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 $C=\frac{1}{8}{h}^2+\frac{1}{4}ef+\frac{1}{4}fe$.

For ${\mathfrak{s}\mathfrak{l}}_2$, the Weyl vector satisfies $\rho (h)=1$, so the scalar from The quadratic Casimir eigenvalue on a highest-weight module is $(\lambda ,\lambda +2\rho )$ is $(\lambda ,\lambda +2\rho )=\lambda (\lambda +2)\mathrm{/}8$. Therefore $C{v}_{\lambda }=\lambda (\lambda +2)v_{\lambda }\mathrm{/}8$.

The symmetric polynomials in $t_1,t_2,t_3$ are generated by the elementary symmetric functions, and the trace-zero relation $t_1+t_2+t_3=0$ removes the degree-one generator. Thus the invariant ring is generated by the degree-two and degree-three symmetric polynomials $s_2$ and $s_3$.

By The center of the enveloping algebra is polynomial on rank-many generators, the center of $U({\mathfrak{s}\mathfrak{l}}_3)$ is polynomial on two homogeneous generators. The Harish-Chandra isomorphism identifies those generators with any algebraically independent pair generating $S(\mathfrak{h}{)}^W$, so the preimages of $s_2$ and $s_3$ give the required quadratic and cubic central elements.

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 $\mu +\rho =s_1(\rho )=(0,1,-1)$.

The degree-two and degree-three symmetric invariants from the $A_2$ Cartan take the same values on $\rho$ and $s_1(\rho )$ because those points are in the same ordinary Weyl orbit. Therefore Central characters are dot-Weyl orbits gives ${\chi }_{\mu }={\chi }_{\lambda }$.

The dot action gives $s\cdot 0=s(\rho )-\rho =-2\omega$, so Central characters are dot-Weyl orbits makes $0$ and $-2\omega$ have the same central character.

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 $w\in W$.

Therefore the entire Weyl group stabilizes $-\rho$, and Central characters are dot-Weyl orbits shows that its central character is singular with full Weyl-group stabilizer.

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.