Note: No lecture next week
Theorem. Let \(\lambda \in \mathfrak{h}^*\), there exists a unique \(1\) dimensional \(\mathfrak{h}\)-module \(\mathbb{C}_\lambda\) such that \(h\cdot v = \lambda(h)v\), furthermore, every simple finite dimensional \(\mathcal{U}(\mathfrak{h})\)-module is one of these modules.
Proof. \(\mathfrak{h}\) is abelian, and the action is commutative. Clearly simple because \(1\)-dimensional.
For a finite dimensional \(\mathfrak{h}\)-module \(V\), take any basis \(h_1,\dots, h_n\) of \(\mathfrak{h}\) and commutativity ensures that we can find a simultaneous diagonalized basis of \(\rho(h_i)\), then \(h_j v_i = \lambda_{v_i}(h_j) v_i\) so \(V = \bigoplus \mathbb{C}_{\lambda_{v_i}}\).
Corollary. For any \(\lambda \in \mathfrak{h}^*\), there exists a unique \(1\) dimensional \(\mathcal{U}(\mathfrak{h})\)-module \(\mathbb{C}_\lambda\) such that \(h\cdot v = \lambda(h)v\). Every \(1\) finite dimensional simple \(\mathcal{U}(\mathfrak{h})\) module arises this way.
Write \(V_\lambda = \{v \in V \mid \lambda(h)v = h\cdot v\}\).
Definition. Let \(V\) be an \(\mathfrak{h}\)-module, we call \(\lambda \in \mathfrak{h}^*\) a weight of \(V\) if \(V_\lambda\neq 0\) and we call \(V_\lambda\) the \(\lambda\)-weight space of \(V\). The set \[\Lambda(V) := \{\lambda \in \mathfrak{h}^* \mid V_\lambda \neq 0\}\] is called the set of weights of \(V\).
Proposition. For any finite dimensional \(\mathfrak{h}\)-module, \[V=\bigoplus_{\lambda \in \Lambda(V)} V_\lambda\] (obvious)
If we have chosen a set of simple roots \(\Pi\), we can define a partial order on \(\mathfrak{h}^*\) as \[\lambda \ge 0 \Leftrightarrow \lambda = \sum_{\alpha\in \Pi} n_\alpha \alpha, n_\alpha\ge 0.\]
A highest weight of \(V\) is any \(\lambda \in \Lambda(V)\) that is maximal with respect to this partial order.