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.