Is this necessary for showing g≅h? (isomorphism)

[Ted123]

• \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}.

If we wanted to show \mathfrak{g} \cong \mathfrak{h} then is it necessary to show that a basis for \mathfrak{h}:

\left\{ E=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} , F=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} , G=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \right\}

satisfies [E,F]=G,\;[E,G]=0,\;[F,G]=0 (i.e. the lie bracket relations in \mathfrak{g}) or is it enough to find a map \varphi : \mathfrak{g} \to\mathfrak{h} and show it is a homomorphism, linear and bijective? (which I have)

 

[HallsofIvy]

It certainly is enough to find a map (that's the definition of "isomorphism") but how did you do that without showing that the first matrix you show maps to E, the second to F and the third to G? That is, if you have, in fact, found such a \phi, what does it map E, F, and G to?

 

[HallsofIvy]

Since the basis vectors define all of g showing that your function maps specific matrices to them is sufficient.

 

[Ted123]

Since the basis vectors define all of g showing that your function maps specific matrices to them is sufficient.

So showing \varphi : \mathfrak{g} \to \mathfrak{h} defined by: \varphi(aE+bF+cG)=\left( \begin{array}{ccc} 0 & a & b\\ 0 & 0 & c\\ 0 & 0 & 0 \end{array} \right) satisfies:
(i) \varphi ([aE+bF+cG,a'E+b'F+c'G])=[\varphi (aE+bF+cG),\varphi (a'E+b'F+c'G)]
(ii) linear transformation
(iii) bijective

is all I need to show to prove \mathfrak{g} \cong \mathfrak{h}?