Can $S = \{D_x, xD_x, x^2 D_x, x^3 D_x\}$ be extended to a set generating a Lie algebra?

Question

Question. Let $D_x := \frac{\mathrm{d}}{\mathrm{d}x}$ be the differentiation operator. Is $S = \{D_x, xD_x, x^2 D_x, x^3 D_x\}$ a basis for a Lie algebra (standard derivation Lie bracket, i.e. $[x,y]=x\circ y - y\circ x$)? If not, can $S$ be extended to a set generating a Lie algebra?

Attempt. $S$ is not a basis, since $\left[x^2 D_x, x^3 D_x\right] = x^4 D_x \notin \mathrm{span} (S)$. However, I'm confused about the next part of the question. As far as I'm aware, any set can generate a Lie algebra, no? The generating set doesn't need to be a basis; the generated algebra automatically includes all Lie brackets and linear combinations.

Consequently, is the question incorrect? Is it meant to be whether $S$ can be extended to a basis for a Lie algebra? Assuming that is the case, my answer would be yes, assuming the Lie algebra can be infinite dimensional: $S = \{D_x, xD_x, x^2 D_x, x^3 D_x, ...\}$. If the Lie algebra has to be finite, then no; adding a new $x^n D_x$ element to the basis would still just get us

$$\left[x^{n-1} D_x, x^n D_x\right] = x^{2n-2} D_x \notin \mathrm{span} (S),$$

for $n \geq 3$.

Is this correct?

Answer 1

We need our algebra to satisfy the Jacobi identity, so for $x^aD_x, x^bD_x$ and $x^cD_x$ we need that: $$[x^aD_x,[x^bD_x,x^cD_x]]+[x^bD_x,[x^cD_x,x^aD_x]]+[x^cD_x,[x^aD_x,x^bD_x]]=0$$ Note that: $$[x^bD_x,x^cD_x]=cx^{b+c-1}D_x-bx^{b+c-1}D_x=(c-b)x^{b+c-1}D_x$$ it follows that: \begin{align} [x^aD_x,[x^bD_x,x^cD_x]]=&(c-b)[x^aD_x,x^{b+c-1}D_x]=(c-b)(c+b-a-1`)x^{b+c+a-2}D_x\\ [x^bD_x,[x^cD_x,x^aD_x]]=&(a-c)(c+a-1-b)x^{a+b+c-2}D_x\\ [x^cD_x,[x^aD_x,x^bD_x]]=&(b-a)(b+a-1-c)x^{a+b+c-2}D_x \end{align} So the only way the Jacobi identity is satisfied is if: $$(c-b)(c+b-a-1`)+(a-c)(c+a-1-b)+(b-a)(b+a-1-c)=0$$ and I'll leave it to you to check that it does :)

It follows that the algebra $\{D_x,xD_x,\dots\}$ is a Lie algebra.