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jdstokes
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[SOLVED] Difficult 3D Lie algebra
Let [itex]\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \in GL_2(\mathbb{C})[/itex].
Consider the Lie algebra [itex]\mathfrak{g}_{(a,b,c,d)}[/itex] with basis {x,y,z} relations given by
[itex][x,y]= ay + cz[/itex]
[itex][x,z] = by + dz[/itex]
[itex][y,z] = 0[/itex]
Show that [itex]\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})[/itex]
Right now I'm struggling to prove this in either direction so let me start with [itex]\implies[/itex]. Since the Lie algebras are isomorphic (via [itex] \psi : \mathfrak{g}_{(a,b,d,c)} \to \mathfrak{g}_{(a',b',c',d')} [/itex] say), the images of the basis vectors satisfy the commutation relations of [itex]\mathfrak{g}_{(a,b,d,c)} [/itex], ie
[itex][\psi x,\psi y]= a\psi y + c\psi z[/itex]
[itex][\psi x,\psi z] = b \psi y + d\psi z[/itex]
[itex][\psi y,\psi z] = 0[/itex]
Moreover, [itex]\mathfrak{g}_{(a',b'c',d')}[/itex] has a basis {x',y',z'}, with the following commutation relations
[itex][x',y']= a'y' + c'z' [/itex]
[itex][x',z'] = b' y' + d' z'[/itex]
[itex][y',z'] = 0[/itex]
My plan was to relate one basis to the other using a change of basis matrix [itex](x',y',z')^T = A \cdot (\psi x,\psi y,\psi z)^T[/itex]. 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
[itex]\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)[/itex]
where B is a matrix involving [itex]a,b,c,d[/itex] and the entries of A. There is one extra relationship from [itex][y',z'] = 0[/itex] 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.
Homework Statement
Let [itex]\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \in GL_2(\mathbb{C})[/itex].
Consider the Lie algebra [itex]\mathfrak{g}_{(a,b,c,d)}[/itex] with basis {x,y,z} relations given by
[itex][x,y]= ay + cz[/itex]
[itex][x,z] = by + dz[/itex]
[itex][y,z] = 0[/itex]
Show that [itex]\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})[/itex]
The Attempt at a Solution
Right now I'm struggling to prove this in either direction so let me start with [itex]\implies[/itex]. Since the Lie algebras are isomorphic (via [itex] \psi : \mathfrak{g}_{(a,b,d,c)} \to \mathfrak{g}_{(a',b',c',d')} [/itex] say), the images of the basis vectors satisfy the commutation relations of [itex]\mathfrak{g}_{(a,b,d,c)} [/itex], ie
[itex][\psi x,\psi y]= a\psi y + c\psi z[/itex]
[itex][\psi x,\psi z] = b \psi y + d\psi z[/itex]
[itex][\psi y,\psi z] = 0[/itex]
Moreover, [itex]\mathfrak{g}_{(a',b'c',d')}[/itex] has a basis {x',y',z'}, with the following commutation relations
[itex][x',y']= a'y' + c'z' [/itex]
[itex][x',z'] = b' y' + d' z'[/itex]
[itex][y',z'] = 0[/itex]
My plan was to relate one basis to the other using a change of basis matrix [itex](x',y',z')^T = A \cdot (\psi x,\psi y,\psi z)^T[/itex]. 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
[itex]\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)[/itex]
where B is a matrix involving [itex]a,b,c,d[/itex] and the entries of A. There is one extra relationship from [itex][y',z'] = 0[/itex] 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.
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