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$V=\left\{a\partial_x+bx+cI:a,b,c\in\mathbb{C}\right\}$ with $[X,Y]:=XY-YX$

Question 1. $V$ with $[X,Y]$ is a Lie álgebra with dimension 3, right?

By Ado's theorem, the Lie Álgebra $V$ is isomorphic to a Lie subalgebra of the square matrix space $M(n,\mathbb{R})$ some $n$ (this is what I have been able to understand)

Question 2. What would be the matrix Lie subalgebra that is isomorphic to $V$?

eraldcoil
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1 Answers1

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We have answered the question together in the comments. You don't need Ado's theorem.

Let $(e_1,e_2,e_3)=(\partial_x,x,I)$ be a basis of $L$. Then $[e_1,e_2]=e_3$ and $[e_1,e_3]=[e_2,e_3]=0$. Hence $L$ is the Heisenberg Lie algebra, with underlying vector space $V$.

It has a faithful matrix representation $\rho\colon L\hookrightarrow \mathfrak{gl}_3(\Bbb C)$, given by $$ \rho(e_1)={\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\\\end{pmatrix}},\quad \rho(e_2)={\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\\\end{pmatrix}},\quad \rho(e_3)={\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\\\end{pmatrix}}. $$ The image is isomorphic to the Lie subalgebra of strictly upper-triangular matrices.

Dietrich Burde
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  • I see. What I want to understand now is: how is that application defined $\rho$ etc. Where does it come from? I'll have to read the theory – eraldcoil Sep 25 '23 at 18:13
  • How is this matrix representation obtained? – eraldcoil Sep 25 '23 at 19:44
  • By knowing the Lie algebra of strictly lower-triangular matrices. It is nilpotent, and it is easy to see what the Lie brackets are. On the other hand, one can just compute matrices $\rho(e_i)$ satisfying $\rho([e_i,e_j])=[\rho(e_i),\rho(e_j)]$. – Dietrich Burde Sep 25 '23 at 20:32
  • Yes. I saw it seconds later, so I deleted the comment. thank you so much – eraldcoil Oct 02 '23 at 20:00