Classify Lie algebra with 1 dimension derived algebra

Question

Problem 3.10 from Erdmann, Wildon ask:

Find, up to isomorphism, all Lie algebras with 1-dimensional derived algebra.

(also, this book assume finite dimension Lie algebra only; I'm not sure whether it's over a general field, or just $\mathbb{C}$, but it's best to do the general field)

I am having a hard time doing this. I got apparently a lot of algebras, and it's not even easy to tell when they are isomorphic or not. Just to illustrate this, consider even this one particular case:

Let $v,a_{1},\ldots,a_{n}$ be a basis where $v$ span the derived algebra, and further assume that $[v,a_{i}]=0$; denote $m_{i,j}$ ($i<j$) to be a constant such that $[a_{i},a_{j}]=m_{i,j}v$. Well, the difficulty here is in the fact it seems like there is no restriction on what $m_{i,j}$ could be (well, aside from the fact that they cannot be all $0$). Yet not every possible set of $m_{i,j}$ produce a different Lie algebra. For example, a Lie algebra with all $m_{i,j}=1$ is isomorphic to one with all $m_{i,j}=2$ because we can scale $v$ be a constant. Other isomorphism between Lie algebras with more complicated set of $m_{i,j}$ are possible by scaling the $a_{i}$ as well. And that's not counting the fact that we can also do addition. So I can't find anyway to classify them.

Any help solving the full question, or this particularly problematic case, is appreciated. Since I only started learning Lie algebra, I don't really have any advanced tools available, and they shouldn't be needed considering this is just chapter 3. Thank you for your help.

Answer 1

Rough outline for an elementary approach.

Let $L$ be a finite-dimensional Lie algebra with $\dim L' = 1$.

If $L'$ is not contained in $Z(L)$, show that in this case $L$ is the direct sum of an abelian Lie algebra and an nonabelian Lie algebra of dimension $2$.

Assume then that $L'$ is contained in $Z(L)$. If $L'$ is properly contained in $Z(L)$, prove that $L$ is the direct sum $K \oplus A$, where $A$ is an abelian Lie algebra and $K$ satisfies $\dim K' = 1$ and $K' = Z(K)$.

We are left with the case where $L' = Z(L) = \operatorname{span} \{z\}$. In this case $\dim L \geq 3$.

Prove that $L$ has a basis $\{z, f_1, g_1, \cdots, f_t, g_t\}$ such that $[f_i, g_i] = z$ for all $i$ and $[f_i, g_j] = [f_i, f_j] = [g_i, g_j] = 0$ for all $i \neq j$. ($L$ is a central product of $t$ copies of the Heisenberg Lie algebra)

To find such a basis, you could proceed by induction. Find $f_1$ and $g_1$ such that $z, f_1, g_1$ span the Heisenberg Lie algebra $H$. Apply induction to a suitable subalgebra (note that any subspace containing $z$ is a subalgebra).

Or you could also consider the bilinear form defined by $(x,y) \mapsto \lambda$, where $[x,y] = \lambda z$. This defines a symplectic bilinear form on $L/Z(L)$.

Finally, prove that for all $t \geq 1$ such an Lie algebra can be constructed.

Answer 2

Here is a proof, assuming that you know what bilinear forms are. I will consider the case when the derived subalgebra is central (this is not be the case in general). We have the following situation: $h$ is a lie algebra (over a field $k$) with the 1-dimensional center $c$ (the derived subalgebra) and the abelian quotient $g=h/c$. Then for every $a, b\in g$, the commutator $[a,b]$ belongs to $c\cong k$. By using axioms of the lie algebra, we immediately see that the map $$ \phi: g\times g\to k, \phi(a,b)=[a,b] $$ is a bilinear antisymmetric ($\phi(a,b)=-\phi(b,a)$) form on the $k$-vector space $g$. Such firms $\phi$ are known as symplectic forms. Using an analogue of Gramm/Schmidt diagonal inaction process one sees that all such forms on the given vector space are isomorphic to each other.

Lastly, since the center of $h$ is exactly 1-dimension, you see that the form $\phi$ is nondegenerate (for every nonzero $a\in g$, $\exists b\in g$, $\phi(a,b)\ne 0$).

Conversely, given a nondegenerate bilinear antisymmetric form $\phi$ on $g$, one defines the Lie algebra structure on the vector space $h:=g\oplus c$: $$ [a,b]=\phi(a,b), [g,c]=0 $$
and extend the bracket by bilinearity to the rest of $g\oplus c$. Now, its your turn to do some work: you sit down and check carefully the axioms of lie algebra for this bracket on $h$ and the fact that its center is exactly $c$. Denote the resulting Lie algebra by $Ext_\phi$.

Suppose that we have two different (nondegenerate) antisymmetric bilinear forms $\phi, \psi$ on $g$ and $\alpha$ is an automorphism of the vector space $g$ sending the form $\phi$ to the form $\psi$: $$ \psi(\alpha(a), \alpha(b))= \phi(a,b). $$ Then you again sit down and check that $\alpha$ defines an isomorphism $$ \alpha_*: Ext_\phi\to Ext_\psi $$ which acts by the identity on $c$ and via $\alpha$ on $g$.

Lastly, you need to check that if $h, h'$ are isomorphic lie algebras with 1-dimensional centers $c, c'$ and commutative quotients $h/c, h'/c'$ which are isomorphic as vector spaces, then there exists an isomorphism $$ \alpha: (h/c, \phi)\to (h'/c', \phi'), $$ where $h=Ext_{\phi}$ and $h'=Ext_{\phi'}$. You prove this by observing that the isomorphism $h\to h'$ has to send center to center.

Therefore, the set of isomorphism classes of your lie algebras is parameterized by the set of isomorphism classes of pairs $(g, \phi)$ where $g$ is a $k$-vector space and $\phi$ is a nondegenerate antisymmetric bilinear form on $g$. Since all such forms are isomorphic, there is a unique one, determined by the dimension of $g$. The corresponding central extension is called Heisenberg Lie algebra.

Answer 3

I am also reading this book for recreation, and just got to Exercise 3.10. I'm not sure if this is "rigorous," but as in the text, I broke down the case where $L' \subset Z(L)$. I created a basis set $\{x_1,\ldots,x_r,z_1,\ldots,z_s\}$ where the $x_i \notin Z(L)$ and $z_j \in Z(L)$. Then you have that, for some index $k$, $[x_i,x_l]=\alpha_{il}z_k$ for all $1\leq i,l \leq r$. Since $[x_l,x_i]=-\alpha_{il}z_k$, the matrix $A$ of the $\alpha_{il}$ (with 0's along the diagonal) is a skew-symmetric matrix -- again, I'm "ignoring" in my counting of necessary rows and columns the center of the Lie algebra, focusing instead on the non-central part. If you encounter another Lie algebra of the same dimension with the same breakdown, but different structure constants, you then use the fact that if you change the basis of a Lie algebra, then in this setting, the new structure constants are represented by $B^TAB$ where $B$ is the change of basis matrix. So now you're looking for equivalence classes of skew-symmetric matrices under conjugacy. This post breaks that down: Two skew symmetric matrices of same rank are congruent..

I think the same reasoning applies in the case where $L' \not\subset Z(L)$. Pick an element that spans $L'$, say $x_1$, pick all your basis vectors that span $Z(L)$, then complete the basis by adding $x_2,x_3,\ldots,x_r$. Here you have $[x_i,x_j]=\alpha_{ij}x_1$. Here again you have an $r\times r$ skew-symmetric matrix.

At least, that's how I'm approaching the problem. Feedback welcome!