BRST cohomology for the dilation action

The operator $x\frac{d}{dx}$ is a derivation; for the one-dimensional abelian $\mathfrak{g}$, its assignment to $e$ is a representation because $[\rho (e),\rho (e)]=0=\rho ([e,e])$. Thus this is an admissible input for [F1].

Since ${\mathrm{\Lambda }}^0\mathfrak{g}=\mathbb{C}$, ${\mathrm{\Lambda }}^1\mathfrak{g}=\mathfrak{g}$, and ${\mathrm{\Lambda }}^n\mathfrak{g}=0$ for $n\ge 2$, the only nonzero cochains are $C^0\cong \mathbb{C}[x]$ and $C^1\cong \mathbb{C}[x]$; [F1] gives $df=x{f}^{\mathrm{\prime }}$ and $d:C^1\to C^2=0$.

For $f={\sum }_{r=0}^N{a}_r{x}^r$, $x{f}^{\mathrm{\prime }}={\sum }_{r=1}^{N}r{a}_r{x}^r$; characteristic zero makes each $r\ge 1$ invertible, so the kernel is $\mathbb{C}$ and $H_{\mathrm{B}\mathrm{R}\mathrm{S}\mathrm{T}}^0\cong \mathbb{C}$ by [F2].

The image of $x\frac{d}{dx}$ is exactly $x\mathbb{C}[x]$: every output has zero constant term and $x^r$ is the image of $x^r\mathrm{/}r$ for $r\ge 1$. Hence $H_{\mathrm{B}\mathrm{R}\mathrm{S}\mathrm{T}}^1\cong \mathbb{C}[x]\mathrm{/}x\mathbb{C}[x]\cong \mathbb{C}$, while $C^n=0$ gives $H^n=0$ for $n\ge 2$.