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Show 2 lie algebras are not isomorphic

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    • [itex]\mathfrak{g}[/itex] is the Lie algebra with basis vectors [itex]E,F,G[/itex] such that the following relations for Lie brackets are satisfied:

    [itex][E,F]=G,\;\;[E,G]=0,\;\;[F,G]=0.[/itex]

    • [itex]\mathfrak{h}[/itex] is the Lie algebra consisting of 3x3 matrices of the form

    [itex]\begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix}[/itex] where [itex]a,b,c[/itex] are any complex numbers. The vector addition and scalar multiplication on [itex]\mathfrak{h}[/itex] are the usual operations on matrices.

    The Lie bracket on [itex]\mathfrak{h}[/itex] is defined as the matrix commutator: [itex][X,Y] = XY - YX[/itex] for any [itex]X,Y \in \mathfrak{h}.[/itex]

    • [itex]\mathfrak{d}_3 \mathbb{C}[/itex] is the Lie algebra consisting of 3x3 diagonal matrices with complex entries with Lie bracket [itex][X,Y]=0[/itex] for all [itex]X,Y \in\mathfrak{d}_3 \mathbb{C}[/itex].

    (We know [itex]\mathfrak{g}\cong \mathfrak{h}[/itex].)

    Show that [itex]\mathfrak{d}_3 \mathbb{C}\ncong\mathfrak{g}[/itex] and [itex]\mathfrak{d}_3 \mathbb{C}\ncong\mathfrak{h}[/itex].

    3. The attempt at a solution

    A basis for [itex]\mathfrak{d}_3 \mathbb{C}[/itex] is [tex]\left\{ E=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , F=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} , G=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \right\}[/tex]

    Is it sufficient to show that [itex][E,F]=0,\;[E,G]=0,\;[F,G]=0[/itex] which doesn't satisfy all the lie bracket relations in [itex]\mathfrak{g}[/itex] so [itex]\mathfrak{d}_3 \mathbb{C}\ncong\mathfrak{g}[/itex]?

    And since [itex]\mathfrak{d}_3 \mathbb{C}\ncong\mathfrak{g}\cong \mathfrak{h},\;\mathfrak{d}_3 \mathbb{C}\ncong\mathfrak{h}[/itex] but to show it:

    [itex]X=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , Y=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\in \mathfrak{h}[/itex]

    but [itex][X,Y]=XY-YX=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \neq \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}[/itex]

    so 2 matrices in [itex]\mathfrak{h}[/itex] don't satisfy the Lie bracket in [itex]\mathfrak{d}_3 \mathbb{C}[/itex] so [itex]\mathfrak{d}_3 \mathbb{C}\ncong\mathfrak{h}[/itex]?

    So is all I need to do to show 2 lie algebras are not isomorphic is to provide a counterexample of how matrices in 1 lie algebra don't satisfy the lie bracket in another lie algebra; therefore the 2 lie algebras can't be isomorphic as isomorphisms preserve lie brackets?
     
    Last edited: Sep 27, 2011
  2. jcsd
  3. Sep 27, 2011 #2

    Office_Shredder

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    Re: Isomorphic

    Just because you picked a map which isn't an isomorphism doesn't mean that no isomorphism exists. To give an example of this argument for inner product spaces:
    A=R3 is not isomorphic to B=R3 because A has a basis of (1,0,0),(1,1,0),(1,1,1), and B has a basis of (1,0,0),(0,1,0),(0,0,1). The inner product of every pair of basis vectors of B is zero, which isn't the case for A, so they are not isomorphic as inner product spaces.

    All we really learned is that different bases can have different properties (for inner products or for lie brackets)
     
  4. Sep 27, 2011 #3
    Re: Isomorphic

    Yeah I see what you're saying.

    Is it true that if [itex]\mathfrak{g}\cong \mathfrak{h}[/itex] and [itex]\mathfrak{g}\ncong \mathfrak{d}_3 \mathbb{C}[/itex] then [itex]\mathfrak{h}\ncong \mathfrak{d}_3 \mathbb{C}[/itex]?
     
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