Lie algebra 3 Dimensional with 2 Dimensional derived lie algebra

i read in mark wildon book , an introduction to lie algebras, in page 22 say that : Suppose that dim $L$ = 3 and dim $L'$ = 2. We shall see that, over $\mathbb{C}$ at least, there are infinitely many non-isomorphic such Lie Algebras. Take a basis of $L'$, say $\{y,z\}$ and extend it to a basis of $L$, say by $x$. To understand the Lie algebra $L$, we need to understand the structure of $L'$ as a Lie algebra in its own right and how the linear map $ad x : L \rightarrow L$ acts on $L'$.

I dont understand this part "To understand the Lie algebra $L$, we need to understand the structure of $L'$ as a Lie algebra in its own right and how the linear map $ad x : L \rightarrow L$ acts on $L'$."

Can someone explain more about this part, how can we understand Lie algebra $L$ with the linear map $ad x : L \rightarrow L$ acts on $L'$? and for the case take in book, only case $x \notin L'$ , why we didnt consider $x \in L'$?

i have problem now to prove part b of lemma 3. $ad x : L' \rightarrow L'$. i stuck to show $ad x : L' \rightarrow L$ is homomorphism. How to show $ad x([y,z]) = [ad x(y), adx(z)]$ where $y,z \in L'$.

• I assume that $L' = [L,L]$. To understand the Lie bracket, we need to know what $[x,y]$, $[x,z]$, and $[y,z]$ are. The first two are encoded in the adjoint map, and the last one comes from the structure of $L'$. – user27126 Nov 8 '12 at 19:07
• @sanchez thanks for your explanation sir,i understand your explanation, what the use we see linear map $adx : L \rightarrow L$ acts on $L′$.? – Ongok Ongoksepen Nov 9 '12 at 5:19
• @OngokOngoksepen If you have a new question, please post as such and don't edit old questions. – Julian Kuelshammer Dec 12 '12 at 12:04
• @JulianKuelshammer ok sir, thanks for your suggestion. here the new link math.stackexchange.com/questions/257147/… – Ongok Ongoksepen Dec 12 '12 at 15:55

We have that $L=L'\oplus \langle x\rangle_k$ as vectorspaces. We want to understand $L$ as a Lie algebra, i.e. we want to know $[\ell,m]$ (for all $\ell, m\in L$). Now write $\ell,m\in L$ as $\ell=\ell'+\lambda x$ with $\ell'\in L'$ and $\lambda\in k$ and similarly $m=m'+\mu x$. Then we have \begin{align} [\ell,x]&=[\ell'+\lambda x,m'+\mu x]\\ &=[\ell',m']+\lambda[x,m']+\mu[\ell',x]+\lambda\mu[x,x] \end{align} Now we understand $[\ell',m']$ if we understand $L'$ as a Lie algebra. We have $[x,m']=\operatorname{ad} x(m')$ and $[\ell',x]=-[x,\ell']=-\operatorname{ad}x(\ell')$ and $[x,x]=0$. The last two equalities follow from anticommutativity. Thus to understand $L$ as a Lie algebra, we have to know how $L'$ is as a Lie algebra and how $\operatorname{ad}x$ acts on $L'$.
• thanks for your answer sir, but what is $<x>_k$? pardon me, if this is low quality question. The case in the book only two , $x \notin L$ and ad $x$ is diagonalizable , case two $x \notin L$ and ad $x$ is not diagonalizable. Why we didnt consider $x \in L$. So the conclusion, if we want to classify 3 dimensional algebra with dim $L'$ = 2 with isomorphism. we just see matrix representasion of $ad x$. if their matrix representation isomorphic , then the lie algebra is isomorphic? – Ongok Ongoksepen Nov 9 '12 at 15:32
• $\langle x\rangle_k$ is the $1$-dimensional vector space spanned by $x$. $x$ is chosen, so that $x\notin L$. That way we have $L=L'\oplus \langle x\rangle_k$ as vector spaces. They determine every Lie algebra $L'$ with $\dim L'=2$ before they determine the $3$-dimensonal ones. – Julian Kuelshammer Nov 9 '12 at 15:36
• The conclusion is if you fix the derived Lie algebra and the action of $\operatorname{ad}x$ up to conjugacy (choosing a different representative of the conjugacy class just results in a different choice of basis for $L'$) you can classify all $3$-dimensional Lie algebras up to isomorphism. – Julian Kuelshammer Nov 9 '12 at 16:16
• Suppose you have two Lie algebras $L$ and $\mathfrak{g}$ with $\dim L=\dim \mathfrak{g}=3$ and $\dim L'=\dim\mathfrak{g}'=2$. Then they are isomorphic iff $L'\cong \mathfrak{g}'$ and $\operatorname{ad} x$ is congruent to $\operatorname{ad} g$, where $x\in L$, but $x\notin L'$ and $g\in \mathfrak{g}$, but $g\notin\mathfrak{g'}$. – Julian Kuelshammer Nov 10 '12 at 17:12
As a first step, show that $L^{\prime}$ is abelian. Then you have $[y,z]=0$ and $[y, x]=a y+bz$, $[z,x]=cy+dz$. So, you have one such Lie algeba for each conjugacy class in $GL(2, \mathbb{C})$.
• what the use we see map $ad x : L \rightarrow L$ acts on $L'$ for understand lie algebra L? and in that book , second step is prove $ad x: L' \rightarrow L'$ is an isomorpishm – Ongok Ongoksepen Nov 9 '12 at 5:22