Lie subalgebra in $Der(\mathbb{C}[z])$ isomorphic to $\mathfrak{sl}_2$

Question

I am to prove that $\{(az^2+bz+c)\frac{\partial}{\partial z}:a,b,c\in\mathbb{C}\}$ regarded as a Lie algebra is isomorphic to $\mathfrak{sl}_2(\mathbb{C})$. I guess it is possible to build a basis $e,f,h$ which satisfies the equations for $\mathfrak{sl}_2$, i.e. $[e,f]=h, [h,e]=2e$ and $[h,f]=-2f$, isn't it?

It can be easily shown that $$[\frac{\partial}{\partial z},z^2\frac{\partial}{\partial z}]=2z\frac{\partial}{\partial z},$$$$[z\frac{\partial}{\partial z},\frac{\partial}{\partial z}]=-\frac{\partial}{\partial z}$$ and $$[z\frac{\partial}{\partial z},z^2\frac{\partial}{\partial z}]=z^2\frac{\partial}{\partial z}.$$ I see that these relations are similar to relations for $\mathfrak{sl}_2$, but I am not getting to fix coefficients to get the relations we need: if we set $h=2z\frac{\partial}{\partial z}$ then $\frac{\partial}{\partial z}$ and $z^2\frac{\partial}{\partial z}$ will be eigenvectors for $\mathrm{ad}h$ with eigenvalues $-2$ and $2$ correspondingly. The last means that $e$ should be $z^2\frac{\partial}{\partial z}$ and $f$ should be $\frac{\partial}{\partial z}$ (remembering the relations for $\mathfrak{sl}_2$). But in $\mathfrak{sl}_2$ we have $[e,f]=h$ but not $[f,e]=h$ as we have just got. Could you please fix it?

Answer 1

Using the notation $\partial_z = \frac{\partial}{\partial z}$, a direct calculation yields: $$ \bigl[ z^m \partial_z, z^n \partial_z \bigr] = (n - m) z^{m+n-1} \partial_z, $$ and in particular, since the bracket is linear over $\Bbb{C}$, $$ \newcommand{\pz}{\,\partial_z} \bigl[ -2z \pz, i \pz \bigr] = 2i \pz, \quad \bigl[ -2z \pz, iz^2 \pz \bigr] = -2iz^2 \pz, \quad \bigl[ i \pz, iz^2 \pz \bigr] = -2z \pz. $$ With $h = -2z \pz$, $e = i \pz$, and $f = iz^2 \pz$, you will recognize the familiar relations for $\mathfrak{sl}_2(\Bbb{C}) = \langle e, h, f \rangle$: $$ \bigl[ h, e \bigr] = 2e, \quad \bigl[ h, f \bigr] = -2f, \quad \bigl[ e, f \bigr] = h. $$