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$\DeclareMathOperator\ad{ad}\DeclareMathOperator\Tr{Tr}$I have been thinking about this for some time but have had no luck. I have found some sources that say higher Casimir elements can be obtained by generalizing the second order Casimir, which is $\sum_{\alpha,\beta} \kappa ^{\alpha \beta} X_{\alpha} X_{\beta}$, where $\kappa ^{\alpha \beta}$ is the inverse of the Killing form, and writing $C_3 = \sum g^{\alpha_1 \alpha_2 \alpha_3} X_{\alpha_1} X_{\alpha_2} X_{\alpha_3}$, where $g^{\alpha_1 \alpha_2 \alpha_3} = \Tr(\ad X^{\alpha_1} \ad X^{\alpha_2} \ad X^{\alpha_3})$ and $X^{\alpha} = \kappa^{\alpha \beta}X_{\beta}$. This definition does not give an element in the center of the universal enveloping algebra.

Is there any text out there where an explicit description of higher Casimir operators is given?

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    $\begingroup$ you may want to look at the book by Humphreys on Lie algebras. The explicit isomorphism between the associated graded of the enveloping algebra of a semi-simple Lie algebra $\mathfrak g$ and the symmetric algebra of $\mathfrak g$ is given there from which you can deduce that the space of invariants of the enveloping algebra (i.e. its centre) is isomorphic to the space of $\mathfrak g$ invariants in the symmetric algebra; the latter is a polynimial algebra with some well chosen generators. In your case, there is one in degree three, namely the determinant of a traceless $3\times 3$ matrix. $\endgroup$
    – Venkataramana
    Commented Nov 28, 2014 at 5:56
  • $\begingroup$ en.wikipedia.org/wiki/Harish-Chandra_isomorphism $\endgroup$
    – Qiaochu Yuan
    Commented Nov 28, 2014 at 9:12

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To answer the question in the title, it's $$(h_2-h_1)(h_2+2h_1+3)(2h_2+h_1+3) - 9f_1(h_1+2h_2+3)e_1 + 9f_2(2h_1+h_2+3)e_2 + 9f_{12}(h_2-h_1)e_{12} -27f_1f_2e_{12} -27f_{12}e_1e_2$$

where $e_{12}=[e_1,e_2]$ and $f_{12}=[f_2,f_1]$. You can find a version of this in a paper by Catoiu called Prime ideals of the universal enveloping algebra $U(\mathfrak{sl}_3)$. I don't know of any source that constructs the higher Casimirs explicitly, but I'm sure it's been discussed here before.

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  • $\begingroup$ Catoiu doesn't have homogeneous expressions for both the second and third Casimir elements. Is this very easy? Am I missing something? $\endgroup$
    – mokim
    Commented Dec 18, 2014 at 3:05
  • $\begingroup$ What do you mean by homogeneous? $U$ isn't graded. The degree two Casimir is relatively easy to calculate: pick a basis $x_i$ of your Lie algebra, then let $x_i^*$ be the dual element with respect to the Killing form. Then $\sum x_i x_i^*$ is the quadratic Casimir. See Fulton and Harris, for example. $\endgroup$
    – M T
    Commented Dec 18, 2014 at 9:34
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    $\begingroup$ What I mean by homogeneous, in the case of the second Casimir is a sum where the summands consist of products of two terms, not one; and for the third Casimir, product of three terms. I do know how to write down the second Casimir. My problem is with the third Casimir. Can you tell me what year this paper was published? Is this paper easily accessible? $\endgroup$
    – mokim
    Commented Dec 18, 2014 at 22:27
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I think $\sum_{i,j,k}X_{ij}X_{jk}X_{ki}$ should work for $\mathfrak{gl}(3)$. Now if you want for $\mathfrak{sl}(3)$, maybe you can change $X_{ii}$ by $X_{ii}-(1/3)\mathrm{Id}$.

The formula above is like the trace of a cube of a generic matrix. I remember a seminar where someone talk about this, and in general he considers

$\sum_{i_1,\dots,i_k}X_{i_1,i_2}X_{i_2,i_3}\cdots X_{i_{k-1},i_k}X_{i_k,i_1}= "\operatorname{tr}(X^k)"$

(for $k=2,3,\dots,n$) and get independent Casimir elements in $\mathfrak{gl}(n)$

Unfortunately I can't remember speaker's name, and I can't google the formula... I actually come to this question because I was trying to find that reference.

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