1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Staff Emeritus
    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] ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Showing a mapping between lie algebras is bijective
  1. Bijective Maps (Replies: 9)

Loading...