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Consider the 3D Lie algebra with the following defined:

$ [x,y] = z $, $ [x,z]=0 $, $ [y,z]=0 $

From this, I want to get a little help how to start finding a 3D representation by $3\times 3$ real matrices. I don't know how to begin.

Andrei
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1 Answers1

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Since the Lie algebra is 3-dimensional, you can consider the adjoint representation.

If we write $L$ your Lie algebra, then for each $x\in L$ there is a linear map $$\operatorname{ad}(x):y\in L\mapsto [x,y]\in L,$$ and this defines a map $$\operatorname{ad}:L\to\operatorname{End}(L)$$ which you can easily check to be a representation.

(Yoy could also consider the trivial representation, but that is slightly less entertaining...)

Mariano Suárez-Álvarez
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  • Thanks, I'm taking a very basic introductory course in this. Could you tell me if this is the trivial (or a correct) representation? $$ad_{e_1}=0$$, $$ad_{e_2}=0$$ and $$ad_{e_3}= \left( {\begin{array}{ccc} 0 & 1 & 0\ -1 & 0 & 0\ 0 & 0 & 0 \end{array} } \right) $$ – Andrei Mar 05 '15 at 22:31
  • It feels very counter-intuitive. – Andrei Mar 05 '15 at 22:33
  • If $e_1$, $e_2$ and $e_3$ are $x$, $y$ and $z$, respectively, that is not right. We have $ad_z=0$, for example. – Mariano Suárez-Álvarez Mar 05 '15 at 23:24
  • Ok, then I don't understand. Because what I have done now is to construct it such as it fulfill: $$[e_i, e_j]=\sum_{k=1}^3 f_{ij}^ke_k$$ When I calculated the matrix elements using this I got the three matrices above. – Andrei Mar 06 '15 at 05:52
  • Well, then you make a mistake. You wrote that $ad_{e_1}=0$, and that means that $[e_1,u]=0$ for all $u\in L$; likewise, your second equation mean sthat $[e_2,u]=0$ for all $u\in L$. This tells us that the center of the Lie algebra has dimension at least two (for $e_1$ and $e_2$ are contained there) Yet the Lie algebra you described in your question has a $1$-dimensional center. – Mariano Suárez-Álvarez Mar 06 '15 at 06:01
  • We have $[x,x]=0$, $[x,y]=z$ and $[x,z]=0$, so with respect to the ordered basis ${x,y,z}$ we have $ad_x=\begin{pmatrix}0&0&0\0&0&0\0&1&0\end{pmatrix}$. Similarly, $[y,x]=-z$, $[y,y]=0$ and $[y,z]=0$, so that $ad_y=\begin{pmatrix}0&0&0\0&0&0\-1&0&0\end{pmatrix}$. Finally, as $[z,x]=[z,y]=[z,z]=0$, $ad_z$ is the zero matrix. – Mariano Suárez-Álvarez Mar 06 '15 at 06:02
  • Thank you Mariano, will have to look at this more when I get home. – Andrei Mar 06 '15 at 06:06