Why $2$-dimensional Lie space $L$ is always solvable?

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Why 2-dimensional Lie space $L$ is always solvable?
For 1-dimensional it is trivial, because $[ax,by]=ab[x,x]=0$ when $a,b in Bbb F$.




abstract-algebra lie-algebras




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    Why 2-dimensional Lie space $L$ is always solvable?
    For 1-dimensional it is trivial, because $[ax,by]=ab[x,x]=0$ when $a,b in Bbb F$.




    abstract-algebra lie-algebras




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      Why 2-dimensional Lie space $L$ is always solvable?
      For 1-dimensional it is trivial, because $[ax,by]=ab[x,x]=0$ when $a,b in Bbb F$.




      abstract-algebra lie-algebras




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      Why 2-dimensional Lie space $L$ is always solvable?
      For 1-dimensional it is trivial, because $[ax,by]=ab[x,x]=0$ when $a,b in Bbb F$.





      abstract-algebra lie-algebras





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      edited Jul 24 at 13:51









      M. Winter

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      asked Jul 24 at 13:36









      Daniel Vainshtein

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          For dimension $2$ it is also trivial, because we have only one Lie bracket between basis vectors $e_1$ and $e_2$, which we may chose arbitrarily non-zero, e.g. $[e_1,e_2]=e_1$. So the commutator ideal is $1$-dimensional, hence abelian, so that the Lie algebra is metabelian, i.e., solvable.






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            For dimension $2$ it is also trivial, because we have only one Lie bracket between basis vectors $e_1$ and $e_2$, which we may chose arbitrarily non-zero, e.g. $[e_1,e_2]=e_1$. So the commutator ideal is $1$-dimensional, hence abelian, so that the Lie algebra is metabelian, i.e., solvable.






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              For dimension $2$ it is also trivial, because we have only one Lie bracket between basis vectors $e_1$ and $e_2$, which we may chose arbitrarily non-zero, e.g. $[e_1,e_2]=e_1$. So the commutator ideal is $1$-dimensional, hence abelian, so that the Lie algebra is metabelian, i.e., solvable.






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                For dimension $2$ it is also trivial, because we have only one Lie bracket between basis vectors $e_1$ and $e_2$, which we may chose arbitrarily non-zero, e.g. $[e_1,e_2]=e_1$. So the commutator ideal is $1$-dimensional, hence abelian, so that the Lie algebra is metabelian, i.e., solvable.






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                For dimension $2$ it is also trivial, because we have only one Lie bracket between basis vectors $e_1$ and $e_2$, which we may chose arbitrarily non-zero, e.g. $[e_1,e_2]=e_1$. So the commutator ideal is $1$-dimensional, hence abelian, so that the Lie algebra is metabelian, i.e., solvable.







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                edited Jul 24 at 13:47


























                answered Jul 24 at 13:40









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