Homework EquationsThe Attempt at a SolutionDifficult 3D Lie algebra [SOLVED]

[jdstokes]

[SOLVED] Difficult 3D Lie algebra

Homework Statement

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})

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.

 

[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.

 

[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.

 

[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.