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can some one give me an example of two-dimensional non-abelian Lie algebra?

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Let $X=\begin{bmatrix} a&b\\0&0 \end{bmatrix} $ and $Y=\begin{bmatrix} x&y\\0&0 \end{bmatrix} $. Furthermore, let the Lie bracket be the matrix commutator: $[X,Y]:=XY-YX$. We get $XY= \begin{bmatrix} ax&ay\\0&0 \end{bmatrix}$, but $YX= \begin{bmatrix} ax&bx\\0&0 \end{bmatrix}$.

(That is the lie algebra of the affine group mentioned by Eric).

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    Thus the Lie algebra aff(1) is spanned by the basis vectors \begin{bmatrix} 1 & 0\\0&0\end{bmatrix} and \begin{bmatrix} 0 & 1\\0&0\end{bmatrix}. Therefore, its element are of the general form \begin{bmatrix} x & y\\0&0\end{bmatrix}. Since Aff(1) is$a$matrix Lie group, the Lie bracket of aff(1) is the common matrix commutator as defined in the answer above. Does this answer your question?2012-11-10
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This is something easy to come up with: take a basis $\{X,Y\}$ of your space. Then to be non-abelian $[X,Y]$ has to be non-zero. So try $[X,Y] = X$. It's straightforward to verify this satisfies the axioms of a Lie algebra. With a little more work you can show this is the unique (up to isomorphism) two dimensional non-abelian Lie algebra.

This Lie algebra has a geometric interpretation as the Lie algebra of affine transformations of the real line, i.e. all maps $\mathbb R \to \mathbb R$ of the form $x \mapsto ax + b$, $a,b \in \mathbb R$.

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    i want ask a little question, how to define the bracket in Lie algebra of the two dimensional affine group in above so we can say it's defines a lie algebra,2012-11-02