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 commutator $[h,f]=-2f$ gives $h{w}_n=(\lambda -2n)w_n$ by induction. Similarly $[e,f]=h$ gives $e{w}_{n+1}=fe{w}_n+h{w}_n$. Starting with $e{w}_0=0$, induction yields $e{w}_n=n(\lambda -n+1)w_{n-1}$ for $n\ge 1$. PBW makes every $w_n$ nonzero, and their $h$-eigenvalues are pairwise distinct.

If $\lambda =m\in {\mathbb{Z}}_{\ge 0}$, step 1.1 gives $e{w}_{m+1}=0$. The tail $W=\mathrm{span}\{w_n:n\ge m+1\}$ is invariant under $f$ and $h$, and under $e$ because $e{w}_{m+1}=0$ and $e{w}_n$ is a scalar multiple of $w_{n-1}$ for larger $n$. It is nonzero and proper by PBW, so $M(m)$ is reducible and $w_{m+1}=f^{m+1}{v}_m$ is singular.

Conversely, suppose $\lambda \mathrm{\notin }{\mathbb{Z}}_{\ge 0}$ and let $N$ be a nonzero submodule. A nonzero vector of $N$ is a finite sum of basis vectors $w_n$. Since their $h$-eigenvalues are distinct, a finite Lagrange interpolation polynomial in $h$ extracts one nonzero $w_n\in N$. Each coefficient $k(\lambda -k+1)$ for $1\le k\le n$ is nonzero, so applying $e^n$ gives a nonzero multiple of $w_0\in N$. Applying powers of $f$ then gives every $w_j\in N$. Thus $N=M(\lambda )$ and $M(\lambda )$ is simple. Together with step 2.1 this proves the asserted classification without a determinant criterion.

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.

Write $w_n=f^n{v}_m$. Induction from $[h,f]=-2f$, $[e,f]=h$, $e{w}_0=0$, and $h{w}_0=m{w}_0$ gives $h{w}_n=(m-2n)w_n$ and $e{w}_n=n(m-n+1)w_{n-1}$ for $n\ge 1$. In particular $e{w}_{m+1}=0$. The tail $W=\mathrm{span}\{w_n:n\ge m+1\}$ is stable under $e,f,h$ and equals $U({\mathfrak{s}\mathfrak{l}}_2)w_{m+1}$: it contains that generated submodule by invariance, while every tail vector is $f^k{w}_{m+1}$. Hence the quotient has the displayed $m+1$ surviving, distinct-weight PBW vectors.

Any nonzero submodule of the quotient contains a nonzero finite combination of its basis vectors. Their $h$-eigenvalues are distinct, so a finite interpolation polynomial in $h$ extracts some ${\bar{w}}_r$. Applying $e^r$ reaches a nonzero multiple of ${\bar{w}}_0$, 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 identifies its kernel with $J(m)$ and the quotient with $L(m)$.