Lie Algebra Cohomology and Kostants Nilradical Theorem — Examples

$1\cdot \lambda =1(\lambda +\rho )-\rho =\lambda =m\omega$, and by [L1] $s\cdot \lambda =s(\lambda +\rho )-\rho =-\lambda -2\rho =-m\omega -\alpha =-(m+2)\omega$; since $\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }=1$ and $\mathrm{\ell }(1)=0$, $\mathrm{\ell }(s)=1$, [L2] gives the displayed cohomology.

In degree one, the generator is ${\epsilon }_{\alpha }\otimes v_{s\lambda }$: its exterior factor has weight $-\alpha$ and $v_{s\lambda }$ has weight $s\lambda =-m\omega$, so the total weight $-\alpha -m\omega =-(m+2)\omega$ is the dot weight, confirming the $\rho$-shift; the degree-zero generator is the highest vector $v_{m\omega }$.

Counting: $\dim H^0=1$, $\dim H^1=1$, and $H^k=0$ for $k\ge 2$, matching the top degree $\mathrm{\mid }{\mathrm{\Phi }}^{+}\mathrm{\mid }=1$; the degree-one weight is $-(m+2)\omega =-m\omega -\alpha$, strictly below $-m\omega$, so the exterior root shift cannot be dropped.

With $s_1\rho =\rho -{\alpha }_1$ and $s_2\rho =\rho -{\alpha }_2$, direct evaluation gives $s_1\cdot 0=-{\alpha }_1$, $s_2\cdot 0=-{\alpha }_2$, $s_1{s}_2\cdot 0=s_1(\rho -{\alpha }_2)-\rho =s_1({\alpha }_1)-\rho =-2{\alpha }_1-{\alpha }_2$, $s_2{s}_1\cdot 0=-2{\alpha }_2-{\alpha }_1$, and $w_0\cdot 0=-\rho -\rho =-2\rho =-2{\alpha }_1-2{\alpha }_2$, while $1\cdot 0=0$.

By [L3] the $k$-th cohomology is the sum of one line for each Weyl element of length $k$, so its dimension is the number of such elements and its weights are precisely those computed in step 1.1 for the length-$k$ elements. For $k>3$ the sum is empty, so $H^k({\mathfrak{n}}^{+},\mathbb{C})=0$.

Counting the elements listed in [L1] gives dimensions $1,2,2,1$ in degrees $0,1,2,3$ and total dimension $6=\mathrm{\mid }W\mathrm{\mid }$; in degree one the two classes have the distinct weights $-{\alpha }_1,-{\alpha }_2$ attached to the two simple roots, and both are nonzero because ${\alpha }_1,{\alpha }_2$ are linearly independent.

By [L1] and [L2], $H^1({\mathfrak{n}}^{+},V)={⨁}_{\mathrm{\ell }(w)=1}{\mathbb{C}}_{w\cdot \lambda }={⨁}_{i=1}^r{\mathbb{C}}_{s_i\cdot \lambda }$, and each summand is generated by the class of ${\gamma }_{s_i}$ by [L3].

By [L4] the weights $s_i\cdot \lambda$ are pairwise distinct, so the displayed sum is direct with one line per simple reflection; and by [L5] the same index set labels the first BGG term, giving $H^1({\mathfrak{n}}^{+},V)\cong {⨁}_{i=1}^r{\mathbb{C}}_{s_i\cdot \lambda }$ against $C_1(\lambda )={⨁}_{i}M(s_i\cdot \lambda )$ as $\mathfrak{h}$-module statements.

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 $H^1({\mathfrak{n}}^{+},\mathbb{C})={\mathbb{C}}_{s\cdot 0}={\mathbb{C}}_{-\alpha }$.

The correct weight is $-\alpha -m\omega =-2\omega -m\omega =-(m+2)\omega$, and by [L1] this equals $s\cdot \lambda$; the exterior factor contributes $s\rho -\rho =-2\rho =-\alpha$ and the vector factor contributes $s\lambda =-m\omega$, so the two shifts are distinct and both are needed.

Dropping the exterior factor leaves the weight $s\lambda =-m\omega$, which differs from the correct weight by $2\omega$ for every $m\ge 0$; since $\omega \mathrm{\ne }0$, the difference never vanishes. Replacing the vector factor by the top weight and moving the whole shift to the exterior part would give $\lambda -\alpha =m\omega -2\omega =(m-2)\omega$, which differs from $-(m+2)\omega$ by $2m\omega$, vanishing exactly when $m=0$, so it differs for every $m\ge 1$.

The true degree-one cohomology weight is therefore $-(m+2)\omega$, and neither of the two listed modifications produces it uniformly in $m$: the exterior root-weight shift $s\rho -\rho =-\alpha$ must be added to the vector shift $s\lambda =-m\omega$, not dropped and not counted twice.