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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 $x \in L'$?

Thanks for your kindness

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 $y,z \in L'$.

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    @JulianKuelshammer ok sir, thanks for your suggestion. here the new link http://math.stackexchange.com/questions/257147/lie-algebra-3-dimensional-with-2-dimensional-derived-lie-algebra-22012-12-12

2 Answers 2

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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 $L'$.

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    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 $y,z \in L'$.2012-12-12
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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 $GL(2, \mathbb{C})$.

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    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 isomorpishm2012-11-09