Showing a mapping between lie algebras is bijective

[Ted123]

Homework Statement

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}.

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?

 

[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.

 

[Ted123]

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.

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?

 

[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} ?