[proofplan] We use the [PBW model of a Verma module](/theorems/9374) to identify $M(\lambda)$, as a [vector space](/page/Vector%20Space), with $U(\mathfrak n^-)$ acting on the highest weight vector. After choosing one nonzero root vector in each negative root space, the Poincare-Birkhoff-Witt basis gives a basis of $M(\lambda)$ indexed by exponent functions on $\Phi^+$. We compute the weight of each PBW monomial by commuting elements of $\mathfrak h$ past the negative root vectors. Grouping the resulting basis vectors by their total root contribution gives the [direct sum](/page/Direct%20Sum) decomposition and the [multiplicity formula](/theorems/2420). [/proofplan] [step:Choose ordered negative root vectors and the corresponding PBW basis] Let \begin{align*} \Phi^+=\{\alpha_1,\dots,\alpha_N\} \end{align*} be an ordering of the finite set of positive roots. For each $i\in\{1,\dots,N\}$, choose a nonzero root vector \begin{align*} f_i\in\mathfrak g_{-\alpha_i}. \end{align*} Since $\mathfrak g$ is finite-dimensional and complex semisimple, every root space $\mathfrak g_\alpha$ is one-dimensional, so $(f_1,\dots,f_N)$ is an ordered basis of \begin{align*} \mathfrak n^-:=\bigoplus_{\alpha\in\Phi^+}\mathfrak g_{-\alpha}. \end{align*} The chosen positive root system gives the triangular decomposition $\mathfrak g=\mathfrak n^-\oplus\mathfrak h\oplus\mathfrak n^+$, with $\mathfrak b=\mathfrak h\oplus\mathfrak n^+$. For an exponent vector \begin{align*} m=(m_1,\dots,m_N)\in\mathbb Z_{\ge 0}^N, \end{align*} define the PBW monomial \begin{align*} f^m:=f_1^{m_1}f_2^{m_2}\cdots f_N^{m_N}\in U(\mathfrak n^-) \end{align*} and the corresponding vector \begin{align*} b_m:=f^m v_\lambda\in M(\lambda). \end{align*} By the Poincare-Birkhoff-Witt theorem [citetheorem:8827] applied to the ordered basis $(f_1,\dots,f_N)$ of $\mathfrak n^-$, the elements $f^m$, with $m\in\mathbb Z_{\ge 0}^N$, form a basis of $U(\mathfrak n^-)$. By the PBW model of a Verma module [citetheorem:9374], the map $U(\mathfrak n^-)\to M(\lambda)$ given by $u\mapsto u v_\lambda$ is a vector-space isomorphism. Hence the vectors $b_m$, with $m\in\mathbb Z_{\ge 0}^N$, form a complex basis of $M(\lambda)$. [/step] [step:Compute the weight of each PBW monomial] For $m=(m_1,\dots,m_N)\in\mathbb Z_{\ge 0}^N$, define its total root contribution by \begin{align*} \beta(m):=\sum_{i=1}^N m_i\alpha_i\in Q_+. \end{align*} We claim that \begin{align*} b_m\in M(\lambda)_{\lambda-\beta(m)}. \end{align*} Let $h\in\mathfrak h$. Since $f_i\in\mathfrak g_{-\alpha_i}$, the root-space relation gives \begin{align*} [h,f_i]=-\alpha_i(h)f_i. \end{align*} Equivalently, in $U(\mathfrak g)$, \begin{align*} h f_i=f_i h-\alpha_i(h)f_i. \end{align*} Commuting $h$ successively past the factors in $f_1^{m_1}\cdots f_N^{m_N}$ gives \begin{align*} h f^m=f^m h-\left(\sum_{i=1}^N m_i\alpha_i(h)\right)f^m. \end{align*} Since $v_\lambda$ is a highest weight vector of weight $\lambda$, we have \begin{align*} h v_\lambda=\lambda(h)v_\lambda. \end{align*} Therefore \begin{align*} h b_m = h f^m v_\lambda = \left(\lambda(h)-\sum_{i=1}^N m_i\alpha_i(h)\right)f^m v_\lambda. \end{align*} By the definition of $\beta(m)$, this is \begin{align*} h b_m=(\lambda-\beta(m))(h)b_m. \end{align*} Since this holds for every $h\in\mathfrak h$, the vector $b_m$ has weight $\lambda-\beta(m)$. [guided] The purpose of this step is to determine exactly which weight is carried by a PBW basis vector. Fix an exponent vector \begin{align*} m=(m_1,\dots,m_N)\in\mathbb Z_{\ge 0}^N \end{align*} and define \begin{align*} \beta(m):=\sum_{i=1}^N m_i\alpha_i\in Q_+. \end{align*} We will prove that the basis vector \begin{align*} b_m=f_1^{m_1}\cdots f_N^{m_N}v_\lambda \end{align*} lies in the weight space of weight $\lambda-\beta(m)$. Let $h\in\mathfrak h$. The defining property of the root space $\mathfrak g_{-\alpha_i}$ is that every element $x\in\mathfrak g_{-\alpha_i}$ satisfies \begin{align*} [h,x]=-\alpha_i(h)x. \end{align*} Since $f_i\in\mathfrak g_{-\alpha_i}$, we get \begin{align*} [h,f_i]=-\alpha_i(h)f_i. \end{align*} Writing the bracket as $[h,f_i]=h f_i-f_i h$ inside the enveloping algebra, this becomes \begin{align*} h f_i=f_i h-\alpha_i(h)f_i. \end{align*} This identity tells us what happens when $h$ is moved past one negative root vector: the cost is subtraction of the scalar $\alpha_i(h)$. Repeating the same commutation through $m_i$ copies of $f_i$ gives a total contribution of $m_i\alpha_i(h)$ from that root. Moving $h$ through all factors of \begin{align*} f^m=f_1^{m_1}f_2^{m_2}\cdots f_N^{m_N} \end{align*} therefore gives \begin{align*} h f^m=f^m h-\left(\sum_{i=1}^N m_i\alpha_i(h)\right)f^m. \end{align*} Now apply this identity to $v_\lambda$. Since $v_\lambda$ has weight $\lambda$, we have \begin{align*} h v_\lambda=\lambda(h)v_\lambda. \end{align*} Thus \begin{align*} h b_m = h f^m v_\lambda = f^m h v_\lambda-\left(\sum_{i=1}^N m_i\alpha_i(h)\right)f^m v_\lambda. \end{align*} Substituting $h v_\lambda=\lambda(h)v_\lambda$ gives \begin{align*} h b_m = \lambda(h)f^m v_\lambda-\left(\sum_{i=1}^N m_i\alpha_i(h)\right)f^m v_\lambda. \end{align*} Hence \begin{align*} h b_m = \left(\lambda(h)-\sum_{i=1}^N m_i\alpha_i(h)\right)b_m. \end{align*} By the definition of $\beta(m)$, the scalar in parentheses is $(\lambda-\beta(m))(h)$. Therefore \begin{align*} h b_m=(\lambda-\beta(m))(h)b_m. \end{align*} Because this equality holds for every $h\in\mathfrak h$, the vector $b_m$ lies in the weight space $M(\lambda)_{\lambda-\beta(m)}$. [/guided] [/step] [step:Group the PBW basis by total root contribution] For each $\beta\in Q_+$, define \begin{align*} B_\beta:=\{b_m:m\in\mathbb Z_{\ge 0}^N\text{ and }\beta(m)=\beta\}. \end{align*} The preceding step shows that every vector in $B_\beta$ lies in $M(\lambda)_{\lambda-\beta}$. Since the full set \begin{align*} \{b_m:m\in\mathbb Z_{\ge 0}^N\} \end{align*} is a basis of $M(\lambda)$, and since every $m\in\mathbb Z_{\ge 0}^N$ has $\beta(m)\in Q_+$, every PBW basis vector belongs to one of the weight spaces $M(\lambda)_{\lambda-\beta}$ with $\beta\in Q_+$. Hence no weights outside the set $\lambda-Q_+$ occur. The sum is direct because if $\beta,\gamma\in Q_+$ and $\lambda-\beta\ne\lambda-\gamma$, then the corresponding simultaneous eigenspaces for the commuting action of $\mathfrak h$ have distinct eigencharacters. Therefore \begin{align*} M(\lambda)=\bigoplus_{\beta\in Q_+} M(\lambda)_{\lambda-\beta}. \end{align*} Moreover, for each $\beta\in Q_+$, the set $B_\beta$ is a basis of $M(\lambda)_{\lambda-\beta}$. To prove spanning, let $w\in M(\lambda)_{\lambda-\beta}$ and expand it in the PBW basis as a finite sum $w=\sum_m c_m b_m$, with $c_m\in\mathbb C$. For every $h\in\mathfrak h$, the preceding step gives $h b_m=(\lambda-\beta(m))(h)b_m$, while the weight condition on $w$ gives $h w=(\lambda-\beta)(h)w$. Hence \begin{align*} 0=\sum_m c_m(\beta-\beta(m))(h)b_m \end{align*} for every $h\in\mathfrak h$. Since the $b_m$ are linearly independent, if $c_m\ne 0$ then $(\beta-\beta(m))(h)=0$ for every $h\in\mathfrak h$, so $\beta(m)=\beta$ as elements of $\mathfrak h^*$. Thus only vectors in $B_\beta$ occur in the expansion of $w$, and $B_\beta$ spans $M(\lambda)_{\lambda-\beta}$. It is linearly independent because it is a subset of the PBW basis of $M(\lambda)$. [/step] [step:Identify the number of basis vectors with Kostant's partition function] For $m=(m_1,\dots,m_N)\in\mathbb Z_{\ge 0}^N$, define the associated function \begin{align*} \widetilde m:\Phi^+\to\mathbb Z_{\ge 0} \end{align*} by \begin{align*} \widetilde m(\alpha_i):=m_i \end{align*} for each $i\in\{1,\dots,N\}$. Then \begin{align*} \beta(m)=\sum_{i=1}^N m_i\alpha_i = \sum_{\alpha\in\Phi^+}\widetilde m(\alpha)\alpha. \end{align*} Thus the exponent vectors $m\in\mathbb Z_{\ge 0}^N$ with $\beta(m)=\beta$ are in bijection with functions $\widetilde m:\Phi^+\to\mathbb Z_{\ge 0}$ satisfying \begin{align*} \beta=\sum_{\alpha\in\Phi^+}\widetilde m(\alpha)\alpha. \end{align*} By the definition of Kostant's partition function, the cardinality of this set is $p_{\Phi^+}(\beta)$. Since $B_\beta$ is a basis of $M(\lambda)_{\lambda-\beta}$, we obtain \begin{align*} \dim M(\lambda)_{\lambda-\beta}=p_{\Phi^+}(\beta). \end{align*} Taking $\beta=0$, the only such function is the zero function, so $B_0=\{v_\lambda\}$. Therefore \begin{align*} M(\lambda)_\lambda=\mathbb C v_\lambda. \end{align*} This proves the theorem. [/step]