7
$\begingroup$

The Wikipedia universal enveloping algebra suggest a way to obtain higher Casimir operators (e.g. generators of the center of $\mathfrak{U(g)}$ for $\mathfrak{g}$ semisimple) by evaluating certain determinant:

$$ \det\, (t\mathrm{I} - \mathrm{ad}_X) = \sum_{i=0}^{\dim \mathfrak{g}} p_i(X) t^i $$

However if one does that for $\mathfrak{sl}_3$ then the resulting polynomial has only degrees (in $t$) 0,1,2,4,6,8 and it's coefficients seem to be just powers of the quadratic Casimir operator. If one tries to do the same for the defining representation $\mathbb{C}^3$ (replacing $\mathrm{ad}_X$ by $\rho(X)$ in the above formula) then one obtains quadratic as well as cubic invariant polynomials.

In the proof of Harish-Chandra isomorphism as presented e.g. in (1) there is construction of elements of $Z(\mathfrak{U(g)}$ using traces of matrices from representations of $\mathfrak{g}$. Something like $$ \sum\mathrm{tr}(\rho(X_{i_1})\rho(X_{i_2})\ldots \rho(X_{i_n}))X_{i_1}^*X_{i_2}^*\ldots X_{i_n}^* $$ where $X_i$ form basis for $\mathfrak{g}$ and $X_i^*$ form dual basis with respect to Killing form.

Q1: What is going on here?

Q2: Is it true that for a semi-simple complex Lie algebra and it's smallest nontrivial representation one obtains in this way all generators of the $Z(\mathfrak{U(g)}$?

Q3: Does the approach through determinant give the same operators that appear in the proof of the H-Ch isomorphism?

(1) Cohomological Induction and Unitary Representations by Knapp, Vogan

$\endgroup$
3

1 Answer 1

3
$\begingroup$

For $\mathfrak{sl}(3)$ the coefficients are not powers of the Casimir, though the polynomial is one in the square of $t$.

$\endgroup$
5
  • $\begingroup$ To be completely honest we haven't tried all coefficients but we checked that the highest degree $p_i$ is indeed a power of Casimir. I've checked again and I discovered I've missed linear term that is there. But all the other odd degrees of $t$ are missing. $\endgroup$ Commented Sep 18, 2018 at 17:15
  • 2
    $\begingroup$ Does "the polynomial is one in the square of $t$" mean "the polynomial has non-$0$ terms only in even degrees"? $\endgroup$
    – LSpice
    Commented Sep 18, 2018 at 19:38
  • $\begingroup$ Yes, that is what I meant $\endgroup$ Commented Sep 18, 2018 at 23:41
  • 1
    $\begingroup$ @LSpice Thank you for the editing. The typos were too many! $\endgroup$ Commented Sep 18, 2018 at 23:56
  • $\begingroup$ The coefficients are naturally elements of $\mathcal S(\mathfrak g^*)$ (and not $\mathcal U(\mathfrak g^*)$). If you symmetrize the coefficient of $t^6$ and evaluate it at the dual basis, you obtain the quadratic Casimir. However, the cubic one is missing, since the coefficient of $t^5$ is $0$. There is a non-zero coefficient next to $t$ (which is of degree $7$), so cubic Casimir might be hidden there? $\endgroup$
    – Ennar
    Commented Sep 19, 2018 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.