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

Asaf Karagila
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user46309
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2 Answers2

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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 $x \mapsto ax + b$, $a,b \in \mathbb R$.

Eric O. Korman
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  • yeah i know this is the unique (up to isomorphism) two dimensional non-abelian Lie algebra. but can u show me example of two dimensional lie algebra(from what vector space to what vector space and lie bracket define in there). like if we talk about three dimensional lie algebra , then sl(2,C) is three dimensional algebra, like that

    btw thanks for your answer

     – user46309 Oct 29 '12 at 15:59
  • It's the Lie algebra of the two-dimensional affine group, which is the semi-direct product of $GL(1,\mathbb R) \simeq \mathbb R^*$ with $(\mathbb R, +)$. – Eric O. Korman Oct 29 '12 at 16:18
  • 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, – user46309 Nov 02 '12 at 08:28
  • Could you please give us the proof of the uniqueness of this algebra or at least mention a source for it(preferably a book not using algebraic geometry terms or commutative algebra ones ) –  Apr 13 '21 at 02:14
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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).

B0rk4
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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, – user46309 Nov 02 '12 at 08:29
  • As Eric said, the 1D (!) affine group consists of transformation of the form $x\rightarrow ax+b$. Written in homogeneous matrix form, these transformation looks like: $\begin{bmatrix} a & b\0&1\end{bmatrix}$. – B0rk4 Nov 10 '12 at 03:37
  • In other words, the Lie group Aff(1) consists of matrices of the form $\begin{bmatrix} a & b\0&1\end{bmatrix}$. The Lie Algebra aff(1) is the tangent space of Aff(1). It is spanned by the derivatives of Aff(1). The partial derivative with respect to $a$ is: $\begin{bmatrix} 1 & 0\0&0\end{bmatrix}$, the one with respect to $b$ is $\begin{bmatrix} 0 & 1\0&0\end{bmatrix}$. – B0rk4 Nov 10 '12 at 03:42
  • 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? – B0rk4 Nov 10 '12 at 03:45