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Let $\cal g$ be a Lie algebra and let $a,b,c\in \cal g$ be such that $ab=ba$ and $[a,b]=c\not =0$. Let $\mathcal h=span\ \{a,b,c\}$. How to prove that $\mathcal h$ is isomorphic to the strictly upper triangular algebra $\mathcal n(3,F)$?

Problem: If $\mathcal h\cong n(3,F)$ then $\exists a',b',c'\in \mathcal n(3,F)$ with $a'b'=b'a'$ and $[a',b']=c'$ as in $h$ But then $c'$ must equal $0$ whereas $c\in h$ is not $0$?

Ronald
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  • Only if the bracket operation is actually given by $ab-ba$. In this case that must not be how the bracket is defined. – Matt Samuel Feb 17 '16 at 03:51
  • Can we define another Lie bracket on $n(3,F)$. For example $[a,b]=ab$? – Ronald Feb 17 '16 at 03:53
  • The bracket just has to satisfy the axioms. You could define $[a,b]=0$ for all $a,b$ in any given vector space to turn it into an abelian Lie algebra. – Matt Samuel Feb 17 '16 at 03:55
  • I thought the definition of $\mathcal n(n,F)$ is the Lie algebra with the commutator bracket – Ronald Feb 17 '16 at 03:56
  • That may be true, but $g$ is a different algebra. – Matt Samuel Feb 17 '16 at 03:57
  • Yes, but if $\phi : g\rightarrow n(3,F)$ we must have $\phi(a)\phi(b)=\phi(b)\phi(a)$, $[\phi (a),\phi(b)]=0 $ – Ronald Feb 17 '16 at 04:01
  • A Lie algebra homomorphism does not have to preserve any other binary operation, so if $ab=ba$ there's no reason for the same to be true in the image. – Matt Samuel Feb 17 '16 at 04:02
  • Sorry, maybe I didn't explain what I mean very well. So the problem is if $h$ and $n(3,F)$ isomorphic. Since in $h$ we have three basis $a,b,c$ with $ab=ba$ and $[a,b]=c\not=0$. Then we must have in $n(3,F)$ a basis $a',b',c'$ with $a'b'=b'a'$ and $[a',b']=c'\not =0$ which is impossible?? – Ronald Feb 17 '16 at 04:08
  • You should put as much relevant information as possible in the post itself. The question is currently not clear. – Matt Samuel Feb 17 '16 at 04:10

1 Answers1

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The Lie algebra with $[a,b]=c$ is the $3$-dimensional Heisenberg Lie algebra $\mathfrak{h}_1$. It has a faithful linear representation given by $$ a = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, \quad b = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix}, \quad c = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}, $$ see here. Obviously this matrix Lie algebra is given by $\mathfrak{n_3}$, so that $\mathfrak{n_3}\cong \mathfrak{h_1}$.

Dietrich Burde
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