(S_3)\cong S_3\cong \operatorname{Aut}(C_2\times C_2), $$ but the symmetric group $S_3$ is not isomorphic to the Kleinian $4$-group $C_2\times C_2$.

For Lie algebras and derivations, the example given by Togo is a standard one (Example $3$ at the end of the paper), which is often cited. But one can easily construct new examples in low dimensions. For example, there are infinitely many solvable complex Lie algebras of dimension $3$, which have only finitely many different derivation algebras. For details see the comments.

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  • $\begingroup$ Thanks for the answer! I agree that the 'somewhat trivial' thing is quite subjective. Indeed, the series of examples by Tôgô are direct sums, which don't fit the concrete problem I'm dealing with, and that's why I used this perhaps unfortunate expression. The last sentence of your answer interests me much more. Can you indicate a reference for that? $\endgroup$ – Dry Bones Jul 26 '20 at 21:42
  • $\begingroup$ Maybe I should have called you using this: @DietrichBurde Still learning how things work here. $\endgroup$ – Dry Bones Jul 26 '20 at 21:59
  • $\begingroup$ Yes, the Lie algebras are $\mathfrak{r}_3(\lambda)$ with brackets $[e_1,e_2]=e_2,\; [e_1,e_3]=\lambda e_3$. See for example here. $\endgroup$ – Dietrich Burde Jul 27 '20 at 8:20
  • $\begingroup$ Ok, I did the calculations for these algebras and hopefully understood the situation a little better. So, if I got the lesson, in that kind of problem, either one has some very special formulae like the one in Tôgô's Example 3, which are probably not very common, or one has to choose a basis with explicit structural constants and calculate, is that right? Thanks again! $\endgroup$ – Dry Bones Jul 27 '20 at 12:58
  • $\begingroup$ Yes, indeed. The second method works almost always, like in the case of finite groups, where we chose some set of generators and then compute. $\endgroup$ – Dietrich Burde Jul 27 '20 at 13:00

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