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We know that not every Lie algebra has commutator $XY-YX$ as its Lie bracket. For example, $R^3$ with vector cross product. However, every Lie derivative $\mathcal{L}_X(Y)=XY-YX$ of a Lie group is a commutator (by differentiating push-forward of a vector field by a flow of another vector field at $0$). Therefore by Lie group-Lie algebra correspondence, I believe every Lie algebra is $isomorphic$ to some Lie algebra with commutator Lie bracket. In above example, $(R^3, \times) \cong (\mathfrak{so_3}, [,])$ as Lie algebras, where the latter has commutator Lie bracket.

My attempt is as follows. Given $\mathfrak{g}$ a Lie algebra, we are done by finding a Lie group which it is tangential to. The idea is to consider the adjoint representation $ad: \mathfrak{g}\to \mathfrak{gl(g)}$ with kernel $Z(\mathfrak{g})$. So we consider $exp(\mathfrak{g}/Z(\mathfrak{g})) \ltimes R^k$ where $k=dim\ Z(\mathfrak{g})$. This $exp(\mathfrak{g}/Z(\mathfrak{g}))$ corresponds to a Lie subalgebra of $\mathfrak{gl(g)}$ so has commutator as Lie bracket. To make the dimension right we just append $R^k$ to compensate the dimension of the kernel of $ad$.

I am new to Lie theory so I am not sure if my intuition on vector field composition or exponential maps are correct. I could not find a proof of this anywhere.

Any suggestions and corrections are very much appreciated.

Andy
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    By Ado's (and Iwasawa's) theorem, every finite-dimensional Lie algebra over a field is isomorphic to a Lie algebra of matrices with Lie bracket $[A,B]=AB-BA$. This is enough for a positive answer to your question. – Dietrich Burde Aug 31 '21 at 10:52
  • @DietrichBurde Thank you a lot! Problem solved! – Andy Aug 31 '21 at 10:55
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    @DietrichBurde What you wrote looks more like an answer than a comment to me. – José Carlos Santos Aug 31 '21 at 10:57
  • See wikipedia for this important theorem. I wrote a wikipedia entry about Ado, if you are interested :) – Dietrich Burde Aug 31 '21 at 10:57
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    @JoséCarlosSantos I can write an answer, but it would be just the statement of Ado's theorem. The moderators were critical lately for such "trivial" answers. – Dietrich Burde Aug 31 '21 at 11:01
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    @DietrichBurde I still think that it is an answer. Let us see what the moderators think about it. – José Carlos Santos Aug 31 '21 at 11:08
  • @DietrichBurde Apart from Ado's theorem, do you have any thoughts about my attempt? Thanks for your help! – Andy Aug 31 '21 at 11:23
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    Yes, I have. Basically you want to prove Ado's theorem, is this correct? – Dietrich Burde Aug 31 '21 at 11:31
  • @DietrichBurde Oh I see. I will go to check the proofs. Thank you! – Andy Aug 31 '21 at 11:37
  • If one wants to avoid Ado's Theorem (and learn about something good), one can alternatively embed a Lie algebra into its universal enveloping algebra. There, the Lie bracket identifies with a commutator of an associative algebra by definition. By the way, this forces upon us to make the question a bit more precise: Is our Lie algebra itself supposed to be closed under the associative product in the commutator? I think neither construction so far achieves that. – Torsten Schoeneberg Aug 31 '21 at 15:44
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    The first sentence sounds unclear. Are you addressing the question which Lie algebras have a bracket of the for $XY-YX$ for some bilinear product $(X,Y)\mapsto XY$? What do you want to assume on the product? associative? unital? without assumption, one can always choose $XY=\frac12[X,Ÿ]$. – YCor Aug 31 '21 at 16:57

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