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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 ?

HK Lee
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2 Answers2

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${\frak sl}_2({\bf R})$ 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.

Noam D. Elkies
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You want to find a basis of the second algebra so that the brackets are as in the first. Express the elements of the basis in terms of the basis you already have, and see what equations you get (you will find that you can solve them over $\mathbb{C}$ but not over $\mathbb{R}.$)

Igor Rivin
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  • I already find an isomorphism ove ${\bf C}$ http://math.stackexchange.com/questions/423419/lie-algebra-isomorphism-between-sl2-bf-c-and-bf-c3?rq=1 But I cannot show that they are different over ${\bf R}$ – HK Lee Dec 07 '13 at 04:47