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Is this necessary for showing g≅h? (isomorphism)

  1. Sep 29, 2011 #1
    • [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]

    If we wanted to show [itex]\mathfrak{g} \cong \mathfrak{h}[/itex] then is it necessary to show that a basis for [itex]\mathfrak{h}[/itex]:

    [itex]\left\{ E=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , F=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} , G=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \right\}[/itex]

    satisfies [itex][E,F]=G,\;[E,G]=0,\;[F,G]=0[/itex] (i.e. the lie bracket relations in [itex]\mathfrak{g}[/itex]) or is it enough to find a map [itex]\varphi : \mathfrak{g} \to\mathfrak{h}[/itex] and show it is a homomorphism, linear and bijective? (which I have)
     
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  3. Sep 29, 2011 #2

    HallsofIvy

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    It certainly is enough to find a map (that's the definition of "isomorphism") but how did you do that without showing that the first matrix you show maps to E, the second to F and the third to G? That is, if you have, in fact, found such a [itex]\phi[/itex], what does it map E, F, and G to?
     
  4. Sep 29, 2011 #3

    HallsofIvy

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    Since the basis vectors define all of g showing that your function maps specific matrices to them is sufficient.
     
  5. Sep 29, 2011 #4
    So showing [itex]\varphi : \mathfrak{g} \to \mathfrak{h}[/itex] defined by: [tex]\varphi(aE+bF+cG)=\left( \begin{array}{ccc} 0 & a & b\\ 0 & 0 & c\\ 0 & 0 & 0 \end{array} \right)[/tex] satisfies:
    (i) [itex]\varphi ([aE+bF+cG,a'E+b'F+c'G])=[\varphi (aE+bF+cG),\varphi (a'E+b'F+c'G)][/itex]
    (ii) linear transformation
    (iii) bijective

    is all I need to show to prove [itex]\mathfrak{g} \cong \mathfrak{h}[/itex]?
     
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