Whitehead vanishing does not apply to the nilpotent radical

In degree zero, $C^0({\mathfrak{n}}^{+},\mathbb{C})=\mathbb{C}$ and $(d^{0}a)(x)=x\cdot a=0$ for every $x\in {\mathfrak{n}}^{+}$ and $a\in \mathbb{C}$ by [L1]; hence $d^0=0$ and $H^0({\mathfrak{n}}^{+},\mathbb{C})=\mathbb{C}$.

In degree one, $C^1({\mathfrak{n}}^{+},\mathbb{C})=({\mathfrak{n}}^{+}{)}^{\ast }$ and for every $1$-cochain $\omega$ and $x,y\in {\mathfrak{n}}^{+}$ one has $(d\omega )(x,y)=x\cdot \omega (y)-y\cdot \omega (x)-\omega ([x,y])=0$ by [L1]; moreover $C^2({\mathfrak{n}}^{+},\mathbb{C})={\mathrm{\Lambda }}^2({\mathfrak{n}}^{+}{)}^{\ast }=0$ because ${\mathfrak{n}}^{+}$ is one dimensional. Hence every $1$-cochain is a cocycle, while $\mathrm{im}{d}^0=0$ by step 1.1, so $H^1({\mathfrak{n}}^{+},\mathbb{C})=({\mathfrak{n}}^{+}{)}^{\ast }\cong \mathbb{C}\mathrm{\ne }0$.

The algebra ${\mathfrak{n}}^{+}$ is nilpotent and $\mathbb{C}$ is finite dimensional, yet $H^1({\mathfrak{n}}^{+},\mathbb{C})\mathrm{\ne }0$; the vanishing statements [L3] have the semisimplicity of the coefficient algebra $\mathfrak{g}$ among their hypotheses, which ${\mathfrak{n}}^{+}$ fails, so they cannot be applied here. Under the inherited choice hypotheses of the Kostant example, the same group appears as the $\lambda =0$ case of Kostant's theorem in Kostant cohomology for sl2, where $H^1({\mathfrak{n}}^{+},\mathbb{C})={\mathbb{C}}_{s\cdot 0}={\mathbb{C}}_{-\alpha }$.