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Casimirs are not completely trivial to compute, and somewhat ill defined (if $C_4$ is a quartic casimir and $C_2$ a quadratic, $C_4'=C_4+p*C_2^2+q*C_2$ is another quartic casimir). With generalized Dynkin indices (here $I_2, I_4$) you remove the latter problem (they are fixed) but you must compute even harder.
Since for my special problem probably any Casimir will do, I went the opposite way and defined a quasimir. Instead of a formal definition, an example should make things clear.
Lie algebra is: $B_2$
Heighest weights are: $x,y$
Positive roots are: $x,y,x+y,x+2y$
Set $x\rightarrow{x+1},y\rightarrow{y+1}$ to get: $x+1,y+1,x+y+2,x+2y+3$
Dimension is: $D=(x+1)/1*(y+1)/1*(x+y+2)/2*(x+2y+3)/3$
Quadratic Quasimir is: $Q_2=(x+1)^2-1^2+(y+1)^2-1^2+(x+y+2)^2-2^2+(x+2y+3)^2-3^2$
Quartic Quasimir is: $Q_4=(x+1)^4-1^4+(y+1)^4-1^4+(x+y+2)^4-2^4+(x+2y+3)^4-3^4$
(Depending on definition you have to multiply them by the dimension. Oh, and odd Quasimirs evidently crash, so hands off $A_2,E_6,...$!)

Clearly the computation is child's play now, but of course first someone (and surely not amateur me :-) should a) prove that Quasimirs are Casimirs, b) relate them to those computed from a fairly standard definition (e.g. Wybourne's SCHUR program) and c) relate them to indices. E.g. here is $I_2=Q_2*D/60,I_4=(-12*Q_4+150*Q_2+5*Q_2^2)*D/9$. (Props to Prof. Schellekens for providing me with the raw data.)

So, first things first: Who can prove that a Quasimir is indeed a Casimir?

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The Harish-Chandra isomorphism tells you all what you want: The radial parts of the "Casimirs" (=elements of the center of the universal enveloping algebra) are of the form $p(\chi+\rho)$ where $p$ is a Weyl group invariant polynomial on a Cartan and $\rho$ is the half-sum of the positive roots. Up to a constant, you seem to take for $p$ the polynomials $$ p_m(\chi)=\sum_{\alpha>0}\langle\chi\mid\alpha^\vee\rangle^m. $$ This is only invariant if $m$ is even. For the $p_{2m}$ to generate the ring of invariants it is necessary for the degrees of the generators to be even and distinct. This excludes $A_n, n\ge2$, $D_n, n\ge4$ and $E_6$. Note, that many other choices for $p$ are possible.

In your example, you seem to have mixed up roots and coroots. Since $$ x(\chi)=\langle\chi\mid\alpha_1^\vee\rangle,\ y(\chi)=\langle\chi\mid\alpha_2^\vee\rangle $$ the other coroots are $x+y$ and $2x+y$. So, after replacing $x+2y$ by $2x+y$ your Quasimirs are Casimirs.

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  • $\begingroup$ Sorry for the mix-up. "Precisely for B and C"...I reproduced the indices of G2 correctly with my quasimirs, and F4 also seems OK (I just didn't test I12). Can you elaborate? $\endgroup$ May 9, 2016 at 8:02
  • $\begingroup$ Ok, I amended my statement. $\endgroup$ May 9, 2016 at 8:20

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