The normalizer acts on Lie algebra cohomology

For fixed $x\in \mathfrak{p}$ the formula of the Statement defines a $k$-linear map ${\theta }_x:C^q(\mathfrak{a},V)\to C^q(\mathfrak{a},V)$: both displayed terms are $k$-linear in $\omega$, and they are alternating in the arguments because $\omega$ is and because the substitution $x_j\mapsto [x,x_j]$ is linear in the slot; the bracket lies in $\mathfrak{a}$ by [F4], so the expression is a legitimate cochain. Moreover $x\mapsto {\theta }_x$ is $k$-linear, since both the action on $V$ and the bracket are.

Write $A_x\omega =x\cdot \omega$ for action on values and $B_x\omega ={\sum }_j\omega (\dots ,[x,x_j],\dots )$ for action on argument slots, so ${\theta }_x=A_x-B_x$. Value and slot operators commute. By [F2], $[A_x,A_y]=A_{[x,y]}$. In $[B_x,B_y]$, substitutions in distinct slots cancel pairwise; the same-slot difference is $\omega (\dots ,[y,[x,x_j]]-[x,[y,x_j]],\dots )=-\omega (\dots ,[[x,y],x_j],\dots )$ by [F3]. Thus $[B_x,B_y]=-B_{[x,y]}$, and $[{\theta }_x,{\theta }_y]=A_{[x,y]}-B_{[x,y]}={\theta }_{[x,y]}$.

Each ${\theta }_x$ commutes with $d$. Evaluate ${\theta }_{x}d\omega -d{\theta }_x\omega$ on $y_0,\dots ,y_q\in \mathfrak{a}$, using the zero-based formula [F1]. Write $\rho (x)$ for the action on $V$. The action terms, after cancelling substitutions in the unacted slots, are ${\sum }_i(-1{)}^i([\rho (x),\rho (y_i)]-\rho ([x,y_i]))\omega (y_0,\dots ,\widehat{y_i},\dots ,y_q)$, which vanish by [F2]. The remaining bracket terms are ${\sum }_{i<j}(-1{)}^{i+j}\omega ([x,[y_i,y_j]]-[[x,y_i],y_j]-[y_i,[x,y_j]],y_0,\dots ,\widehat{y_i},\dots ,\widehat{y_j},\dots ,y_q)$, which vanish by Jacobi [F3]. Thus ${\theta }_{x}d=d{\theta }_x$.

For $x\in \mathfrak{a}$ the Cartan formula ${\theta }_x=d{i}_x+i_{x}d$ holds. Evaluate at $(x_1,\dots ,x_q)$. In $i_{x}d$, the terms of $d\omega$ in which the leading argument $x$ acts give $x\omega (x_1,\dots ,x_q)$, and the terms in which $x$ is the bracket argument give ${\sum }_j(-1{)}^j\omega ([x,x_j],x_1,\dots ,\widehat{x_j},\dots ,x_q)$; all terms in which one of the $x_i$ acts, and all bracket terms with two entries among the $x_i$, cancel against the corresponding terms of $d{i}_x$, because $\omega$ is alternating: $\omega (x,[x_i,x_j],\dots )=-\omega ([x_i,x_j],x,\dots )$ and the two appear with the same coefficient $(-1{)}^{i+j}$, while $x_i\omega (x,\dots )$ and $x_i\omega (x,\dots )$ appear with opposite coefficients $(-1{)}^i$ and $(-1{)}^{i-1}$. Since moving the bracket $[x,x_j]$ from the first slot into the $j$-th slot costs the sign $(-1{)}^{j-1}$, the surviving sum equals $-{\sum }_j\omega (x_1,\dots ,[x,x_j],\dots ,x_q)$, so $d{i}_x\omega +i_{x}d\omega =x\omega -{\sum }_j\omega (\dots ,[x,x_j],\dots )={\theta }_x\omega$.

For $x\in \mathfrak{a}$, ${\theta }_x$ induces the zero map on every cohomology group. If $\omega$ is a cocycle, then by step 2.3 ${\theta }_x\omega =d(i_x\omega )+i_x(d\omega )=d(i_x\omega )$ is a coboundary; and by step 2.2 ${\theta }_x$ maps coboundaries to coboundaries, since ${\theta }_x(d\alpha )=d({\theta }_x\alpha )$. Hence the induced endomorphism of $H^q(\mathfrak{a},V)=\ker d^q\mathrm{/}\mathrm{im}{d}^{q-1}$ is zero for every $q$.

The action descends. By steps 2.1 and 2.2, $\theta :\mathfrak{p}\to {\mathrm{End}}_k(C^{\bullet }(\mathfrak{a},V))$ is a representation preserving the differential, so it induces a representation of $\mathfrak{p}$ on each cohomology space $H^q(\mathfrak{a},V)$; by step 3.1 this induced representation kills $\mathfrak{a}$. Since $\mathfrak{a}$ is an ideal [F4] and the quotient map $q:\mathfrak{p}\twoheadrightarrow \mathfrak{p}\mathrm{/}\mathfrak{a}$ is a surjective Lie algebra homomorphism, there is a unique Lie algebra homomorphism $\bar{\theta }:\mathfrak{p}\mathrm{/}\mathfrak{a}\to {\mathrm{End}}_k(H^q)$ with $\bar{\theta }\circ q=\theta$: it is well defined because $\theta$ vanishes on $\mathfrak{a}$, and it is a homomorphism because $\theta$ is and $q$ is surjective. By [F4] the representation $\bar{\theta }$ makes $H^q(\mathfrak{a},V)$ a unital $U(\mathfrak{p}\mathrm{/}\mathfrak{a})$-module. For $\mathfrak{a}={\mathfrak{n}}^{+}◃\mathfrak{b}$ the quotient is $\mathfrak{h}$ by [F5], so $H^{\bullet }({\mathfrak{n}}^{+},V)$ is an $\mathfrak{h}$-module for every $\mathfrak{b}$-module $V$; and every $\mathfrak{g}$-module is a $\mathfrak{b}$-module by restriction. The same conclusion holds for $\mathfrak{p}=N_{\mathfrak{g}}(\mathfrak{a})$, which makes $\mathfrak{a}$ an ideal by [F4] applied to the normalizer definition, so the statement covers every subalgebra whose normalizer acts.