(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Showing a mapping between lie algebras is bijective

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