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Let L be a Lie algebra of dimension 3.

suppose $ad_z = 0$ with $z\ne 0$. Show that there are $x$ and $y$ in $L$ such that $[x,y] = z$ and realize L in $\mathfrak{b}_3$ the Borel algebra of upper triangular matrices.

What I did:

I proved the existence of x and y. I am stuck in realizing L in $\mathfrak{b}_3$: i.e. finding a morphism $f:L\to \mathfrak{b}_3$ by giving to $x,y$ and $z$ and image in $\mathfrak{b}_3$.

$ad_z = 0$ means z is in the center of L so clearly $f(z)=I_3$. (This is wrong as pointed in comments)

But what about $f(x)$ and $f(y)$. I tried to look for solutions in the form $f(x)=I_3+E_{i,j}$ and $f(y)=I_3+E_{k,l}$ but I can't find a a solution such that $[f(x),f(y)] = f(z)=I_3$

Please note that [,] is the Lie Bracket.

Thanks for any help

Conjecture
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  • I doubt that you want to realise it in some Borel group; presumably the task is to find it in the Borel subalgebra of, say, upper triangular $3\times3$-matrices. Please clarify. – Torsten Schoeneberg Mar 27 '19 at 09:51
  • It is exacltly what is asked from me: To realize it in the Borel algebra of upper triangular 3x3 matrices. Could you please advise what's not clear? A Borel subgroup is the set of upper triangular matrices from what I see on https://en.wikipedia.org/wiki/Borel_subgroup – Conjecture Mar 27 '19 at 09:53
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    Then why do you write "group"? Also, your claim about $f(z)$ having to be $I_3$ is wrong, because $f(z)$ only needs to be in the centre of the image $f(L)$, which certainly is not all upper triangular matrices (as these form a six-dimensional space). – Torsten Schoeneberg Mar 27 '19 at 10:00
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    No, a Borel subgroup consists e.g. of invertible upper triangular matrices. You should pay close attention to the fundamental difference between matrix groups and matrix Lie algebras. – Torsten Schoeneberg Mar 27 '19 at 10:03
  • Thank you @TorstenSchoeneberg fixed that! For f(z) indeed you are right too, I was mistakenly thinking about the center of $\mathfrak{b}_3$ being in $Z(GL_3)$ – Conjecture Mar 27 '19 at 10:07

1 Answers1

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The classification of $3$-dimensional Lie algebras with $1$-dimensional derived algebra has been explained at this duplicate.

For completing the classification in general in dimension $3$ is a bit more work necessary, even over the complex numbers. For the classification of $3$-dimensional Lie algebras over an arbitrary field $K$ see for example this thesis. It has $58$ pages and the classification is quite complicated, in particular for characteristic $2$. Over the complex numbers the classification is much easier. This can be found, for example, in Jacobson's book on Lie algebras. Bianchi classified $3$-dimensional real and complex Lie algebras $115$ years ago. There are still new articles on this classification, e.g., to put it in representation theoretic context, see for example here.

Dietrich Burde
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