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

1. Sep 29, 2011

### 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)

2. Sep 29, 2011

### HallsofIvy

Staff Emeritus
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?

3. Sep 29, 2011

### HallsofIvy

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

4. Sep 29, 2011

### Ted123

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 $\mathfrak{g} \cong \mathfrak{h}$?