Verma module

Remarks

The vector $v_\lambda = 1 \otimes 1$ has weight $\lambda$, is killed by $\mathfrak{n}^+$, and generates $M(\lambda)$. With $\mathfrak{n}^- = \bigoplus_{\alpha \in \Phi^+} \mathfrak{g}_{-\alpha}$, the map $u \mapsto u v_\lambda$ is a linear isomorphism $U(\mathfrak{n}^-) \to M(\lambda)$, by the Poincaré–Birkhoff–Witt theorem. Every module generated by a vector of weight $\lambda$ killed by $\mathfrak{n}^+$ is a quotient of $M(\lambda)$. $M(\lambda)$ has a unique maximal submodule, and the quotient is the irreducible module $L(\lambda)$. Example: for $\mathfrak{sl}_2(\C)$ with basis $e, f, h$, put $v_k = f^k v_\lambda$ and write $\lambda$ for $\lambda(h)$. Then $h v_k = (\lambda - 2k) v_k$, $f v_k = v_{k+1}$ and $e v_k = k(\lambda - k + 1) v_{k-1}$. If $\lambda = n \in \Z_{\ge 0}$, then $e v_{n+1} = 0$, the vectors $v_k$ with $k > n$ span a submodule isomorphic to $M(-n-2)$, and the quotient is $L(n)$ of dimension $n + 1$. Non-example: $L(n)$ itself is generated by a highest weight vector $v$, but $f^{n+1} v = 0$, so $u \mapsto uv$ is not injective on $U(\mathfrak{n}^-)$ and $L(n)$ is not a Verma module.