Homework Help: Showing a mapping between lie algebras is bijective

1. Sep 15, 2011

Ted123

1. The problem statement, all variables and given/known data

Show the map $\varphi : \mathfrak{g} \to \mathfrak{h}$ defined by

$\varphi (aE + bF + cG) = \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix}$

is bijective.

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

$[E,F]=G,\;\;[E,G]=0,\;\;[F,G]=0.$

$\mathfrak{h}$ is the Lie algebra consisting of 3x3 matrices of the form

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

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

3. The attempt at a solution

For showing $\varphi$ is 1-1 (injective) is this proof OK:

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

$\varphi (x) = \varphi (y) \Rightarrow \begin{bmatrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & a' & c' \\ 0 & 0 & b' \\ 0 & 0 & 0 \end{bmatrix}$

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

And $\varphi$ is onto (surjective) since $\text{Im}(\varphi) = \mathfrak{h}$ - how do you explictly show this?

Last edited: Sep 15, 2011
2. Sep 15, 2011

HallsofIvy

It is sufficient to show that the kernel of $\phi$ is just the 0 vector. That is, if
$$\phi(x)= \begin{bmatrix}0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0\end{bmatrix}= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$
then a= b= c= 0.

3. Sep 15, 2011

Ted123

Is that for 1-1 or onto?

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

From linear algebra that proves phi is 1-1. How about onto?

4. Sep 15, 2011

Ted123

To give a reason why $\varphi$ is surjective, can I just say that the images of any 2 of the basis vectors in $\mathfrak{g}$ are clearly linearly independent, hence the image of $\mathfrak{g}$ is all of $\mathfrak{h}$ ?