Verma Modules and Shapovalov Forms — Examples

The first two identities follow by $[h,f]=-2f$ and multiplication; at $n=0$, $e{v}_{\lambda }=0$.

Induction using $e{f}^{n+1}=fe{f}^n+h{f}^n$ transforms the $n$th coefficient into $(n+1)(\lambda -n)$, which is the stated formula for $n+1$.

For $n=0$ the formula is the normalization. For $n>0$, contravariance gives $S(f^{n}v,f^{n}v)=S(f^{n-1}v,e{f}^{n}v)=n(\lambda -n+1)S(f^{n-1}v,f^{n-1}v)$.

Iterating that recurrence from $0$ to $n$ gives exactly the displayed factorial product, including its empty-product boundary.

The only positive root has coroot pairing $⟨\lambda +\rho ,{\alpha }^{\vee }⟩=\lambda +1$, so the criterion says reducible precisely for $\lambda \in {\mathbb{Z}}_{\ge 0}$.

For $\lambda =m$, the action formula gives $e{f}^{m+1}{v}_m=(m+1)(m-(m+1)+1)f^m{v}_m=0$; its weight is below $m$, so it is the indicated nonzero singular vector.

PBW gives exactly the two displayed monomials. The Chevalley relations give $e_1{f}_1{f}_2{v}_{\lambda }=({\lambda }_1+1)f_2{v}_{\lambda }$, $e_1{f}_{12}{v}_{\lambda }=-f_2{v}_{\lambda }$, $e_{12}{f}_1{f}_2{v}_{\lambda }=-{\lambda }_2{v}_{\lambda }$, and $e_{12}{f}_{12}{v}_{\lambda }=({\lambda }_1+{\lambda }_2)v_{\lambda }$. Contravariance, together with $S_{\lambda }(f_2{v}_{\lambda },f_2{v}_{\lambda })={\lambda }_2$, gives the four displayed matrix entries.

Taking the $2\times 2$ determinant gives ${\lambda }_2({\lambda }_1+1)({\lambda }_1+{\lambda }_2)-{\lambda }_2^2={\lambda }_1{\lambda }_2({\lambda }_1+{\lambda }_2+1)$.

At $n=1$ the product formula gives $S_{-1}(f{v}_{-1},f{v}_{-1})=1!(-1)=-1$.

A positive-definite real bilinear form cannot take a negative value on a nonzero vector; hence this refutes positivity on the specified real PBW span, without asserting degeneracy or turning the complex-bilinear form into a Hermitian form.

The action formula gives $e{f}^{m+1}{v}_m=0$, and $f$ and $h$ preserve the span of $f^n{v}_m$ for $n\ge m+1$; this is therefore the submodule generated by $f^{m+1}{v}_m$. Its quotient has the displayed $m+1$ surviving, distinct-weight PBW vectors.

Any nonzero submodule of the quotient contains a weight vector $\overline{f^r{v}_m}$; applying $e^r$ reaches ${\bar{v}}_m$ because all coefficients $k(m-k+1)$ for $1\le k\le r\le m$ are nonzero. Applying $f$ then gives every displayed basis vector, so the quotient is simple. The unique-simple-quotient theorem therefore identifies its kernel with $J(m)$ and the quotient with $L(m)$.