Lie algebra over $F$ of dimension $3$ whose derived algebra has dimension $1$ and lies in $Z(L)$

Question

Show that (up to isomorphism) there is a unique Lie algebra over $F$ of dimension $3$ whose derived algebra has dimension $1$ and lies in $Z(L)$.

I think that I must construct a basis for $L$ satisfying such conditions. My first idea is to consider any canonical basis $x,y,h$ and calculate $[x,y],[x,h]$ and $[y,h]$. Then finding some relation between those brackets to define a new basis $x',y',h'$ such that whenever I compute the brackets of the new basis elements I will always lie in the subspace $[LL]$.

But unforunatly i don't know how. Any help would be much appreciated!

2

See this question. – Dietrich Burde Dec 15 '17 at 19:12

score 2 · Accepted Answer · answered Dec 15 '17 at 18:48

2

There must be $x$ and $y$ linearly independent, with $h=[x,y]$ nonzero. As $h$ lies in the centre, $[x,h]=[y,h]=0$. This means that $x$, $y$ and $h$ are linearly independent, so form a basis for $L$.

answered Dec 15 '17 at 18:48

Angina Seng

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How do you know that $h$ lies in the centre? – user2345678 Dec 15 '17 at 19:14
2

@math.h Since you said so in the title. – Angina Seng Dec 15 '17 at 19:26
Am I missing something? Isn't it part of what i must show? I suppose that i have to show existence of such an algebra, then prove its uniqueness, right? – user2345678 Dec 15 '17 at 19:31
@math.h Exactly, see my answer. Your text says "whose derived algebra has dimension $1$ and lies in $Z(L)$". This implies $[L,L]\subseteq Z(L)$ so that $[L,[L,L]]=0$. Then you are done. – Dietrich Burde Dec 19 '17 at 10:35
How can we know that such algebra is unique (up to isomorhpism) ? – learner Jan 20 '21 at 07:05

score 1 · Answer 2 · answered Dec 15 '17 at 19:17

Since $[L,L]$ is $1$-dimensional, there is a basis $(x,y,z)$ of $L$ such that $[x,y]=\lambda z,\; [x,z]=\mu z $ and $[y,z]=\nu z$ for some $\lambda,\mu,\nu\in F$, not all zero. Since $[L,[L,L]]\subseteq [L,Z(L)]=0$, $L$ is nilpotent, so that all adjoint operators are nilpotent, i.e., $\mu=\nu=0$. Now $L$ is isomorphic to the Heisenberg Lie algebra, given by $[x,y]=z$.

Lie algebra over $F$ of dimension $3$ whose derived algebra has dimension $1$ and lies in $Z(L)$

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