TITLE: How am I supposed to create a multiplication table for the commutator/Lie-Bracket operaton if $\left[X_1,X_2\right]=X_1X_2-X_2X_1=0$?
QUESTION [0 upvotes]: Let's say we have the Lie-group of
$$\left\{X_1=\frac{\partial }{\partial x},\:X_2=x\frac{\partial }{\partial x},\:X_3=x^2\frac{\partial }{\partial x}\right\}$$
We have to create a "multiplication table" where the operation is the Lie-Bracket operation (also known as commutator)
But what I don't get is: Why should we get anything different than zero in the first place?
Let's say:
$$\left[X_1,\:X_2\right]=X_1X_2-X_2X_1=\frac{\partial }{\partial x}\left(x\frac{\partial }{\partial x}\right)-x\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}\left(x\right)\right)-\frac{\partial }{\partial x}\left(x\right)\frac{\partial }{\partial x}=\frac{\partial }{\partial x}-\frac{\partial }{\partial x}=0$$
And the funny thing is that I can repeat this for any arbitrary IFG
So where's my mistake?

REPLY [0 votes]: In fact, one can establish relationship
$$[X_1,X_2]=X_1\tag{1}$$
in a somewhat weaker sense using matrices.
Indeed, let $(P_n)$ be the (n+1)-dimensional vector spaces of polynomials with degree $ \le n$ with $$(P_0) \subset (P_1) \subset (P_2) \subset ...\subset (P_n) \subset ...$$
We are going to obtain (1) on any of the $(P_n)$. For the sake of simplicity, let us see it for $n=3$.
In this space, operators $X_1$ and $X_2$ are resp.:
$$\begin{cases}X_1: \ ax^3+bx^2+cx+d \ \mapsto 0x^3+3ax^2+2bx+c\\
               X_2: \ ax^3+bx^2+cx+d \ \mapsto 3ax^3+2bx^2+cx+0\end{cases}$$
whose matrices wrt basis $\{x^3,x^2,x,1\}$ are resp.
$$M_1=\begin{pmatrix}0&0&0&0\\3&0&0&0\\0&2&0&0\\0&0&1&0\end{pmatrix} \ \text{and} \ 
M_2=\begin{pmatrix}3&0&0&0\\0&2&0&0\\0&0&1&0\\0&0&0&0\end{pmatrix}$$
It remains to check that:
$$[M_1,M_2]=M_1M_2-M_2M_1=M_1$$
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