The central reduction of U(sl2) is simple away from the finite-dimensional central characters

The element $\mathrm{\Omega }=ef+fe+\frac{1}{2}{h}^2$ is central. Indeed, using $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$: $[h,\mathrm{\Omega }]=2ef-2ef-2fe+2fe=0$; $[e,ef]=[e,e]f+e[e,f]=eh$, $[e,fe]=[e,f]e+f[e,e]=he$, and $[e,h^2]=[e,h]h+h[e,h]=-2eh-2he$, so $[e,\mathrm{\Omega }]=eh+he-eh-he=0$; and $[f,ef]=[f,e]f+e[f,f]=-hf$, $[f,fe]=[f,f]e+f[f,e]=-fh$, $[f,h^2]=[f,h]h+h[f,h]=2fh+2hf$, so $[f,\mathrm{\Omega }]=-hf-fh+fh+hf=0$. The Killing form in the basis $e,f,h$ has $B(h,h)=8$ and $B(e,f)=B(f,e)=4$ with all other pairings $0$ (the adjoint matrices have columns $a{d}_h(e)=2e$, $a{d}_h(f)=-2f$, $a{d}_h(h)=0$ and $a{d}_e(f)=h$, $a{d}_e(h)=-2e$, $a{d}_f(e)=-h$, $a{d}_f(h)=2f$, giving traces $8$ and $4$), so dual bases are $h\mathrm{/}8$, $f\mathrm{/}4$, $e\mathrm{/}4$ and the library Casimir is $C=\frac{1}{8}{h}^2+\frac{1}{4}(ef+fe)=\frac{1}{4}\mathrm{\Omega }$. To prove that this Casimir generates the center, first $\mathbb{C}h$ is a Cartan subalgebra: it is abelian, hence nilpotent, and the displayed brackets show $N_{\mathfrak{g}}(\mathbb{C}h)=\mathbb{C}h$ by Cartan subalgebra and Normalizer of a Lie subalgebra. The root spaces are $\mathbb{C}e$, $\mathbb{C}f$ with roots $\alpha ,-\alpha$, where $\alpha (h)=2$; therefore the unique root reflection acts by sign on ${\mathfrak{h}}^{\ast }$, and $S(\mathfrak{h}{)}^W=\mathbb{C}[h^2]$ (Root and root space, Root reflections and the Weyl group action). By Chevalley restriction, $S(\mathfrak{g}{)}^{\mathfrak{g}}\cong S(\mathfrak{h}{)}^W$, and the PBW symbol ${\sigma }_2(C)=\frac{1}{8}{h}^2+\frac{1}{2}ef$ restricts to $h^2\mathrm{/}8$, so it generates $S(\mathfrak{g}{)}^{\mathfrak{g}}$ (Chevalley restriction for symmetric invariants). If $z\in Z(U(\mathfrak{g}))$ has PBW degree $d>0$, the adjoint-symbol identity in The adjoint action preserves the associated graded of a two-sided ideal shows $\sigma (z)$ is invariant. It is homogeneous; since the invariant ring is the polynomial ring generated by the degree-two element $\sigma (C)$, one has $d=2k$ and $\sigma (z)=a\sigma (C{)}^k$ for some $a\in {\mathbb{C}}^{\times }$. Thus $z-a{C}^k$ is central of strictly smaller PBW degree. Induction on degree, with degree-zero elements being scalars and $C$ central by the direct commutator calculation above, yields $Z(U(\mathfrak{g}))=\mathbb{C}[C]=\mathbb{C}[\mathrm{\Omega }]$. Consequently $\ker \chi =(\mathrm{\Omega }-c)$. Since $ef=fe+h$, one has $\mathrm{\Omega }=2fe+h+\frac{1}{2}{h}^2$ in $U(\mathfrak{g})$, hence $fe=\frac{1}{2}(\mathrm{\Omega }-h-\frac{1}{2}{h}^2)$ exactly; in $U_{\chi }$, where $\mathrm{\Omega }$ acts by $c$, this reads $fe=p(h):=\frac{1}{2}(c-h-\frac{1}{2}{h}^2)$ and $ef=p(h)+h=:q(h)$. Equivalently $p(h)=-\frac{1}{4}(h^2+2h-2c)$ and $q(h)=p(h-2)=-\frac{1}{4}(h^2-2h-2c)$; also $he=e(h+2)$ and $hf=f(h-2)$.

The associated graded of $U_{\chi }$ for the induced PBW filtration is $\mathbb{C}[e,f,h]\mathrm{/}(\sigma )$, where $\sigma$ is the symbol of $\mathrm{\Omega }-c$, namely $2ef+\frac{1}{2}{h}^2$ up to a nonzero scalar: the kernel of $U(\mathfrak{g})\twoheadrightarrow U_{\chi }$ is generated by $\mathrm{\Omega }-c$, whose symbol is a nonzerodivisor of the polynomial algebra $\mathbb{C}[e,f,h]$, so an element of that ideal has top-degree part in the ideal $(\sigma )$ and conversely every element of $(\sigma )$ occurs as the top-degree part of a product $u(\mathrm{\Omega }-c)v$. Multiplying $\sigma$ by $2$ identifies the quotient with $\mathbb{C}[e,f,h]\mathrm{/}(h^2+4ef)$; multiplication by $h^2+4ef$ from degree $d-2$ to degree $d$ is injective, so the degree-$d$ part has dimension $\left(\genfrac{}{}{0px}{}{d+2}{2}\right)-\left(\genfrac{}{}{0px}{}{d}{2}\right)=2d+1$.

Let $N$ be the $\mathbb{C}$-span of the normal forms $f^i{h}^a$ $(i,a\ge 0)$ and $e^j{h}^a$ $(j\ge 1,a\ge 0)$, filtered by degree $i+a$, respectively $j+a$. Left multiplication by the generators preserves $N$ and raises degree by at most one: using $hf=f(h-2)$ and $he=e(h+2)$ from step 1.1 gives $h\cdot f^i{h}^a=f^i{h}^{a+1}-2i{f}^i{h}^a$ and $h\cdot e^j{h}^a=e^j{h}^{a+1}+2j{e}^j{h}^a$; $e\cdot f^i{h}^a=f^{i-1}q(h-2i+2)h^a$ for $i\ge 1$ by step 1.1 and $e\cdot h^a=e{h}^a$; $e\cdot e^j{h}^a=e^{j+1}{h}^a$; $f\cdot f^i{h}^a=f^{i+1}{h}^a$ and $f\cdot h^a=f{h}^a$; $f\cdot e^j{h}^a=e^{j-1}p(h+2j-2)h^a$ for $j\ge 1$. Since $F^n{U}_{\chi }$ is spanned by words of length at most $n$ in $e,f,h$, induction on $n$ gives $F^n{U}_{\chi }\subseteq N\cap F^n{U}_{\chi }$, so the normal forms span $U_{\chi }$. Exactly $(n+1{)}^2$ of them have degree at most $n$, and their symbols span $(\mathrm{gr}{U}_{\chi }{)}_{\le n}$, whose dimension is ${\sum }_{d\le n}(2d+1)=(n+1{)}^2$ by step 2.1; hence those symbols are a basis of $(\mathrm{gr}{U}_{\chi }{)}_{\le n}$ and, comparing top degrees, the normal forms are linearly independent. They therefore form a $\mathbb{C}$-basis of $U_{\chi }$, so $U_{\chi }$ is infinite-dimensional, and the basis exhibits the ${\mathrm{ad}}_h$-weight decomposition $W_{2k}=e^k\mathbb{C}[h]$ for $k\ge 0$, $W_{-2k}=f^k\mathbb{C}[h]$ for $k\ge 1$.

Let $J\mathrm{\ne }0$ be a two-sided ideal of $U_{\chi }$. Since $[h,u]\in J$ for $u\in J$, and the weight components of $u$ are polynomial combinations of the elements $({\mathrm{ad}}_h{)}^{i}u$, the ideal $J$ contains a nonzero weight vector. If it contains $u=e^{k}Q(h)$ with $k\ge 1$ and $Q\mathrm{\ne }0$, then $u{f}^k=Q(h-2k){\prod }_{i=0}^{k-1}q(h-2i)\in J\cap \mathbb{C}[h]$ is nonzero, because $e^k{f}^k={\prod }_{i=0}^{k-1}q(h-2i)$ and $\mathbb{C}[h]$ has no zero divisors and $Q(h-2k)\mathrm{\ne }0$; if it contains $u=f^{k}Q(h)$ with $k\ge 1$, then $e^{k}u=Q(h){\prod }_{i=0}^{k-1}q(h-2i)$ is such a nonzero polynomial; and if it contains a nonzero element of weight $0$, that element already lies in $\mathbb{C}[h]$. Hence $J\cap \mathbb{C}[h]\mathrm{\ne }0$.

By [F6] the nonzero ideal $J\cap \mathbb{C}[h]$ is generated by a polynomial $P$, chosen of least degree; if $J\mathrm{\ne }{U}_{\chi }$ then $P$ is nonconstant, since a nonzero constant in $J\cap \mathbb{C}[h]$ would put $1$ in $J$ and force $J=U_{\chi }$.

For $P\in J\cap \mathbb{C}[h]$ one has $[e,P]=eP-Pe=(P(h-2)-P(h))e\in J$ and $[f,P]=(P(h+2)-P(h))f\in J$, because $eP(h)=P(h-2)e$ and $fP(h)=P(h+2)f$. Multiplying by $f$ respectively $e$ and using $fe=p$, $ef=q=p+h$ gives the divisibilities $P\mid p(P(h+2)-P(h))$ and $P\mid q(P(h-2)-P(h))$ in $\mathbb{C}[h]$.

Let $r$ be a root of $P$. If $r+2$ is not a root of $P$, then $(P(h+2)-P(h))(r)=P(r+2)-P(r)=P(r+2)\mathrm{\ne }0$, so the first divisibility of step 6.1 forces $p(r)=0$. If $r-2$ is not a root of $P$, then $(P(h-2)-P(h))(r)=P(r-2)\mathrm{\ne }0$, so the second divisibility forces $q(r)=p(r)+r=0$.

Since $P$ has finitely many roots, for every root $r$ the upward chain $r,r+2,r+4,\dots$ has a last member $s=r+2k$ with $s$ a root of $P$ and $s+2$ not a root, and the downward chain has a last member $s^{\mathrm{\prime }}=r-2l$ with $s^{\mathrm{\prime }}-2$ not a root. Step 7.1 then gives $p(s)=0$ and $q(s^{\mathrm{\prime }})=0$, so $s\in \{r_{+},r_{-}\}$ and $s^{\mathrm{\prime }}\in \{q_{+},q_{-}\}=\{r_{+}+2,r_{-}+2\}$, where by step 1.1 $r_{\pm }=-1\pm t$ with $t^2=1+2c$ are the roots of $p$, and $r=-2k+s=s^{\mathrm{\prime }}+2l$ for some $k,l\ge 0$.

If $t\mathrm{\notin }\mathbb{Z}$ then the two roots $r_{\pm }$ are distinct and not congruent modulo $2$ (their difference is $2t$). Writing any root $r$ as in step 8.1, the identity $s-s^{\mathrm{\prime }}=2(k+l)\ge 0$ forces $s-s^{\mathrm{\prime }}\in \{-2,2t-2,-2t-2\}$ to be nonnegative and even; the first value gives $k+l=-1$; the second gives $t=1+k+l\in \mathbb{Z}$; the third gives $t=-(1+k+l)\in \mathbb{Z}$. All are impossible, so $P$ has no roots and [F7] makes it constant, hence $J=U_{\chi }$ by step 5.1. Thus $U_{\chi }$ is simple whenever $t\mathrm{\notin }\mathbb{Z}$, in particular whenever $1+2c$ is not the square of an integer.

If $t=0$ then $p(h)=-\frac{1}{4}(h+1{)}^2$ has the single root $r_{+}=-1$, while $q(h)=p(h-2)=-\frac{1}{4}(h-1{)}^2$ has the single root $q_{+}=1$; step 8.1 would give $-1-2k=r=1+2l$, i.e. $2(k+l)=-2$, which is impossible. So $P$ has no roots and [F7] makes it constant, whence $J=U_{\chi }$; $U_{\chi }$ is simple for $c=-\frac{1}{2}$ as well.

Conversely let $t\in \mathbb{Z}\setminus \{0\}$, and put $n:=\mathrm{\mid }t\mathrm{\mid }-1\ge 0$; then $c=(t^2-1)\mathrm{/}2=n(n+2)\mathrm{/}2$ is the $\mathrm{\Omega }$-eigenvalue of the nonzero finite-dimensional simple module $L(n)$ by [F4], so $L(n)$ is a nonzero $U_{\chi }$-module. The action map $\phi \text{\hspace{0.17em}⁣}:U_{\chi }\to {\mathrm{End}}_{\mathbb{C}}L(n)$ is a nonzero algebra homomorphism (it sends $1$ to the identity), and its image is finite-dimensional while $U_{\chi }$ is infinite-dimensional by step 3.1, so $\ker \phi \mathrm{\ne }0$ and $\ker \phi \mathrm{\ne }{U}_{\chi }$: $U_{\chi }$ has a nonzero proper two-sided ideal and is not simple. Conversely, if a nonzero finite-dimensional $U(\mathfrak{g})$-module has central character $\chi$, choose a nonzero submodule of least positive dimension; it is simple and $\mathrm{\Omega }$ acts there by $c$, so [F4] makes it some $L(m)$ and $c=m(m+2)\mathrm{/}2$. Thus $\chi$ is a finite-dimensional central character exactly when $1+2\chi (\mathrm{\Omega })$ is a positive square, equivalently when $\chi (\mathrm{\Omega })=n(n+2)\mathrm{/}2$ for some $n\ge 0$ (the integer parameter is $n=\mathrm{\mid }t\mathrm{\mid }-1$, since $t^2-1=(\mathrm{\mid }t\mathrm{\mid }-1)(\mathrm{\mid }t\mathrm{\mid }+1)$). Together with steps 9.1 and 9.2 this is the asserted criterion, and the eigenvalues $n(n+2)\mathrm{/}2$ are pairwise distinct by [F4].

Finally suppose $U_{\chi }$ is simple, let $N\mathrm{\ne }0$ be a $U_{\chi }$-module, and let $\psi \text{\hspace{0.17em}⁣}:U_{\chi }\to {\mathrm{End}}_{\mathbb{C}}(N)$ be the action. Its kernel is a two-sided ideal and is proper because $N\mathrm{\ne }0$ makes $\psi (1)={\mathrm{id}}_N\mathrm{\ne }0$; simplicity forces $\ker \psi =0$, so $N$ is faithful over $U_{\chi }$. If ${\chi }_{\lambda }=\chi$ for a weight $\lambda$ then $\lambda \mathrm{\notin }{\mathbb{Z}}_{\ge 0}$, because otherwise ${\chi }_{\lambda }$ would be the central character of the finite-dimensional module $L(\lambda )$ by [F5], contrary to the established criterion; hence $M(\lambda )$ is simple by [F5]. The annihilator of $M(\lambda )$ in $U(\mathfrak{g})$ is the preimage of $\ker \psi$ for $N=M(\lambda )$, which is $\ker (U(\mathfrak{g})\to U_{\chi })=U(\mathfrak{g})\ker \chi$ by the definition of the annihilator as a kernel; therefore ${\mathrm{Ann}}_{U(\mathfrak{g})}M(\lambda )=U(\mathfrak{g})\ker \chi$, as claimed.