TITLE: Lie algebras ${\rm sl}(2,{\bf R})$ and $({\bf R}^3,\wedge)$ are not isomorphic
QUESTION [8 upvotes]: As I said in the title,
I want distinguish algebras between ${\rm sl}(2,{\bf R})$ and $({\bf R}^3,\wedge)$ :
On ${\rm sl}(2,{\bf R})$ $$ e=\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix},\ f= \begin{bmatrix} 0 & 0 \\ -1 & 0\end{bmatrix},\ h=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$$  so that $$ [e,f]=h,\ [f,h]=2f,\ [e,h]=-2e $$ 
(Note that this algebra has no proper ideal)
On ${\bf R}^3$ $$ e=(1,0,0),\ f=(0,1,0),\ h=(0,0,1) $$ so that $$ 
[e,f]=h,\ [f,h]=e,\ [h,e]=f$$
How can we prove that there exist no isomorphism ?

REPLY [7 votes]: ${\frak sl}_2({\bf R})$ has elements like $e$ that are nonzero but nilpotent
(in fact $[e,[e,[e,a]]]=0$ for all $a$).  The cross product on ${\bf R}^3$
has no such elements: for any nonzero $e$, you can choose for $a$ any
nonzero vector orthogonal to $e$, and then repeated cross-product with $e$
takes $a$ to a vector of length $|e|^n |a|$ and thus never yields zero.
Hence the two Lie algebras are not isomorphic, QED.