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Difficult 3D Lie algebra

  1. Feb 6, 2008 #1
    [SOLVED] Difficult 3D Lie algebra

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

    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]

    3. 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.
    Last edited: Feb 6, 2008
  2. jcsd
  3. Feb 6, 2008 #2
    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.
  4. Feb 6, 2008 #3
    Let the Lie brackets [,] and [,]' work over a common vector space V.

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

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

    [itex]\varphi_{\mathbb{C}\{ a,b \}} \circ \varphi = \varphi' \circ \varphi_{\mathbb{C}}[/itex].

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

    [itex]g A = \alpha A'g[/itex]. Which is what we set out to prove.
  5. Feb 6, 2008 #4


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