Harish Chandra Isomorphism Casimir and Central Characters - Examples

The Killing form values are $B(h,h)=8$ and $B(e,f)=B(f,e)=4$, so the dual basis is $h\mathrm{/}8,f\mathrm{/}4,e\mathrm{/}4$. Substituting into the definition of the Casimir gives $C=\frac{1}{8}{h}^2+\frac{1}{4}ef+\frac{1}{4}fe$.

Since $e{v}_{\lambda }=0$ and $[e,f]=h$, one has $fe{v}_{\lambda }=0$ and $ef{v}_{\lambda }=(fe+h)v_{\lambda }=\lambda v_{\lambda }$. The formula in step 1.1 therefore gives $C{v}_{\lambda }=({\lambda }^2\mathrm{/}8+\lambda \mathrm{/}4)v_{\lambda }=\lambda (\lambda +2)v_{\lambda }\mathrm{/}8$.

For diagonal $h=\mathrm{diag}(t_1,t_2,t_3)$, the root vectors satisfy $[h,E_{ij}]=(t_i-t_j)E_{ij}$. The corresponding root reflections from Root reflections and the Weyl group action interchange $t_i$ and $t_j$, so $W=S_3$ on the trace-zero Cartan. The symmetric polynomials in $t_1,t_2,t_3$ are generated freely by the elementary symmetric functions $e_1,e_2,e_3$. The trace-zero relation sets $e_1=0$. Every invariant class modulo this relation has an invariant representative, obtained by averaging any representative over the finite group $S_3$. If an invariant polynomial is a multiple of $e_1$ in the full polynomial ring, averaging its quotient makes it a multiple of $e_1$ already in the invariant ring. Thus the quotient invariant ring is exactly $\mathbb{C}[e_2,e_3]$. Newton's identities give $s_2=-2e_2$ and $s_3=3e_3$ there. Hence $s_2,s_3$ are algebraically independent generators.

Under Choice, The center of the enveloping algebra is polynomial on rank-many generators says the center of $U({\mathfrak{s}\mathfrak{l}}_3)$ is polynomial on two generators of definite PBW degrees. The filtered isomorphism Harish-Chandra isomorphism for the center and step 1.1 show that the preimages of $s_2$ and $s_3$ are algebraically independent generators of respective PBW degrees two and three.

In the $A_2$ realization, $\rho =(1,0,-1)$ and $s_1$ swaps the first two coordinates. Hence $\mu =s_1\cdot 0=s_1(\rho )-\rho =-{\alpha }_1$, and $\mu +\rho =s_1(\rho )=(0,1,-1)$.

Here $⟨\rho ,{\alpha }_1^{\vee }⟩=1$. By The simple-root singular vector in a Verma module, $f_1{v}_0$ is a nonzero highest vector of weight $\mu =-{\alpha }_1$ in $M(0)$. The universal property The universal property of Verma modules gives a nonzero map $M(\mu )\to M(0)$. Every central element acts by a scalar on either cyclic highest-weight module by The Harish-Chandra projection computes the highest-weight scalar, and the map intertwines these actions. Hence ${\chi }_{\mu }={\chi }_{\lambda }$. The displayed degree-two and degree-three symmetric invariants also agree on $\rho$ and $s_1(\rho )$, as the coordinates show.

The dot action gives $s\cdot 0=s(\rho )-\rho =-2\omega =-\alpha$. Since $⟨\rho ,{\alpha }^{\vee }⟩=1$, the singular-vector lemma The simple-root singular vector in a Verma module gives a nonzero highest vector $f{v}_0$ of weight $-2\omega$ in $M(0)$. By The universal property of Verma modules it induces a nonzero map $M(-2\omega )\to M(0)$. Every central element acts by a scalar on either Verma module by The Harish-Chandra projection computes the highest-weight scalar; the nonzero map intertwines these actions, so their central characters are equal.

Under the ordinary Weyl action, $s(0)=0$, so the ordinary orbit of $0$ is just $\{0\}$, while the orbit of $-2\omega$ is $\{\pm 2\omega \}$. Thus the two weights are not ordinarily conjugate.

Therefore ordinary Weyl orbits are too fine here: they separate weights that have the same central character. The $\rho$-shifted dot action is essential.

By definition of the dot action, $w\cdot (-\rho )=w((-\rho )+\rho )-\rho =w(0)-\rho =-\rho$ for every $w\in W$.

The cyclic highest-weight module $M(-\rho )$ has a central character ${\chi }_{-\rho }$ because The Harish-Chandra projection computes the highest-weight scalar makes every central element act by a scalar. Step 1.1 shows that the label $-\rho$ has the full Weyl group as its dot-action stabilizer, which is the asserted singularity.

Using the relations $[f,e]=-h$ and $he-eh=2e$, one computes $[f,e^2]=[f,e]e+e[f,e]=-he-eh$.

The element $-he-eh$ is nonzero in the PBW basis, so step 1.1 shows that $e^2$ does not commute with $f$. Hence $e^2$ is not central.