# Difficult 3D Lie algebra

1. Feb 6, 2008

### jdstokes

[SOLVED] Difficult 3D Lie algebra

1. The problem statement, all variables and given/known data

Let $\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \in GL_2(\mathbb{C})$.

Consider the Lie algebra $\mathfrak{g}_{(a,b,c,d)}$ with basis {x,y,z} relations given by
$[x,y]= ay + cz$
$[x,z] = by + dz$
$[y,z] = 0$

Show that $\mathfrak{g}_{(a,b,c,d)} \cong \mathfrak{g}_{(a',b',c',d')}; \iff \left(\begin{array}{cc} a' & b' \\ c' & d' \end{array}\right) = \lambda g \left(\begin{array}{cc} a & b\\ c & d \end{array}\right) g^{-1}\; \exists \lambda \in \mathbb{C}^\times, g \in GL_2(\mathbb{C})$

3. The attempt at a solution

Right now I'm struggling to prove this in either direction so let me start with $\implies$. Since the Lie algebras are isomorphic (via $\psi : \mathfrak{g}_{(a,b,d,c)} \to \mathfrak{g}_{(a',b',c',d')}$ say), the images of the basis vectors satisfy the commutation relations of $\mathfrak{g}_{(a,b,d,c)}$, ie

$[\psi x,\psi y]= a\psi y + c\psi z$
$[\psi x,\psi z] = b \psi y + d\psi z$
$[\psi y,\psi z] = 0$

Moreover, $\mathfrak{g}_{(a',b'c',d')}$ has a basis {x',y',z'}, with the following commutation relations

$[x',y']= a'y' + c'z'$
$[x',z'] = b' y' + d' z'$
$[y',z'] = 0$

My plan was to relate one basis to the other using a change of basis matrix $(x',y',z')^T = A \cdot (\psi x,\psi y,\psi z)^T$. Then plugging each equation for x',y',z' into the commutators in the above commutation relations I hoped to obtain a relationship amongst (a,b,c,d) and (a',b',c',d').

The result was something like

$\left(\begin{array}{cc} a' & c' \\ b' & d' \end{array}\right)\left(\begin{array}{c} y' \\ z'\end{array} \right)= B\left(\begin{array}{c}\psi y\\\psi z\end{array} \right)$

where B is a matrix involving $a,b,c,d$ and the entries of A. There is one extra relationship from $[y',z'] = 0$ which connects a,b,c,d and the entries of A, but I don't know how to use this.

Any help would be greatly appreciated.

Last edited: Feb 6, 2008
2. Feb 6, 2008

### jdstokes

Could anyone suggest if this is even the correct plan of attack?

I've been thinking about this for hours but am getting nowhere. Somehow I need to find a relationship amongst the matrices (a,b,c,d) and (a',b',c',d') without those column vectors of basis vectors in there.

3. Feb 6, 2008

### jdstokes

Let the Lie brackets [,] and [,]' work over a common vector space V.

A is invertible then $\{ay + cz, by + dz \}$ is a basis for $\mathbb{C}\{y,z\}$. Thus $\{[x,y],[x,z] \}$ is a basis so $\mathbb{C}\{x,y\}$ is composed entirely of Lie brackets. This shows that $\psi([y,z]) = [\psi y, \psi z]$.

Define two linear maps $\varphi : \mathbb{C}{y,z} \to V ; w \mapsto [x,w]$ and $\varphi' : \mathbb{C}{y,z} \to V ; w \mapsto [\varphi(x),w]$. Then the isomorphism condition becomes for the first two commutation relations that

$\varphi_{\mathbb{C}\{ a,b \}} \circ \varphi = \varphi' \circ \varphi_{\mathbb{C}}$.

The matrix for $\varphi$ is A. If the coefficient of x in $\varphi(x)$ is alpha, say, then the matrix for $\varphi'$ is A', because all commutators of y with z vanish. Thus in terms of matrices we have

$g A = \alpha A'g$. Which is what we set out to prove.

4. Feb 6, 2008

### quasar987

Note that for grad level HW questions, it is okay to post in the non HW sections of PF, where you have more chance of getting a reply.