Differentiation and integration of highest weights

By [L4], $\pi$ is smooth and $\pi (\exp X)=\exp (d\pi (X))$. An exponential neighborhood generates a connected group: the generated subgroup is open and its other cosets are open, so it is also closed and must be the whole group. Therefore a complex subspace invariant under the differential is group invariant, since the matrix exponential preserves it; the converse follows by differentiating. A commuting endomorphism of a nonzero irreducible complex representation is scalar: choose an eigenvalue, whose nonzero eigenspace is invariant and hence is the entire space. In particular $Z(G{)}^0$ acts by scalars. Differentiating the covering in [L5] gives $\mathfrak{g}=\mathrm{Lie}(Z(G{)}^0)\oplus \mathfrak{s}$, with the first summand central. A subspace invariant under ${\mathfrak{s}}_{\mathbb{C}}$ is consequently invariant under all of $d\pi$ and under $G$. Thus restriction to ${\mathfrak{s}}_{\mathbb{C}}$ is irreducible.

Let $v$ be a highest vector of $\pi$, of character $\lambda$ from [L1]. Differentiating $\pi (t)v=\lambda (t)v$ for $t\in T$ gives $d\pi (X)v=d\lambda (X)v$ for $X\in \mathfrak{t}$, and complex-linear extension gives the same identity on ${\mathfrak{t}}_{\mathbb{C}}$. Positive root operators annihilate $v$: such an operator takes a $T$-weight vector of character $\lambda$ to one of character $\lambda \alpha$, by conjugating the differentiated action with $t\in T$, and a nonzero such weight would lie strictly above the highest weight. The derived Cartan weight is therefore the restriction of this complex-linear $d\lambda$ to $(\mathfrak{t}\cap \mathfrak{s}{)}_{\mathbb{C}}$.

Conversely let $M$ be a finite-dimensional complex ${\mathfrak{s}}_{\mathbb{C}}$-module. Restrict its action to the real algebra $\mathfrak{s}$ and apply [L3] with source $S$ and target ${\mathrm{GL}}_{\mathbb{C}}(M)$ regarded as a real Lie group. This integrates the action uniquely to $\rho :S\to {\mathrm{GL}}_{\mathbb{C}}(M)$. Choose a continuous representation $\zeta :Z(G{)}^0\to {\mathrm{GL}}_{\mathbb{C}}(M)$ commuting with $\rho$. Then $R(z,s)=\zeta (z)\rho (s)$ is a representation of $H$. The derived-algebra data contain no prescribed action of the central factor; when that factor is trivial there is of course no extra choice. For $M=0$ every action and the resulting descent are the unique zero-dimensional ones.

Write $K=\ker p$. To justify the torus language, let $U$ be the identity component of the closed Lie subgroup $p^{-1}(T)$. The covering charts imply $p(U)$ contains an identity neighborhood of $T$, hence equals connected $T$. For $u,v\in U$, their commutator lies in finite $K$; continuity on connected $U\times U$ makes it identity. Thus $U$ is a compact connected abelian subgroup, hence a torus. It is maximal: a torus containing it maps to a torus containing $T$, so maps into $T$ and lies in $U$. By [L7] central $K$ lies in $C_H(U)=U$. Since $p(U)=T$, every element of $p^{-1}(T)$ differs from an element of $U$ by one in $K$. Consequently $p^{-1}(T)=U$ is a torus and $U\mathrm{/}K=T$.

By [L6] the restriction $R{\mathrm{\mid }}_U$ is unitary. A finite-dimensional commuting family of unitary operators has a common orthonormal eigenbasis: if some operator is not scalar, its mutually orthogonal eigenspaces are preserved by every other operator, and induction on dimension diagonalizes the restrictions; if all are scalar any orthonormal basis suffices. The resulting diagonal entries are continuous characters $\chi \in X^{\ast }(U)$. Since $K\subseteq U$, it acts trivially on $M$ exactly when every occurring character is trivial on $K$. Such a character factors uniquely through $U\mathrm{/}K=T$, and the factor is continuous because the compact-to-Hausdorff surjection $U\to T$ is a quotient map. Thus this is precisely the condition that all weights lie in $p^{\ast }{X}^{\ast }(T)$. Conversely a pulled-back character is trivial on $K$. For the zero module the character family is empty and both conditions hold.

The product representation $R$ descends exactly when $R(hk)=R(h)$ for every $h\in H,k\in K$, equivalently when $R(k)=I$. In that case define $\pi (p(h))=R(h)$; this is well defined and is a homomorphism. Local inverse sheets of the covering show it is continuous and smooth. Necessity follows by pulling back any representation of $G$. Step 2.1 proves the equivalent character-lattice condition. In the forward direction, the irreducibility established in step 1.1 means the nonzero highest vector of step 1.2 generates the whole derived-algebra module, not merely a submodule. If $\mathfrak{s}=0$, irreducibility forces dimension one, its derived highest weight is zero, and the independent datum is exactly a torus character. These arguments prove the Statement including the central-action qualification.