My questions are the following.

  • Prove that the center of the universal enveloping algebra of a nilpotent Lie algebra is generated by the center of the Lie algebra.
  • Give a solvable Lie algebra such that the center of its universal enveloping algebra is not generated by the center of the Lie algebra.

I heard these two are classical results, however I finally could not find the proof. Thank you.


I know that there is a counterexample in the semi-simple Lie algebra case and I already calculated the center of universal enveloping algebras of several nilpotent Lie algebras (Heisenberg algebras, ladder algebras and so on). I want to know a general proof in the nilpotent case and I could not find such a question in the sugested.

  • 2
    $\begingroup$ Have a look at other posts here for this topic, e.g. here, or here. $\endgroup$ – Dietrich Burde Jun 3 '18 at 14:19
  • $\begingroup$ > Dietrich For the 2-dimentional non-abelian Lie algebra $L$, the center of $L$ is zero and the center of $U(L)$ is $\mathbf{C} \cdot 1$. This is not a counterexample which I am seeking. $\endgroup$ – ShyGuy Jun 3 '18 at 14:37

Excuse me but I got an answer by myself. In fact, I got a mistake but the ladder Lie algebra is a counterexample of my first question.

The ladder Lie algebra is a Lie algebra $\mathfrak{g} := \langle X_0, X_1, X_2, X_3 \rangle$ whose Lie bracket is defined by the following:

\begin{align} [X_0,X_1] = X_2, [X_0,X_2] = X_3, [X_0,X_3] = [X_1,X_2] = [X_1,X_3] = [X_2,X_3] = 0. \end{align}

In this case, the center of $\mathfrak{g}$ is $\langle X_3 \rangle$, however $X_2^2 - 2 X_1 X_3$ is contained in the center of $U(\mathfrak{g})$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.