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Homework Help: Showing a mapping between lie algebras is bijective

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Show the map [itex]\varphi : \mathfrak{g} \to \mathfrak{h}[/itex] defined by

    [itex]\varphi (aE + bF + cG) = \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix}[/itex]

    is bijective.

    [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]

    3. The attempt at a solution

    For showing [itex]\varphi[/itex] is 1-1 (injective) is this proof OK:

    Letting [itex]x = aE+bF+cG \in \mathfrak{g}[/itex] and [itex]y = a'E+b'F+c'G \in \mathfrak{g}[/itex],

    [itex]\varphi (x) = \varphi (y) \Rightarrow \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & a' & c' \\ 0 & 0 & b' \\ 0 & 0 & 0 \end{bmatrix}[/itex]

    [itex]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\, \Rightarrow aE +bF+cG = a'E + b'F + c'G[/itex] [i.e. [itex]x=y[/itex]]

    And [itex]\varphi[/itex] is onto (surjective) since [itex]\text{Im}(\varphi) = \mathfrak{h}[/itex] - how do you explictly show this?
     
    Last edited: Sep 15, 2011
  2. jcsd
  3. Sep 15, 2011 #2

    HallsofIvy

    User Avatar
    Science Advisor

    It is sufficient to show that the kernel of [itex]\phi[/itex] is just the 0 vector. That is, if
    [tex]\phi(x)= \begin{bmatrix}0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}[/tex]
    then a= b= c= 0.
     
  4. Sep 15, 2011 #3
    Is that for 1-1 or onto?

    Doesn't [itex]\begin{bmatrix}0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} \Rightarrow a= b= c= 0\;?[/itex] (just from looking at it)

    From linear algebra that proves phi is 1-1. How about onto?
     
  5. Sep 15, 2011 #4
    To give a reason why [itex]\varphi[/itex] is surjective, can I just say that the images of any 2 of the basis vectors in [itex]\mathfrak{g}[/itex] are clearly linearly independent, hence the image of [itex]\mathfrak{g}[/itex] is all of [itex]\mathfrak{h}[/itex] ?
     
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